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2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.

2.2 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words, ∆ x v = lim (2.1a) ∆ t → 0 ∆ t Fig. 2.1 Determining velocity from position-time d x = (2.1b) graph. Velocity at t = 4 s is the slope of the dt tangent to the graph at that instant. lim where the symbol stands for the operation ∆→t 0 Now, we decrease the value of ∆t from 2 s to 1 of taking limit as ∆tg0 of the quantity on its s. Then line P1P2 becomes Q1Q2 and its slope right. In the language of calculus, the quantity gives the value of the average velocity over on the right hand side of Eq. (2.1a) is the the interval 3.5 s to 4.5 s. In the limit ∆t → 0, differential coefficient of x with respect to t and the line P1P2 becomes tangent to the position- d x time curve at the point P and the velocity at t is denoted by (see Appendix 2.1). It is the d t = 4 s is given by the slope of the tangent at rate of change of position with respect to time, that point. It is difficult to show this process graphically. But if we useat that instant. numerical method to obtain the value of We can use Eq. (2.1a) for obtaining the the velocity, the meaning of the limiting value of velocity at an instant either process becomes clear. For the graph shown graphically or numerically. Suppose that we in Fig. 2.1, x = 0.08 t3. Table 2.1 gives the want to obtain graphically the value of value of ∆x/∆t calculated for ∆t equal to 2.0 s, velocity at time t = 4 s (point P) for the motion 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = of the car represented in Fig.2.1 calculation. 4.0 s. The second and third columns give theLet us take ∆t = 2 s centred at t = 4 s. Then, t  t    ∆ ∆by the definition of the average velocity, the t + t −     and t 2 = and the value of t1= 2 2slope of ( Fig. 2.1) gives the value of     line P1P2 average velocity over the interval 3 s to 5 s. fourth and the fifth columns give the ∆x Table 2.1 Limiting value of at t = 4 s ∆ t Reprint 2025-26 MOTION IN A STRAIGHT LINE 15 3 a + 16b – a – 4b corresponding values of x, i.e. x (t1) = 0.08 t1 = = 6.0 × b 2.0 -1 ⊳and x (t2) = 0.08 t23. The sixth column lists the = 6.0 × 2.5 =15 m s difference ∆x = x (t2) – x (t1) and the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting + 24.0 m s–1 and a velocity of – 24.0 m s–1 — value 3.84 m s–1 which is the value of velocity at both have an associated speed of 24.0 m s-1. It should be noted that though average speed over dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt magnitude of the average velocity, manner, we can calculate velocity at each instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a 2.3 ACCELERATIONconvenient method. For this, we must carefully plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the ⊳ increasing distance of fall. This led to the Example 2.1 The position of an object concept of acceleration as the rate of change moving along x-axis is given by x = a + bt2 of velocity with time. where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time interval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? v 2 – v1 ∆v (2.2)Answer In notation of differential calculus, the a = = t 2 – t1 ∆tvelocity is where v2 and v1 are the instantaneous velocities dx d 2 -1 2b t = 5.0 t m s or simply velocities at time t2 and t1 . It is thev = = ( a + bt ) = dt dt average change of velocity per unit time. The SI At t = 0 s, v = 0 m s–1 and at t = 2.0 s, unit of acceleration is m s–2 . v = 10 m s-1 . On a plot of velocity versus time, the average acceleration is the slope of the straight line x ( 4.0 ) − x ( 2.0 )Average velocity = connecting the points corresponding to (v2, t2) 4.0 − 2.0 and (v1, t1). Reprint 2025-26 16 PHYSICS Instantaneous acceleration is defined in the (c) An object is moving in negative direction same way as the instantaneous velocity : with a negative acceleration. ∆v d v (d) An object is moving in positive direction lim a = = (2.3) till time t1, and then turns back with the d t ∆→ t 0 ∆ t same negative acceleration. The acceleration at an instant is the slope of the tangent to the v–t curve at that An interesting feature of a velocity-time instant. graph for any moving object is that the area Since velocity is a quantity having both under the curve represents the magnitude and direction, a change in displacement over a given time interval. A velocity may involve either or both of these general proof of this statement requires use of factors. Acceleration, therefore, may result calculus. We can, however, see that it is true from a change in speed (magnitude), a for the simple case of an object moving with change in direction or changes in both. Like constant velocity u. Its velocity-time graph is velocity, acceleration can also be positive, as shown in Fig. 2.4. negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vo at t = 0 and v at time t, we have v − v 0 a = or, v = v 0 + a t (2.4) t − 0 Fig. 2.3 Velocity–time graph for motions with Fig. 2.2 Position-time graph for motion with constant acceleration. (a) Motion in positive (a) positive acceleration; (b) negative direction with positive acceleration, acceleration, and (c) zero acceleration. (b) Motion in positive direction with Let us see how velocity-time graph looks like negative acceleration, (c) Motion in for some simple cases. Fig. 2.3 shows velocity- negative direction with negative acceleration, (d) Motion of an object withtime graph for motion with constant acceleration negative acceleration that changesfor the following cases : direction at time t1. Between times 0 to (a) An object is moving in a positive direction t1, it moves in positive x - direction with a positive acceleration. and between t1 and t2 it moves in the (b) An object is moving in positive direction opposite direction. with a negative acceleration. Reprint 2025-26 MOTION IN A STRAIGHT LINE 17 Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which Fig. 2.5 Area under v-t curve for an object with is the displacement in this time interval. How uniform acceleration. come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at As explained in the previous section, the area the answer. under v-t curve represents the displacement. Therefore, the displacement x of the object is : Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp 1 x = ( v – v 0 ) t + v 0 t (2.5)kinks at some points implying that the 2 functions are not differentiable at these But v − v 0 = a tpoints. In any realistic situation, the functions will be differentiable at all points 1 2 Therefore, x = a t + v 0 tand the graphs will be smooth. 2 What this means physically is that 1 2 or, x = v 0 t + at (2.6)acceleration and velocity cannot change 2 values abruptly at an instant. Changes are Equation (2.5) can also be written as always continuous. v + v 0 x = t = v t (2.7a)2.4 KINEMATIC EQUATIONS FOR 2 UNIFORMLY ACCELERATED MOTION where, For uniformly accelerated motion, we can derive some simple equations that relate displacement v + v 0 v = (constant acceleration only)(x), time taken (t), initial velocity (v0), final 2 velocity (v) and acceleration (a). Equation (2.4) (2.7b) already obtained gives a relation between final and initial velocities v and v0 of an object moving Equations (2.7a) and (2.7b) mean that the object with uniform acceleration a : has undergone displacement x with an average velocity equal to the arithmetic average of the v = v0 + at (2.4) initial and final velocities. From Eq. (2.4), t = (v – v0)/a. Substituting this in This relation is graphically represented in Fig. 2.5. Eq. (2.7a), we get The area under this curve is : Area between instants 0 and t = Area of triangle  v + v 0   v − v 0  v 2 − v 02 x = v t =ABC + Area of rectangle OACD  2   a = 2a 1 2 2 = (v – v 0 ) t + v 0 t v = v 0 + 2ax (2.8) 2 Reprint 2025-26 18 PHYSICS This equation can also be obtained by t v 0 + at ) d tsubstituting the value of t from Eq. (2.4) into Eq. = ∫ 0 ( (2.6). Thus, we have obtained three important equations : 1 2 x – x 0 = v 0 t + a t 2 v = v 0 + at 1 2 1 2 x = x 0 + v 0 t + a t x = v 0t + at 2 2 We can write v 2 = v 02 + 2ax (2.9a) d v d v d x d v a = = = v d t d x d t d x connecting five quantities v0, v, a, t and x. These or, v dv = a dxare kinematic equations of rectilinear motion for Integrating both sides,constant acceleration. The set of Eq. (2.9a) were obtained by v x v d v = a d xassuming that at t = 0, the position of the particle, ∫ v 0 ∫ x 0 x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non- v 2 – v 02 = a ( x – x 0 ) zero, say x0. Then Eqs. (2.9a) are modified 2 (replacing x by x – x0 ) to : 2 2 v = v 0 + 2a ( x – x 0 ) v = v 0 + at The advantage of this method is that it can be used 1 2 for motion with non-uniform acceleration x = x 0 + v 0t + at (2.9b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (2.9c) important cases. ⊳ ⊳ ⊳ Example 2.3 A ball is thrown vertically Example 2.2 Obtain equations of motion upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how longAnswer By definition will it be before the ball hits the ground? d v Take g = 10 m s–2. a = d t dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the v t ∫ v 0 d v = ∫ 0 a d t ground, as shown in Fig. 2.6. Now vo = + 20 m s–1, t d t (a is a = – g = –10 m s–2, = a ∫ 0 v = 0 m s–1 constant) If the ball rises to height y from the point of v – v 0 = at launch, then using the equation 2 + 2 a 0 ( y – y 0 ) v = v 0 + at v 2 = v we get d x Further, v = 0 = (20)2 + 2(–10)(y – y0) d t Solving, we get, (y – y0) = 20 m. dx = v dt Integrating both sides (b) We can solve this part of the problem in two x t ways. Note carefully the methods used. ∫ x 0 dx = ∫v0 d t Reprint 2025-26 MOTION IN A STRAIGHT LINE 19 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. ⊳ ⊳ Example 2.4 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 2.6 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use v = 0 – g t = –9.8 t m s–1equation y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y 0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t2 2 the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, with time have been plotted in Fig. 2.7(a), (b)Therefore, the total time taken by the ball before and (c). it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y = 0 m vo = 20 m s-1, a = –10m s–2, t = ? (a) Reprint 2025-26 20 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 2 y = −1 gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 2.2. If we take (–1/ 2) gτ2 as y0 — the position coordinate after first time interval τ, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. ⊳ ⊳ Example 2.6 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and theFig. 2.7 Motion of an object under free fall. braking capacity, or deceleration, –a that (a) Variation of acceleration with time. (b) Variation of velocity with time. is caused by the braking. Derive an (c) Variation of distance with time ⊳ expression for stopping distance of a vehicle in terms of vo and a. ⊳ Example 2.5 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling 2 motion v2 = vo + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. – v 02 d s = 2aAnswer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals τ and find out the distances the square of the initial velocity. Doubling the Table 2.2 Reprint 2025-26 MOTION IN A STRAIGHT LINE 21 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example, in school zones. ⊳ ⊳ Example 2.7 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think Fig. 2.8 Measuring the reaction time. and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before Answer The ruler drops under free fall. he slams the brakes of the car is the Therefore, vo = 0, and a = – g = –9.8 m s–2. The reaction time. Reaction time depends distance travelled d and the reaction time tr are on complexity of the situation and on related by an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between Or, your thumb and forefinger (Fig. 2.8). Given d = 21.0 cm and g = 9.8 m s–2 the reaction After you catch it, find the distance d time is travelled by the ruler. In a particular case, d was found to be 21.0 cm. ⊳ Estimate reaction time. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small : ∆ x d x v = lim v = lim = ∆→t 0 ∆→t 0 ∆t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. Reprint 2025-26 22 PHYSICS 3. Average acceleration is the change in velocity divided by the time interval during which the change occurs : ∆ v a = ∆t 4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆v d v a = lim a = lim = ∆→t 0 ∆→t 0 ∆ t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 5. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at 1 2 x = v0 t + at 2 v 2 = v 02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x – x0). Reprint 2025-26 MOTION IN A STRAIGHT LINE 23 POINTS TO PONDER 1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. Reprint 2025-26 24 PHYSICS EXERCISES 2.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice). Fig. 2.9 2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? 2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). 2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. Reprint 2025-26 MOTION IN A STRAIGHT LINE 25 2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. 2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. 2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] Fig. 2.10 2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. 2.13 Figure 2.11shows the x-t plot of one- dimensional motion of a particle. Is it correct to say from the graph that the particle moves Fig. 2.11 in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car). Reprint 2025-26 26 PHYSICS 2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12): Fig. 2.12 2.16 Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s. Fig. 2.13 2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 2.14 2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Fig. 2.15 Reprint 2025-26 CHAPTER THREE MOTION IN A PLANE 3.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 3.1 Introduction describe the motion of an object along a straight line. We 3.2 Scalars and vectors found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two3.3 Multiplication of vectors by real numbers directions are possible. But in order to describe motion of an 3.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 3.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to3.6 Vector addition — analytical method add, subtract and multiply vectors ? What is the result of 3.7 Motion in a plane multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and3.8 Motion in a plane with constant acceleration acceleration in a plane. We then discuss motion of an object 3.9 Projectile motion in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail3.10 Uniform circular motion the projectile motion. Circular motion is a familiar class of Summary motion that has a special significance in daily-life situations. Points to ponder We shall discuss uniform circular motion in some detail. Exercises The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 3.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided Reprint 2025-26 28 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

2.7 kg, then its volume is 10–3 m3 (a scalar) and its density is 2.7×103 kg m–3 (a scalar). A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the Fig. 3.1 (a) Position and displacement vectors. parallelogram law of addition. So, a vector is (b) Displacement vector PQ and different specified by giving its magnitude by a number courses of motion. and its direction. Some physical quantities that It is important to note that displacement are represented by vectors are displacement, vector is the straight line joining the initial and velocity, acceleration and force. final positions and does not depend on the actual To represent a vector, we use a bold face type path undertaken by the object between the two in this book. Thus, a velocity vector can be positions. For example, in Fig. 3.1(b), given the represented by a symbol v. Since bold face is initial and final positions as P and Q, the difficult to produce, when written by hand, a displacement vector is the same PQ for different vector is often representedrv by an arrow placedrv paths of journey, say PABCQ, PDQ, and PBEFQ.over a letter, say . Thus, both v and Therefore, the magnitude of displacement is represent the velocity vector. The magnitude of either less or equal to the path length of an a vector is often called its absolute value, object between two points. This fact was indicated by |v| = v. Thus, a vector is emphasised in the previous chapter also whilerepresented by a bold face, e.g. by A, a, p, q, r, ... discussing motion along a straight line.x, y, with respective magnitudes denoted by light face A, a, p, q, r, ... x, y. 3.2.2 Equality of Vectors 3.2.1 Position and Displacement Vectors Two vectors A and B are said to be equal if, and only if, they have the same magnitude and theTo describe the position of an object moving in same direction.**a plane, we need to choose a convenient point, say O as origin. Let P and P′ be the positions of Figure 3.2(a) shows two equal vectors A and the object at time t and t′, respectively [Fig. 3.1(a)]. B. We can easily check their equality. Shift B We join O and P by a straight line. Then, OP is parallel to itself until its tail Q coincides with that the position vector of the object at time t. An of A, i.e. Q coincides with O. Then, since their arrow is marked at the head of this line. It is tips S and P also coincide, the two vectors are represented by a symbol r, i.e. OP = r. Point P′ is said to be equal. In general, equality is indicated * Addition and subtraction of scalars make sense only for quantities with same units. However, you can multiply and divide scalars of different units. ** In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves the vector unchanged. Such vectors are called free vectors. However, in some physical applications, location or line of application of a vector is important. Such vectors are called localised vectors. Reprint 2025-26 MOTION IN A PLANE 29 The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector. 3.4 ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD Fig. 3.2 (a) Two equal vectors A and B. (b) Two As mentioned in section 4.2, vectors, by vectors A′ and B′ are unequal though they definition, obey the triangle law or equivalently, are of the same length. the parallelogram law of addition. We shall now describe this law of addition using the graphical as A = B. Note that in Fig. 3.2(b), vectors A′ and method. Let us consider two vectors A and B that B′ have the same magnitude but they are not lie in a plane as shown in Fig. 3.4(a). The lengths equal because they have different directions. of the line segments representing these vectors Even if we shift B′ parallel to itself so that its tail are proportional to the magnitude of the vectors. Q′ coincides with the tail O′ of A′, the tip S′ of B′ To find the sum A + B, we place vector B so that does not coincide with the tip P′ of A′. its tail is at the head of the vector A, as in 3.3 MULTIPLICATION OF VECTORS BY REAL Fig. 3.4(b). Then, we join the tail of A to the head NUMBERS of B. This line OQ represents a vector R, that is, the sum of the vectors A and B. Since, in thisMultiplying a vector A with a positive number λ procedure of vector addition, vectors aregives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A : λ A = λ A if λ > 0. For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. 3.3(a). Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|. Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig 3.3(b). (c) (d) Fig. 3.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. Fig. 3.4 (a) Vectors A and B. (b) Vectors A and B (b) Vector A and resultant vectors after added graphically. (c) Vectors B and A multiplying it by a negative number –1 added graphically. (d) Illustrating the and –1.5. associative law of vector addition. Reprint 2025-26 30 PHYSICS arranged head to tail, this graphical method is What is the physical meaning of a zero vector? called the head-to-tail method. The two vectors Consider the position and displacement vectors and their resultant form three sides of a triangle, in a plane as shown in Fig. 3.1(a). Now suppose so this method is also known as triangle method that an object which is at P at time t, moves to of vector addition. If we find the resultant of P′ and then comes back to P. Then, what is its B + A as in Fig. 3.4(c), the same vector R is displacement? Since the initial and final obtained. Thus, vector addition is commutative: positions coincide, the displacement is a “null vector”. A + B = B + A (3.1) Subtraction of vectors can be defined in termsThe addition of vectors also obeys the associative of addition of vectors. We define the differencelaw as illustrated in Fig. 3.4(d). The result of of two vectors A and B as the sum of two vectorsadding vectors A and B first and then adding A and –B :vector C is the same as the result of adding B and C first and then adding vector A : A – B = A + (–B) (3.5) (A + B) + C = A + (B + C) (3.2) It is shown in Fig 3.5. The vector –B is added to vector A to get R2 = (A – B). The vector R1 = A + BWhat is the result of adding two equal and is also shown in the same figure for comparison.opposite vectors ? Consider two vectors A and We can also use the parallelogram method to–A shown in Fig. 3.3(b). Their sum is A + (–A). find the sum of two vectors. Suppose we haveSince the magnitudes of the two vectors are the same, but the directions are opposite, the two vectors A and B. To add these vectors, we resultant vector has zero magnitude and is bring their tails to a common origin O as represented by 0 called a null vector or a zero shown in Fig. 3.6(a). Then we draw a line from vector : the head of A parallel to B and another line from the head of B parallel to A to complete a A – A = 0 |0|= 0 (3.3) parallelogram OQSP. Now we join the point of Since the magnitude of a null vector is zero, its the intersection of these two lines to the origin direction cannot be specified. O. The resultant vector R is directed from the The null vector also results when we multiply common origin O along the diagonal (OS) of the a vector A by the number zero. The main parallelogram [Fig. 3.6(b)]. In Fig.3.6(c), the properties of 0 are : triangle law is used to obtain the resultant of A A + 0 = A and B and we see that the two methods yield the λ 0 = 0 same result. Thus, the two methods are 0 A = 0 (3.4) equivalent. Fig. 3.5 (a) Two vectors A and B, – B is also shown. (b) Subtracting vector B from vector A – the result is R2. For comparison, addition of vectors A and B, i.e. R1 is also shown. Reprint 2025-26 MOTION IN A PLANE 31 Fig. 3.6 (a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method. ⊳ Example 3.1 Rain is falling vertically with 3.5 RESOLUTION OF VECTORS a speed of 35 m s–1. Winds starts blowing Let a and b be any two non-zero vectors in a after sometime with a speed of 12 m s–1 in plane with different directions and let A be east to west direction. In which direction another vector in the same plane (Fig. 3.8). A should a boy waiting at a bus stop hold can be expressed as a sum of two vectors — one his umbrella ? obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have A = OP = OQ + QP (3.6) But since OQ is parallel to a, and QP is parallel to b, we can write : OQ = λ a, and QP = µ b (3.7) Fig. 3.7 where λ and µ are real numbers. Answer The velocity of the rain and the wind Therefore, A = λ a + µ b (3.8)are represented by the vectors vr and vw in Fig.

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…, rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge q1 is 1 q1 V1 = 4 πε0 r1P where r1P is the distance between q1 and P. Similarly, the potential V2 at P due to q2 and V3 due to q3 are given by 1 q 2 1 q 3 V 2 = , V 3 = 4 πε0 r2P 4 πε0 r3P where r2P and r3P are the distances of P from charges q2 and q3, respectively; and so on for the potential due to other charges. By the FIGURE 2.6 Potential at a point due to a superposition principle, the potential V at P due system of charges is the sum of potentials to the total charge configuration is the algebraic due to individual charges. sum of the potentials due to the individual charges V = V1 + V2 + ... + Vn (2.17) 51 Reprint 2025-26 Physics 1  q1 q 2 q n  = + + ...... + (2.18) 4 πε0  r1P r2 P rnP  If we have a continuous charge distribution characterised by a charge density r (r), we divide it, as before, into small volume elements each of size Dv and carrying a charge rDv. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution. We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by 1 q V = (r ≥ R ) [2.19(a)] 4 πε0 r where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is 1 q V = [2.19(b)] 4 πε0 R Example 2.2 Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Solution Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7). FIGURE 2.7 Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have 1  3 × 10 – 8 2 × 10 –8  − x × 10 –2 4 πε0  (15 − x ) × 10 –2  = 0 where x is in cm. That is, 3 2 − = 0 2.2 x 15 − x which gives x = 9 cm. If x lies on the extended line OA, the required condition is 3 2 − = 0 EXAMPLE x x − 15 Reprint 2025-26 Electrostatic Potential and Capacitance which gives x = 45 cm Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the EXAMPLE formula for potential used in the calculation required choosing potential to be zero at infinity. 2.2 Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively. Electric potential, equipotential-sufaces-12584/ FIGURE 2.8 equipotential (a) Give the signs of the potential difference VP – VQ; VB – VA. (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B. surfaces: (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Solution 1 (a) As V ∝ , VP > VQ. Thus, (VP – VQ) is positive. Also VB is less negative r than VA . Thus, VB > VA or (VB – VA) is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive. http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field- (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. EXAMPLE (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A. 2.3 53 Reprint 2025-26 Physics 2.6 EQUIPOTENTIAL SURFACES An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8): 1 q V = 4 πεo r This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential FIGURE 2.9 For a surface: there is no potential difference between any two points on the single charge q surface and no work is required to move a test charge on the surface. (a) equipotential The electric field must, therefore, be normal to the equipotential surface surfaces are at every point. Equipotential surfaces offer an alternative visual picture spherical surfaces in addition to the picture of electric field lines around a charge centred at the configuration. charge, and (b) electric field lines are radial, starting from the charge if q > 0. FIGURE 2.10 Equipotential surfaces for a uniform electric field. For a uniform electric field E, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11. FIGURE 2.11 Some equipotential surfaces for (a) a dipole, 54 (b) two identical positive charges. Reprint 2025-26 Electrostatic Potential and Capacitance 2.6.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + dV, where dV is the change in V in the direction of the electric field E. Let P be a point on the surface B. d l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|dl. This work equals the potential difference VA–VB. Thus, |E|d l = V – (V + dV)= – dV V i.e., |E|= −δ (2.20) δl Since dV is negative, dV = – |dV|. we can rewrite FIGURE 2.12 From the Eq (2.20) as potential to the field. δV δV E = − = + (2.21) δl δl We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

2.3 Two charges 2 mC and –2 mC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? 79 Reprint 2025-26 Physics 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. 2.10 A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? 2.11 A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Reprint 2025-26 Chapter Three CURRENT ELECTRICITY 3.1 INTRODUCTION In Chapter 1, all charges whether free or bound, were considered to be at rest. Charges in motion constitute an electric current. Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we see many devices where charges flow in a steady manner, like water flowing smoothly in a river. A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents. 3.2 ELECTRIC CURRENT Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let q– be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q–. This is proportional to t for steady current Reprint 2025-26 Physics and the quotient q I = (3.1) t is defined to be the current across the area in the forward direction. (If it turn out to be a negative number, it implies a current in the backward direction.) Currents are not always steady and hence more generally, we define the current as follows. Let DQ be the net charge flowing across a cross- section of a conductor during the time interval Dt [i.e., between times t and (t + Dt)]. Then, the current at time t across the cross-section of the conductor is defined as the value of the ratio of DQ to Dt in the limit of Dt tending to zero, ∆Q lim (3.2) I (t ) ≡ t 0 ∆→ ∆t In SI units, the unit of current is ampere. An ampere is defined through magnetic effects of currents that we will study in the following chapter. An ampere is typically the order of magnitude of currents in domestic appliances. An average lightning carries currents of the order of tens of thousands of amperes and at the other extreme, currents in our nerves are in microamperes. 3.3 ELECTRIC CURRENTS IN CONDUCTORS An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, some of the electrons are practically free to move within the bulk material. These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons. There are, however, other types of conductors like electrolytic solutions where positive and negative charges both can move. In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions. Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential 82 direction for the velocities of the electrons. Thus on the average, the Reprint 2025-26 Current Electricity number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current. Let us now see what happens to such a piece of conductor if an electric field is applied. To focus our thoughts, imagine the conductor in the shape of a cylinder of radius R (Fig. 3.1). Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift positive charge +Q distributed over one disc because of the electric field created to and similarly –Q at the other disc. We attach neutralise the charges. The current thus the two discs on the two flat surfaces of the will stop after a while unless the charges +Q cylinder. An electric field will be created and and –Q are continuously replenished. is directed from the positive towards the negative charge. The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter. We can also imagine a mechanism where the ends of the cylinder are supplied with fresh charges to make up for any charges neutralised by electrons moving inside the conductor. In that case, there will be a steady electric field in the body of the conductor. This will result in a continuous current rather than a current for a short period of time. Mechanisms, which maintain a steady electric field are cells or batteries that we shall study later in this chapter. In the next sections, we shall study the steady current that results from a steady electric field in conductors.

2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE As we learnt in the last chapter, an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5). We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r and p. Further, 49 Reprint 2025-26 Physics the field falls off, at large distance, not as 1/r 2 (typical of field due to a single charge) but as 1/r3. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge. As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q 1  q q  V = − (2.9)FIGURE 2.5 Quantities involved in the calculation 4 πε0  r1 r2  of potential due to a dipole. where r1 and r2 are the distances of the point P from q and –q, respectively. Now, by geometry, r12 = r 2 + a 2 − 2ar cosq r22 = r 2 + a 2 + 2ar cosq (2.10) We take r much greater than a ( r  a ) and retain terms only upto the first order in a/r 2 2  2a cosθ a 2  r1 = r 1 − + 2  r r  2  2a cosθ (2.11) ≅ r  1 − r  Similarly, 2 2  2a cosθ (2.12) r2 ≅ r 1 + r  Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain, 1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 − ≅ 1 + [2.13(a)] r1 r  r  r  r  1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 + ≅ 1 − [2.13(b)] r2 r  r  r  r  Using Eqs. (2.9) and (2.13) and p = 2qa, we get q 2 acosθ p cos θ V = = 4 πε0 r 2 4 πε0r 2 (2.14) 50 Now, p cos q = p.rˆ Reprint 2025-26 Electrostatic Potential and Capacitance where ˆr is the unit vector along the position vector OP. The electric potential of a dipole is then given by 1 p.rˆ V = 2 ; (r >> a) (2.15) 4 πε0 r Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however, exact. From Eq. (2.15), potential on the dipole axis (q = 0, p ) is given by 1 p V = ± 2 (2.16) 4 πε0 r (Positive sign for q = 0, negative sign for q = p.) The potential in the equatorial plane (q = p/2) is zero. The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (2.8) and (2.15): (i) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping q fixed, the points corresponding to P on the cone so generated will have the same potential as at P.) (ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig. 2.5 for graphs of 1/r 2 versus r and 1/r versus r, drawn there in another context.)

2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of C1 has charge –Q and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance C¢ of these three capacitors is given by 1 1 1 1 = + + C ′ C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is  10  C = C¢ + C4 =  3 + 10 mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, Q¢/C4 = 500 V. This gives for the given value of the capacitances, 10 −3 Q = 500 V × µ F = 1.7 × 10 C and EXAMPLE 3 Q ′ = 500 V × 10 µ F = 5.0 × 10 −3 C 2.9

2.5 Relative velocity within its local group of galaxies. Motion is change in position of an object with time. How Summary does the position change with time ? In this chapter, we shall Points to ponder learn how to describe motion. For this, we develop the Exercises concepts of velocity and acceleration. We shall confine ourselves to the study of motion of objects along a straight line, also known as rectilinear motion. For the case of rectilinear motion with uniform acceleration, a set of simple equations can be obtained. Finally, to understand the relative nature of motion, we introduce the concept of relative velocity. In our discussions, we shall treat the objects in motion as point objects. This approximation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time. In a good number of situations in real-life, the size of objects can be neglected and they can be considered as point-like objects without much error. In Kinematics, we study ways to describe motion without going into the causes of motion. What causes motion described in this chapter and the next chapter forms the subject matter of Chapter 4. Reprint 2025-26 14 PHYSICS

2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

2.13 EFFECT OF DIELECTRIC ON CAPACITANCE With the understanding of the behaviour of dielectrics in an external field developed in Section 2.10, let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present. As before, we have two large plates, each of area A, separated by a distance d. The charge on the plates is ±Q, corresponding to the charge density ±s (with s = Q/A). When there is vacuum between the plates, σ E 0 = ε0 69 Reprint 2025-26 Physics and the potential difference V0 is V0 = E0d The capacitance C0 in this case is Q A C 0 = = ε0 (2.46) V 0 d Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and, as explained in Section 2.10, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities sp and –sp. The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(s – sp). That is, σ − σP E = (2.47) ε0 so that the potential difference across the plates is σ − σP V = E d = d (2.48) ε0 For linear dielectrics, we expect sp to be proportional to E0, i.e., to s. Thus, (s – sp) is proportional to s and we can write σ σ − σP = (2.49) K where K is a constant characteristic of the dielectric. Clearly, K > 1. We then have σd Qd V = = (2.50) ε0 K Aε0 K The capacitance C, with dielectric between the plates, is then Q ε0KA C = = (2.51) V d The product e0K is called the permittivity of the medium and is denoted by e e = e0 K (2.52) For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum. The dimensionless ratio ε K = (2.53) ε0 is called the dielectric constant of the substance. As remarked before, from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and (2. 51) C K = (2.54) C 0 Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is 70 inserted fully between the plates of a capacitor. Though we arrived at Reprint 2025-26 Electrostatic Potential and Capacitance Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any type of capacitor and can, in fact, be viewed in general as a definition of the dielectric constant of a substance. Example 2.8 A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? Solution Let E0 = V0/d be the electric field between the plates when there is no dielectric and the potential difference is V0. If the dielectric is now inserted, the electric field in the dielectric will be E = E0/K. The potential difference will then be 1 E 0 3 V = E 0 ( d ) + ( d ) 4 K 4 1 3 K + 3 = E 0 d ( + ) = V 0 4 4 K 4 K The potential difference decreases by the factor (K + 3)/4K while the free charge Q0 on the plates remains unchanged. The capacitance thus increases EXAMPLE Q 0 4 K Q 0 4 K C = = = C 0 V K + 3 V 0 K + 3 2.8

2.6 Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

2.15 ENERGY STORED IN A CAPACITOR A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by 73 Reprint 2025-26 Physics bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Fig 2.30 ). In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges Q¢ and FIGURE 2.30 (a) Work done in a small –Q¢ respectively. At this stage, the potential difference step of building charge on conductor 1 V¢ between conductors 1 to 2 is Q¢/C, where C is thefrom Q¢ to Q¢ + d Q¢. (b) Total work done in charging the capacitor may be capacitance of the system. Next imagine that a small viewed as stored in the energy of charge d Q¢ is transferred from conductor 2 to 1. Work electric field between the plates. done in this step (d W), resulting in charge Q¢ on conductor 1 increasing to Q¢+ d Q¢, is given by Q ′ δW = V ′δQ ′ = δQ ′ (2.68) C Integrating eq. (2.68) Q 2 Q 2 Q ′ 1 Q ′ Q δ Q ’ = = W = ∫ C 2 2C 0 C 0 We can write the final result, in different ways Q 2 1 2 1 W = = CV = QV (2.69) 2C 2 2 Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (2.69)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area A (of each plate) and separation d between the plates]. Energy stored in the capacitor 1 Q 2 ( Aσ)2 d = = × (2.70) 2 C 2 ε0 A The surface charge density s is related to the electric field E between the plates, σ E = (2.71) ε0 From Eqs. (2.70) and (2.71) , we get Energy stored in the capacitor 2 × A d (2.72) U = (1/2 ) ε0 E 74 Reprint 2025-26 Electrostatic Potential and Capacitance Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.72) shows that Energy density of electric field, u =(1/2)e0E 2 (2.73) Though we derived Eq. (2.73) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges. Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system? FIGURE 2.31 Solution (a) The charge on the capacitor is Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C The energy stored by the capacitor is = (1/2) CV 2 = (1/2) QV = (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J (b) In the steady situation, the two capacitors have their positive EXAMPLE plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V¢. The 2.10 75 Reprint 2025-26 Physics charge on each capacitor is then Q¢ = CV¢. By charge conservation, Q¢ = Q/2. This implies V¢ = V/2. The total energy of the system is 1 1 − 6 = 2 × Q ' V ' = QV = 2.25 × 10 J 2 4 Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone? 2.10 There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in EXAMPLE the form of heat and electromagnetic radiation. SUMMARY 1. Electrostatic force is a conservative force. Work done by an external force (equal and opposite to the electrostatic force) in bringing a charge q from a point R to a point P is q(VP–VR), which is the difference in potential energy of charge q between the final and initial points. 2. Potential at a point is the work done per unit charge (by an external agency) in bringing a charge from infinity to that point. Potential at a point is arbitrary to within an additive constant, since it is the potential difference between two points which is physically significant. If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by 1 Q V ( r ) = 4 πεo r 3. The electrostatic potential at a point with position vector r due to a point dipole of dipole moment p placed at the origin is 1 p.rˆ V ( r ) = 2 4 πεo r The result is true also for a dipole (with charges –q and q separated by 2a) for r >> a. 4. For a charge configuration q1, q2, ..., qn with position vectors r1, r2, ... rn, the potential at a point P is given by the superposition principle 1 q1 q 2 qn V = ( + + ... + ) 4 πε0 r1P r2P rnP where r1P is the distance between q1 and P, as and so on. 5. An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential. Reprint 2025-26 Electrostatic Potential and Capacitance 6. Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by 1 q1 q 2 U = 4 πε0 r12 where r12 is distance between q1 and q2. 7. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E. 8. Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by σ E = nˆ where ˆn is the unit vector along the outward normal to the ε0 surface and s is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero. 9. A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V –1. For a parallel plate capacitor (with vacuum between the plates), A C = ε0 d where A is the area of each plate and d the separation between them. 10. If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), C = KC0 where K is the dielectric constant of the insulating substance. 11. For capacitors in the series combination, the total capacitance C is given by 1 1 1 1 = + + + ... C C1 C 2 C 3 In the parallel combination, the total capacitance C is: C = C1 + C2 + C3 + ... where C1, C2, C3... are individual capacitances. 77 Reprint 2025-26 Physics 12. The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is 1 1 2 1 Q 2 U = QV = CV = 2 2 2 C The electric energy density (energy per unit volume) in a region with electric field is (1/2)e0E2. Physical quantity Symbol Dimensions Unit Remark Potential or V [M1 L2 T–3 A–1] V Potential difference is physically significant Capacitance C [M–1 L–2 T–4 A2] F Polarisation P [L–2 AT] C m-2 Dipole moment per unit volume Dielectric constant K [Dimensionless] POINTS TO PONDER 1. Electrostatics deals with forces between charges at rest. But if there is a force on a charge, how can it be at rest? Thus, when we are talking of electrostatic force between charges, it should be understood that each charge is being kept at rest by some unspecified force that opposes the net Coulomb force on the charge. 2. A capacitor is so configured that it confines the electric field lines within a small region of space. Thus, even though field may have considerable strength, the potential difference between the two conductors of a capacitor is small. 3. Electric field is discontinuous across the surface of a spherical charged ˆn outside. Electric potential is, however shell. It is zero inside and σε0 continuous across the surface, equal to q/4pe0R at the surface. 4. The torque p × E on a dipole causes it to oscillate about E. Only if there is a dissipative mechanism, the oscillations are damped and the dipole eventually aligns with E. 5. Potential due to a charge q at its own location is not defined – it is infinite. 6. In the expression qV (r) for potential energy of a charge q, V (r) is the potential due to external charges and not the potential due to q. As seen in point 5, this expression will be ill-defined if V (r) includes potential 78 due to a charge q itself. Reprint 2025-26 Electrostatic Potential and Capacitance 7. A cavity inside a conductor is shielded from outside electrical influences. It is worth noting that electrostatic shielding does not work the other way round; that is, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. EXERCISES

2.10 DIELECTRICS AND POLARISATION Dielectrics are non-conducting substances. In contrast to conductors, they have no (or negligible number of ) charge carriers. Recall from Section

2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES Consider first the simple case of two charges q1and q2 with position vector r1 and r2 relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1 is zero. This charge produces a potential in space given by 1 q1 V1 = 4 πε0 r1P where r1P is the distance of a point P in space from the location of q1. From the definition of potential, work done in bringing charge q2 from infinity to the point r2 is q2 times the potential at r2 due to q1: 1 q1q 2 work done on q2 = 4 πε0 r12 55 Reprint 2025-26 Physics where r12 is the distance between points 1 and 2. Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q1 and q2 is FIGURE 2.13 Potential energy of a 1 q1q 2 U = system of charges q1 and q2 is 4 πε0 r12 (2.22) directly proportional to the product of charges and inversely to the Obviously, if q2 was brought first to its present location and distance between them. q1 brought later, the potential energy U would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force. Equation (2.22) is true for any sign of q1and q2. If q1q2 > 0, potential energy is positive. This is as expected, since for like charges (q1q2 > 0), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (q1 q2 < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative. Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first from infinity to r1, no work is required. Next we bring q2 from infinity to r2. As before, work done in this step is 1 q1q 2 q 2 V1 ( r2 ) = (2.23) 4 πε0 r12 The charges q1 and q2 produce a potential, which at any point P is given by 1  q1 q 2  V1, 2 = + (2.24) 4 πε0  r1P r2 P  Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3 1  q1q 3 q 2 q 3  q 3 V1, 2 ( r3 ) = + (2.25) 4 πε0  r13 r23  The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)], 1  q1q 2 q1q 3 q 2 q 3  U = + + (2.26) FIGURE 2.14 Potential energy of a 4 πε0  r12 r13 r23  system of three charges is given by Again, because of the conservative nature of the Eq. (2.26), with the notation given electrostatic force (or equivalently, the path in the figure. independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which 56 the configuration is assembled. The potential energy Reprint 2025-26 Electrostatic Potential and Capacitance is characteristic of the present state of configuration, and not the way the state is achieved. Example 2.4 Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge q0 is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this? FIGURE 2.15 Solution (a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps: (i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero. (ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)  q  q 2 = −q × − ε  4 πε0 d = 4 π 0 d (iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)  + q −q  = + q + ε 0 d   4 πε0 d 2 4 π −q 2  1  = 4 πε0 d 1 − 2  (iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)  + q −q q  = −q + + ε  4 π 0 d 4 πε0 d 2 4 πε0 d  EXAMPLE −q 2  1  = 4 πε0 d  2 − 2  2.4 57 Reprint 2025-26 Physics Add the work done in steps (i), (ii), (iii) and (iv). The total work required is −q 2   1   1   =  4 πε0 d ( 0 ) + (1) + 1 − 2 +  2 − 2    −q 2 = 4 − 2 ( ) 4 πε0 d The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.) 2.4 the(b) Thefourextrachargesworkarenecessaryat A, B, CtoandbringD ais chargeq0 × (electrostaticq0 to the pointpotentialE whenat E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to EXAMPLE point E. 2.8 POTENTIAL ENERGY IN AN EXTERNAL FIELD 2.8.1 Potential energy of a single charge In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined. In this section, we ask a related but a distinct question. What is the potential energy of a charge q in a given field? This question was, in fact, the starting point that led us to the notion of the electrostatic potential (Sections 2.1 and 2.2). But here we address this question again to clarify in what way it is different from the discussion in Section 2.7. The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field. The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest. Note again that we are interested in determining the potential energy of a given charge q (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field. The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work 58 done in bringing a unit positive charge from infinity to the point P. Reprint 2025-26 Electrostatic Potential and Capacitance (We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write: Potential energy of q at r in an external field = qV(r) (2.27) where V(r) is the external potential at the point r. Thus, if an electron with charge q = e = 1.6×10–19 C is accelerated by a potential difference of DV = 1 volt, it would gain energy of qDV = 1.6 × 10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV=1.6 × 10–19J. The units based on eV are most commonly used in atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV = 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV = 1.6 × 10–7J). [This has already been defined on Page 117, XI 2.8.2 Potential energy of a system of two charges in an external field Next, we ask: what is the potential energy of a system of two charges q1 and q2 located at r1and r2, respectively, in an external field? First, we calculate the work done in bringing the charge q1 from infinity to r1. Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the work done in bringing q2 to r2. In this step, work is done not only against the external field E but also against the field due to q1. Work done on q2 against the external field = q2 V (r2) Work done on q2 against the field due to q1 q1q 2 = 4 πεo r12 where r12 is the distance between q1 and q2. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on q2 against the two fields (E and that due to q1): Work done in bringing q2 to r2 q1q 2 = q 2 V ( r2 ) + (2.28) 4 πεo r12 Thus, Potential energy of the system = the total work done in assembling the configuration q1q 2 = q1V ( r1 ) + q 2 V ( r2 ) + (2.29) 4 πε0r12 Example 2.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 mC and –2 mC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. EXAMPLE (b) How much work is required to separate the two charges infinitely 2.5 away from each other? 59 Reprint 2025-26 Physics (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2. What would the electrostatic energy of the configuration be? Solution 1 q1q 2 9 7 × ( −2) × 10 −12 (a) U = = 9 × 10 × = –0.7 J. 4 πε0 r 0.18 (b) W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J. (c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find, 7 µ C −µ2 C q1V ( r1 ) + q 2 V ( r2 ) = A + A 0.09m 0.09m 2.5 and the net electrostatic energy is q1q 2 7 µC −µ2 C q1V ( r1 ) + q 2 V ( r2 ) + = A + A − 0.7 J 4 πε0r12 0.09 m 0.09 m EXAMPLE = 70 − 20 − 0.7 = 49.3 J 2.8.3 Potential energy of a dipole in an external field Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig. 2.16. As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque ttttt given by ttttt ===== p × E (2.30) which will tend to rotate it (unless p is parallel or antiparallel to E). Suppose an external torque text is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle q0 to angle q1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by FIGURE 2.16 Potential energy of a dipole in a uniform external field. = pE ( cosθ0 − cosθ1 ) (2.31) This work is stored as the potential energy of the system. We can then associate potential energy U(q) with an inclination q of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take q0 = p / 2. (An explanation for it is provided towards the end of discussion.) We can then write, (2.32) 60 Reprint 2025-26 Electrostatic Potential and Capacitance This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges +q and –q. The potential energy expression then reads q 2 U ′ (θ) = q [V ( r1 ) − V ( r2 )] − (2.33) 4 πε0 × 2a Here, r1 and r2 denote the position vectors of +q and –q. Now, the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1. The displacement parallel to the force is 2a cosq. Thus, [V(r1)–V (r2)] = –E × 2a cosq . We thus obtain, q 2 q 2 U ′ (θ) = − pE cosθ− = − p.E − (2.34) 4 πε0 × 2a 4 πε0 × 2a We note that U¢(q) differs from U(q ) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32). We can now understand why we took q0=p/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0. Example 2.6 A molecule of a substance has a permanent electric dipole moment of magnitude 10–29 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample. Solution Here, dipole moment of each molecules = 10–29 C m As 1 mole of the substance contains 6 × 1023 molecules, total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m = 6 × 10–6 C m Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J Change in potential energy = –3 J – (–6J) = 3 J EXAMPLE So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles. 2.6

3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. R = A + B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos θ SN = B sin θ OS2 = (A + B cos θ)2 + (B sin θ)2 or, R2 = A2 + B2 + 2AB cos θ R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 The average acceleration for three time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and the acceleration of an object are always along -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  direction or in the opposite direction). ⊳ However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (b) The time taken to return to the same level is Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r′ be the position vectors and v and r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 3.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 3.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 3.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? Reprint 2025-26 48 PHYSICS 3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 3.16 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 3.17 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 2.0 j m s-2. (a) At what time is ( ) the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t1 to t2) 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is 2 2 2 2 −1 −1 R = v r + v w = 35 + 12 m s = 37 m s The direction θ that R makes with the vertical is given by v w 12 tan θ = = = 0.343 Fig. 3.8 (a) Two non-colinear vectors a and b. v r 35 (b) Resolving a vector A in terms of vectors Or, θ = tan-1 ( 0.343 ) = 19° a and b. Therefore, the boy should hold his umbrella We say that A has been resolved into two in the vertical plane at an angle of about 19o component vectors λ a and µ b along a and b with the vertical towards the east. ⊳ Reprint 2025-26 32 PHYSICS respectively. Using this method one can resolve and A2 is parallel to ɵj, we have :a given vector into two component vectors along a set of two vectors – all the three lie in the same A1= Ax ɵi , A2 = Ay ɵj (3.11) plane. It is convenient to resolve a general vector where Ax and Ay are real numbers.along the axes of a rectangular coordinate system using vectors of unit magnitude. These Thus, A = Ax ɵi + Ay ɵj (3.12) are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and This is represented in Fig. 3.9(c). The quantities points in a particular direction. It has no Ax and Ay are called x-, and y- components of the dimension and unit. It is used to specify a vector A. Note that Ax is itself not a vector, but direction only. Unit vectors along the x-, y- and ɵi is a vector, and so is Ay ɵj. Using simplez-axes of a rectangular coordinate system are Ax trigonometry, we can express Ax and Ay in terms denoted by ɵi , ɵj and ˆk , respectively, as shown of the magnitude of A and the angle θ it makes in Fig. 3.9(a). with the x-axis : Since these are unit vectors, we have Ax = A cos θ Ay = A sin θ (3.13) ˆi  = ˆj  = ˆk =1 (3.9) As is clear from Eq. (3.13), a component of a These unit vectors are perpendicular to each vector can be positive, negative or zero other. In this text, they are printed in bold face depending on the value of θ. with a cap (^) to distinguish them from other Now, we have two ways to specify a vector A vectors. Since we are dealing with motion in two in a plane. It can be specified by : dimensions in this chapter, we require use of (i) its magnitude A and the direction θ it makes only two unit vectors. If we multiply a unit vector, with the x-axis; or say ˆn by a scalar, the result is a vector (ii) its components Ax and Ay λ = λ ˆn. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained using Eq. (3.13). If Ax and Ay are given, A and θ A = |A| ˆn (3.10) can be obtained as follows : where ˆn is a unit vector along A. 2 2 2 2 2 2 A x + A y = A cos θ + A sin θ We can now resolve a vector A in terms = A2 of component vectors that lie along unit vectors iˆ and ɵj. Consider a vector A that lies in x-y Or, A = A 2x + Ay2 (3.14) plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate Ay A y tan θ = , θ = tan− 1 axes as in Fig. 3.9(b), and get vectors A1 and A2 And A x A x (3.15) such that A1 + A2 = A. Since A1 is parallel to ɵi Fig. 3.9 (a) Unit vectors ɵi , ɵj and ɵk lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of ɵi and ɵj . Reprint 2025-26 MOTION IN A PLANE 33 So far we have considered a vector lying in ɵ ɵ B = B x i + B y jan x-y plane. The same procedure can be used Let R be their sum. We haveto resolve a general vector A into three components along x-, y-, and z-axes in three R = A + B dimensions. If α, β, and γ are the angles* ɵ ɵ ɵ ɵ = + (3.19a) ( A x i + A y j ) ( B x i + B y j )between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : ɵ ɵ j (3.19b) R = ( A x + B x ) i + ( A y + B y ) ɵ ɵ SinceR = R x i + R y j (3.20) we have, R x = A x + B x , R y = A y + B y (3.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have ɵ ɵ ɵ A = A x i + Ay j + A z k ɵ ɵ ɵ B = B x i + B y j + B z k (d) ɵ ɵ ɵ R = A + B = R x i + R y j + R z kFig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes with R x = A x + B x y = A y + B yA x = A cos α, A y = A cos β, A z = A cos γ (3.16a) R In general, we have R z = A z + B z (3.22) A = Ax ˆi + Ay ˆj + Az kˆ (3.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = A x2 + Ay2 + A z2 (3.16c) example, if vectors a, b and c are given as ɵ ɵ ɵ A position vector r can be expressed as a = a x i + a y j + a z k ɵ ɵ ɵ r = x i + y j + z k (3.17) ɵ ɵ ɵ b = b x i + b y j + b z k where x, y, and z are the components of r along ɵ ɵ ɵ x-, y-, z-axes, respectively. c = c x i + c y j + c z k (3.23a)

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1. 105 Reprint 2025-26 Physics 3.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment? 3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver. 3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1. 3.7 Determine the current in each branch of the network shown in Fig. 3.20: FIGURE 3.20 3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? 3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Reprint 2025-26 Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related. During a lecture demonstration in the summer of 1820, Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Following this, there was intense experimentation. In 1864, the laws obeyed by electricity and magnetism were unified and formulated by Reprint 2025-26 Physics James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress took place in the 20th century. This was due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer.(1777–1851) In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot (¤). A current or a field going into the plane of the paper is depicted by a cross ()*. Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these twoOERSTED (1777–1851) Danish situations, respectively. physicist and chemist, professor at Copenhagen. 4.2 MAGNETIC FORCE He observed that a compass needle suffers a 4.2.1 Sources and fields deflection when placed Before we introduce the concept of a magnetic field B, weCHRISTIAN near a wire carrying an electric current. This shall recapitulate what we have learnt in Chapter 1 about discovery gave the first the electric field E. We have seen that the interaction empirical evidence of a between two charges can be considered in two stages.HANS connection between electric The charge Q, the source of the field, produces an electric and magnetic phenomena. field E, where * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered 108 tail of an arrow moving away from you. Reprint 2025-26 Moving Charges and Magnetism E = Q ˆr / (4pe0)r2 (4.1) where ˆr is unit vector along r, and the field E is a vector field. A charge q interacts with this field and experiences a force F given by F = q E = q Q ˆr / (4pe0) r 2 (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a function of time. In our Hendrik Antoon Lorentz discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated themore charges. If there are more charges the fields add HENDRIK relationship between vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and is called the principle of superposition. Once the field is mechanics. In order to known, the force on a test charge is given by Eq. (4.2). explain the observed effect Just as static charges produce an electric field, the of magnetic fields on ANTOONcurrents or moving charges produce (in addition) a emitters of light (Zeeman magnetic field, denoted by B (r), again a vector field. It effect), he postulated the existence of electric chargeshas several basic properties identical to the electric field. in the atom, for which he It is defined at each point in space (and can in addition was awarded the Nobel Prize depend on time). Experimentally, it is found to obey the in 1902. He derived a set of LORENTZprinciple of superposition: the magnetic field of several transformation equations sources is the vector addition of magnetic field of each (known after him, as individual source. Lorentz transformation equations) by some tangled (1853 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, – but he was not aware thatLet us suppose that there is a point charge q (moving these equations hinge on a with a velocity v and, located at r at a given time t) in new concept of space andpresence of both the electric field E (r) and the magnetic 1928) time. field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] º Felectric +Fmagnetic (4.3) This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity 109 and magnetic field. The vector product makes the force due to magnetic Reprint 2025-26 Physics field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q ˆn , where q is force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when magnetic field B is given by the right-hand the force acting on a unit charge (1 C), moving rule. (b) A moving charged particle q is perpendicular to B with a speed 1m/s, is one deflected in an opposite sense to –q in the newton. presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit of B are Newton second / (coulomb metre). This unit is called tesla (T) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nlA)q vd ´ B where q is the value of the charge on a carrier. Now nq vd is the current density j and |(nq vd)|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nq vd )lA] × B = [ jAl ] ´ B = Il ´ B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F  Idl j × B j This summation can be converted to an integral in most cases. Reprint 2025-26 Moving Charges and Magnetism Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB m g B = I l Interactive Charged 0.2 × 9.8 = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per EXAMPLE particles unit length of the wire. The earth’s 4 × 10–5 T and we have ignored it. magnetic field is approximately 4.1 moving demonstration: in a Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative magnetic charge), (b) a proton (positive charge). field. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html FIGURE 4.4 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z EXAMPLE axis. (b) for a positive charge (proton) the force is along +z axis. 4.2 111 Reprint 2025-26 Physics 4.3 MOTION IN A MAGNETIC FIELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 5) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, qE, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 3) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the FIGURE 4.5 Circular motion velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB (4.5) for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If w is the angular frequency, then v = w r. So, w = 2p n = q B/ m [4.6(a)] which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. The time taken for one revolution is T= 2p/ w º 1/n. If there is a component of the velocity FIGURE 4.6 Helical motion parallel to the magnetic field (denoted by v||), 112 it will make the particle move along the field and the path of the Reprint 2025-26 Moving Charges and Magnetism particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v||T = 2pm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T) = 28 × 10–2 m = 28 cm n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz. 2 EXAMPLE E = (½ )mv = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J ≈ 4×10–16 J = 2.5 keV. 4.3

3.10 CELLS, EMF, INTERNAL RESISTANCE We have already mentioned that a simple device to maintain a steady current in an electric circuit is the electrolytic cell. Basically a cell has two electrodes, called the positive (P) and the negative (N), as shown in 93 Reprint 2025-26 Physics Fig. 3.12. They are immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte. The positive electrode has a potential difference V+ (V+ > 0) between itself and the electrolyte solution immediately adjacent to it marked A in the figure. Similarly, the negative electrode develops a negative potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it, marked as B in the figure. When there is no current, the electrolyte has the same potential throughout, so that the potential difference between P and N is V+ – (–V–) = V+ + V– . This difference is called the electromotive force (emf) of the cell and is denoted by e. Thus e = V++V– > 0 (3.36) Note that e is, actually, a potential difference and not a force. The name emf, however, is used because of historical reasons, and was given at a time when the phenomenon was not understood properly. To understand the significance of e, consider a resistor R connected across the cell (Fig. 3.12). A current I flows across R FIGURE 3.12 (a) Sketch of from C to D. As explained before, a steady current is maintained an electrolyte cell with because current flows from N to P through the electrolyte. Clearly, positive terminal P and across the electrolyte the same current flows through the electrolyte negative terminal N. The but from N to P, whereas through R, it flows from P to N. gap between the electrodes The electrolyte through which a current flows has a finite is exaggerated for clarity. A resistance r, called the internal resistance. Consider first the and B are points in the situation when R is infinite so that I = V/R = 0, where V is the electrolyte typically close to potential difference between P and N. Now, P and N. (b) the symbol for V = Potential difference between P and A a cell, + referring to P and – referring to the N + Potential difference between A and B electrode. Electrical + Potential difference between B and N connections to the cell are = e (3.37) made at P and N. Thus, emf e is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell. If however R is finite, I is not zero. In that case the potential difference between P and N is V = V++ V– – I r = e – I r (3.38) Note the negative sign in the expression (I r) for the potential difference between A and B. This is because the current I flows from B to A in the electrolyte. In practical calculations, internal resistances of cells in the circuit may be neglected when the current I is such that e >> I r. The actual values of the internal resistances of cells vary from cell to cell. The internal resistance of dry cells, however, is much higher than the common electrolytic cells. We also observe that since V is the potential difference across R, we have from Ohm’s law V = I R (3.39) 94 Combining Eqs. (3.38) and (3.39), we get Reprint 2025-26 Current Electricity I R = e – I r ε Or, I = (3.40) R + r The maximum current that can be drawn from a cell is for R = 0 and it is Imax = e/r. However, in most cells the maximum allowed current is much lower than this to prevent permanent damage to the cell. 3.113.113.113.113.11 CELLSELLSELLSELLSELLS INININININ SSSERIESSSERIESERIESERIESERIES ANDANDANDANDAND INININININ PPPARALLELPPARALLELARALLELARALLELARALLEL Like resistors, cells can be combined together in an electric circuit. And like resistors, one can, for calculating currents and voltages in a circuit, replace a combination of cells by an equivalent cell. FIGUREFIGUREFIGUREFIGUREFIGURE 3.133.133.133.133.13 Two cells of emf’s e1 and e2 in the series. r1, r2 are their internal resistances. For connections across A and C, the combination can be considered as one cell of emf eeq and an internal resistance req. Consider first two cells in series (Fig. 3.13), where one terminal of the two cells is joined together leaving the other terminal in either cell free. e1, e2 are the emf’s of the two cells and r1, r2 their internal resistances, respectively. Let V (A), V (B), V (C) be the potentials at points A, B and C shown in Fig. 3.13. Then V (A) – V (B) is the potential difference between the positive and negative terminals of the first cell. We have already calculated it in Eq. (3.38) and hence, (3.41) V AB ≡ V ( A ) – V ( B ) = ε1 – I r1 Similarly, V BC ≡ V ( B ) – V ( C ) = ε2 – I r2 (3.42) Hence, the potential difference between the terminals A and C of the combination is V AC ≡ V (A)– V (C) = V ( A ) – V ( B ) + V ( B ) – V ( C ) = (ε1 + ε2 ) – I (r1 + r2 ) (3.43) If we wish to replace the combination by a single cell between A and C of emf eeq and internal resistance req, we would have VAC = eeq– I req (3.44) Comparing the last two equations, we get eeq = e1 + e2 (3.45) and req = r1 + r2 (3.46) In Fig.3.13, we had connected the negative electrode of the first to the 95 positive electrode of the second. If instead we connect the two negatives, Reprint 2025-26 Physics Eq. (3.42) would change to VBC = –e2–Ir2 and we will get eeq = e1 – e2 (e1 > e2) (3.47) The rule for series combination clearly can be extended to any number of cells: (i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and (ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances. This is so, when the current leaves each cell from the positive electrode. If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for eeq with a negative sign, as in Eq. (3.47). Next, consider a parallel combination of the cells (Fig. 3.14). I1 and I2 are the currents leaving the positive electrodes of the cells. At the point B1, I1 and I2 flow in whereas the current I flows out. Since as much charge flows in as out, we have FIGURE 3.14 Two cells in I = I1 + I2 (3.48) parallel. For connections Let V (B1) and V (B2) be the potentials at B1 and B2, respectively. across A and C, the Then, considering the first cell, the potential difference across its combination can be terminals is V (B1) – V (B2). Hence, from Eq. (3.38) replaced by one cell of emf eeq and internal resistances V ≡ V ( B1 ) – V ( B 2 ) = ε1 – I 1r1 (3.49) req whose values are given in Points B1 and B2 are connected exactly similarly to the second Eqs. (3.54) and (3.55). cell. Hence considering the second cell, we also have V ≡ V ( B1 ) – V ( B 2 ) = ε2 – I 2r2 (3.50) Combining the last three equations I = I 1 + I 2 ε 1 – V ε2 – V  ε1 ε2   1 1  = + = + – V + (3.51) r1 r2  r1 r2   r1 r2  Hence, V is given by, ε1r2 + ε2r1 r1r2 V = – I (3.52) r1 + r2 r1 + r2 If we want to replace the combination by a single cell, between B1 and B2, of emf eeq and internal resistance req, we would have V = eeq – I req (3.53) The last two equations should be the same and hence ε1r2 + ε2r1 εeq = (3.54) r1 + r2 r1r2 req = (3.55) r1 + r2 We can put these equations in a simpler way, 96 Reprint 2025-26 Current Electricity 1 1 1 = + (3.56) req r1 r2 εeq ε1 ε2 = + (3.57) req r1 r2 In Fig. (3.14), we had joined the positive terminals together and similarly the two negative ones, so that the currents I1, I2 flow out of positive terminals. If the negative GUSTAVterminal of the second is connected to positive terminal of the first, Eqs. (3.56) and (3.57) would still be valid with e ® 2 –e2 Equations (3.56) and (3.57) can be extended easily. If there are n cells of emf e1, . . . en and of internal ROBERT resistances r1,... rn respectively, connected in parallel, the Gustav Robert Kirchhoff combination is equivalent to a single cell of emf eeq and (1824 – 1887) German internal resistance req, such that physicist, professor at Heidelberg and at 1 1 1 = + ... + Berlin. Mainly known for (3.58) req r1 rn his development of spectroscopy, he also KIRCHHOFF ε eq ε1 εn made many important = + ... + (3.59) contributions to mathe- req r1 rn matical physics, among them, his first and (1824 –3.12 KIRCHHOFF’S RULES second rules for circuits. Electric circuits generally consist of a number of resistors and cells interconnected sometimes in a complicated way. The formulae we have derived earlier for series and parallel combinations of resistors are not always sufficient to determine all the currents and potential differences in the circuit. Two rules, called Kirchhoff’s rules, are very useful for analysis of electric circuits. Given a circuit, we start by labelling currents in each resistor by a symbol, say I, and a directed arrow to indicate that a current I flows along the resistor in the direction indicated. If ultimately I is determined to be positive, the actual current in the resistor is in the direction of the arrow. If I turns out to be negative, the current actually flows in a direction opposite to the arrow. Similarly, for each source (i.e., cell or some other source of electrical power) the positive and negative electrodes are labelled, as well as, a directed arrow with a symbol for the current flowing through the cell. This will tell us the potential difference, V = V (P) – V (N) = e – I r [Eq. (3.38) between the positive terminal P and the negative terminal N; I here is the current flowing from N to P through the cell]. If, while labelling the current I through the cell one goes from P to N, then of course V = e + I r (3.60) Having clarified labelling, we now state the rules and the proof: (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction (Fig. 3.15). 97 Reprint 2025-26 Physics This applies equally well if instead of a junction of several lines, we consider a point in a line. The proof of this rule follows from the fact that when currents are steady, there is no accumulation of charges at any junction or at any point in a line. Thus, the total current flowing in, (which is the rate at which charge flows into the junction), must equal the total current flowing out. (b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero FIGURE 3.15 At junction a the current (Fig. 3.15). leaving is I1 + I2 and current entering is I3. This rule is also obvious, since electric potential is The junction rule says I3 = I1 + I2. At point dependent on the location of the point. Thus h current entering is I1. There is only one starting with any point if we come back to the same current leaving h and by junction rule point, the total change must be zero. In a closed that will also be I1. For the loops ‘ahdcba’ loop, we do come back to the starting point and and ‘ahdefga’, the loop rules give –30I1 – hence the rule. 41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0. Example 3.5 A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 W (Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube. Z 3.5 FIGURE 3.16 EXAMPLE 98 Reprint 2025-26 Current Electricity Solution The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. The paths AA¢, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A¢, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second rule: –IR – (1/2)IR – IR + e = 0 where R is the resistance of each edge and e the emf of battery. Thus, 5 e = I R 2 The equivalent resistance Req of the network is ε 5 R eq = = R Similation 3 I 6 for For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in the network is 3I = 10 V/(5/6) W = 12 A, i.e., I = 4 A EXAMPLE The current flowing in each edge can now be read off from the Fig. 3.16. 3.5 application of It should be noted that because of the symmetry of the network, the great power of Kirchhoff’s rules has not been very apparent in Example 3.5. In a general network, there will be no such simplification due to symmetry,and only by application of Kirchhoff’s rules to junctions and closed loops Kirchhoff’s (as many as necessary to solve the unknowns in the network) can we handle the problem. This will be illustrated in Example 3.6. rules: Example 3.6 Determine the current in each branch of the network http://www.phys.hawaii.edu/~teb/optics/java/kirch3/ shown in Fig. 3.17. EXAMPLE FIGURE 3.17 3.6 99 Reprint 2025-26 Physics Solution Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns I1, I2 and I3 which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop ADCA gives, 10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0 [3.61(a)] that is, 7I1– 6I2 – 2I3 = 10 For the closed loop ABCA, we get 10 – 4I2– 2 (I2 + I3) – I1 = 0 that is, I1 + 6I2 + 2I3 =10 [3.61(b)] For the closed loop BCDEB, we get 5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0 that is, 2I1 – 4I2 – 4I3 = –5 [3.61(c)] Equations (3.61 a, b, c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give 5 7 I1 = 2.5A, I2 = A, I3 = A 8 18 The currents in the various branches of the network are 5 1 7 AB : A, CA : 2 A, DEB : 1 A 8 2 8 7 1 AD : 18 A, CD : 0 A, BC : 2 2 A It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage 3.6 drop over the closed loop BADEB  5   15  5 V +  8 × 4  V −  8 × 4  V EXAMPLE equal to zero, as required by Kirchhoff’s second rule. 3.13 WHEATSTONE BRIDGE As an application of Kirchhoff’s rules consider the circuit shown in Fig. 3.18, which is called the Wheatstone bridge. The bridge has four resistors R1, R2, R3 and R4. Across one pair of diagonally opposite points (A and C in the figure) a source is connected. This (i.e., AC) is called the battery arm. Between the other two vertices, B and D, a galvanometer G (which is a device to detect currents) is connected. This line, shown as BD in the figure, is called the galvanometer arm. For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current Ig through G. Of special interest, is the case of a balanced bridge where the resistors are such that Ig = 0. We can easily get the balance condition, such that there is no current through G. In this case, the 100 Kirchhoff’s junction rule applied to junctions D and B (see the figure) Reprint 2025-26 Current Electricity immediately gives us the relations I1 = I3 and I2 = I4. Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives –I1 R1 + 0 + I2 R2 = 0 (Ig = 0) (3.62) and the second loop gives, upon using I3 = I1, I4 = I2 I2 R4 + 0 – I1 R3 = 0 (3.63) From Eq. (3.62), we obtain, I1 R 2 = I 2 R1 whereas from Eq. (3.63), we obtain, I1 R 4 = I 2 R 3 Hence, we obtain the condition R 2 R 4 FIGURE 3.18 = [3.64(a)] R 1 R 3 This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection. The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R4 is thus not known. Keeping known resistances R1 and R2 in the first and second arm of the bridge, we go on varying R3 till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R4 is given by, R 2 R 4 = R 3 [3.64(b)] R1 A practical device using this principle is called the meter bridge. Example 3.7 The four arms of a Wheatstone bridge (Fig. 3.19) have the following resistances: AB = 100W, BC = 10W, CD = 5W, and DA = 60W. EXAMPLE FIGURE 3.19 3.7 101 Reprint 2025-26 Physics A galvanometer of 15W resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. Solution Considering the mesh BADB, we have 100I1 + 15Ig – 60I2 = 0 or 20I1 + 3Ig – 12I2= 0 [3.65(a)] Considering the mesh BCDB, we have 10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0 10I1 – 30Ig –5I2 = 0 2I1 – 6Ig – I2 = 0 [3.65(b)] Considering the mesh ADCEA, 60I2 + 5 (I2 + Ig) = 10 65I2 + 5Ig = 10 13I2 + Ig = 2 [3.65(c)] Multiplying Eq. (3.65b) by 10 20I1 – 60Ig – 10I2 = 0 [3.65(d)] From Eqs. (3.65d) and (3.65a) we have 63Ig – 2I2 = 0 I2 = 31.5Ig [3.65(e)] 3.7 Substituting the value of I2 into Eq. [3.65(c)], we get 13 (31.5Ig ) + Ig = 2 410.5 Ig = 2 EXAMPLE Ig = 4.87 mA. SUMMARY 1. Current through a given area of a conductor is the net charge passing per unit time through the area. 2. To maintain a steady current, we must have a closed circuit in which an external agency moves electric charge from lower to higher potential energy. The work done per unit charge by the source in taking the charge from lower to higher potential energy (i.e., from one terminal of the source to the other) is called the electromotive force, or emf, of the source. Note that the emf is not a force; it is the voltage difference between the two terminals of a source in open circuit. 3. Ohm’s law: The electric current I flowing through a substance is proportional to the voltage V across its ends, i.e., V µ I or V = RI, where R is called the resistance of the substance. The unit of resistance is ohm: 1W = 1 V A–1. Reprint 2025-26 Current Electricity 4. The resistance R of a conductor depends on its length l and cross-sectional area A through the relation, ρl R = A where r, called resistivity is a property of the material and depends on temperature and pressure. 5. Electrical resistivity of substances varies over a very wide range. Metals have low resistivity, in the range of 10–8 W m to 10–6 W m. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale. 6. In most substances, the carriers of current are electrons; in some cases, for example, ionic crystals and electrolytic liquids, positive and negative ions carry the electric current. 7. Current density j gives the amount of charge flowing per second per unit area normal to the flow, j = nq vd where n is the number density (number per unit volume) of charge carriers each of charge q, and vd is the drift velocity of the charge carriers. For electrons q = – e. If j is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nevd A. 8. Using E = V/l, I = nevd A, and Ohm’s law, one obtains eE ne 2 = ρ v d m m The proportionality between the force eE on the electrons in a metal due to the external field E and the drift velocity vd (not acceleration) can be understood, if we assume that the electrons suffer collisions with ions in the metal, which deflect them randomly. If such collisions occur on an average at a time interval t, vd = at = eEt/m where a is the acceleration of the electron. This gives m ρ= 2 ne τ 9. In the temperature range in which resistivity increases linearly with temperature, the temperature coefficient of resistivity a is defined as the fractional increase in resistivity per unit increase in temperature. 10. Ohm’s law is obeyed by many substances, but it is not a fundamental law of nature. It fails if (a) V depends on I non-linearly. (b) the relation between V and I depends on the sign of V for the same absolute value of V. (c) The relation between V and I is non-unique. An example of (a) is when r increases with I (even if temperature is kept fixed). A rectifier combines features (a) and (b). GaAs shows the feature (c). 11. When a source of emf e is connected to an external resistance R, the voltage Vext across R is given by ε R Vext = IR = R + r where r is the internal resistance of the source. 103 Reprint 2025-26 Physics 12. Kirchhoff’s Rules – (a) Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. (b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero. 13. The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4 as shown in the text. The null-point condition is given by R1 R 3 = R 2 R 4 using which the value of one resistance can be determined, knowing the other three resistances. Physical Quantity Symbol Dimensions Unit Remark Electric current I [A] A SI base unit Charge Q, q [T A] C Voltage, Electric V [M L2 T –3 A –1] V Work/charge potential difference Electromotive force e [M L2 T –3 A –1] V Work/charge Resistance R [M L2 T –3 A –2] W R = V/I Resistivity r [M L3 T –3 A –2] W m R = rl/A Electrical s [M –1 L–3 T 3 A 2] S s = 1/r conductivity –3 –1 Electric force Electric field E [M L T A–1] V m charge –1 e E τ Drift speed vd [L T –1] m s vd = m Relaxation time t [T] s Current density j [L–2 A] A m–2 current/area Mobility m [M L3 T –4 A –1] m 2 V –1s –1 v d / E POINTS TO PONDER 1. Current is a scalar although we represent current with an arrow. Currents do not obey the law of vector addition. That current is a scalar also follows from it’s definition. The current I through an area of cross-section is given by the scalar product of two vectors: I = j . DS 104 where j and DS are vectors. Reprint 2025-26 Current Electricity 2. Refer to V-I curves of a resistor and a diode as drawn in the text. A resistor obeys Ohm’s law while a diode does not. The assertion that V = IR is a statement of Ohm’s law is not true. This equation defines resistance and it may be applied to all conducting devices whether they obey Ohm’s law or not. The Ohm’s law asserts that the plot of I versus V is linear i.e., R is independent of V. Equation E = ρ j leads to another statement of Ohm’s law, i.e., a conducting material obeys Ohm’s law when the resistivity of the material does not depend on the magnitude and direction of applied electric field. 3. Homogeneous conductors like silver or semiconductors like pure germanium or germanium containing impurities obey Ohm’s law within some range of electric field values. If the field becomes too strong, there are departures from Ohm’s law in all cases. 4. Motion of conduction electrons in electric field E is the sum of (i) motion due to random collisions and (ii) that due to E. The motion due to random collisions averages to zero and does not contribute to vd (Chapter 10, Textbook of Class XI). vd , thus is only due to applied electric field on the electron. 5. The relation j = ρ v should be applied to each type of charge carriers separately. In a conducting wire, the total current and charge density arises from both positive and negative charges: j = ρ+ v+ + ρ– v– ρ = ρ+ + ρ– Now in a neutral wire carrying electric current, ρ+ = – ρ– Further, v+ ~ 0 which gives ρ = 0 j = ρ– v Thus, the relation j = ρ v does not apply to the total current charge density. 6. Kirchhoff’s junction rule is based on conservation of charge and the outgoing currents add up and are equal to incoming current at a junction. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule. EXERCISES

3.4 OHM’S LAW A basic law regarding flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for flow of currents was discovered. Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then Ohm’s law states that V µ I or, V = R I (3.3) where the constant of proportionality R is called the resistance of the conductor. The SI units of resistance is ohm, and is denoted by the symbol W. The resistance R not only depends on the material of the conductor but also on the dimensions of the conductor. The dependence of R on the dimensions of the conductor can easily be determined as follows. FIGURE 3.2 Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of Illustrating the length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such relation R = rl/A for a rectangular slabidentical slabs side by side [Fig. 3.2(b)], so that the length of the of length l and areacombination is 2l. The current flowing through the combination is the of cross-section A. same as that flowing through either of the slabs. If V is the potential difference across the ends of the first slab, then V is also the potential 83difference across the ends of the second slab since the second slab is Reprint 2025-26 Physics identical to the first and the same current I flows through both. The potential difference across the ends of the combination is clearly sum of the potential difference across the two individual slabs and hence equals 2V. The current through the combination is I and the resistance of the combination RC is [from Eq. (3.3)], 2V R C = = 2 R (3.4) I since V/I = R, the resistance of either of the slabs. Thus, doubling the length of a conductor doubles the(1787–1854) resistance. In general, then resistance is proportional to length, R ∝ l (3.5)OHM Georg Simon Ohm (1787– Next, imagine dividing the slab into two by cutting it 1854) German physicist, lengthwise so that the slab can be considered as a professor at Munich. Ohm combination of two identical slabs of length l, but each was led to his law by anSIMON having a cross sectional area of A/2 [Fig. 3.2(c)]. analogy between the For a given voltage V across the slab, if I is the current conduction of heat: the through the entire slab, then clearly the current flowing electric field is analogous to the temperature gradient, through each of the two half-slabs is I/2. Since theGEORG and the electric current is potential difference across the ends of the half-slabs is V, analogous to the heat flow. i.e., the same as across the full slab, the resistance of each of the half-slabs R1 is V V R1 = = 2 = 2 R. (3.6) ( I /2) I Thus, halving the area of the cross-section of a conductor doubles the resistance. In general, then the resistance R is inversely proportional to the cross-sectional area, 1 R ∝ (3.7) A Combining Eqs. (3.5) and (3.7), we have l R ∝ (3.8) A and hence for a given conductor l R = ρ (3.9) A where the constant of proportionality r depends on the material of the conductor but not on its dimensions. r is called resistivity. Using the last equation, Ohm’s law reads I ρl V = I × R = (3.10) A Current per unit area (taken normal to the current), I/A, is called current density and is denoted by j. The SI units of the current density are A/m2. Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its 84 ends is El. Using these, the last equation reads Reprint 2025-26 Current Electricity E l = j r l or, E = j r (3.11) The above relation for magnitudes E and j can indeed be cast in a vector form. The current density, (which we have defined as the current through unit area normal to the current) is also directed along E, and is also a vector j (º j E/E). Thus, the last equation can be written as, E = jr (3.12) or, j = s E (3.13) where s º1/r is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we will try to understand the origin of the Ohm’s law as arising from the characteristics of the drift of electrons.

3.7 RESISTIVITY OF VARIOUS MATERIALS The materials are classified as conductors, semiconductors and insulators 89depending on their resistivities, in an increasing order of their values. Reprint 2025-26 Physics Metals have low resistivities in the range of 10–8 Wm to 10–6 Wm. At the other end are insulators like ceramic, rubber and plastics having resistivities 1018 times greater than metals or more. In between the two are the semiconductors. These, however, have resistivities characteristically decreasing with a rise in temperature. The resistivities of semiconductors can be decreased by adding small amount of suitable impurities. This last feature is exploited in use of semiconductors for electronic devices. 3.8 TEMPERATURE DEPENDENCE OF RESISTIVITY The resistivity of a material is found to be dependent on the temperature. Different materials do not exhibit the same dependence on temperatures. Over a limited range of temperatures, that is not too large, the resistivity of a metallic conductor is approximately given by, rT = r0 [1 + a (T–T0)] (3.26) where rT is the resistivity at a temperature T and r0 is the same at a reference temperature T0. a is called the temperature co-efficient of resistivity, and from Eq. (3.26), the dimension of a is (Temperature)–1. For metals, a is positive. The relation of Eq. (3.26) implies that a graph of rT plotted against T would be a straight line. At temperatures much lower than 0°C, the graph, however, deviates considerably from a straight line (Fig. 3.8). Equation (3.26) thus, can be used approximately over a limited range of T around any reference temperature T0, where the graph can be approximated as a straight line.  FIGURE 3.8 FIGURE 3.9 Resistivity FIGURE 3.10 Resistivity rT of rT of nichrome as a Temperature dependence copper as a function function of absolute of resistivity for a typical of temperature T. temperature T. semiconductor. Some materials like Nichrome (which is an alloy of nickel, iron and chromium) exhibit a very weak dependence of resistivity with temperature (Fig. 3.9). Manganin and constantan have similar properties. These materials are thus widely used in wire bound standard resistors since 90 their resistance values would change very little with temperatures. Reprint 2025-26 Current Electricity Unlike metals, the resistivities of semiconductors decrease with increasing temperatures. A typical dependence is shown in Fig. 3.10. We can qualitatively understand the temperature dependence of resistivity, in the light of our derivation of Eq. (3.23). From this equation, resistivity of a material is given by 1 m ρ= = 2 (3.27) σ n e τ r thus depends inversely both on the number n of free electrons per unit volume and on the average time t between collisions. As we increase temperature, average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of collisions t, thus decreases with temperature. In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of t with rise in temperature causes r to increase as we have observed. For insulators and semiconductors, however, n increases with temperature. This increase more than compensates any decrease in t in Eq.(3.23) so that for such materials, r decreases with temperature. Example 3.3 An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 W. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10–4 °C–1. Solution When the current through the element is very small, heating effects can be ignored and the temperature T1 of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R2 at the steady temperature T2 is 230 V R2 = 2.68 A = 85.8 Ω Using the relation R2 = R1 [1 + a (T2 – T1)] with a = 1.70 × 10–4 °C–1, we get (85.8 – 75.3) T2 – T1 = –4 = 820 °C (75.3) × 1.70 × 10 that is, T2 = (820 + 27.0) °C = 847 °C Thus, the steady temperature of the heating element (when heating EXAMPLE effect due to the current equals heat loss to the surroundings) is 3.3 91 847 °C. Reprint 2025-26 Physics Example 3.4 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 W and at steam point is 5.23 W. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 W. Calculate the temperature of the bath. Solution R0 = 5 W, R100 = 5.23 W and Rt = 5.795 W Rt − R 0 Now, t = × 100, Rt = R 0 (1 + αt ) R100 − R 0 3.4 5.795 − 5 = × 100 5.23 − 5 0.795 EXAMPLE = × 100 = 345.65 °C 0.23 3.9 ELECTRICAL ENERGY, POWER Consider a conductor with end points A and B, in which a current I is flowing from A to B. The electric potential at A and B are denoted by V(A) and V(B) respectively. Since current is flowing from A to B, V(A) > V(B) and the potential difference across AB is V = V(A) – V(B) > 0. In a time interval Dt, an amount of charge DQ = I Dt travels from A to B. The potential energy of the charge at A, by definition, was Q V(A) and similarly at B, it is Q V(B). Thus, change in its potential energy DUpot is DUpot = Final potential energy – Initial potential energy = DQ[(V (B) – V (A)] = –DQ V = –I VDt < 0 (3.28) If charges moved without collisions through the conductor, their kinetic energy would also change so that the total energy is unchanged. Conservation of total energy would then imply that, DK = –DUpot (3.29) that is, DK = I VDt > 0 (3.30) Thus, in case charges were moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move. We have, however, seen earlier that on the average, charge carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The atoms vibrate more vigorously, i.e., the conductor heats up. Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval Dt is, DW = I VDt (3.31) The energy dissipated per unit time is the power dissipated P = DW/Dt and we have, 92 P = I V (3.32) Reprint 2025-26 Current Electricity Using Ohm’s law V = IR, we get P = I 2 R = V 2/R (3.33) as the power loss (“ohmic loss”) in a conductor of resistance R carrying a current I. It is this power which heats up, for example, the coil of an electric bulb to incandescence, radiating out heat and light. Where does the power come from? As we have reasoned before, we need an external source to keep a steady current through the conductor. It is clearly this source which must supply this power. In the simple circuit shown with a cell (Fig.3.11), it is the chemical energy of the cell which supplies this power for as long as it can. The expressions for power, Eqs. (3.32) and (3.33), show the dependence of the power dissipated in a resistor R on the current through it and the voltage FIGURE 3.11 Heat is produced in the across it. resistor R which is connected across Equation (3.33) has an important application to the terminals of a cell. The energy power transmission. Electrical power is transmitted dissipated in the resistor R comes from from power stations to homes and factories, which the chemical energy of the electrolyte. may be hundreds of miles away, via transmission cables. One obviously wants to minimise the power loss in the transmission cables connecting the power stations to homes and factories. We shall see now how this can be achieved. Consider a device R, to which a power P is to be delivered via transmission cables having a resistance Rc to be dissipated by it finally. If V is the voltage across R and I the current through it, then P = V I (3.34) The connecting wires from the power station to the device has a finite resistance Rc. The power dissipated in the connecting wires, which is wasted is Pc with Pc = I 2 Rc P 2 R c = 2 (3.35) V from Eq. (3.32). Thus, to drive a device of power P, the power wasted in the connecting wires is inversely proportional to V 2. The transmission cables from power stations are hundreds of miles long and their resistance Rc is considerable. To reduce Pc, these wires carry current at enormous values of V and this is the reason for the high voltage danger signs on transmission lines — a common sight as we move away from populated areas. Using electricity at such voltages is not safe and hence at the other end, a device called a transformer lowers the voltage to a value suitable for use.

4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

4.10 Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

4.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 4.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 4.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 4.10 Circular motion agency is needed to provide force to move a body from rest. 4.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion. Summary In these examples, the external agency of force (hands, Points to ponder wind, stream, etc) is in contact with the object. This is not Exercises always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? Reprint 2025-26 50 PHYSICS 4.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion isThe question posed above appears to be simple. possible with no frictional forces opposing. ThisHowever, it took ages to answer it. Indeed, the is what Galileo did.correct answer to this question given by Galileo in the seventeenth century was the foundation 4.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an interme- something external is required to keep it moving. diate situation. Galileo concluded that an object According to this view, for example, an arrow moving on a frictionless horizontal plane must shot from a bow keeps flying since the air behind neither have acceleration nor retardation, i.e. it the arrow keeps pushing it. The view was part of should move with constant velocity (Fig. 4.1(a)).an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force (i) (ii) (iii)is required to keep a body in motion. Fig. 4.1(a) Aristotelian law of motion is flawed, as we shall Another experiment by Galileo leading to thesee. However, it is a natural view that anyone same conclusion involves a double inclined plane.would hold from common experience. Even a A ball released from rest on one of the planes rollssmall child playing with a simple (non-electric) down and climbs up the other. If the planes are toy-car on a floor knows intuitively that it needs smooth, the final height of the ball is nearly the to constantly drag the string attached to the toy- same as the initial height (a little less but never car with some force to keep it going. If it releases greater). In the ideal situation, when friction is the string, it comes to rest. This experience is absent, the final height of the ball is the same common to most terrestrial motion. External as its initial height. forces seem to be needed to keep bodies in If the slope of the second plane is decreased motion. Left to themselves, all bodies eventually and the experiment repeated, the ball will still come to rest. reach the same height, but in doing so, it will What is the flaw in Aristotle’s argument? The travel a longer distance. In the limiting case, when answer is: a moving toy car comes to rest because the slope of the second plane is zero (i.e. is a the external force of friction on the car by the floor horizontal) the ball travels an infinite distance. opposes its motion. To counter this force, the child In other words, its motion never ceases. This is, has to apply an external force on the car in the of course, an idealised situation (Fig. 4.1(b)). direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle Fig. 4.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. Reprint 2025-26 LAWS OF MOTION 51 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he formu- Galileo thus, arrived at a new insight on lated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, Two kinds of situations are encountered in thea body at rest continues to remain at rest and a application of this law in practice. In somebody in motion continues to move with a uniform examples, we know that the net external forcevelocity. This property of the body is called on the object is zero. In that case we caninertia. Inertia means ‘resistance to change’. A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 4.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. Reprint 2025-26 52 PHYSICS More often, however, we do not know all the The acceleration of the car cannot be accounted forces to begin with. In that case, if we know for by any internal force. This might sound that an object is unaccelerated (i.e. it is either surprising, but it is true. The only conceivable at rest or in uniform linear motion), we can infer external force along the road is the force of from the first law that the net external force on friction. It is the frictional force that accelerates the object must be zero. Gravity is everywhere. the car as a whole. (You will learn about friction For terrestrial phenomena, in particular, every in section 4.9). When the car moves with object experiences gravitational force due to the constant velocity, there is no net external force. earth. Also objects in motion generally experience The property of inertia contained in the First friction, viscous drag, etc. If then, on earth, an law is evident in many situations. Suppose we object is at rest or in uniform linear motion, it is are standing in a stationary bus and the driver not because there are no forces acting on it, but starts the bus suddenly. We get thrown because the various external forces cancel out backward with a jerk. Why ? Our feet are in touch i.e. add up to zero net external force. with the floor. If there were no friction, we would Consider a book at rest on a horizontal surface remain where we were, while the floor of the bus Fig. (4.2(a)). It is subject to two external forces : would simply slip forward under our feet and the the force due to gravity (i.e. its weight W) acting back of the bus would hit us. However, downward and the upward force on the book by fortunately, there is some friction between the the table, the normal force R . R is a self-adjusting feet and the floor. If the start is not too sudden, force. This is an example of the kind of situation i.e. if the acceleration is moderate, the frictional mentioned above. The forces are not quite known force would be enough to accelerate our feet fully but the state of motion is known. We observe along with the bus. But our body is not strictly the book to be at rest. Therefore, we conclude a rigid body. It is deformable, i.e. it allows some from the first law that the magnitude of R equals relative displacement between different parts. that of W. A statement often encountered is : What this means is that while our feet go with “Since W = R, forces cancel and, therefore, the book the bus, the rest of the body remains where it is is at rest”. This is incorrect reasoning. The correct due to inertia. Relative to the bus, therefore, we statement is : “Since the book is observed to be at are thrown backward. As soon as that happens, rest, the net external force on it must be zero, however, the muscular forces on the rest of the according to the first law. This implies that the body (by the feet) come into play to move the body normal force R must be equal and opposite to the along with the bus. A similar thing happens weight W ”. when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest. ⊳ Example 4.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a Fig. 4.2 (a) a book at rest on the table, and (b) a car constant rate of 100 m s–2. What is the moving with uniform velocity. The net force acceleration of the astronaut the instant after is zero in each case. he is outside the spaceship ? (Assume that Consider the motion of a car starting from there are no nearby stars to exert rest, picking up speed and then moving on a gravitational force on him.) smooth straight road with uniform speed (Fig. Answer Since there are no nearby stars to exert(4.2(b)). When the car is stationary, there is no gravitational force on him and the smallnet force acting on it. During pick-up, it spaceship exerts negligible gravitationalaccelerates. This must happen due to a net attraction on him, the net force acting on theexternal force. Note, it has to be an external force. Reprint 2025-26 LAWS OF MOTION 53 astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. ⊳ you may have noticed, he draws in the hands backward in the act of catching the ball4.5 NEWTON’S SECOND LAW OF MOTION (Fig. 4.3). The novice, on the other hand, The first law refers to the simple case when the keeps his hands fixed and tries to catch the net external force on a body is zero. The second ball almost instantly. He needs to provide a law of motion refers to the general situation when much greater force to stop the ball instantly, there is a net external force acting on the body. and this hurts. The conclusion is clear: force It relates the net external force to the not only depends on the change in momentum, acceleration of the body. but also on how fast the change is brought Momentum about. The same change in momentum Momentum of a body is defined to be the product brought about in a shorter time needs a of its mass m and velocity v, and is denoted greater applied force. In short, the greater the by p: rate of change of momentum, the greater is p = m v (4.1) the force. Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. • If two stones, one light and the other heavy, are dropped from the top of a building, a Fig. 4.3 Force not only depends on the change in person on the ground will find it easier to catch momentum but also on how fast the change is brought about. A seasoned cricketer draws the light stone than the heavy stone. The in his hands during a catch, allowing greater mass of a body is thus an important time for the ball to stop and hence requires a parameter that determines the effect of force smaller force. on its motion. • Speed is another important parameter to consider. A bullet fired by a gun can easily • Observations confirm that the product of pierce human tissue before it stops, resulting mass and velocity (i.e. momentum) is basic to in casualty. The same bullet fired with the effect of force on motion. Suppose a fixed moderate speed will not cause much damage. force is applied for a certain interval of time Thus for a given mass, the greater the speed, on two bodies of different masses, initially at the greater is the opposing force needed to stop rest, the lighter body picks up a greater speed the body in a certain time. Taken together, than the heavier body. However, at the end of the product of mass and velocity, that is the time interval, observations show that each momentum, is evidently a relevant variable body acquires the same momentum. Thus of motion. The greater the change in the the same force for the same time causes momentum in a given time, the greater is the the same change in momentum for force that needs to be applied. • A seasoned cricketer catches a cricket ball different bodies. This is a crucial clue to the second law of motion. coming in with great speed far more easily • In the preceding observations, the vector than a novice, who can hurt his hands in the Reprint 2025-26 54 PHYSICS character of momentum has not been evident. ∆p ∆p In the examples so far, momentum and change F ∝ or F = k ∆t ∆ t in momentum both have the same direction. But this is not always the case. Suppose a where k is a constant of proportionality. Taking stone is rotated with uniform speed in a ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 4.4). A force is needed d p to cause this change in momentum vector. respect to t, denoted by . Thus dt This force is provided by our hand through the string. Experience suggests that our hand d p F = k (4.2) needs to exert a greater force if the stone is d t rotated at greater speed or in a circle of For a body of fixed mass m, smaller radius, or both. This corresponds to greater acceleration or equivalently a greater d p d d v = (m v ) = m = m a (4.3) rate of change in momentum vector. This d t d t d t suggests that the greater the rate of change i.e the Second Law can also be written as in momentum vector the greater is the force F = k m a (4.4) applied. which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (4.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp F = = m a (4.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Let us note at this stage some important points Fig. 4.4 Force is necessary for changing the direction about the second law : of momentum, even if its magnitude is constant. We can feel this while rotating a 1. In the second law, F = 0 implies a = 0. The second stone in a horizontal circle with uniform speed law is obviously consistent with the first law. by means of a string. 2. The second law of motion is a vector law. It is These qualitative observations lead to the equivalent to three equations, one for each second law of motion expressed by Newton as component of the vectors : follows : d p x F x = = ma xThe rate of change of momentum of a body is d t directly proportional to the applied force and d p ytakes place in the direction in which the force F y = = ma y acts. d t dp zThus, if under the action of a force F for time F z = =m a z (4.6) dtinterval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum This means that if a force is not parallel to the velocity of the body, but makes some anglep = m v changes by ∆ p = m ∆v . According to the with it, it changes only the component of Second Law, velocity along the direction of force. The Reprint 2025-26 LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical – 90 × 90 – u 2 gravitational force, the horizontal component m s −2 = – 6750 m s −2 a = = of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force = 0.04 kg × 6750 m s-2 = 270 N on the particle and a stands for acceleration of the particle. It turns out, however, that the The actual resistive force, and therefore, law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The answer therefore, only indicates the average even more generally, to a system of particles. resistive force. ⊳ In that case, F refers to the total external force ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is 1 2 the acceleration of the centre of mass of the mass m is described by y = ut + gt . Find 2 system about which we shall study in detail in the force acting on the particle. Chapter 6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y = ut + gt 2 Now, d y v = = u + gt d t dv acceleration, a = = g d t Fig. 4.5 Acceleration at an instant is determined by Then the force is given by Eq. (4.5) the force at that instant. The moment after a F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, if of a particle under acceleration due to gravity air resistance is neglected. The stone carries no memory of its acceleration with the train and y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces determined by the force here and now, not by back, the force on the ball by the wall acts for a any history of the motion of the particle very short time when the two are in contact, yet the force is large enough to reverse the momentum (See Fig. 4.5). of the ball. Often, in these situations, the force ⊳ and the time duration are difficult to ascertain Example 4.2 A bullet of mass 0.04 kg separately. However, the product of force and time, moving with a speed of 90 m s–1 enters a which is the change in momentum of the body heavy wooden block and is stopped after a remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2025-26 56 PHYSICS A large force acting for a short time to produce a Thus, according to Newtonian mechanics, finite change in momentum is called an impulsive force never occurs singly in nature. Force is the force. In the history of science, impulsive forces were mutual interaction between two bodies. Forces put in a conceptually different category from always occur in pairs. Further, the mutual forces ordinary forces. Newtonian mechanics has no such between two bodies are always equal and distinction. Impulsive force is like any other force – opposite. This idea was expressed by Newton in except that it is large and acts for a short time. the form of the third law of motion. ⊳ To every action, there is always an equal and Example 4.4 A batsman hits back a ball opposite reaction. straight in the direction of the bowler without changing its initial speed of 12 m s–1. Newton’s wording of the third law is so crisp and If the mass of the ball is 0.15 kg, determine beautiful that it has become a part of common the impulse imparted to the ball. (Assume language. For the same reason perhaps, linear motion of the ball) misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of theAnswer Change in momentum terms : action and reaction. = 0.15 × 12–(–0.15×12) 1. The terms action and reaction in the third law = 3.6 N s, mean nothing else but ‘force’. Using different Impulse = 3.6 N s, terms for the same physical concept in the direction from the batsman to the bowler. can sometimes be confusing. A simple and clear way of stating the third law is as This is an example where the force on the ball follows :by the batsman and the time of contact of the ball and the bat are difficult to know, but the Forces always occur in pairs. Force on a impulse is readily calculated. ⊳ body A by B is equal and opposite to the force on the body B by A.

4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10–19 C, me = 9.1×10–31 kg)

4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

4.5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT LOOP In this section, we shall evaluate the magnetic field due to a circular coil along its axis. The evaluation entails summing up the effect of infinitesimal current elements (I dl) mentioned in the previous section. We assume that the current I is steady and that the evaluation is carried out in free space (i.e., vacuum). Fig. 4.9 depicts a circular loop carrying a steady current I. The loop is placed in the plane with its centre at the origin O and has a radius R. The x-axis is the axis of the loop. We wish to calculate the magnetic field at the point P on this axis. Let x be the distance of P from the centre O of the loop. Consider a conducting element dl of the loop. This is FIGURE 4.9 Magnetic field on the shown in Fig. 4.9. The magnitude dB of the magnetic axis of a current carrying circular field due to dl is given by the Biot-Savart law [Eq. 4.7(a)], loop of radius R. Shown are the µ0 I d l × r magnetic field dB (due to a line dB = 3 (4.8) element dl ) and its 4π r components along and Now r2 = x2 + R2 . Further, any element of the loop perpendicular to the axis. will be perpendicular to the displacement vector from the element to the axial point. For example, the element dl in Fig. 4.9 is in the plane, whereas, the displacement vector r from dl to the axial point P is in the plane. Hence |dl × r|=r dl. Thus, µ0 Idl d B = 4 π x 2 + R 2 (4.9) 115 ( ) Reprint 2025-26 Physics The direction of dB is shown in Fig. 4.9. It is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis, dB⊥. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result. For example, the dB⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element, shown in Fig. 4.9. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop. For Fig. 4.9, R cosθ= 2 2 1/2 (4.10) ( x + R ) From Eqs. (4.9) and (4.10), µ0 Idl R d B x = 4 π x 2 + R 2 3/2 ( ) The summation of elements dl over the loop yields 2πR, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is 2 µ 0 I R ˆ i B = B x ˆi = 3/2 (4.11) 2 2 2 x + R ( ) As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain, B 0 = µ0 I ˆi (4.12) 2 R The magnetic field lines due to a circular wire form closed loops and are shown in Fig. 4.10. The direction of the magnetic field is given by (another) right-hand thumb rule stated below: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field. FIGURE 4.10 The magnetic field lines for a current loop. The direction of the field is given by the right-hand thumb rule described in the text. The upper side of the loop may be thought of as the north pole and the lower 116 side as the south pole of a magnet. Reprint 2025-26 Moving Charges and Magnetism Example 4.5 A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. 4.11(a). Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. 4.11(b)? FIGURE 4.11 Solution (a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|. (b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 × 10–4 T normal to the plane of the EXAMPLE paper going into it. (c) Same magnitude of B but opposite in direction to that in (b). 4.5 Example 4.6 Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? Solution Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, EXAMPLE –7 NI µ 4 π × 10 × 10 2 × 1 0 B = = –1 = 2 π × 10−4 = 6.28 × 10 −4 T 4.6 R 2 10 2 ×

4.6 NEWTON’S THIRD LAW OF MOTION 2. The terms action and reaction in the third law The second law relates the external force on a may give a wrong impression that action body to its acceleration. What is the origin of the comes before reaction i.e action is the cause external force on the body ? What agency and reaction the effect. There is no cause- provides the external force ? The simple answer effect relation implied in the third law. The in Newtonian mechanics is that the external force on A by B and the force on B by A act force on a body always arises due to some other at the same instant. By the same reasoning, body. Consider a pair of bodies A and B. B gives any one of them may be called action and the rise to an external force on A. A natural question other reaction. is: Does A in turn give rise to an external force 3. Action and reaction forces act on different on B ? In some examples, the answer seems bodies, not on the same body. Consider a pair clear. If you press a coiled spring, the spring is of bodies A and B. According to the third law, compressed by the force of your hand. The FAB = – FBA (4.8) compressed spring in turn exerts a force on your (force on A by B) = – (force on B by A)hand and you can feel it. But what if the bodies are not in contact ? The earth pulls a stone Thus if we are considering the motion of any downwards due to gravity. Does the stone exert one body (A or B), only one of the two forces is a force on the earth ? The answer is not obvious relevant. It is an error to add up the two forces since we hardly see the effect of the stone on the and claim that the net force is zero. earth. The answer according to Newton is: Yes, However, if you are considering the system the stone does exert an equal and opposite force of two bodies as a whole, FAB and FBA are on the earth. We do not notice it since the earth internal forces of the system (A + B). They add is very massive and the effect of a small force on up to give a null force. Internal forces in a its motion is negligible. body or a system of particles thus cancel away Reprint 2025-26 LAWS OF MOTION 57 in pairs. This is an important fact that the force on the wall due to the ball is normal to enables the second law to be applicable to a the wall along the positive x-direction. The body or a system of particles (See Chapter 6). magnitude of force cannot be ascertained since the small time taken for the collision has not ⊳ Example 4.5 Two identical billiard balls been specified in the problem. strike a rigid wall with the same speed but Case (b) at different angles, and get reflected without = − m u sin 30 any change in speed, as shown in Fig. 4.6. ( p x ) initial = m u cos 30 , ( p y ) initial What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the = − m u sin 30 p x ) final = – m u cos 30 , ( p y ) final magnitudes of impulses imparted to the ( balls by the wall ? Note, while px changes sign after collision, py does not. Therefore, x-component of impulse = –2 m u cos 30° y-component of impulse = 0 The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction. The ratio of the magnitudes of the impulses Fig. 4.6 imparted to the balls in (a) and (b) is 2 Answer An instinctive answer to (i) might be 2 m u/ 2 m u cos30 = ≈ 1.2 ⊳ ( ) 3 that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° 4.7 CONSERVATION OF MOMENTUM to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. The second and third laws of motion lead to How to find the force on the wall? The trick is an important consequence: the law of to consider the force (or impulse) on the ball conservation of momentum. Take a familiar due to the wall using the second law, and then example. A bullet is fired from a gun. If the force use the third law to answer (i). Let u be the speed on the bullet by the gun is F, the force on the gun of each ball before and after collision with the by the bullet is – F, according to the third law. wall, and m the mass of each ball. Choose the x The two forces act for a common interval of time and y axes as shown in the figure, and consider ∆t. According to the second law, F ∆t is the change the change in momentum of the ball in each in momentum of the bullet and – F ∆t is the case : change in momentum of the gun. Since initially, both are at rest, the change in momentum equals Case (a) the final momentum for each. Thus if pb is the initial = mu initial = 0 momentum of the bullet after firing and pg is the ( p x ) ( p y ) recoil momentum of the gun, pg = – pb i.e. pb + pg = 0 = 0. That is, the total momentum of the (bullet + ( p x )final = −mu ( p y )final gun) system is conserved. Thus in an isolated system (i.e. a system withImpulse is the change in momentum vector. no external force), mutual forces between pairsTherefore, of particles in the system can cause momentum x-component of impulse = – 2 m u change in individual particles, but since the y-component of impulse = 0 mutual forces for each pair are equal and Impulse and force are in the same direction. opposite, the momentum changes cancel in pairs Clearly, from above, the force on the ball due to and the total momentum remains unchanged. the wall is normal to the wall, along the negative This fact is known as the law of conservation x-direction. Using Newton’s third law of motion, of momentum : Reprint 2025-26 58 PHYSICS The total momentum of an isolated system of interacting particles is conserved. An important example of the application of the law of conservation of momentum is the collision of two bodies. Consider two bodies A and B, with initial momenta pA and pB. The bodies collide, get apart, with final momenta p′A and p′B Fig. 4.7 Equilibrium under concurrent forces. respectively. By the Second Law In other words, the resultant of any two forces say F1 and F2, obtained by the parallelogram FAB ∆=t p ′A − p A and law of forces must be equal and opposite to the FBA ∆=t p ′B − p B third force, F3. As seen in Fig. 4.7, the three forces in equilibrium can be represented by the (where we have taken a common interval of time sides of a triangle with the vector arrows taken for both forces i.e. the time for which the two in the same sense. The result can be bodies are in contact.) generalised to any number of forces. A particle is in equilibrium under the action of forces F1,Since F AB = − FBA by the third law, F2,... Fn if they can be represented by the sides of a closed n-sided polygon with arrows directed p ′A − p A = − ( p ′B − p B ) in the same sense. i.e. p ′A + p ′B = p A + p B (4.9) Equation (4.11) implies that which shows that the total final momentum of F1x + F2x + F3x = 0 the isolated system equals its initial momentum. F1y + F2y + F3y = 0 Notice that this is true whether the collision is F1z + F2z + F3z = 0 (4.12) elastic or inelastic. In elastic collisions, there is where F1x, F1y and F1z are the components of F1a second condition that the total initial kinetic along x, y and z directions respectively. energy of the system equals the total final kinetic ⊳ energy (See Chapter 5). Example 4.6 See Fig. 4.8. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the 4.8 EQUILIBRIUM OF A PARTICLE horizontal direction is applied at the mid- point P of the rope, as shown. What is theEquilibrium of a particle in mechanics refers to angle the rope makes with the vertical inthe situation when the net external force on the equilibrium ? (Take g = 10 m s-2). Neglectparticle is zero.* According to the first law, this the mass of the rope. means that, the particle is either at rest or in uniform motion. If two forces F1 and F2, act on a particle, equilibrium requires F1 = − F2 (4.10) i.e. the two forces on the particle must be equal and opposite. Equilibrium under three concurrent forces F1, F2 and F3 requires that (a) (b) (c) the vector sum of the three forces is zero. Fig. 4.8 F1 + F2 + F3 = 0 (4.11) * Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque), as we shall see in Chapter 6. Reprint 2025-26 LAWS OF MOTION 59 Answer Figures 4.8(b) and 4.8(c) are known as other types of supports), there are mutual free-body diagrams. Figure 4.8(b) is the free-body contact forces (for each pair of bodies) satisfying diagram of W and Fig. 4.8(c) is the free-body the third law. The component of contact force diagram of point P. normal to the surfaces in contact is called Consider the equilibrium of the weight W. normal reaction. The component parallel to the Clearly,T2 = 6 × 10 = 60 N. surfaces in contact is called friction. Contact forces arise also when solids are in contact with Consider the equilibrium of the point P under fluids. For example, for a solid immersed in a the action of three forces - the tensions T1 and fluid, there is an upward bouyant force equal toT2, and the horizontal force 50 N. The horizontal the weight of the fluid displaced. The viscousand vertical components of the resultant force must vanish separately : force, air resistance, etc are also examples of contact forces (Fig. 4.9). T1 cos θ = T2 = 60 N Two other common forces are tension in a T1 sin θ = 50 N string and the force due to spring. When a spring which gives that is compressed or extended by an external force, a restoring force is generated. This force is usually proportional to the compression or Note the answer does not depend on the length elongation (for small displacements). The spring of the rope (assumed massless) nor on the point force F is written as F = – k x where x is the at which the horizontal force is applied. ⊳ displacement and k is the force constant. The negative sign denotes that the force is opposite 4.9 COMMON FORCES IN MECHANICS to the displacement from the unstretched state. In mechanics, we encounter several kinds of For an inextensible string, the force constant is forces. The gravitational force is, of course, very high. The restoring force in a string is called pervasive. Every object on the earth experiences tension. It is customary to use a constant tension the force of gravity due to the earth. Gravity also T throughout the string. This assumption is true governs the motion of celestial bodies. The for a string of negligible mass. gravitational force can act at a distance without We learnt that there are four fundamental the need of any intervening medium. forces in nature. Of these, the weak and strong All the other forces common in mechanics are forces appear in domains that do not concern contact forces.* As the name suggests, a contact us here. Only the gravitational and electrical force on an object arises due to contact with some forces are relevant in the context of mechanics. other object: solid or fluid. When bodies are in The different contact forces of mechanics contact (e.g. a book resting on a table, a system mentioned above fundamentally arise from of rigid bodies connected by rods, hinges and electrical forces. This may seem surprising Fig. 4.9 Some examples of contact forces in mechanics. * We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are electrical and magnetic non-contact forces. Reprint 2025-26 60 PHYSICS since we are talking of uncharged and non- exist by itself. When there is no applied force, magnetic bodies in mechanics. At the microscopic there is no static friction. It comes into play the level, all bodies are made of charged constituents moment there is an applied force. As the applied (nuclei and electrons) and the various contact force F increases, fs also increases, remaining forces arising due to elasticity of bodies, molecular equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, itcollisions and impacts, etc. can ultimately be is called static friction. Static friction opposestraced to the electrical forces between the charged impending motion. The term impending motionconstituents of different bodies. The detailed means motion that would take place (but does microscopic origin of these forces is, however, not actually take place) under the applied force, complex and not useful for handling problems in if friction were absent. mechanics at the macroscopic scale. This is why We know from experience that as the applied they are treated as different types of forces with force exceeds a certain limit, the body begins to their characteristic properties determined move. It is found experimentally that the limiting empirically. value of static friction ( sf )max is independent of 4.9.1 Friction the area of contact and varies with the normal Let us return to the example of a body of mass m force(N) approximately as : at rest on a horizontal table. The force of gravity = µs N (4.13) ( f s )max (mg) is cancelled by the normal reaction force where µs is a constant of proportionality(N) of the table. Now suppose a force F is applied depending only on the nature of the surfaces in horizontally to the body. We know from contact. The constant µs is called the coefficientexperience that a small applied force may not of static friction. The law of static friction may be enough to move the body. But if the applied thus be written as force F were the only external force on the body, fs ≤ µs N (4.14) it must move with acceleration F/m, however small. Clearly, the body remains at rest because If the applied force F exceeds ( sf )max the body some other force comes into play in the begins to slide on the surface. It is found experi- horizontal direction and opposes the applied mentally that when relative motion has started, force F, resulting in zero net force on the body. the frictional force decreases from the static This force fs parallel to the surface of the body in maximum value ( sf )max . Frictional force thatcontact with the table is known as frictional opposes relative motion between surfaces inforce, or simply friction (Fig. 4.10(a)). The contact is called kinetic or sliding friction and issubscript stands for static friction to distinguish denoted by fk . Kinetic friction, like static fric-it from kinetic friction fk that we consider later tion, is found to be independent of the area of (Fig. 4.10(b)). Note that static friction does not contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: f k = µk N (4.15) where µk′ the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that µk is Fig. 4.10 Static and sliding friction: (a) Impending less than µs. When relative motion has begun, motion of the body is opposed by static the acceleration of the body according to the friction. When external force exceeds the second law is ( F – fk )/m. For a body moving with maximum limit of static friction, the body constant velocity, F = fk. If the applied force on begins to move. (b) Once the body is in the body is removed, its acceleration is – fk /m motion, it is subject to sliding or kinetic friction and it eventually comes to a stop. which opposes relative motion between the The laws of friction given above do not have two surfaces in contact. Kinetic friction is the status of fundamental laws like those for usually less than the maximum value of static gravitational, electric and magnetic forces. They friction. are empirical relations that are only Reprint 2025-26 LAWS OF MOTION 61 approximately true. Yet they are very useful in Answer The forces acting on a block of mass m practical calculations in mechanics. at rest on an inclined plane are (i) the weight Thus, when two bodies are in contact, each mg acting vertically downwards (ii) the normal experiences a contact force by the other. Friction, force N of the plane on the block, and (iii) the by definition, is the component of the contact force static frictional force fs opposing the impending parallel to the surfaces in contact, which opposes motion. In equilibrium, the resultant of these impending or actual relative motion between the forces must be zero. Resolving the weight mg two surfaces. Note that it is not motion, but along the two directions shown, we have relative motion that the frictional force opposes. m g sin θ = fs , m g cos θ = N Consider a box lying in the compartment of a train As θ increases, the self-adjusting frictional force that is accelerating. If the box is stationary fs increases until at θ = θmax, fs achieves itsrelative to the train, it is in fact accelerating along with the train. What forces cause the acceleration maximum value, ( sf )max = µs N. of the box? Clearly, the only conceivable force in the horizontal direction is the force of friction. If Therefore, there were no friction, the floor of the train would tan θmax = µs or θmax = tan–1 µsslip by and the box would remain at its initial position due to inertia (and hit the back side of When θ becomes just a little more than θmax , the train). This impending relative motion is there is a small net force on the block and it opposed by the static friction fs. Static friction begins to slide. Note that θmax depends only on provides the same acceleration to the box as that µs and is independent of the mass of the block. of the train, keeping it stationary relative to the For θmax = 15°,train. ⊳ µs = tan 15° Example 4.7 Determine the maximum = 0.27 ⊳ acceleration of the train in which a box ⊳ Example 4.9 What is the acceleration of lying on its floor will remain stationary, the block and trolley system shown in a given that the co-efficient of static friction Fig. 4.12(a), if the coefficient of kinetic friction between the box and the train’s floor is between the trolley and the surface is 0.04? 0.15. What is the tension in the string? (Take g = Answer Since the acceleration of the box is due 10 m s-2). Neglect the mass of the string. to the static friction, ma = fs ≤ µs N = µs m g i.e. a ≤ µs g ∴ amax = µs g = 0.15 x 10 m s–2 = 1.5 m s–2 ⊳ ⊳ Example 4.8 See Fig. 4.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ? (a) (b) (c) Fig. 4.11 Fig. 4.12 Reprint 2025-26 62 PHYSICS Answer As the string is inextensible, and the is the reason why discovery of the wheel has pully is smooth, the 3 kg block and the 20 kg been a major milestone in human history. trolley both have same magnitude of Rolling friction again has a complex origin, acceleration. Applying second law to motion of though somewhat different from that of static the block (Fig. 4.12(b)), and sliding friction. During rolling, the surfaces in contact get momentarily deformed a little, and 30 – T = 3a this results in a finite area (not a point) of theApply the second law to motion of the trolley (Fig. body being in contact with the surface. The net4.12(c)), effect is that the component of the contact force T – fk = 20 a. parallel to the surface opposes motion. Now fk = µk N, We often regard friction as something Here µk = 0.04, undesirable. In many situations, like in a N = 20 x 10 machine with different moving parts, friction = 200 N. does have a negative role. It opposes relative Thus the equation for the motion of the trolley is motion and thereby dissipates power in the form T – 0.04 x 200 = 20 a Or T – 8 = 20a. of heat, etc. Lubricants are a way of reducing kinetic friction in a machine. Another way is to 22 –2 These equations give a = m s = 0.96 m s-2 use ball bearings between two moving parts of a 23 and T = 27.1 N. ⊳ machine [Fig. 4.13(a)]. Since the rolling friction between ball bearings and the surfaces in Rolling friction contact is very small, power dissipation is reduced. A thin cushion of air maintainedA body like a ring or a sphere rolling without between solid surfaces in relative motion isslipping over a horizontal plane will suffer no another effective way of reducing friction friction, in principle. At every instant, there is (Fig. 4.13(a)). just one point of contact between the body and In many practical situations, however, friction the plane and this point has no motion relative is critically needed. Kinetic friction that to the plane. In this ideal situation, kinetic or dissipates power is nevertheless important for static friction is zero and the body should quickly stopping relative motion. It is made use continue to roll with constant velocity. We know, of by brakes in machines and automobiles. in practice, this will not happen and some Similarly, static friction is important in daily resistance to motion (rolling friction) does occur, life. We are able to walk because of friction. It i.e. to keep the body rolling, some applied force is impossible for a car to move on a very slippery is needed. For the same weight, rolling friction road. On an ordinary road, the friction between is much smaller (even by 2 or 3 orders of the tyres and the road provides the necessary magnitude) than static or sliding friction. This external force to accelerate the car. Fig. 4.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion. Reprint 2025-26 LAWS OF MOTION 63 4.10 CIRCULAR MOTION is the static friction that provides the centripetal acceleration. Static friction opposes the We have seen in Chapter 4 that acceleration of impending motion of the car moving away from a body moving in a circle of radius R with uniform the circle. Using equation (4.14) & (4.16) we get speed v is v2/R directed towards the centre. the result According to the second law, the force f providing this acceleration is : c mv 2 f = ≤ µs N 2 mv R f c = (4.16) R 2 µs RN v ≤ = µs Rg [∵N = mg] where m is the mass of the body. This force m directed forwards the centre is called the which is independent of the mass of the car. centripetal force. For a stone rotated in a circle This shows that for a given value of µs and R, by a string, the centripetal force is provided by there is a maximum speed of circular motion of the tension in the string. The centripetal force the car possible, namely for motion of a planet around the sun is the v max = µs Rg (4.18) (a) (b) Fig. 4.14 Circular motion of a car on (a) a level road, (b) a banked road. gravitational force on the planet due to the sun. Motion of a car on a banked road For a car taking a circular turn on a horizontal We can reduce the contribution of friction to the road, the centripetal force is the force of friction. circular motion of the car if the road is banked The circular motion of a car on a flat and (Fig. 4.14(b)). Since there is no acceleration along banked road give interesting application of the the vertical direction, the net force along this laws of motion. direction must be zero. Hence, Motion of a car on a level road N cos θ = mg + f sin θ (4.19a) Three forces act on the car (Fig. 4.14(a): The centripetal force is provided by the horizontal(i) The weight of the car, mg components of N and f.(ii) Normal reaction, N (iii) Frictional force, f 2 mv As there is no acceleration in the vertical N sin θ + f cos θ = (4.19b) R direction N – mg = 0 But f ≤ µsN N = mg (4.17) Thus to obtain vmax we putThe centripetal force required for circular motion is along the surface of the road, and is provided f = µs N . by the component of the contact force between Then Eqs. (4.19a) and (4.19b) become road and the car tyres along the surface. This by definition is the frictional force. Note that it N cos θ = mg + µsN sin θ (4.20a) Reprint 2025-26 64 PHYSICS N sin θ + µsN cos θ = mv2/R (4.20b) ⊳ Example 4.11 A circular racetrack of From Eq. (4.20a), we obtain radius 300 m is banked at an angle of 15°. mg If the coefficient of friction between the N = wheels of a race-car and the road is 0.2, cosθ – µs sinθ what is the (a) optimum speed of the race- Substituting value of N in Eq. (4.20b), we get car to avoid wear and tear on its tyres, and 2 (b) maximum permissible speed to avoidmg ( sinθ+ µs cosθ) mv max = slipping ? cosθ– µs sinθ R 1 Answer On a banked road, the horizontal 2 component of the normal force and the frictional  µs + tanθ  or v max =  Rg  (4.21) force contribute to provide centripetal force to  1 – µs tanθ keep the car moving on a circular turn without Comparing this with Eq. (4.18) we see that slipping. At the optimum speed, the normal maximum possible speed of a car on a banked reaction’s component is enough to provide the road is greater than that on a flat road. needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by For µs = 0 in Eq. (4.21 ), Eq. (4.22): vo = ( R g tan θ ) ½ (4.22) vO = (R g tan θ)1/2 At this speed, frictional force is not needed at all Here R = 300 m, θ = 15°, g = 9.8 m s-2; we to provide the necessary centripetal force. have Driving at this speed on a banked road will cause vO = 28.1 m s-1. little wear and tear of the tyres. The same The maximum permissible speed vmax is given by equation also tells you that for v < vo, frictional Eq. (4.21): force will be up the slope and that a car can be parked only if tan θ ≤ µs. ⊳ ⊳ Example 4.10 A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction 4.11 SOLVING PROBLEMS IN MECHANICS between the tyres and the road is 0.1. Will The three laws of motion that you have learnt in the cyclist slip while taking the turn? this chapter are the foundation of mechanics. You should now be able to handle a large variety Answer On an unbanked road, frictional force of problems in mechanics. A typical problem in alone can provide the centripetal force needed mechanics usually does not merely involve a to keep the cyclist moving on a circular turn single body under the action of given forces. without slipping. If the speed is too large, or if More often, we will need to consider an assembly the turn is too sharp (i.e. of too small a radius) of different bodies exerting forces on each other. or both, the frictional force is not sufficient to Besides, each body in the assembly experiences provide the necessary centripetal force, and the the force of gravity. When trying to solve a cyclist slips. The condition for the cyclist not to problem of this type, it is useful to remember slip is given by Eq. (4.18) : the fact that we can choose any part of the assembly and apply the laws of motion to that R g v2 ≤ µs part provided we include all forces on the chosen Now, R = 3 m, g = 9.8 m s-2, µs = 0.1. That is, part due to the remaining parts of the assembly. R g = 2.94 m2 s-2. v = 18 km/h = 5 m s-1; i.e., We may call the chosen part of the assembly asµs v2 = 25 m2 s-2. The condition is not obeyed. the system and the remaining part of the The cyclist will slip while taking the assembly (plus any other agencies of forces) as circular turn. ⊳ the environment. We have followed the same Reprint 2025-26 LAWS OF MOTION 65 method in solved examples. To handle a typical the net force on the block must be zero i.e., problem in mechanics systematically, one R = 20 N. Using third law the action of the should use the following steps : block (i.e. the force exerted on the floor by (i) Draw a diagram showing schematically the the block) is equal to 20 N and directed various parts of the assembly of bodies, the vertically downwards. links, supports, etc. (b) The system (block + cylinder) accelerates (ii) Choose a convenient part of the assembly downwards with 0.1 m s-2. The free-body as one system. diagram of the system shows two forces on (iii) Draw a separate diagram which shows this the system : the force of gravity due to the system and all the forces on the system by earth (270 N); and the normal force R′ by the the remaining part of the assembly. Include floor. Note, the free-body diagram of the also the forces on the system by other system does not show the internal forces agencies. Do not include the forces on the between the block and the cylinder. Applying environment by the system. A diagram of the second law to the system, this type is known as ‘a free-body diagram’. 270 – R′ = 27 × 0.1N (Note this does not imply that the system ie. R′ = 267.3 N under consideration is without a net force). (iv) In a free-body diagram, include information about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion. (v) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F. The following example illustrates the above procedure : ⊳ Example 4.12 See Fig. 4.15. A wooden Fig. 4.15 block of mass 2 kg rests on a soft horizontal By the third law, the action of the system on floor. When an iron cylinder of mass 25 kg the floor is equal to 267.3 N vertically downward. is placed on top of the block, the floor yields steadily and the block and the cylinder Action-reaction pairs together go down with an acceleration of For (a): (i) the force of gravity (20 N) on the block 0.1 m s–2. What is the action of the block by the earth (say, action); the force of on the floor (a) before and (b) after the floor gravity on the earth by the block yields ? Take g = 10 m s–2. Identify the (reaction) equal to 20 N directed action-reaction pairs in the problem. upwards (not shown in the figure). (ii) the force on the floor by the block Answer (action); the force on the block by the (a) The block is at rest on the floor. Its free-body floor (reaction). diagram shows two forces on the block, the For (b): (i) the force of gravity (270 N) on the force of gravitational attraction by the earth system by the earth (say, action); the equal to 2 × 10 = 20 N; and the normal force force of gravity on the earth by the R of the floor on the block. By the First Law, system (reaction), equal to 270 N, Reprint 2025-26 66 PHYSICS directed upwards (not shown in the gravity on the mass in (a) or (b) and the normal figure). force on the mass by the floor are not action- (ii) the force on the floor by the system reaction pairs. These forces happen to be equal (action); the force on the system by the and opposite for (a) since the mass is at rest. floor (reaction). In addition, for (b), the They are not so for case (b), as seen already. force on the block by the cylinder and The weight of the system is 270 N, while the the force on the cylinder by the block normal force R′ is 267.3 N. ⊳ also constitute an action-reaction pair. The practice of drawing free-body diagrams is The important thing to remember is that an of great help in solving problems in mechanics. action-reaction pair consists of mutual forces It allows you to clearly define your system and which are always equal and opposite between consider all forces on the system due to objects two bodies. Two forces on the same body which that are not part of the system itself. A number happen to be equal and opposite can never of exercises in this and subsequent chapters will constitute an action-reaction pair. The force of help you cultivate this practice. SUMMARY 1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A force is necessary in practice to counter the opposing force of friction. 2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and arrived at the law of inertia. Newton’s first law of motion is the same law rephrased thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise”. In simple terms, the First Law is “If external force on a body is zero, its acceleration is zero”. 3. Momentum (p ) of a body is the product of its mass (m) and velocity (v) : p = m v 4. Newton’s second law of motion : The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts. Thus d p F = k = k m a d t where F is the net external force on the body and a its acceleration. We set the constant of proportionality k = 1 in SI units. Then dp F = = ma dt The SI unit of force is newton : 1 N = 1 kg m s-2. (a) The second law is consistent with the First Law (F = 0 implies a = 0) (b) It is a vector equation (c) It is applicable to a particle, and also to a body or a system of particles, provided F is the total external force on the system and a is the acceleration of the system as a whole. (d) F at a point at a certain instant determines a at the same point at that instant. That is the Second Law is a local law; a at an instant does not depend on the history of motion. 4. Impulse is the product of force and time which equals change in momentum. The notion of impulse is useful when a large force acts for a short time to produce a measurable change in momentum. Since the time of action of the force is very short, one can assume that there is no appreciable change in the position of the body during the action of the impulsive force. 6. Newton’s third law of motion: To every action, there is always an equal and opposite reaction Reprint 2025-26 LAWS OF MOTION 67 In simple terms, the law can be stated thus : Forces in nature always occur between pairs of bodies. Force on a body A by body B is equal and opposite to the force on the body B by A. Action and reaction forces are simultaneous forces. There is no cause-effect relation between action and reaction. Any of the two mutual forces can be called action and the other reaction. Action and reaction act on different bodies and so they cannot be cancelled out. The internal action and reaction forces between different parts of a body do, however, sum to zero. 7. Law of Conservation of Momentum The total momentum of an isolated system of particles is conserved. The law follows from the second and third law of motion. 8. Friction Frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surface in contact. Static friction fs opposes impending relative motion; kinetic friction fk opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws : = µs R f s ≤ ( f s )max f = µ R k k µs (co-efficient of static friction) and µk (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that µk is less than µs . POINTS TO PONDER 1. Force is not always in the direction of motion. Depending on the situation, F may be along v, opposite to v, normal to v or may make some other angle with v. In every case, it is parallel to acceleration. 2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v = 0 but the force continues to be its weight mg and the acceleration is not zero but g. 3. Force on a body at a given time is determined by the situation at the location of the body at that time. Force is not ‘carried’ by the body from its earlier history of motion. The moment after a stone is released out of an accelerated train, there is no horizontal force (or acceleration) on the stone, if the effects of the surrounding air are neglected. The stone then has only the vertical force of gravity. 4. In the second law of motion F = m a, F stands for the net force due to all material agencies external to the body. a is the effect of the force. ma should not be regarded as yet another force, besides F. Reprint 2025-26 68 PHYSICS 5. The centripetal force should not be regarded as yet another kind of force. It is simply a name given to the force that provides inward radial acceleration to a body in circular motion. We should always look for some material force like tension, gravitational force, electrical force, friction, etc as the centripetal force in any circular motion. 6. Static friction is a self-adjusting force up to its limit µs N (fs ≤ µs N). Do not put fs= µs N without being sure that the maximum value of static friction is coming into play. 7. The familiar equation mg = R for a body on a table is true only if the body is in equilibrium. The two forces mg and R can be different (e.g. a body in an accelerated lift). The equality of mg and R has no connection with the third law. 8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for simultaneous mutual forces between a pair of bodies. Unlike their meaning in ordinary language, action does not precede or cause reaction. Action and reaction act on different bodies. 9. The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’, ‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’ in different contexts. For clarity, every force and its equivalent terms encountered in mechanics should be reduced to the phrase ‘force on A by B’. 10. For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground. 11. The objective concept of force in physics should not be confused with the subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of our body are subject to an inward force, but we have a feeling of being pushed outward – the direction of impending motion. EXERCISES (For simplicity in numerical calculations, take g = 10 m s-2) 4.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance. 4.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c ) just after it is dropped from the window of a train accelerating with 1 m s-2, (d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout. Reprint 2025-26 LAWS OF MOTION 69 4.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is : mv2 mv 2 (i) T, (ii) T − , (iii) T + , (iv) 0 l l T is the tension in the string. [Choose the correct alternative]. 4.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop ? 4.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ? 4.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body. 4.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. 4.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast. 4.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s. 4.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.) 4.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position. 4.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s-1, (b) downwards with a uniform acceleration of 5 m s-2, (c) upwards with a uniform acceleration of 5 m s-2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ? 4.14 Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only). Fig. 4.16 4.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case? Reprint 2025-26 70 PHYSICS 4.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released. 4.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions. 4.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ? 4.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun ? 4.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.) 4.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ? 4.22 If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks : (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ? 4.23 Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch. Reprint 2025-26 CHAPTER FIVE WORK, ENERGY AND POWER 5.1 INTRODUCTION The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a 5.1 Introduction construction worker carrying bricks, a student studying for a competitive examination, an artist painting a beautiful

4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

4.10 THE MOVING COIL GALVANOMETER Currents and voltages in circuits have been discussed extensively in Chapters 3. But how do we measure them? How do we claim that current in a circuit is 1.5 A or the voltage drop across a resistor is 1.2 V? Figure 4.20 exhibits a very useful instrument for this purpose: the moving coil galvanometer (MCG). It is a device whose principle can be understood on the basis of our discussion in Section 4.9. The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. 4.20), in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it. This torque is given by Eq. (4.20) to be 129 τ = NI AB Reprint 2025-26 Physics where the symbols have their usual meaning. Since the field is radial by design, we have taken sin θ = 1 in the above expression for the torque. The magnetic torque NIAB tends to rotate the coil. A spring Sp provides a counter torque kφ that balances the magnetic torque NIAB; resulting in a steady angular deflection φ. In equilibrium kφ = NI AB where k is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection φ is indicated on the scale by a pointer attached to the spring. We have  NAB  φ =  k  I (4.26) The quantity in brackets is a constant for a given galvanometer. The galvanometer can be used in a number of ways. It can be used as a detector to check if a current is FIGURE 4.20 The moving coil flowing in the circuit. We have come across this usage galvanometer. Its elements are in the Wheatstone’s bridge arrangement. In this usage described in the text. Depending on the neutral position of the pointer (when no current is the requirement, this device can be flowing through the galvanometer) is in the middle of used as a current detector or for the scale and not at the left end as shown in Fig.4.20. measuring the value of the current Depending on the direction of the current, the pointer’s (ammeter) or voltage (voltmeter). deflection is either to the right or the left. The galvanometer cannot as such be used as an ammeter to measure the value of the current in a given circuit. This is for two reasons: (i) Galvanometer is a very sensitive device, it gives a full- scale deflection for a current of the order of µA. (ii) For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit. To overcome these difficulties, one attaches a small resistance rs, called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt. The resistance of this arrangement is, ≃ RG rs / (RG + rs) rs if RG >> rs If rs has small value, in relation to the resistance of the rest of the circuit Rc, the effect of introducing the measuring instrument is also small and negligible. This arrangement is schematically shown in Fig. 4.21. FIGURE 4.21 The scale of this ammeter is calibrated and then graduated to read off Conversion of a the current value with ease. We define the current sensitivity of the galvanometer (G) to galvanometer as the deflection per unit current. From Eq. (4.26) this an ammeter by the current sensitivity is, introduction of a NAB φshunt resistance rs of = (4.27) very small value in I k parallel. A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N. We choose galvanometers having 130 sensitivities of value, required by our experiment. Reprint 2025-26 Moving Charges and Magnetism The galvanometer can also be used as a voltmeter to measure the voltage across a given section of the circuit. For this it must be connected in parallel with that section of the circuit. Further, it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Usually we like to keep the disturbance due to the measuring device below one per cent. To ensure this, a large resistance R is connected in series with the galvanometer. This arrangement is schematically depicted in Fig.4.22. Note that the resistance of the voltmeter is now, FIGURE 4.22 RG + R ≃ R : large Conversion of a The scale of the voltmeter is calibrated to read off the voltage value galvanometer (G) to a with ease. We define the voltage sensitivity as the deflection per unit voltmeter by the introduction of avoltage. From Eq. (4.26), resistance R of large φ  NAB  I  NAB  1 value in series. = = (4.28) V  k  V  k  R An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.27) which provides a measure of current sensitivity. If N → 2N, i.e., we double the number of turns, then φ φ → 2 I I Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N →2N, and R →2R, thus the voltage sensitivity, φ φ → V V remains unchanged. So in general, the modification needed for conversion of a galvanometer to an ammeter will be different from what is needed for converting it into a voltmeter. Example 4.12 In the circuit (Fig. 4.23) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance RG = 60.00 Ω; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance rs = 0.02 Ω; (c) is an ideal ammeter with zero resistance? EXAMPLE FIGURE 4.23 4.12 131 Reprint 2025-26 Physics Solution (a) Total resistance in the circuit is, RG + 3 = 63 Ω. Hence, I = 3/63 = 0.048 A. (b) Resistance of the galvanometer converted to an ammeter is, R G rs 60 Ω × 0. 02 Ω = ≃ 0.02Ω R G + rs ( 60 + 0 .02 )Ω 4.12 Total resistance in the circuit is, 0.02 Ω+ 3 Ω= 3.02 Ω. Hence, I = 3/3.02 = 0.99 A. (c) For the ideal ammeter with zero resistance, EXAMPLE I = 3/3 = 1.00 A SUMMARY 1. The total force on a charge q moving with velocity v in the presence of magnetic and electric fields B and E, respectively is called the Lorentz force. It is given by the expression: F = q (v × B + E) The magnetic force q (v × B) is normal to v and work done by it is zero. 2. A straight conductor of length l and carrying a steady current I experiences a force F in a uniform external magnetic field B, F = I l × B where|l| = l and the direction of l is given by the direction of the current. 3. In a uniform magnetic field B, a charge q executes a circular orbit in a plane normal to B. Its frequency of uniform circular motion is called the cyclotron frequency and is given by: q B νc = 2 π m This frequency is independent of the particle’s speed and radius. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles. 4. The Biot-Savart law asserts that the magnetic field dB due to an element dl carrying a steady current I at a point P at a distance r from the current element is: r d l µ × 0 I dB = 3 r 4 π To obtain the total field at P, we must integrate this vector expression over the entire length of the conductor. 5. The magnitude of the magnetic field due to a circular coil of radius R carrying a current I at an axial distance x from the centre is Reprint 2025-26 Moving Charges and Magnetism 2 IR µ 0 B = 2 2 2( x + R )3/2 At the centre this reduces to µ0 I B = 2 R 6. Ampere’s Circuital Law: Let an open surface S be bounded by a loop B.d l = µ0 I where I refers to C. Then the Ampere’s law states that ∫CÑ the current passing through S. The sign of I is determined from the right-hand rule. We have discussed a simplified form of this law. If B is directed along the tangent to every point on the perimeter L of a closed curve and is constant in magnitude along perimeter then, BL = µ0 Ie where Ie is the net current enclosed by the closed circuit. 7. The magnitude of the magnetic field at a distance R from a long, straight wire carrying a current I is given by: µ0 I B = 2 π R The field lines are circles concentric with the wire. 8. The magnitude of the field B inside a long solenoid carrying a current I is B = µ0nI where n is the number of turns per unit length. 9. Parallel currents attract and anti-parallel currents repel. 10. A planar loop carrying a current I, having N closely wound turns, and an area A possesses a magnetic moment m where, m = N I A and the direction of m is given by the right-hand thumb rule : curl the palm of your right hand along the loop with the fingers pointing in the direction of the current. The thumb sticking out gives the direction of m (and A) When this loop is placed in a uniform magnetic field B, the force F on it is: F = 0 And the torque on it is, τ = m × B In a moving coil galvanometer, this torque is balanced by a counter- torque due to a spring, yielding kφ = NI AB where φ is the equilibrium deflection and k the torsion constant of the spring. 11. A moving coil galvanometer can be converted into a ammeter by introducing a shunt resistance rs, of small value in parallel. It can be converted into a voltmeter by introducing a resistance of a large value in series. 133 Reprint 2025-26 Physics Physical Quantity Symbol Nature Dimensions Units Remarks Permeability of free µ0 Scalar [MLT –2A–2] T m A–1 4π × 10–7 T m A–1 space Magnetic Field B Vector [M T –2A–1] T (telsa) Magnetic Moment m Vector [L2A] A m2 or J/T Torsion Constant k Scalar [M L2T –2] N m rad–1 Appears in MCG POINTS TO PONDER 1. Electrostatic field lines originate at a positive charge and terminate at a negative charge or fade at infinity. Magnetic field lines always form closed loops. 2. The discussion in this Chapter holds only for steady currents which do not vary with time. When currents vary with time Newton’s third law is valid only if momentum carried by the electromagnetic field is taken into account. 3. Recall the expression for the Lorentz force, F = q (v × B + E) This velocity dependent force has occupied the attention of some of the greatest scientific thinkers. If one switches to a frame with instantaneous velocity v, the magnetic part of the force vanishes. The motion of the charged particle is then explained by arguing that there exists an appropriate electric field in the new frame. We shall not discuss the details of this mechanism. However, we stress that the resolution of this paradox implies that electricity and magnetism are linked phenomena (electromagnetism) and that the Lorentz force expression does not imply a universal preferred frame of reference in nature. 4. Ampere’s Circuital law is not independent of the Biot-Savart law. It can be derived from the Biot-Savart law. Its relationship to the Biot-Savart law is similar to the relationship between Gauss’s law and Coulomb’s law. EXERCISES 4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? 4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? 4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B 134 at a point 2.5 m east of the wire. Reprint 2025-26 Moving Charges and Magnetism

4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.) 135 Reprint 2025-26 Physics Chapter Five MAGNETISM AND MATTER 5.1 INTRODUCTION Magnetic phenomena are universal in nature. Vast, distant galaxies, the tiny invisible atoms, humans and beasts all are permeated through and through with a host of magnetic fields from a variety of sources. The earth’s magnetism predates human evolution. The word magnet is derived from the name of an island in Greece called magnesia where magnetic ore deposits were found, as early as 600 BC. In the previous chapter we have learned that moving charges or electric currents produce magnetic fields. This discovery, which was made in the early part of the nineteenth century is credited to Oersted, Ampere, Biot and Savart, among others. In the present chapter, we take a look at magnetism as a subject in its own right. Some of the commonly known ideas regarding magnetism are: (i) The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north. (ii) When a bar magnet is freely suspended, it points in the north-south direction. The tip which points to the geographic north is called the north pole and the tip which points to the geographic south is called 136 the south pole of the magnet. Reprint 2025-26 Magnetism and Matter (iii) There is a repulsive force when north poles ( or south poles ) of two magnets are brought close together. Conversely, there is an attractive force between the north pole of one magnet and the south pole of the other. (iv) We cannot isolate the north, or south pole of a magnet. If a bar magnet is broken into two halves, we get two similar bar magnets with somewhat weaker properties. Unlike electric charges, isolated magnetic north and south poles known as magnetic monopoles do not exist. (v) It is possible to make magnets out of iron and its alloys. We begin with a description of a bar magnet and its behaviour in an external magnetic field. We describe Gauss’s law of magnetism. We next describe how materials can be classified on the basis of their magnetic properties. We describe para-, dia-, and ferromagnetism. 5.25.25.25.25.2 TTTHETTHEHEHEHE BBBARBBARARARAR MMMAGNETMMAGNETAGNETAGNETAGNET We begin our study by examining iron filings sprinkled on a sheet of glass placed over a short bar magnet. The arrangement of iron filings is shown in Fig. 5.1. The pattern of iron filings suggests that the magnet has two poles similar to the positive and negative charge of an electric dipole. As mentioned in the introductory section, one pole is designated the North pole and the other, the South pole. When suspended freely, these poles point approximately towards the geographic north and south poles, respectively. A similar pattern of iron filings is observed around a current carrying solenoid. FIGUREFIGUREFIGUREFIGUREFIGURE 5.15.15.15.15.1 The arrangement 5.2.15.2.15.2.15.2.15.2.1 TheTheTheTheThe magneticmagneticmagneticmagneticmagnetic fieldfieldfieldfieldfield lineslineslineslineslines of iron filings surrounding a bar magnet. The pattern mimics The pattern of iron filings permits us to plot magnetic field lines. The pattern the magnetic field lines*. This is shown both suggests that the bar magnet is for the bar-magnet and the current- a magnetic dipole. carrying solenoid in Fig. 5.2. For comparison refer to the Chapter 1, Figure 1.14(d). Electric field lines of an electric dipole are also displayed in Fig. 5.2(c). The magnetic field lines are a visual and intuitive realisation of the magnetic field. Their properties are: (i) The magnetic field lines of a magnet (or a solenoid) form continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity. * In some textbooks the magnetic field lines are called magnetic lines of force. This nomenclature is avoided since it can be confusing. Unlike electrostatics the field lines in magnetism do not indicate the direction of the force on a 137 (moving) charge. Reprint 2025-26 Physics FIGURE 5.2 The field lines of (a) a bar magnet, (b) a current-carrying finite solenoid and (c) electric dipole. At large distances, the field lines are very similar. The curves labelled i and ii are closed Gaussian surfaces. (ii) The tangent to the field line at a given point represents the direction of the net magnetic field B at that point. (iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B. In Fig. 5.2(a), B is larger around region ii than in region i . (iv) The magnetic field lines do not intersect, for if they did, the direction of the magnetic field would not be unique at the point of intersection. One can plot the magnetic field lines in a variety of ways. One way is to place a small magnetic compass needle at various positions and note its orientation. This gives us an idea of the magnetic field direction at various points in space. 5.2.2 Bar magnet as an FIGURE 5.3 Calculation of (a) The axial field of a equivalent solenoid finite solenoid in order to demonstrate its In the previous chapter, we have similarity to that of a bar magnet. (b) A magnetic explained how a current loop acts as a needle in a uniform magnetic field B. The arrangement may be used to determine either B magnetic dipole (Section 4.9). We or the magnetic moment m of the needle. mentioned Ampere’s hypothesis that all magnetic phenomena can be explained in 138 terms of circulating currents. Reprint 2025-26 Magnetism and Matter The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smaller solenoids with weaker magnetic properties. The field lines remain continuous, emerging from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases. To make this analogy more firm we may calculate the axial field of a finite solenoid depicted in Fig. 5.3 (a). We can demonstrate that at large distances this axial field resembles that of a bar magnet. The magnitude of the field at point P due to the solenoid is µ0 2 m B = 3 (5.1) 4π r This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field. 5.2.3 The dipole in a uniform magnetic field Let’s place a small compass needle of known magnetic moment m allowing it to oscillate in the magnetic field. This arrangement is shown in Fig. 5.3(b). The torque on the needle is [see Eq. (4.23)], τ = m × B (5.2) In magnitude τ = mB sinθ Here τ is restoring torque and θ is the angle between m and B. An expression for magnetic potential energy can be obtained on lines similar to electrostatic potential energy. The magnetic potential energy Um is given by (θ)d θ U m = ∫τ sin θ d θ = −mB cos θ = ∫mB = –m.B (5.3) We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one’s convenience. Taking the constant of integration to be zero means fixing the zero of potential energy at θ = 90°, i.e., when the needle is perpendicular to the field. Equation (5.3) shows that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position). Example 5.1 (a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a EXAMPLE torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? 5.1 139 Reprint 2025-26 Physics (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.] Solution (a) In either case, one gets two magnets, each with a north and south pole. (b) No force if the field is uniform. The iron nail experiences a non- uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole. (c) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor. (d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the 5.1 other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, EXAMPLE then A is magnetised. 5.2.4 The electrostatic analog Comparison of Eqs. (5.1), (5.2) and (5.3) with the corresponding equations for electric dipole (Chapter 1), suggests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p, by making the following replacements: 1 µ0 E → B , p → m , → 4 πε0 4 π In particular, we can write down the equatorial field (BE) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet: µ0 m B E = − 4 π r 3 (5.4) Likewise, the axial field (BA) of a bar magnet for r >> l is: µ0 2 m B A =140 4 π r 3 (5.5) Reprint 2025-26 Magnetism and Matter Equation (5.5) is just Eq. (5.1) in the vector form. Table 5.1 summarises the analogy between electric and magnetic dipoles. TABLE 5.1 THE DIPOLE ANALOGY Electrostatics Magnetism 1/ε0 µ0 Dipole moment p m Equatorial Field for a short dipole –p/4πε0r 3 – µ0 m / 4π r 3 Axial Field for a short dipole 2p/4πε0r 3 µ0 2m / 4π r 3 External Field: torque p × E m × B External Field: Energy –p.E –m.B Example 5.2 Figure 5.4 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown? FIGURE 5.4 Solution Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression [Eqs. (5.4) and (5.5)]: µ0 m P B P = − 3 (on the normal bisector) 4π r µ0 2 m P B P = 3 (on the axis) 4π r where mP is the magnetic moment of the dipole P. EXAMPLE Equilibrium is stable when mQ is parallel to BP, and unstable when it is anti-parallel to BP. 5.2 141 Reprint 2025-26 Physics For instance for the configuration Q3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q3 is stable. 5.2 Thus, (a) PQ1 and PQ2 (b) (i) PQ3, PQ6 (stable); (ii) PQ5, PQ4 (unstable) EXAMPLE (c) PQ6 5.3 MAGNETISM AND GAUSS’S LAW In Chapter 1, we studied Gauss’s law for electrostatics. In Fig 5.2(c), we see that for a closed surface represented by i , the number of lines leaving the surface is equal to the number of lines entering it. This is consistent with the fact that no net charge is enclosed by the surface. However, in the same figure, for the closed surface ii , there is a net outward flux, since it does include a net (positive) charge.1855) The situation is radically different for magnetic fields – which are continuous and form closed loops. Examine the Gaussian surfaces represented by i or ii in Fig 5.2(a) or Karl Friedrich Gauss Fig. 5.2(b). Both cases visually demonstrate that the(1777 (1777 – 1855) He was a number of magnetic field lines leaving the surface is child prodigy and was gifted balanced by the number of lines entering it. The net in mathematics, physics, magnetic flux is zero for both the surfaces. This is true engineering, astronomy for any closed surface.GAUSS and even land surveying. The properties of numbers fascinated him, and in his work he anticipated major mathematical development of later times. Along with Wilhelm Welser, he built theFRIEDRICH first electric telegraph in 1833. His mathematical theory of curved surface laid the foundation for theKARL later work of Riemann. FIGURE 5.5 Consider a small vector area element ∆S of a closed surface S as in Fig. 5.5. The magnetic flux through ÄS is defined as ∆φB = B.∆S, where B is the field at ∆S. We divide S into many small area elements and calculate the individual flux through each. Then, the net flux φB is, B.∆S = 0 (5.6) φB = ∑ ∆ φB = ∑ ’ all ’ ’ all ’ where ‘all’ stands for ‘all area elements ∆S′. Compare this with the Gauss’s law of electrostatics. The flux through a closed surface in that case is given by E.∆S =142 ∑ q ε0 Reprint 2025-26 Magnetism and Matter where q is the electric charge enclosed by the surface. The difference between the Gauss’s law of magnetism and that for electrostatics is a reflection of the fact that isolated magnetic poles (also called monopoles) are not known to exist. There are no sources or sinks of B; the simplest magnetic element is a dipole or a current loop. All magnetic phenomena can be explained in terms of an arrangement of dipoles and/or current loops. Thus, Gauss’s law for magnetism is: The net magnetic flux through any closed surface is zero. Example 5.3 Many of the diagrams given in Fig. 5.6 show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones. EXAMPLE FIGURE 5.6 5.3 143 Reprint 2025-26 Physics Solution (a) Wrong. Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor, as described in Chapter 4. (b) Wrong. Magnetic field lines (like electric field lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges. (c) Right. Magnetic lines are completely confined within a toroid. Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field. (d) Wrong. Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops. (e) Right. These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the net flux of the field is zero. (f ) Wrong. These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged 5.3 carefullylower plate.grasped.The difference between Fig. [5.6(e) and (f)] should be (g) Wrong. Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric EXAMPLE field lines. Example 5.4 (a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) If magnetic monopoles existed, how would the Gauss’s law of 5.4 magnetism be modified? (c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another EXAMPLE element of the same wire? Reprint 2025-26 Magnetism and Matter (d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero? Solution (a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force. (b) Gauss’s law of magnetism states that the flux of B through any closed surface is always zero ∫s B .∆s = 0 . If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, B.∆s = µ0qm where qm is the ∫S (monopole) magnetic charge enclosed by S.] (c) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.) (d) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection EXAMPLE with paramagnetic material where atoms have net dipole moment 5.4 through their net charge is zero.

4.4 MAGNETIC FIELD DUE TO A CURRENT ELEMENT, BIOT-SAVART LAW All magnetic fields that we know are due to currents (or moving charges) and due to intrinsic magnetic moments of particles. Here, we shall study the relation between current and the magnetic field it produces. It is given by the Biot-Savart’s law. Fig. 4.7 shows a finite conductor XY carrying current I. Consider an infinitesimal element dl of the conductor. The magnetic field dB due to this element is to be determined at a point P which is at a distance r from it. Let q be the angle between dl and the displacement vector r. According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length |dl|, and inversely proportional to the square of the distance r. Its direction* is perpendicular to the plane containing dl and r . Thus, in vector notation, I d l × r FIGURE 4.7 Illustration of d B ∝ r 3 the Biot-Savart law. The current element I dl µ0 I d l × r produces a field dB at a = 3 [4.7(a)] distance r. The Ä sign 4π r indicates that the where m0/4p is a constant of proportionality. The above expression field is perpendicular holds when the medium is vacuum. to the plane of this page and directed into it. * The sense of dl × r is also given by the Right Hand Screw rule : Look at the plane containing vectors dl and r. Imagine moving from the first vector towards second vector. If the movement is anticlockwise, the resultant is towards you. 113 If it is clockwise, the resultant is away from you. Reprint 2025-26 Physics The magnitude of this field is, θ µ0 I d l sin d B = [4.7(b)] 2 4 π r where we have used the property of cross-product. Equation [4.7 (a)] constitutes our basic equation for the magnetic field. The proportionality constant in SI units has the exact value, µ0 − 7 = 10 Tm/A [4.7(c)] 4 π We call µ0 the permeability of free space (or vacuum). The Biot-Savart law for the magnetic field has certain similarities, as well as, differences with the Coulomb’s law for the electrostatic field. Some of these are: (i) Both are long range, since both depend inversely on the square of distance from the source to the point of interest. The principle of superposition applies to both fields. [In this connection, note that the magnetic field is linear in the source I dl just as the electrostatic field is linear in its source: the electric charge.] (ii) The electrostatic field is produced by a scalar source, namely, the electric charge. The magnetic field is produced by a vector source I dl. (iii) The electrostatic field is along the displacement vector joining the source and the field point. The magnetic field is perpendicular to the plane containing the displacement vector r and the current element I dl. (iv) There is an angle dependence in the Biot-Savart law which is not present in the electrostatic case. In Fig. 4.7, the magnetic field at any point in the direction of dl (the dashed line) is zero. Along this line, θ = 0, sin θ = 0 and from Eq. [4.7(a)], |dB| = 0. There is an interesting relation between ε0, the permittivity of free space; µ0, the permeability of free space; and c, the speed of light in vacuum: µ0 1 −7 1 1 10 = = = ε0µ0 = ( 4 πε0 ) ( ) c 2 4 π (3 × 108 )2 9 × 10 9 We will discuss this connection further in Chapter 8 on the electromagnetic waves. Since the speed of light in vacuum is constant, the product µ0ε0 is fixed in magnitude. Choosing the value of either ε0 or µ0, fixes the value of the other. In SI units, µ0 is fixed to be equal to 4π × 10–7 in magnitude. Example 4.4 An element ∆=l ∆x ˆi is placed at the origin and carries a large current I = 10 A (Fig. 4.8). What is the magnetic field on the y- axis at a distance of 0.5 m. ∆x = 1 cm. 4.4 EXAMPLE 114 FIGURE 4.8 Reprint 2025-26 Moving Charges and Magnetism Solution µ0 I d l sin θ |d B | = 2 [using Eq. (4.7)] 4 π r −2 − 7 T m dl = ∆ x = 10 m , I = 10 A, r = 0.5 m = y, µ0 /4 π = 10 A θ = 90° ; sin θ = 1 10 − 7 × 10 × 10 −2 d B = − 2 = 4 × 10–8 T 25 × 10 The direction of the field is in the +z-direction. This is so since, ˆ ˆi × ˆj dl × r = ∆x ˆi × y ˆj = y ∆x ( ) = y ∆kx We remind you of the following cyclic property of cross-products, EXAMPLE ˆi × ˆj = kˆ ; ˆj × kˆ = ˆi ; kˆ × ˆi = ˆj Note that the field is small in magnitude. 4.4 In the next section, we shall use the Biot-Savart law to calculate the magnetic field due to a circular loop.

4.7 THE SOLENOID We shall discuss a long solenoid. By long solenoid we mean that the solenoid’s length is large compared to its radius. It consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced. So each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enamelled wires are used for winding so that turns are insulated from each other. FIGURE 4.15 (a) The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the exterior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. (b) The magnetic field of a finite solenoid. Figure 4.15 displays the magnetic field lines for a finite solenoid. We show a section of this solenoid in an enlarged manner in Fig. 4.15(a). Figure 4.15(b) shows the entire finite solenoid with its magnetic field. In Fig. 4.15(a), it is clear from the circular loops that the field between two neighbouring turns vanishes. In Fig. 4.15(b), we see that the field at the interior mid-point P is uniform, strong and along the axis of the solenoid. The field at the exterior mid-point Q is weak and moreover is along the axis of the solenoid with no perpendicular or normal component. As the FIGURE 4.16 The magnetic field of a very long solenoid. We consider a 121 rectangular Amperian loop abcd to determine the field. Reprint 2025-26 Physics solenoid is made longer it appears like a long cylindrical metal sheet. Figure 4.16 represents this idealised picture. The field outside the solenoid approaches zero. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis. Consider a rectangular Amperian loop abcd. Along cd the field is zero as argued above. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h. Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law [Eq. 4.13 (b)] BL = µ0Ie, B h = µ0I (n h) B = µ0 n I (4.16) The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field. We shall see in the next chapter that a large field is possible by inserting a soft iron core inside the solenoid. Example 4.8 A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? Solution The number of turns per unit length is, 500 n = = 1000 turns/m 4.8 0.5 The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a. Hence, we can use the long solenoid formula, namely, Eq. (4.20) B = µ0n I = 4π × 10–7 × 103 × 5 EXAMPLE = 6.28 × 10–3 T 4.8 FORCE BETWEEN TWO PARALLEL CURRENTS, THE AMPERE We have learnt that there exists a magnetic field due to a conductor carrying a current which obeys the Biot-Savart law. Further, we have learnt that an external magnetic field will exert a force on a current-carrying conductor. This follows from the Lorentz force formula. Thus, it is logical to expect that two current-carrying conductors placed near each other will exert (magnetic) forces on each other. In the period 1820-25, Ampere studied the nature of this FIGURE 4.17 Two long straight magnetic force and its dependence on the magnitude of parallel conductors carrying steady the current, on the shape and size of the conductors, as currents Ia and Ib and separated by a well as, the distances between the conductors. In this distance d. Ba is the magnetic field section, we shall take the simple example of two parallelset up by conductor ‘a’ at conductor ‘b’. current- carrying conductors, which will perhaps help us 122 to appreciate Ampere’s painstaking work. Reprint 2025-26 Moving Charges and Magnetism Figure 4.17 shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents Ia and Ib, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). Its magnitude is given by Eq. [4.15(a)] or from Ampere’s circuital law, µ0 I a Ba = 2 π d The conductor ‘b’ carrying a current Ib will experience a sideways force due to the field Ba. The direction of this force is towards the conductor ‘a’ (Verify this). We label this force as Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by Eq. (4.4), Fba = I b LB a µ0 I a I b = L (4.17) 2πd It is of course possible to compute the force on ‘a’ due to ‘b’. From considerations similar to above we can find the force Fab, on a segment of length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to Fba, and directed towards ‘b’. Thus, Fba = –Fab (4.18) Note that this is consistent with Newton’s third Law. Thus, at least for parallel conductors and steady currents, we have shown that the Biot-Savart law and the Lorentz force yield results in accordance with Newton’s third Law*. We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other. Thus, Parallel currents attract, and antiparallel currents repel. This rule is the opposite of what we find in electrostatics. Like (same sign) charges repel each other, but like (parallel) currents attract each other. Let fba represent the magnitude of the force Fba per unit length. Then, from Eq. (4.17), µ0 I a I b f ba = 2 π d (4.19) The above expression is used to define the ampere (A), which is one of the seven SI base units. * It turns out that when we have time-dependent currents and/or charges in motion, Newton’s third law may not hold for forces between charges and/or conductors. An essential consequence of the Newton’s third law in mechanics is conservation of momentum of an isolated system. This, however, holds even for the case of time-dependent situations with electromagnetic fields, provided 123 the momentum carried by fields is also taken into account. Reprint 2025-26 Physics The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. This definition of the ampere was adopted in 1946. It is a theoretical definition. In practice, one must eliminate the effect of the earth’s magnetic field and substitute very long wires by multiturn coils of appropriate geometries. An instrument called the current balance is used to measure this mechanical force. The SI unit of charge, namely, the coulomb, can now be defined in terms of the ampere. When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C). Example 4.9 The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Solution F = Il × B F = IlB sinθ The force per unit length is f = F/l = I B sinθ (a) When the current is flowing from east to west, θ = 90° Hence, f = I B = 1 × 3 × 10–5 = 3 × 10–5 N m–1 This is larger than the value 2×10–7 Nm–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere. 4.9 The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors. (b) When the current is flowing from south to north, θ = 0o f = 0 EXAMPLE Hence there is no force on the conductor. 4.9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE 4.9.1 Torque on a rectangular current loop in a uniform magnetic field We now show that a rectangular loop carrying a steady current I and placed in a uniform magnetic field experiences a torque. It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.11).124 Reprint 2025-26 Moving Charges and Magnetism We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in Fig. 4.18(a). The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop. Its magnitude is, F1 = I b B Similarly, it exerts a force F2 on the arm CD and F2 is directed out of the plane of the paper. F2 = I b B = F1 Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces F1 and F2. Figure 4.18(b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anticlockwise. This torque is (in magnitude), a a τ = F1 + F2 2 2 a a = IbB + IbB = I (ab ) B 2 2 FIGURE 4.18 (a) A rectangular = I A B (4.20) current-carrying coil in uniform where A = ab is the area of the rectangle. magnetic field. The magnetic moment We next consider the case when the plane of the loop, m points downwards. The torque τ is is not along the magnetic field, but makes an angle with along the axis and tends to rotate the it. We take the angle between the field and the normal to coil anticlockwise. (b) The couple acting on the coil.the coil to be angle θ (The previous case corresponds to θ = π/2). Figure 4.19 illustrates this general case. The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are F1 and F2. They too are equal and opposite, with magnitude, F1 = F2 = I b B But they are not collinear! This results in a couple as before. The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.19(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is, a a τ = F1 sin θ+ F2 sin θ 2 2 = I ab B sin θ = I A B sin θ (4.21) 125 Reprint 2025-26 Physics As θ à 0, the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. The torques in Eqs. (4.20) and (4.21) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as, m = I A (4.22) where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.18. Then as the angle between m and B is θ , Eqs. (4.20) and (4.21) can be expressed by one expression (4.23) This is analogous to the electrostatic case (Electric dipole of dipole moment pe in an electric field E). τ = p e × E As is clear from Eq. (4.22), the dimensions of the magnetic moment are [A][L2] and its unit is Am2. FIGURE 4.19 (a) The area vector of the loop From Eq. (4.23), we see that the torque τ ABCD makes an arbitrary angle θ with vanishes when m is either parallel or antiparallel the magnetic field. (b) Top view of to the magnetic field B. This indicates a state of the loop. The forces F1 and F2 acting equilibrium as there is no torque on the coil (this on the arms AB and CD also applies to any object with a magnetic moment are indicated. m). When m and B are parallel the equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position. When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. If the loop has N closely wound turns, the expression for torque, Eq. (4.23), still holds, with m = N I A (4.24) Example 4.10 A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis 4.10 of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated EXAMPLE 126 by 90°? The moment of inertia of the coil is 0.1 kg m2. Reprint 2025-26 Moving Charges and Magnetism Solution (a) From Eq. (4.12) µ0 NI B = 2R Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence, 4 × 10 −5 × 10 = − 1 (using π × 3.2 = 10) 2 × 10 = 2 × 10–3 T The direction is given by the right-hand thumb rule. (b) The magnetic moment is given by Eq. (4.24), m = N I A = N I π r2 = 100 × 3.2 × 3.14 × 10–2 = 10 A m2 The direction is once again given by the right-hand thumb rule. (c) τ = m × B [from Eq. (4.23)] = m B sin θ Initially, θ = 0. Thus, initial torque τi = 0. Finally, θ = π/2 (or 90º). Thus, final torque τf = m B = 10 × 2 = 20 N m. (d) From Newton’s second law, I where I is the moment of inertia of the coil. From chain rule, d ω d ω d θ d ω = = ω d t d θ d t d θ Using this, I ωd ω = m B sin θ d θ Integrating from θ = 0 to θ = π/2, EXAMPLE 4.10 Example 4.11 (a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis). EXAMPLE(b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of 4.11 127 Reprint 2025-26 Physics the total field (external field + field produced by the loop) is maximum. (c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape? Solution (a) No, because that would require τ to be in the vertical direction. But τ = I A × B, and since A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B. (b) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this 4.11 orientation,direction as theexternalmagneticfield,fieldbothproducednormal byto thethe loopplaneis inof thethe sameloop, thus giving rise to maximum flux of the total field. (c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater EXAMPLE area than any other shape. 4.9.2 Circular current loop as a magnetic dipole In this section, we shall consider the elementary magnetic element: the current loop. We shall show that the magnetic field (at large distances) due to current in a circular current loop is very similar in behaviour to the electric field of an electric dipole. In Section 4.5, we have evaluated the magnetic field on the axis of a circular loop, of a radius R, carrying a steady current I. The magnitude of this field is [(Eq. (4.11)], µ0 I R 2 B = 2 2 3/2 2 x + R ( ) and its direction is along the axis and given by the right-hand thumb rule (Fig. 4.10). Here, x is the distance along the axis from the centre of the loop. For x >> R, we may drop the R2 term in the denominator. Thus, µ0 IR 2 B = 3 2x Note that the area of the loop A = πR2. Thus, µ0 IA B = 3 2 πx As earlier, we define the magnetic moment m to have a magnitude IA, m = I A. Hence, B ≃µ0 m3 2 πx µ0 2 m = 3 [4.25(a)] 4 π x The expression of Eq. [4.25(a)] is very similar to an expression obtained earlier for the electric field of a dipole. The similarity may be seen if we substitute,128 µ0 → 1/ε0 Reprint 2025-26 Moving Charges and Magnetism m → p e (electrostatic dipole) B → E (electrostatic field) We then obtain, 2 pe E = 3 4 π ε0 x which is precisely the field for an electric dipole at a point on its axis. considered in Chapter 1, Section 1.9 [Eq. (1.20)]. It can be shown that the above analogy can be carried further. We had found in Chapter 1 that the electric field on the perpendicular bisector of the dipole is given by [See Eq.(1.21)], pe E ≃ 4 πε0 x 3 where x is the distance from the dipole. If we replace p à m and µ0 → 1/ ε0 in the above expression, we obtain the result for B for a point in the plane of the loop at a distance x from the centre. For x >>R, m x >> R B ≃µ0 3 ; [4.25(b)] 4π x The results given by Eqs. [4.25(a)] and [4.25(b)] become exact for a point magnetic dipole. The results obtained above can be shown to apply to any planar loop: a planar current loop is equivalent to a magnetic dipole of dipole moment m = I A, which is the analogue of electric dipole moment p. Note, however, a fundamental difference: an electric dipole is built up of two elementary units — the charges (or electric monopoles). In magnetism, a magnetic dipole (or a current loop) is the most elementary element. The equivalent of electric charges, i.e., magnetic monopoles, are not known to exist. We have shown that a current loop (i) produces a magnetic field (see Fig. 4.10) and behaves like a magnetic dipole at large distances, and (ii) is subject to torque like a magnetic needle. This led Ampere to suggest that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment, not accounted by circulating currents.

6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be

6.12 ANGULAR MOMENTUM IN CASE OF For computing the total angular momentum ROTATION ABOUT A FIXED AXIS of the whole rigid body, we add up the contribution of each particle of the body. We have studied in section 6.7, the angular momentum of a system of particles. We already Thus know from there that the time rate of total We denote by L ⊥ and L z the components of angular momentum of a system of particles L respectively perpendicular to the z-axis andabout a point is equal to the total external torque along the z-axis;on the system taken about the same point. When OC i × m i v i (6.42a)the total external torque is zero, the total angular L ⊥= ∑ momentum of the system is conserved. where mi and vi are respectively the mass and We now wish to study the angular momentum the velocity of the ith particle and Ci is the centrein the special case of rotation about a fixed axis. of the circle described by the particle; The general expression for the total angular momentum of the system of n particles is N and ˆ (6.42b) L = =∑i 1 ri × p i (6.25b) or L z = Iωk We first consider the angular momentum of The last step follows since the perpendicular a typical particle of the rotating rigid body. We distance of the ith particle from the axis is ri; andthen sum up the contributions of individual by definition the moment of inertia of the body particles to get L of the whole body. m i ri2 . For a typical particle l = r × p. As seen in the about the axis of rotation is I =∑ last section r = OP = OC + CP [Fig. 6.17(b)]. With Note L = L z + L ⊥ (6.42c)p = m v , l = ( OC × m v ) + ( CP × m v ) The rigid bodies which we have mainly considered in this chapter are symmetric about The magnitude of the linear velocity v of the the axis of rotation, i.e. the axis of rotation is particle at P is given by v = ωr⊥ where r⊥ is the one of their symmetry axes. For such bodies, for length of CP or the perpendicular distance of P a given OCi, for every particle which has a from the axis of rotation. Further, v is tangential velocity vi , there is another particle of velocity at P to the circle which the particle describes. –vi located diametrically opposite on the circle Using the right-hand rule one can check that with centre Ci described by the particle. TogetherCP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L ⊥ and as a is ˆk . Hence result for symmetric bodies L ⊥ is zero, and CP × m v = r⊥ (mv ) kˆ hence z = Iωkˆ (6.42d) = mr⊥2ω kˆ (since υ = ωr⊥ ) L = L Similarly, we can check that OC × v is For bodies, which are not symmetric about perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to Lz and hence the part of l along the fixed axis (i.e. the z-axis) L does not lie along the axis of rotation. by lz, then Referring to Table 6.1, can you tell in which l z = CP × m v = mr⊥2ωkˆ cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since ˆk is a and l = l z + OC × m v fixed (constant) vector, we get We note that lz is parallel to the fixed axis, ˆbut l is not. In general, for a particle, the angular I ω) k d ( L z ) =  d (  d t momentum l is not along the axis of rotation, d t i.e. for a particle, l and ω are not necessarily Now, Eq. (6.28b) states parallel. Compare this with the corresponding dL fact in translation. For a particle, p and v are = τ dtalways parallel to each other. Reprint 2025-26 122 PHYSICS As we have seen in the last section, only We have already derived this equation using those components of the external torques which the work - kinetic energy route. are along the axis of rotation, need to be taken into account, when we discuss rotation about a 6.12.1 Conservation of angular momentum fixed axis. This means we can take τ = τkˆ . We are now in a position to revisit the principle of conservation of angular momentum in the Since L = L z + L ⊥ and the direction of Lz (vector context of rotation about a fixed axis. From Eq. ˆk ) is fixed, it follows that for rotation about a (6.43c), if the external torque is zero, fixed axis, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz d L z = τkˆ (6.43a) may be replaced by L .(L and Lz are respectively d t the magnitudes of L and Lz.) This then is the required form, for fixed axis d L rotation, of Eq. (6.29a), which expresses theand ⊥= 0 (6.43b) dt general law of conservation of angular momentum Thus, for rotation about a fixed axis, the of a system of particles. Eq. (6.44) applies to many component of angular momentum perpendicular situations that we come across in daily life. You may do this experiment with your friend. Sit on a to the fixed axis is constant. As L z = Iωkˆ , we swivel chair (a chair with a seat, free to rotate get from Eq. (6.43a), about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your d ( Iω) = τ (6.43c) friend to rotate the chair rapidly. While the chair d t is rotating with considerable angular speed If the moment of inertia I does not change with stretch your arms horizontally. What happens? time, Your angular speed is reduced. If you bring back d dω your arms closer to your body, the angular speed ( Iω) = I = Iα increases again. This is a situation where thed t d t principle of conservation of angular momentumand we get from Eq. (6.43c), is applicable. If friction in the rotational τ = I α (6.41) Fig 6.32 (a) A demonstration of conservation of Fig 6.32 (b) An acrobat employing the principle of angular momentum. A girl sits on a conservation of angular momentum in swivel chair and stretches her arms/ her performance. brings her arms closer to the body. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 mechanism is neglected, there is no external A circus acrobat and a diver take advantage torque about the axis of rotation of the chair and of this principle. Also, skaters and classical, hence Iω is constant. Stretching the arms Indian or western, dancers performing a increases I about the axis of rotation, resulting in pirouette (a spinning about a tip–top) on the toes decreasing the angular speed ω. Bringing the of one foot display ‘mastery’ over this principle. arms closer to the body has the opposite effect. Can you explain? SUMMARY 1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. 2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. 3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. 4. In pure translation, every particle of the body moves with the same velocity at any instant of time. 5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. 6. The vector or cross product of two vector a and b is a vector written as a×b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule. 7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin. 8. The centre of mass of a system of n particles is defined as the point whose position vector is ri ∑m i R = M 9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant. 10. The angular momentum of a system of n particles about the origin is n L = ri × pi i =∑1 The torque or moment of force on a system of n particles about the origin is τ = ∑ri × Fi 1 The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0 and Reprint 2025-26 124 PHYSICS dL = τ ext dt 11. A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : Fi = 0 , ∑ and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : Fi = 0 . ∑ τi = ∑ri × 12. The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero. 13. The moment of intertia of a rigid body about an axis is defined by the formula I m i ri2 =∑ where ri is the perpendicular distance of the ith point of the body from the axis. The 1 2 kinetic energy of rotation is K = Iω . 2 POINTS TO PONDER 1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body. 2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV2/2, K = K′ + MV2/2 3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles. 4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles. 5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero. 6. The total torque on a system is independent of the origin if the total external force is zero. 7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 125 8. The angular momentum L and the angular velocity ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis. EXERCISES 6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. Fig. 6.33 6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. Reprint 2025-26 126 PHYSICS 6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping. 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Reprint 2025-26 CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything

6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero the velocity of the centre of mass remainsLet us recall that the linear momentum of a constant. (We assume throughout the particle is defined as discussion on systems of particles in this p = m v (6.12) chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F = (6.13) another, the individual particles may have dt Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 101 complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle. The vector Eq. (6.18a) is equivalent to three scalar equations, Px = c1, Py = c2 and Pz = c3 (6.18 b) (a) (b) Here Px, Py and Pz are the components of the total linear momentum vector P along the x–, y– Fig. 6.14 (a) Trajectories of two stars, S1 (dotted line) and z–axes respectively; c1, c2 and c3 are and S2 (solid line) forming a binary constants. system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest. move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b). In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference. (a) (b) In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star Fig. 6.13 (a) A heavy nucleus radium (Ra) splits into moves like a free particle, as shown in Fig.6.14 a lighter nucleus radon (Rn) and an alpha (a). The trajectories of the two stars of equal particle (nucleus of helium atom). The CM mass are also shown in the figure; they look of the system is in uniform motion. complicated. If we go to the centre of mass (b) The same spliting of the heavy nucleus radium (Ra) with the centre of mass at frame, then we find that there the two stars rest. The two product particles fly back are moving in a circle, about the centre of to back. mass, which is at rest. Note that the position of the stars have to be diametrically opposite As an example, let us consider the to each other [Fig. 6.14(b)]. Thus in our frame radioactive decay of a moving unstable particle, of reference, the trajectories of the stars are a combination of (i) uniform motion in a straightlike the nucleus of radium. A radium nucleus line of the centre of mass and (ii) circulardisintegrates into a nucleus of radon and an orbits of the stars about the centre of mass.alpha particle. The forces leading to the decay As can be seen from the two examples,are internal to the system and the external separating the motion of different parts of aforces on the system are negligible. So the total system into motion of the centre of mass andlinear momentum of the system is the same motion about the centre of mass is a verybefore and after decay. The two particles useful technique that helps in understanding produced in the decay, the radon nucleus and the motion of the system. the alpha particle, move in different directions in such a way that their centre of mass moves 6.5 VECTOR PRODUCT OF TWO VECTORS along the same path along which the original decaying radium nucleus was moving We are already familiar with vectors and their [Fig. 6.13(a)]. use in physics. In chapter 5 (Work, Energy, Power) If we observe the decay from the frame of we defined the scalar product of two vectors. An reference in which the centre of mass is at rest, important physical quantity, work, is defined as the motion of the particles involved in the decay a scalar product of two vector quantities, force looks particularly simple; the product particles and displacement. Reprint 2025-26 102 PHYSICS We shall now define another product of two A simpler version of the right hand rule is vectors. This product is a vector. Two important the following : Open up your right hand palm quantities in the study of rotational motion, and curl the fingers pointing from a to b. Your namely, moment of a force and angular stretched thumb points in the direction of c. momentum, are defined as vector products. It should be remembered that there are two angles between any two vectors a and b . In Definition of Vector Product Fig. 6.15 (a) or (b) they correspond to θ(as shown) A vector product of two vectors a and b is a and (3600– θ). While applying either of the above vector c such that rules, the rotation should be taken through the (i) magnitude of c = c = ab sinθ where a and b smaller angle (<1800) between a and b. It is θ are magnitudes of a and b and θ is the here. angle between the two vectors. Because of the cross (×) used to denote the (ii) c is perpendicular to the plane containing vector product, it is also referred to as cross product. a and b. • Note that scalar product of two vectors is (iii) if we take a right handed screw with its head commutative as said earlier, a.b = b.a lying in the plane of a and b and the screw The vector product, however, is not perpendicular to this plane, and if we turn commutative, i.e. a × b ≠ b × a the head in the direction from a to b, then The magnitude of both a × b and b × a is the the tip of the screw advances in the direction same ( ab sin θ ); also, both of them are of c. This right handed screw rule is perpendicular to the plane of a and b. But the illustrated in Fig. 6.15a. rotation of the right-handed screw in case of Alternately, if one curls up the fingers of a × b is from a to b, whereas in case of b × a it right hand around a line perpendicular to the is from b to a. This means the two vectors are plane of the vectors a and b and if the fingers in opposite directions. We have are curled up in the direction from a to b, then a × b = − b × a the stretched thumb points in the direction of • Another interesting property of a vector c, as shown in Fig. 6.15b. product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have x →− x , y →−y and z →− z . As a result all the components of a vector change sign and thus a →−a , b →−b . What happens to a × b under reflection? a × b →−( a ) × ( − b ) = a × b Thus, a × b does not change sign under reflection. • Both scalar and vector products are distributive with respect to vector addition. Thus, a.( b + c ) = a.b + a.c a × ( b + c ) = a × b + a × c (a) (b) • We may write c = a × b in the component form. For this we first need to obtain some elementary cross products: Fig. 6.15 (a) Rule of the right handed screw for (i) a × a = 0 (0 is a null vector, i.e. a vector defining the direction of the vector with zero magnitude) product of two vectors. This follows since magnitude of a × a is (b) Rule of the right hand for defining the direction of the vector product. a 2 sin0° = 0 . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 103 From this follow the results ˆ ˆ ˆ i j k (i) ˆi × ˆi = 0, ˆj × ˆj = 0, kˆ × kˆ = 0 a × b = 3 − 4 5 = 7 ˆi − ˆj − 5 kˆ (ii) ˆi × ˆj = kˆ − 2 1 − 3 Note that the magnitude of ˆi × ˆj is sin900 Note b × a = −7ˆi + ˆj + 5 kˆ ⊳ or 1, since ˆi and ˆj both have unit magnitude and the angle between them is 900. 6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY Thus, ˆi × ˆj is a unit vector. A unit vector In this section we shall study what is angular perpendicular to the plane of ˆi and ˆj and velocity and its role in rotational motion. We related to them by the right hand screw rule is have seen that every particle of a rotating body moves in a circle. The linear velocity of the ˆk . Hence, the above result. You may verify particle is related to the angular velocity. The similarly, relation between these two quantities involves ˆ j × kˆ = ˆi and kˆ × ˆi = ˆj a vector product which we learnt about in the last section. From the rule for commutation of the cross Let us go back to Fig. 6.4. As said above, inproduct, it follows: rotational motion of a rigid body about a fixed ˆ j × ˆi = − kˆ , kˆ × ˆj = − ˆi, ˆi × kˆ = − ˆj axis, every particle of the body moves in a circle, Note if ˆi, ˆj, kˆ occur cyclically in the above vector product relation, the vector product is positive. If ˆi, ˆj, kˆ do not occur in cyclic order, the vector product is negative. Now, a × b = (a x ˆi + a y ˆj + a z kˆ ) × (b x ˆi + b y ˆj + b z kˆ ) = a x b y kˆ − a x b z ˆj − a y b x kˆ + a y b z ˆi + a z b x ˆj − a z b y ˆi = (a y b z − a z b y )i + (a z b x − a x b z ) j + (a x b y − a y b x )k We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember. ˆ ˆ ˆ i j k a × b = a x a y a z b x b y b z u Example 6.4 Find the scalar and vector Fig. 6.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed products of two vectors. a = (z-) axis moves in a circle with centre (C) and b = on the axis.) Answer which lies in a plane perpendicular to the axis a i b = (3ˆi − 4 ˆj + 5 kˆ )i( − 2ˆi + ˆj − 3 kˆ ) and has its centre on the axis. In Fig. 6.16 we redraw Fig. 6.4, showing a typical particle (at a = −6 − 4 − 15 point P) of the rigid body rotating about a fixed = −25 axis (taken as the z-axis). The particle describes Reprint 2025-26 104 PHYSICS a circle with a centre C on the axis. The radius and points out in the direction in which a right of the circle is r, the perpendicular distance of handed screw would advance, if the head of the the point P from the axis. We also show the screw is rotated with the body. (See Fig. 6.17a). linear velocity vector v of the particle at P. It is The magnitude of this vector is ω = d θ dt along the tangent at P to the circle. referred as above. Let P′ be the position of the particle after an interval of time ∆t (Fig. 6.16). The angle PCP′ describes the angular displacement ∆θ of the particle in time ∆t. The average angular velocity of the particle over the interval ∆t is ∆θ/∆t. As ∆t tends to zero (i.e. takes smaller and smaller values), the ratio ∆θ/∆t approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocity by ω (the Greek letter omega). We know from our study Fig. 6.17 (a) If the head of a right handed screw of circular motion that the magnitude of linear rotates with the body, the screw velocity v of a particle moving in a circle is advances in the direction of the angular related to the angular velocity of the particle ω velocity ω. If the sense (clockwise or by the simple relation υ = ωr , where r is the anticlockwise) of rotation of the body changes, so does the direction of ω.radius of the circle. We observe that at any given instant the relation v = ωr applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by v i = ωri (6.19) The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, r = 0 , and hence v = ω r = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed. Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body. We have characterised pure translation of a body by all parts of the body having the same Fig. 6.17 (b) The angular velocity vector ω is directed velocity at any instant of time. Similarly, we along the fixed axis as shown. The linear may characterise pure rotation by all parts of velocity of the particle at P is v = ω × r. the body having the same angular velocity at It is perpendicular to both ωωωωω and r and any instant of time. Note that this is directed along the tangent to the circle described by the particle. characterisation of the rotation of a rigid body about a fixed axis is just another way of saying We shall now look at what the vector as in Sec. 6.1 that each particle of the body moves product ω × r corresponds to. Refer to Fig. in a circle, which lies in a plane perpendicular 6.17(b) which is a part of Fig. 6.16 reproduced to the axis and has the centre on the axis. to show the path of the particle P. The figure In our discussion so far the angular velocity shows the vector ω directed along the fixed (z–) appears to be a scalar. In fact, it is a vector. We axis and also the position vector r = OP of the shall not justify this fact, but we shall accept particle at P of the rigid body with respect to it. For rotation about a fixed axis, the angular the origin O. Note that the origin is chosen to velocity vector lies along the axis of rotation, be on the axis of rotation. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 105 Now ω × r = ω × OP = ω × (OC + CP) If the axis of rotation is fixed, the direction But ω × OC = 00000 as ωωωωω is along OC of ωωωωω and hence, that of α is fixed. In this case Hence ω × r = ω × CP the vector equation reduces to a scalar equation dω α = (6.22) The vector ω × CP is perpendicular to ω, i.e. dt to the z-axis and also to CP, the radius of the circle described by the particle at P. It is 6.7 TORQUE AND ANGULAR MOMENTUM therefore, along the tangent to the circle at P. In this section, we shall acquaint ourselves with Also, the magnitude of ω × CP is ω (CP) since two physical quantities (torque and angular ω and CP are perpendicular to each other. We momentum) which are defined as vector products shall denote CP by ⊥r and not by r, as we did of two vectors. These as we shall see, are earlier. especially important in the discussion of motion Thus, ω × r is a vector of magnitude ωr⊥ of systems of particles, particularly rigid bodies. and is along the tangent to the circle described by the particle at P. The linear velocity vector v 6.7.1 Moment of force (Torque) at P has the same magnitude and direction. We have learnt that the motion of a rigid body, Thus, in general, is a combination of rotation and v = ωωωωω × r (6.20) translation. If the body is fixed at a point or along In fact, the relation, Eq. (6.20), holds good a line, it has only rotational motion. We know even for rotation of a rigid body with one point that force is needed to change the translationalfixed, such as the rotation of the top [Fig. 6.6(a)]. In this case r represents the position vector of state of a body, i.e. to produce linear the particle with respect to the fixed point taken acceleration. We may then ask, what is the as the origin. analogue of force in the case of rotational We note that for rotation about a fixed motion? To look into the question in a concrete axis, the direction of the vector ω does not situation let us take the example of opening or change with time. Its magnitude may, closing of a door. A door is a rigid body which however, change from instant to instant. For can rotate about a fixed vertical axis passing the more general rotation, both the magnitude and the direction of ωωωωω may change through the hinges. What makes the door from instant to instant. rotate? It is clear that unless a force is applied the door does not rotate. But any force does not 6.6.1 Angular acceleration do the job. A force applied to the hinge line You may have noticed that we are developing cannot produce any rotation at all, whereas a the study of rotational motion along the lines force of given magnitude applied at right angles of the study of translational motion with which to the door at its outer edge is most effective in we are already familiar. Analogous to the kinetic producing rotation. It is not the force alone, but variables of linear displacement (s) and velocity how and where the force is applied is important (v) in translational motion, we have angular in rotational motion. displacement (θ) and angular velocity (ω) in The rotational analogue of force in linear rotational motion. It is then natural to define motion is moment of force. It is also referred to in rotational motion the concept of angular as torque or couple. (We shall use the words acceleration in analogy with linear acceleration moment of force and torque interchangeably.) defined as the time rate of change of velocity in We shall first define the moment of force for the translational motion. We define angular special case of a single particle. Later on we acceleration α as the time rate of change of shall extend the concept to systems of particles angular velocity. Thus, including rigid bodies. We shall also relate it to d ω a change in the state of rotational motion, i.e. is α = (6.21) dt angular acceleration of a rigid body. Reprint 2025-26 106 PHYSICS of the line of action of F from the origin and F⊥=( F sin θ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin. One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same. 6.7.2 Angular momentum of a particle Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue Fig. 6.18 τττττ ===== r × F, τττττ is perpendicular to the plane of linear momentum. We shall first define containing r and F, and its direction is angular momentum for the special case of a given by the right handed screw rule. single particle and look at its usefulness in the context of single particle motion. We shall then If a force acts on a single particle at a point extend the definition of angular momentum to P whose position with respect to the origin O is systems of particles including rigid bodies. given by the position vector r (Fig. 6.18), the Like moment of a force, angular momentum moment of the force acting on the particle with is also a vector product. It could also be referred respect to the origin O is defined as the vector to as moment of (linear) momentum. From this product term one could guess how angular momentum τ = r × F (6.23) is defined. The moment of force (or torque) is a vector Consider a particle of mass m and linear quantity. The symbol τττττ stands for the Greek momentum p at a position r relative to the origin letter tau. The magnitude of τττττ is O. The angular momentum l of the particle with τ = r F sinθ (6.24a) respect to the origin O is defined to be l = r × p (6.25a)where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force The magnitude of the angular momentum F and θ is the angle between r and F as vector is shown. l = r p sinθ (6.26a) Moment of force has dimensions M L2 T -2. where p is the magnitude of p and θ is the angle Its dimensions are the same as those of work between r and p. We may write or energy. It is, however, a very different physical l = r p⊥ or r ⊥ p (6.26b)quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of where r⊥ (= r sinθ) is the perpendicular distance moment of force is newton metre (N m). The of the directional line of p from the origin and magnitude of the moment of force may be p ⊥=( p sin θ) is the component of p in a directionwritten perpendicular to r. We expect the angular τ = (r sin θ)F = r⊥ F (6.24b) momentum to be zero (l = 0), if the linear or τ = r F sin θ = rF ⊥ (6.24c) momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p where r⊥ = r sinθ is the perpendicular distance passes through the origin θ = 00 or 1800. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 107 The physical quantities, moment of a force and angular momentum, have an important An experiment with the bicycle rim relation between them. It is the rotational Take a analogue of the relation between force and linear bicycle rim momentum. For deriving the relation in the and extend context of a single particle, we differentiate its axle on l = r × p with respect to time, both sides. Tie two d l d = ( r × p ) s t r i n g s d t d t at both ends Applying the product rule for differentiation A and B, to the right hand side, as shown in the d d r d p ( r × p ) = × p + r × a d j o i n i n g d t d t d t figure. Hold Now, the velocity of the particle is v = dr/dt both the and p = m v Initially After s t r i n g s together in dr one hand such that the rim is vertical. If you Because of this × p = v × m v = 0, dt leave one string, the rim will tilt. Now keeping the rim in vertical position with both the stringsas the vector product of two parallel vectors in one hand, put the wheel in fast rotation vanishes. Further, since dp / dt = F, around the axle with the other hand. Then leave d p one string, say B, from your hand, and observe r × = r × F = t dt what happens. The rim keeps rotating in a vertical plane d and the plane of rotation turns around the string Hence ( r × p ) = τ A which you are holding. We say that the axis dt of rotation of the rim or equivalently or (6.27) its angular momentum precesses about the string A. Thus, the time rate of change of the angular The rotating rim gives rise to an angular momentum of a particle is equal to the torque momentum. Determine the direction of this acting on it. This is the rotational analogue of angular momentum. When you are holding the the equation F = dp/dt, which expresses rotating rim with string A, a torque is generated. (We leave it to you to find out how the torque isNewton’s second law for the translational motion generated and what its direction is.) The effect of a single particle. of the torque on the angular momentum is to make it precess around an axis perpendicular Torque and angular momentum for a system to both the angular momentum and the torque. of particles Verify all these statements. To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles, particle has mass mi and velocity vi) We may write the total angular momentum of a system of particles as (6.25b) The angular momentum of the ith particle is given by li = ri × pi This is a generalisation of the definition of angular momentum (Eq. 6.25a) for a singlewhere ri is the position vector of the ith particle particle to a system of particles.with respect to a given origin and p = (mivi) is Using Eqs. (6.23) and (6.25b), we getthe linear momentum of the particle. (The Reprint 2025-26 108 PHYSICS d L d d l Note that like Eq.(6.17), Eq.(6.28b) holds = ( l ) = ∑ = ∑ τ (6.28a) good for any system of particles, whether it is a d t d t i d t i rigid body or its individual particles have all where τi is the torque acting on the ith particle; kinds of internal motion. τi = ri × Fi Conservation of angular momentum The force Fi on the ith particle is the vector ext If τext = 0, Eq. (6.28b) reduces to Fi sum of external forces acting on the particle d L = 0 and the internal forces iFint exerted on it by the dt other particles of the system. We may therefore or L = constant. (6.29a) separate the contribution of the external and Thus, if the total external torque on a system the internal forces to the total torque of particles is zero, then the total angular momentum of the system is conserved, i.e. τ = ∑ τ i = ∑ ri × Fi as remains constant. Eq. (6.29a) is equivalent to i i three scalar equations, τ = τext + τ int , Lx = K1, Ly = K2 and Lz = K3 (6.29 b) Here K1, K2 and K3 are constants; Lx, Ly and τ ext = ∑ri × Fi ext Lz are the components of the total angular where i momentum vector L along the x,y and z axes respectively. The statement that the total i × Fiint τ int = ∑r and angular momentum is conserved means that i each of these three components is conserved. We shall assume not only Newton’s third law Eq. (6.29a) is the rotational analogue of of motion, i.e. the forces between any two particles Eq. (6.18a), i.e. the conservation law of the total of the system are equal and opposite, but also that linear momentum for a system of particles. these forces are directed along the line joining the Like Eq. (6.18a), it has applications in many two particles. In this case the contribution of the practical situations. We shall look at a few of internal forces to the total torque on the system is the interesting applications later on in zero, since the torque resulting from each action- this chapter. reaction pair of forces is zero. We thus have, τint = 0 and therefore τ = τττext.ττ u Example 6.5 Find the torque of a force Since τ = ∑ τ i , it follows from Eq. (6.28a) + – about the origin. The force acts on a particle whose position vector is .that d L = τ ext (6.28 b) Answer Here r = ˆi − ˆj + kˆ d t and F = 7 ˆi + 3 ˆj − 5 kˆ . Thus, the time rate of the total angular We shall use the determinant rule to find themomentum of a system of particles about a τ = r × Fpoint (taken as the origin of our frame of torque reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (6.28 b) is the generalisation of the single particle case of Eq. (6.23) to a system of particles. Note that when we have only one or ⊳ particle, there are no internal forces or torques. Eq.(6.28 b) is the rotational analogue of Example 6.6 Show that the angular u momentum about any point of a single d P = Fext (6.17) particle moving with constant velocity d t remains constant throughout the motion. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 109 Answer Let the particle with velocity v be at acceleration nor angular acceleration. This means point P at some instant t. We want to calculate (1) the total force, i.e. the vector sum of the the angular momentum of the particle about an forces, on the rigid body is zero; arbitrary point O. n F1 + F2 + ... + Fn = =∑i 1 Fi = 0 (6.30a) If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body. (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero, n τ + τ i = 0 (6.30b) 1 2 + ... + τ n = =∑i 1 τ Fig 6.19 If the total torque on the rigid body is zero, The angular momentum is l = r × mv. Its the total angular momentum of the body does magnitude is mvr sinθ, where θ is the angle not change with time. Eq. (6.30 b) gives the between r and v as shown in Fig. 6.19. Although condition for the rotational equilibrium of the the particle changes position with time, the line body. of direction of v remains the same and hence One may raise a question, whether theOM = r sin θ. is a constant. rotational equilibrium condition [Eq. 6.30(b)] Further, the direction of l is perpendicular remains valid, if the origin with respect to whichto the plane of r and v. It is into the page of the the torques are taken is shifted. One can showfigure.This direction does not change with time. Thus, l remains the same in magnitude and that if the translational equilibrium condition direction and is therefore conserved. Is there [Eq. 6.30(a)] holds for a rigid body, then such a any external torque on the particle? ⊳ shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the 6.8 EQUILIBRIUM OF A RIGID BODY location of the origin about which the torques are taken. Example 6.7 gives a proof of this result We are now going to concentrate on the motion in a special case of a couple, i.e. two forcesof rigid bodies rather than on the motion of acting on a rigid body in translationalgeneral systems of particles. equilibrium. The generalisation of this result to We shall recapitulate what effect the external forces have on a rigid body. (Henceforth n forces is left as an exercise. we shall omit the adjective ‘external’ because Eq. (6.30a) and Eq. (6.30b), both, are vector unless stated otherwise, we shall deal with only equations. They are equivalent to three scalar external forces and torques.) The forces change equations each. Eq. (6.30a) corresponds to the translational state of the motion of the rigid n n n body, i.e. they change its total linear =∑i 1 Fix = 0 , =∑i 1 Fiy = 0 and =∑i 1 Fiz = 0 (6.31a)momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The where Fix, Fiy and Fiz are respectively the x, y and total torque on the body may not vanish. Such z components of the forces Fi. Similarly, Eq. a torque changes the rotational state of motion (6.30b) is equivalent to three scalar equations of the rigid body, i.e. it changes the total angular n n 0momentum of the body in accordance with τix = 0 , τiy = and (6.31b) =∑i 1 =∑i 1 Eq. (6.28 b). where τix, τiy and τiz are respectively the x, y and A rigid body is said to be in mechanical z components of the torque τi .equilibrium, if both its linear momentum and Eq. (6.31a) and (6.31b) give six independentangular momentum are not changing with time, conditions to be satisfied for mechanicalor equivalently, the body has neither linear Reprint 2025-26 110 PHYSICS equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis Fig. 6.20 (b) perpendicular to the plane of the forces must be zero. The force at B in Fig. 6.20(a) is reversed in Fig. 6.20(b). Thus, we have the same rod with The conditions of equilibrium of a rigid body two forces of equal magnitude but acting inmay be compared with those for a particle, which opposite diretions applied perpendicular to the we considered in earlier chapters. Since rod, one at end A and the other at end B. Here consideration of rotational motion does not apply the moments of both the forces are equal, but to a particle, only the conditions for translational they are not opposite; they act in the same sense equilibrium (Eq. 6.30 a) apply to a particle. Thus, and cause anticlockwise rotation of the rod. The for equilibrium of a particle the vector sum of total force on the body is zero; so the body is in all the forces on it must be zero. Since all these translational equilibrium; but it is not in forces act on the single particle, they must be rotational equilibrium. Although the rod is not fixed in any way, it undergoes pure rotation (i.e.concurrent. Equilibrium under concurrent rotation without translation).forces was discussed in the earlier chapters. A pair of forces of equal magnitude but acting A body may be in partial equilibrium, i.e., it in opposite directions with different lines of may be in translational equilibrium and not in action is known as a couple or torque. A couple rotational equilibrium, or it may be in rotational produces rotation without translation. equilibrium and not in translational When we open the lid of a bottle by turning equilibrium. it, our fingers are applying a couple to the lid Consider a light (i.e. of negligible mass) rod [Fig. 6.21(a)]. Another known example is a compass needle in the earth’s magnetic field as(AB) as shown in Fig. 6.20(a). At the two ends (A shown in the Fig. 6.21(b). The earth’s magneticand B) of which two parallel forces, both equal field exerts equal forces on the north and southin magnitude and acting along same direction poles. The force on the North Pole is towards are applied perpendicular to the rod. the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field. Fig. 6.20 (a) Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF ), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational fingers apply a couple to turnequilibrium; F ≠ 0 Fig. 6.21(a) Our ∑ the lid. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 111 length. This point is called the fulcrum. A see- saw on the children’s playground is a typical example of a lever. Two forces F1 and F2, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances d1 and d2 respectively from the fulcrum as shown in Fig. 6.23. Fig. 6.21(b) The Earth’s magnetic field exerts equal and opposite forces on the poles of a Fig. 6.23 compass needle. These two forces form a couple. The lever is a system in mechanical equilibrium. Let R be the reaction of the supportu Example 6.7 Show that moment of a at the fulcrum; R is directed opposite to the couple does not depend on the point about forces F1 and F2. For translational equilibrium, which you take the moments. R – F1 – F2 = 0 (i) Answer For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero, d1F1 – d2F2 = 0 (ii) Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum. Fig. 6.22 In the case of the lever force F1 is usually Consider a couple as shown in Fig. 6.22 some weight to be lifted. It is called the load and acting on a rigid body. The forces F and -F act its distance from the fulcrum d1 is called the respectively at points B and A. These points have load arm. Force F2 is the effort applied to lift the position vectors r1 and r2 with respect to origin load; distance d2 of the effort from the fulcrum O. Let us take the moments of the forces about is the effort arm. the origin. Eq. (ii) can be written as The moment of the couple = sum of the d1F1 = d2 F2 (6.32a) moments of the two forces making the couple or load arm × load = effort arm × effort = r1 × (–F) + r2 × F The above equation expresses the principle = r2 × F – r1 × F of moments for a lever. Incidentally the ratio = (r2–r1) × F F1/F2 is called the Mechanical Advantage (M.A.); But r1 + AB = r2, and hence AB = r2 – r1. F1 d 2 The moment of the couple, therefore, is M.A. = = (6.32b) F2 d1AB × F. Clearly this is independent of the origin, the If the effort arm d2 is larger than the load point about which we took the moments of the arm, the mechanical advantage is greater than forces. ⊳ one. Mechanical advantage greater than one means that a small effort can be used to lift a 6.8.1 Principle of moments large load. There are several examples of a lever An ideal lever is essentially a light (i.e. of around you besides the see-saw. The beam of a negligible mass) rod pivoted at a point along its balance is a lever. Try to find more such Reprint 2025-26 112 PHYSICS examples and identify the fulcrum, the effort and The CG of the cardboard is so located that effort arm, and the load and the load arm of the the total torque on it due to the forces m1g, m2g lever in each case. …. etc. is zero. You may easily show that the principle of If ri is the position vector of the ith particle moment holds even when the parallel forces F1 of an extended body with respect to its CG, then and F2 are not perpendicular, but act at some the torque about the CG, due to the force of angle, to the lever. gravity on the particle is τi = ri × mi g. The total gravitational torque about the CG is zero, i.e. 6.8.2 Centre of gravity i × m i g = 0 (6.33) τ g = ∑ τ i = ∑r Many of you may have the experience of We may therefore, define the CG of a body balancing your notebook on the tip of a finger. as that point where the total gravitational torque Figure 6.24 illustrates a similar experiment that on the body is zero. you can easily perform. Take an irregular- We notice that in Eq. (6.33), g is the same shaped cardboard having mass M and a narrow for all particles, and hence it comes out of the tipped object like a pencil. You can locate by trial summation. This gives, since g is non-zero, and error a point G on the cardboard where it ir = 0. Remember that the position vectorscan be balanced on the tip of the pencil. (The ∑mi cardboard remains horizontal in this position.) (ri) are taken with respect to the CG. Now, in This point of balance is the centre of gravity (CG) accordance with the reasoning given below of the cardboard. The tip of the pencil provides Eq. (6.4a) in Sec. 6.2, if the sum is zero, the origin a vertically upward force due to which the must be the centre of mass of the body. Thus, cardboard is in mechanical equilibrium. As the centre of gravity of the body coincides with shown in the Fig. 6.24, the reaction of the tip is the centre of mass in uniform gravity or gravity- equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the card board due to the forces of gravity like m1g, m2g …. etc, acting on the individual particles that make up the cardboard. Fig. 6.25 Determining the centre of gravity of a body Fig. 6.24 Balancing a cardboard on the tip of a of irregular shape. The centre of gravity G pencil. The point of support, G, is the lies on the vertical AA1 through the point centre of gravity. of suspension of the body A. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 113 free space. We note that this is true because = 30 cm, PG = 5 cm, AK1= BK2 = 10 cm and K1G = the body being small, g does not K2G = 25 cm. Also, W= weight of the rod = 4.00 vary from one point of the body to the other. If kg and W 1= suspended load = 6.00 kg; the body is so extended that g varies from part R1 and R2 are the normal reactions of the to part of the body, then the centre of gravity support at the knife edges. and centre of mass will not coincide. Basically, For translational equilibrium of the rod, the two are different concepts. The centre of R1+R2 –W1 –W = 0 (i) mass has nothing to do with gravity. It depends Note W1 and W act vertically down and R1 only on the distribution of mass of the body. and R2 act vertically up. In Sec. 6.2 we found out the position of the For considering rotational equilibrium, we centre of mass of several regular, homogeneous take moments of the forces. A convenient point objects. Obviously the method used there gives to take moments about is G. The moments of us also the centre of gravity of these bodies, if R2 and W1 are anticlockwise (+ve), whereas the they are small enough. moment of R1 is clockwise (-ve). Figure 6.25 illustrates another way of For rotational equilibrium, determining the CG of an irregular shaped body –R1 (K1G) + W1 (PG) + R2 (K2G) = 0 (ii) like a cardboard. If you suspend the body from It is given that W = 4.00g N and W1 = 6.00g some point like A, the vertical line through A N, where g = acceleration due to gravity. We passes through the CG. We mark the vertical take g = 9.8 m/s2. AA1. We then suspend the body through other With numerical values inserted, from (i) points like B and C. The intersection of the R1 + R2 – 4.00g – 6.00g = 0 verticals gives the CG. Explain why the method or R1 + R2 = 10.00g N (iii) works. Since the body is small enough, the = 98.00 N method allows us to determine also its centre From (ii), – 0.25 R1 + 0.05 W1 + 0.25 R2 = 0 of mass. or R1 – R2 = 1.2g N = 11.76 N (iv) From (iii) and (iv), R1 = 54.88 N, u Example 6.8 A metal bar 70 cm long and R2 = 43.12 N 4.00 kg in mass supported on two knife- Thus the reactions of the support are about edges placed 10 cm from each end. A 6.00 55 N at K1 and 43 N at K2. ⊳ kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. u Example 6.9 A 3m long ladder weighing (Assume the bar to be of uniform cross 20 kg leans on a frictionless wall. Its feet section and homogeneous.) rest on the floor 1 m from the wall as shown in Fig.6.27. Find the reaction forces of the Answer wall and the floor. Answer Fig. 6.26 Figure 6.26 shows the rod AB, the positions of the knife edges K1 and K2 , the centre of gravity of the rod at G and the suspended load at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP Fig. 6.27 Reprint 2025-26 114 PHYSICS The ladder AB is 3 m long, its foot A is at from the axis, the linear velocity is υi = ir ω. The distance AC = 1 m from the wall. From kinetic energy of motion of this particle is Pythagoras theorem, BC = 2 2 m. The forces 1 2 1 2 2 on the ladder are its weight W acting at its centre k i = m i υi = m i ri ω 2 2 of gravity D, reaction forces F1 and F2 of the wall where mi is the mass of the particle. The totaland the floor respectively. Force F1 is kinetic energy K of the body is then given byperpendicular to the wall, since the wall is the sum of the kinetic energies of individualfrictionless. Force F2 is resolved into two particles,components, the normal reaction N and the force of friction F. Note that F prevents the ladder n 1 n 2 2 from sliding away from the wall and is therefore K = ∑ k i = ∑ (m i ri ω ) i =1 2 i =1 directed toward the wall. For translational equilibrium, taking the Here n is the number of particles in the body. forces in the vertical direction, Note ωis the same for all particles. Hence, taking N – W = 0 (i) ω out of the sum, Taking the forces in the horizontal direction, n 1 2 2 i ri ) F – F1 = 0 (ii) K = 2 ω ( ∑i =1 m For rotational equilibrium, taking the We define a new parameter characterisingmoments of the forces about A, the rigid body, called the moment of inertia I , 2 2 F1 −(1/2) W = 0 (iii) given by Now W = 20 g = 20 × 9.8 N = 196.0 N n 2 I = ∑ m i ri (6.34)From (i) N = 196.0 N i =1 With this definition,From (iii) F1 = W 4 2 = 196.0/4 2 = 34.6 N 1 2 From (ii) F = F1 = 34.6 N K = Iω (6.35) 2 2 2 Note that the parameter I is independent of F2 = F + N = 199.0 N the magnitude of the angular velocity. It is a The force F2 makes an angle α with the characteristic of the rigid body and the axis horizontal, about which it rotates. −1 Compare Eq. (6.35) for the kinetic energy oftan α = N F = 4 2 , α = tan (4 2) ≈ 80 ⊳ a rotating body with the expression for the kinetic energy of a body in linear (translational)6.9 MOMENT OF INERTIA motion, We have already mentioned that we are 1 2developing the study of rotational motion parallel K = m υ 2to the study of translational motion with which Here, m is the mass of the body and v is itswe are familiar. We have yet to answer one major velocity. We have already noted the analogy question in this connection. What is the between angular velocity ω (in respect of analogue of mass in rotational motion? We shall rotational motion about a fixed axis) and linear attempt to answer this question in the present velocity v (in respect of linear motion). It is then section. To keep the discussion simple, we shall evident that the parameter, moment of inertia consider rotation about a fixed axis only. Let us I, is the desired rotational analogue of mass in try to get an expression for the kinetic energy of linear motion. In rotation (about a fixed axis), a rotating body. We know that for a body rotating the moment of inertia plays a similar role as about a fixed axis, each particle of the body moves mass does in linear motion. We now apply the definition Eq. (6.34), toin a circle with linear velocity given by Eq. (6.19). calculate the moment of inertia in two simple cases.(Refer to Fig. 6.16). For a particle at a distance Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 115 (a) Consider a thin ring of radius R and mass change in its rotational motion, it can be M, rotating in its own plane around its centre regarded as a measure of rotational inertia of with angular velocity ω. Each mass element the body; it is a measure of the way in which of the ring is at a distance R from the axis, different parts of the body are distributed at and moves with a speed Rω. The kinetic different distances from the axis. Unlike the energy is therefore, mass of a body, the moment of inertia is not a fixed quantity but depends on distribution of 1 2 1 2 2 K = M υ = MR ω mass about the axis of rotation, and the 2 2 orientation and position of the axis of rotation Comparing with Eq. (6.35) we get I = MR 2 with respect to the body as a whole. As a for the ring. measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body. Notice from the Table 6.1 that in all cases, we can write I = Mk2, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, k 2 = L2 12, i.e. k = L 12 . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of Fig. 6.28 A light rod of length l with a pair of gyration. The radius of gyration of a body masses rotating about an axis through about an axis may be defined as the distance the centre of mass of the system and perpendicular to the rod. The total mass from the axis of a mass point whose mass is of the system is M. equal to the mass of the whole body and whose moment of inertia is equal to the moment of (b) Next, take a rigid rod of negligible mass of inertia of the body about the axis. length of length l with a pair of small masses, Thus, the moment of inertia of a rigid body rotating about an axis through the centre of depends on the mass of the body, its shape and mass perpendicular to the rod (Fig. 6.28). size; distribution of mass about the axis of Each mass M/2 is at a distance l/2 from rotation, and the position and orientation of the the axis. The moment of inertia of the masses axis of rotation. is therefore given by From the definition, Eq. (6.34), we can infer (M/2) (l/2)2 + (M/2)(l/2)2 that the dimensions of moments of inertia are Thus, for the pair of masses, rotating about ML2 and its SI units are kg m2. the axis through the centre of mass The property of this extremely important perpendicular to the rod 2 quantity I, as a measure of rotational inertia of I = Ml / 4 the body, has been put to a great practical use. Table 6.1 simply gives the moment of inertia of The machines, such as steam engine and thevarious familiar regular shaped bodies about automobile engine, etc., that produce rotationalspecific axes. (The derivations of these motion have a disc with a large moment ofexpressions are beyond the scope of this inertia, called a flywheel. Because of its largetextbook and you will study them in higher classes.) moment of inertia, the flywheel resists the As the mass of a body resists a change in its sudden increase or decrease of the speed of the state of linear motion, it is a measure of its inertia vehicle. It allows a gradual change in the speed in linear motion. Similarly, as the moment of and prevents jerky motions, thereby ensuring inertia about a given axis of rotation resists a a smooth ride for the passengers on the vehicle. Reprint 2025-26 116 PHYSICS Table 6.1 Moments of inertia of some regular shaped bodies about specific axes Z Body Axis Figure I (1) Thin circular Perpendicular to M R 2 ring, radius R plane, at centre (2) Thin circular Diameter M R2/2 ring, radius R (3) Thin rod, Perpendicular to M L2/12 length L rod, at mid point (4) Circular disc, Perpendicular to M R2/2 radius R disc at centre (5) Circular disc, Diameter M R2/4 radius R (6) Hollow cylinder, Axis of cylinder M R2 radius R (7) Solid cylinder, Axis of cylinder M R2/2 radius R (8) Solid sphere, Diameter 2 M R2/5 radius R 6.10 KINEMATICS OF ROTATIONAL MOTION translation. We wish to take this analogy further. ABOUT A FIXED AXIS In doing so we shall restrict the discussion only We have already indicated the analogy between to rotation about fixed axis. This case of motion rotational motion and translational motion. For involves only one degree of freedom, i.e., needs example, the angular velocity ω plays the same only one independent variable to describe the role in rotation as the linear velocity v in motion. This in translation corresponds to linear Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 117 motion. This section is limited only to kinematics. We shall turn to dynamics in later sections. We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.6.29) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 6.29 also shows θ0, the angular displacement at t = 0. We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a Fig.6.29 Specifying the angular position of a rigid vector. Further, the angular acceleration, α = body. dω/dt. u Example 6.10 Obtain Eq. (6.36) from first The kinematical quantities in rotational principles. motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) Answer The angular acceleration is uniform, respectively are analogous to kinematic hence quantities in linear motion, displacement (x), dω velocity (v) and acceleration (a). We know the = α = constant (i) kinematical equations of linear motion with d t uniform (i.e. constant) acceleration: Integrating this equation, α dt + c v = v0 + at (a) ω = ∫ 1 2 x = x 0 + υ0t + at (b) = αt + c (as α is constant) 2 At t = 0, ω = ω0 (given) 2 2 υ = υ0 + 2ax (c) From (i) we get at t = 0, ω = c = ω0 Thus, ω = αt + ω0 as required. where x0 = initial displacement and v0= initial With the definition of ω = dθ/dt we may velocity. The word ‘initial’ refers to values of the integrate Eq. (6.36) to get Eq. (6.37). This quantities at t = 0 derivation and the derivation of Eq. (6.38) is left The corresponding kinematic equations for as an exercise. rotational motion with uniform angular acceleration are: u Example 6.11 The angular speed of a motor wheel is increased from 1200 rpm to ω= ω0 + αt (6.36) 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the 1 2 θ = θ0 + ω0t + αt (6.37) acceleration to be uniform? (ii) How many 2 revolutions does the engine make during and ω2 = ω0 2 + 2α(θ– θ0 ) (6.38) this time? Answer where θ0= initial angular displacement of the (i) We shall use ω = ω0 + αt rotating body, and ω0 = initial angular velocity ω0 = initial angular speed in rad/s of the body. Reprint 2025-26 118 PHYSICS = 2π × angular speed in rev/s It is, however, necessary that these correspondences are established on sound 2π × angular speed in rev/min dynamical considerations. This is what we now = 60 s/min turn to. Before we begin, we note a simplification 2π × 1200 that arises in the case of rotational motion = rad/s 60 about a fixed axis. Since the axis is fixed, only those components of torques, which are along = 40π rad/s the direction of the fixed axis need to be Similarly ω = final angular speed in rad/s considered in our discussion. Only these 2π × 3120 components can cause the body to rotate about = rad/s the axis. A component of the torque 60 perpendicular to the axis of rotation will tend to = 2π × 52 rad/s turn the axis from its position. We specifically = 104 π rad/s assume that there will arise necessary forces of constraint to cancel the effect of the ∴Angular acceleration perpendicular components of the (external) torques, so that the fixed position of the axis ω − ω will be maintained. The perpendicular α = 0 = 4 π rad/s2 t components of the torques, therefore need not be taken into account. This means that for our The angular acceleration of the engine calculation of torques on a rigid body: = 4π rad/s2 (1) We need to consider only those forces that (ii) The angular displacement in time t is lie in planes perpendicular to the axis. given by Forces which are parallel to the axis will give torques perpendicular to the axis and need 1 2 θ = ω0 t + αt not be taken into account. 2 (2) We need to consider only those components 1 2 of the position vectors which are = (40π × 16 + × 4π × 16 ) rad 2 perpendicular to the axis. Components of position vectors along the axis will result in = (640π + 512π) rad torques perpendicular to the axis and need = 1152π rad not be taken into account. 1152π = 576 ⊳ Work done by a torqueNumber of revolutions = 2π

6.12 Angular momentum in case solved by considering them to be rigid bodies. Ideally a of rotation about a fixed axis rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of Summary particles of such a body do not change. It is evident from Points to Ponder this definition of a rigid body that no real body is truly rigid, Exercises since real bodies deform under the influence of forces. But in many situations the deformations are negligible. In a number of situations involving bodies such as wheels, tops, steel beams, molecules and planets on the other hand, we can ignore that they warp (twist out of shape), bend or vibrate and treat them as rigid. 6.1.1 What kind of motion can a rigid body have? Let us try to explore this question by taking some examples of the motion of rigid bodies. Let us begin with a rectangular Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 93 most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on (Fig Fig 6.1 Translational (sliding) motion of a block down 6.3(a) and (b)). an inclined plane. (Any point like P1 or P2 of the block moves with the same velocity at any instant of time.) block sliding down an inclined plane without any sidewise movement. The block is taken as a rigid body. Its motion down the plane is such that all the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion (Fig. 6.1). In pure translational motion at any instant of time, all particles of the body have the same velocity. Consider now the rolling motion of a solid metallic or wooden cylinder down the same (a) inclined plane (Fig. 6.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, seems to have translational motion. But as Fig.

6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current

6.11 DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS Table 6.2 lists quantities associated with linear motion and their analogues in rotational motion. We have already compared kinematics of the two motions. Also, we know that in rotational motion moment of inertia and torque play the same role as mass and force respectively in linear motion. Given this we should be able to guess what the other analogues indicated in the table are. For Fig. 6.30 Work done by a force F1 acting on a example, we know that in linear motion, work particle of a body rotating about a fixed done is given by F dx, in rotational motion about axis; the particle describes a circular path a fixed axis it should be τdθ , since we already with centre C on the axis; arc P1P′1(ds1) gives the displacement of the particle. know the correspondence d x → dθ and F → τ . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 119 Table 6.2 Comparison of Translational and Rotational Motion Linear Motion Rotational Motion about a Fixed Axis 1 Displacement x Angular displacement θ 2 Velocity v = dx/dt Angular velocity ω = dθ/dt 3 Acceleration a = dv/dt Angular acceleration α = dω/dt 4 Mass M Moment of inertia I 5 Force F = Ma Torque τ = I α 6 Work dW = F ds Work W = τ dθ 7 Kinetic energy K = Mv2/2 Kinetic energy K = Iω2/2 8 Power P = F v Power P = τω 9 Linear momentum p = Mv Angular momentum L = Iω If there are more than one forces acting on Figure 6.30 shows a cross-section of a rigid the body, the work done by all of them can bebody rotating about a fixed axis, which is taken added to give the total work done on the body. as the z-axis (perpendicular to the plane of the Denoting the magnitudes of the torques due to page; see Fig. 6.29). As said above we need to the different forces as τ1, τ2, … etc,consider only those forces which lie in planes perpendicular to the axis. Let F1 be one such d W = (τ1 + τ2 + ...)dθ typical force acting as shown on a particle of Remember, the forces giving rise to the the body at point P1 with its line of action in a torques act on different particles, but the plane perpendicular to the axis. For convenience angular displacement dθ is the same for all we call this to be the x′–y′ plane (coincident particles. Since all the torques considered are with the plane of the page). The particle at P1 parallel to the fixed axis, the magnitude τ of the describes a circular path of radius r1 with centre total torque is just the algebraic sum of the C on the axis; CP1 = r1. magnitudes of the torques, i.e., τ = τ1 + τ2 + ..... In time ∆t, the point moves to the position We, therefore, have P1′. The displacement of the particle ds1, d W = τdθ (6.39) therefore, has magnitude ds1 = r1dθ and This expression gives the work done by the direction tangential at P1 to the circular path total (external) torque τ which acts on the bodyas shown. Here dθ is the angular displacement rotating about a fixed axis. Its similarity withof the particle, dθ = ∠P1CP1′ .The work done by the corresponding expression the force on the particle is dW= F ds dW1 = F1. ds1= F1ds1 cosφ1= F1(r1 dθ)sinα1 for linear (translational) motion is obvious. where φ1 is the angle between F1 and the tangent Dividing both sides of Eq. (6.39) by dt gives at P1, and α1 is the angle between F1 and the d W dθ P = = τ = τωradius vector OP1; φ1 + α1 = 90°. d t d t The torque due to F1 about the origin is or P = τω (6.40) OP1 × F1. Now OP1 = OC + OP1. [Refer to This is the instantaneous power. Compare Fig. 6.17(b).] Since OC is along the axis, the torque this expression for power in the case of rotational resulting from it is excluded from our motion about a fixed axis with that of power in consideration. The effective torque due to F1 is the case of linear motion, τ1= CP × F1; it is directed along the axis of rotation P = Fv and has a magnitude τ1= r1F1 sinα , Therefore, In a perfectly rigid body there is no internal dW1 = τ1dθ motion. The work done by external torques is Reprint 2025-26 120 PHYSICS therefore, not dissipated and goes on to increase Answer the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (6.40). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is d  I ω2  ( 2ω) d ω I d t  2 = 2 d t We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body. Since α= dω/d,t we get d  I ω2  I ω α Fig. 6.31 dt  2 = (a) We use I α = τ Equating rates of work done and of increase the torque τ = F Rin kinetic energy, = 25 × 0.20 Nm (as R = 0.20m) τω= I ωα = 5.0 Nm τ = I α (6.41) I = Moment of inertia of flywheel about its Eq. (6.41) is similar to Newton’s second law 2 for linear motion expressed symbolically as axis = MR F = ma 2 Just as force produces acceleration, torque 2 20.0 × (0.2) produces angular acceleration in a body. The = = 0.4 kg m2 2 angular acceleration is directly proportional to α = angular acceleration the applied torque and is inversely proportional = 5.0 N m/0.4 kg m2 = 12.5 s–2 to the moment of inertia of the body. In this (b) Work done by the pull unwinding 2m of the respect, Eq.(6.41) can be called Newton’s second cord law for rotational motion about a fixed axis. = 25 N × 2m = 50 J u (c) Let ω be the final angular velocity. The Example 6.12 A cord of negligible mass is 1 2 kinetic energy gained = Iω , wound round the rim of a fly wheel of mass 2 20 kg and radius 20 cm. A steady pull of since the wheel starts from rest. Now, 25 N is applied on the cord as shown in 2 2 ω = ω0 + 2αθ, ω0 = 0 Fig. 6.31. The flywheel is mounted on a horizontal axle with frictionless bearings. The angular displacement θ = length of unwound string / radius of wheel (a) Compute the angular acceleration of = 2m/0.2 m = 10 rad the wheel. (b) Find the work done by the pull, when ω2 = 2 × 12 .5 × 10 .0 = 250 (rad/s )2 2m of the cord is unwound. ∴ (c) Find also the kinetic energy of the wheel at this point. Assume that the (d) The answers are the same, i.e. the kinetic energy wheel starts from rest. gained by the wheel = work done by the force. (d) Compare answers to parts (b) and (c). There is no loss of energy due to friction. ⊳ Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 121

6.4 FARADAY’S LAW OF INDUCTION From the experimental observations, Faraday arrived at a conclusion that an emf is induced in a coil when magnetic flux through the coil changes with time. Experimental observations discussed in Section 6.2 can be explained using this concept. The motion of a magnet towards or away from coil C1 in Experiment 6.1 and moving a current-carrying coil C2 towards or away from coil C1 in Experiment 6.2, change the magnetic flux associated with coil C1. The change in magnetic flux induces emf in coil C1. It was this induced emf which caused electric current to flow in coil C1 and through the galvanometer. A FIGURE 6.5 Magnetic field Bi plausible explanation for the observations of Experiment 6.3 is at the ith area element. dAi represents area vector of theas follows: When the tapping key K is pressed, the current in ith area element. coil C2 (and the resulting magnetic field) rises from zero to a maximum value in a short time. Consequently, the magnetic flux through the neighbouring coil C1 also increases. It is the change in magnetic flux through coil C1 that produces an induced emf in coil C1. When the key is held pressed, current in coil C2 is constant. Therefore, there is no change in the magnetic flux through coil C1 and the current in coil C1 drops to zero. When the key is released, the current in C2 and the resulting magnetic field decreases from the maximum value to zero in a short time. This results in a decrease in magnetic flux through coil C1 and hence again induces an electric current in coil C1*. The common point in all these observations is that the time rate of change of magnetic flux through a circuit induces emf in it. Faraday stated experimental observations in the form of a law called Faraday’s law of electromagnetic induction. The law is stated below. * Note that sensitive electrical instruments in the vicinity of an electromagnet can be damaged due to the induced emfs (and the resulting currents) when the electromagnet is turned on or off. 157 Reprint 2025-26 Physics The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically, the induced emf is given by dΦB ε = – (6.3) d t The negative sign indicates the direction of e and hence the direction of current in a closed loop. This will be discussed in detail in the next section. In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, Michael Faraday [1791– the expression for the total induced emf is given by 1867] Faraday made numerous contributions to dΦB ε = – N (6.4)(1791–1867) science, viz., the discovery d t of electromagnetic induction, the laws of The induced emf can be increased by increasing the electrolysis, benzene, and number of turns N of a closed coil. the fact that the plane of From Eqs. (6.1) and (6.2), we see that the flux can be polarisation is rotated in an varied by changing any one or more of the terms B, A andFARADAY electric field. He is also credited with the invention q. In Experiments 6.1 and 6.2 in Section 6.2, the flux is of the electric motor, the changed by varying B. The flux can also be altered by electric generator and the changing the shape of a coil (that is, by shrinking it or transformer. He is widely stretching it) in a magnetic field, or rotating a coil in a regarded as the greatest magnetic field such that the angle q between B and AMICHAEL experimental scientist of changes. In these cases too, an emf is induced in the the nineteenth century. respective coils. Example 6.1 Consider Experiment 6.2. (a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer? Solution (a) To obtain a large deflection, one or more of the following steps can be taken: (i) Use a rod made of soft iron inside the coil C2, (ii) Connect the coil to a powerful battery, and (iii) Move the arrangement rapidly towards the test coil C1. (b) Replace the galvanometer by a small bulb, the kind one finds in a 6.1 small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current. In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary EXAMPLE for his innovative skills. 6.2 Example 6.2 A square loop of side 10 cm and resistance 0.5 W is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine EXAMPLE the magnitudes of induced emf and current during this time-interval.158 Reprint 2025-26 Electromagnetic Induction Solution The angle q made by the area vector of the coil with the magnetic field is 45°. From Eq. (6.1), the initial magnetic flux is F = BA cos q 0.1 × 10 –2 = Wb 2 Final flux, Fmin = 0 The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by –3 ∆ΦB (Φ – 0 ) 10 ε = = = = 1.0 mV ∆ t ∆ t 2 × 0.7 And the magnitude of the current is ε 10 –3 V I = = = 2 mA R 0.5 Ω Note that the earth’s magnetic field also produces a flux through the EXAMPLE loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf. 6.2 Example 6.3 A circular coil of radius 10 cm, 500 turns and resistance 2 W is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is 3.0 × 10–5 T. Solution Initial flux through the coil, FB (initial) = BA cos q = 3.0 × 10–5 × (p ×10–2) × cos 0° = 3p × 10–7 Wb Final flux after the rotation, FB (final) = 3.0 × 10–5 × (p ×10–2) × cos 180° = –3p × 10–7 Wb Therefore, estimated value of the induced emf is, ∆Φ ε = N ∆t = 500 × (6p × 10–7)/0.25 = 3.8 × 10–3 V I = e/R = 1.9 × 10–3 A Note that the magnitudes of e and I are the estimated values. Their EXAMPLE instantaneous values are different and depend upon the speed of rotation at the particular instant. 6.3 159 Reprint 2025-26 Physics 6.5 LENZ’S LAW AND CONSERVATION OF ENERGY In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule, known as Lenz’s law which gives the polarity of the induced emf in a clear and concise fashion. The statement of the law is: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The negative sign shown in Eq. (6.3) represents this effect. We can understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1. In Fig. 6.1, we see that the North-pole of a bar magnet is being pushed towards the closed coil. As the North-pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux. This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. Similarly, if the North- pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in clockwise direction and its South- pole faces the receding North-pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux. What will happen if an open circuit is used in place of the closed loop in the above example? In this case too, an emf is induced across the open ends of the circuit. The direction of the induced emf can be found using Lenz’s law. Consider Figs. 6.6 (a) and (b). They provide an easier way to understand the direction of induced currents. Note that the direction shown by and indicate the directions of the induced currents. A little reflection on this matter should convince us on the correctness of Lenz’s law. Suppose that the induced current was in the direction opposite to the one depicted in Fig. 6.6(a). In that case, the South-pole due to the induced current will face the approaching North-pole of the magnet. The bar magnet will then be attracted towards the coil at an ever increasing acceleration. A gentle push on the magnet will initiate the process and its velocity and kinetic energy will continuously increase without expending any energy. If this can happen, one could construct a perpetual-motion machine by a suitable arrangement. This violates the law of conservation of energy and hence can not happen. FIGURE 6.6 Now consider the correct case shown in Fig. 6.6(a). In this situation, Illustration of the bar magnet experiences a repulsive force due to the induced Lenz’s law. current. Therefore, a person has to do work in moving the magnet. Where does the energy spent by the person go? This energy is 160 dissipated by Joule heating produced by the induced current. Reprint 2025-26 Electromagnetic Induction Example 6.4 Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law. FIGURE 6.7 Solution (i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux. (ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux. (iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. EXAMPLE Note that there are no induced current as long as the loops are 6.4 completely inside or outside the region of the magnetic field. Example 6.5 (a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets? (b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig. 6.8) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be EXAMPLE during constant the passage out of the field region? The field is 6.5 normal to the loops. 161 Reprint 2025-26 Physics FIGURE 6.8 (d) Predict the polarity of the capacitor in the situation described by Fig. 6.9. FIGURE 6.9 Solution (a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop. (b) No current is induced in either case. Current can not be induced by changing the electric flux. 6.5 (c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly. (d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in EXAMPLE the capacitor. 6.6 MOTIONAL ELECTROMOTIVE FORCE Let us consider a straight conductor moving in a uniform and time- independent magnetic field. Figure 6.10 shows a rectangular conductor PQRS in which the conductor PQ is free to move. The rod PQ is moved towards the left with a constant velocity v as shown in the figure. Assume that there is no loss of energy due to friction. PQRS forms a closed circuit enclosing an area that changes as PQ moves. It is placed in a uniform magnetic field B which is perpendicular to the plane of this system. If the length RQ = x and RS = l, the magnetic flux FB enclosed by the loop PQRS will be FB = Blx Since x is changing with time, the rate of change of flux FB will induce an emf given by: FIGURE 6.10 The arm PQ is moved to the left – dΦB d side, thus decreasing the area of the ε= = – ( Blx ) d t d t rectangular loop. This movement induces a current I as shown. d x 162 = – Bl = Blv (6.5) d t Reprint 2025-26 Electromagnetic Induction where we have used dx/dt = –v which is the speed of the conductor PQ. The induced emf Blv is called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit. It is also possible to explain the motional emf expression in Eq. (6.5) by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ. The work done in moving the charge from P to Q is, W = qvBl Since emf is the work done per unit charge, W ε = q = Blv This equation gives emf induced across the rod PQ and is identical to Eq. (6.5). We stress that our presentation is not wholly rigorous. But it does help us to understand the basis of Faraday’s law when the conductor is moving in a uniform and time-independent magnetic field. On the other hand, it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing – a fact which Faraday verified by numerous experiments. In the case of a stationary conductor, the force on its charges is given by F = q (E + v ´ B) = qE (6.6) since v = 0. Thus, any force on the charge must arise from the electric field term E alone. Therefore, to explain the existence of induced emf or induced current, we must assume that a time-varying magnetic field generates an electric field. However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields. In Chapter 4, we learnt that charges in motion (current) can exert force/torque on a stationary magnet. Conversely, a bar magnet in motion (or more generally, a changing magnetic field) can exert a force on the stationary charge. This is the fundamental significance of the Faraday’s discovery. Electricity and magnetism are related. Example 6.6 A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. 6.11). A constant and uniform magnetic field of 1 T parallel to the EXAMPLE axis is present everywhere. What is the emf between the centre and the metallic ring? 6.6 163 Reprint 2025-26 Physics FIGURE 6.11 Solution Method I As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by dε = Bv dr . Hence, R R B ωR 2 ωr d r = Bv d r = ∫B ε =∫ d ε = ∫ 2 0 0 Note that we have used v = w r. This gives 1 2 e = × 1.0 × 2 π × 50 × (1 ) 2 = 157 V Method II To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If q is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by 2 θ 1 2 π R × = R θ 2 π 2 where R is the radius of the circle. Hence, the induced emf is d  1  1 2 dθ BωR 2 e = B × R 2θ = BR = d t  2  2 d t 2 6.6 dθ [Note: = ω = 2 π ν] dt This expression is identical to the expression obtained by Method I EXAMPLE and we get the same value of e. Reprint 2025-26 Electromagnetic Induction Example 6.7 A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field HE at a place. If HE = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1 G = 10–4 T. Solution Induced emf = (1/2) ω B R2 = (1/2) × 4π × 0.4 × 10–4 × (0.5)2 = 6.28 × 10–5 V EXAMPLE The number of spokes is immaterial because the emf’s across the 6.7 spokes are in parallel.

6.7 INDUCTANCE An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. These two situations are described separately in the next two sub-sections. However, in both the cases, the flux through a coil is proportional to the current. That is, ΦB α I. Further, if the geometry of the coil does not vary with time then, dΦB d I ∝ d t d t For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. When the flux ΦB through the coil changes, each turn contributes to the induced emf. Therefore, a term called flux linkage is used which is equal to NΦB for a closely wound coil and in such a case NΦB∝ I The constant of proportionality, in this relation, is called inductance. We shall see that inductance depends only on the geometry of the coil and intrinsic material properties. This aspect is akin to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the dielectric constant K of the intervening medium (intrinsic material property). Inductance is a scalar quantity. It has the dimensions of [M L2 T–2 A–2] given by the dimensions of flux divided by the dimensions of current. The SI unit of inductance is henry and is denoted by H. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA, independently of Faraday in England. 6.7.1 Mutual inductance Consider Fig. 6.12 which shows two long co-axial solenoids each of length l. We denote the radius of the inner solenoid S1 by r1 and the number of turns per unit length by n1. The corresponding quantities for the outer solenoid S2 are r2 and n2, respectively. Let N1 and N2 be the total number 165 of turns of coils S1 and S2, respectively. Reprint 2025-26 Physics When a current I2 is set up through S2, it in turn sets up a magnetic flux through S1. Let us denote it by Φ1. The corresponding flux linkage with solenoid S1 is N1 Φ1 = M 12 I 2 (6.7) M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2. It is also referred to as the coefficient of mutual induction. For these simple co-axial solenoids it is possible to calculate M12. The magnetic field due to the current I2 in S2 is µ0n2I2. The resulting flux linkage with coil S1 is, πr12 N 1Φ1 = ( µ0n 2 I 2 ) n1l ) ( ) ( = µ0n 1n 2 πr12l I 2 (6.8) where n1l is the total number of turns in solenoid S1. FIGURE 6.12 Two long co-axial Thus, from Eq. (6.7) and Eq. (6.8), solenoids of same 2 M12 = µ0n1n2πr 1l (6.9) length l. Note that we neglected the edge effects and considered the magnetic field µ0n2I2 to be uniform throughout the length and width of the solenoid S2. This is a good approximation keeping in mind that the solenoid is long, implying l >> r2. We now consider the reverse case. A current I1 is passed through the solenoid S1 and the flux linkage with coil S2 is, N2Φ2 = M21 I1 (6.10) M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1. The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoids are very long. Thus, flux linkage with solenoid S2 is πr12 N 2Φ2 = ( µ0n1 I1 ) n 2l ) ( ) ( where n2l is the total number of turns of S2. From Eq. (6.10), M21 = µ0n1n2πr 1l2 (6.11) Using Eq. (6.9) and Eq. (6.10), we get M12 = M21= M (say) (6.12) We have demonstrated this equality for long co-axial solenoids. However, the relation is far more general. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage N1Φ1 because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. In this case, the calculation of M12 would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. Therefore, the calculation of M21 would also be extremely difficult in this case. The 166 equality M12=M21 is very useful in such situations. Reprint 2025-26 Electromagnetic Induction We explained the above example with air as the medium within the solenoids. Instead, if a medium of relative permeability µr had been present, the mutual inductance would be M =µr µ0 n1n2π r21 l It is also important to know that the mutual inductance of a pair of coils, solenoids, etc., depends on their separation as well as their relative orientation. Example 6.8 Two concentric circular coils, one of small radius r1 and the other of large radius r2, such that r1 << r2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Solution Let a current I2 flow through the outer circular coil. The field at the centre of the coil is B2 = µ0I2 / 2r2. Since the other co-axially placed coil has a very small radius, B2 may be considered constant over its cross-sectional area. Hence, Φ1 = πr 1B22 µ0 π r12 = I 2 2r2 = M12 I2 Thus, µ0 πr12 M 12 = 2r2 From Eq. (6.12) µ0 πr12 M 12 = M 21 = 2 r2 Note that we calculated M12 from an approximate value of Φ1, assuming EXAMPLE the magnetic field B2 to be uniform over the area π r12. However, we can accept this value because r1 << r2. 6.8 Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil C1 wherever there was any change in current through coil C2. Let Φ1 be the flux through coil C1 (say of N1 turns) when current in coil C2 is I2. Then, from Eq. (6.7), we have N1Φ1 = MI2 For currents varrying with time, d d ( N 1Φ1 ) ( MI 2 ) = d t d t Since induced emf in coil C1 is given by d N 1Φ1 ) ( ε1 = – d t We get, d I 2 ε1 = – M d t 167 Reprint 2025-26 Physics It shows that varying current in a coil can induce emf in a neighbouring coil. The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils. 6.7.2 Self-inductance In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as NΦB ∝ I NΦB = L I (6.13) where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.13), the induced emf is given by d ( NΦB ) ε = – dt d I ε = – L (6.14) d t Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross- sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ0 n I (neglecting edge effects, as before). The total flux linked with the solenoid is n I A ) NΦB = (nl )(µ0 )( = µ0n2 Al I where nl is the total number of turns. Thus, the self-inductance is, ΝΦΒ L = I = µ0n 2 Al (6.15) If we fill the inside of the solenoid with a material of relative permeability µr (for example soft iron, which has a high value of relative permeability), then, L = µr µ0 n 2 Al (6.16) The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any 168 change in the current in a circuit. Physically, the self-inductance plays Reprint 2025-26 Electromagnetic Induction the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is d W = ε I d t If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.14), d W d I = L I d t d t Total amount of work done in establishing the current I is I W = ∫ d W = ∫ L I d I 0 Thus, the energy required to build up the current I is, 1 2 W = LI (6.17) 2 This expression reminds us of mv 2/2 for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit). Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. Equation (6.7) would be modified into N1 Φ1 = M 11 I 1 + M 12 I 2 where M11 represents inductance due to the same coil. Therefore, using Faraday’s law, d I 1 d I 2 ε1 = − M 11 − M 12 d t d t M11 is the self-inductance and is written as L1. Therefore, d I 1 d I 2 ε1 = − L 1 − M 12 d t d t Example 6.9 (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? Solution (a) From Eq. (6.17), the magnetic energy is 1 2 U B = LI 2 2 EXAMPLE 1  B  L = ( since B = µ0 nI , for a solenoid ) 2  µ0 n  6.9 169 Reprint 2025-26 Physics 1 2  B  2 = ( µ0 n Al ) [from Eq. (6.15)] 2  µ0n  1 2 = B Al 2µ0 (b) The magnetic energy per unit volume is, U B u B = (where V is volume that contains flux) V U B = Al B 2 generator: = (6.18) 2µ0 ac We have already obtained the relation for the electrostatic energy on stored per unit volume in a parallel plate capacitor (refer to Chapter 2, Eq. 2.73), 1 2 u Ε = ε0 E (2.73) animation 2 6.9 In both the cases energy is proportional to the square of the field strength. Equations (6.18) and (2.73) have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they Interactive http://micro.magnet.fsu.edu/electromag/java/generator/ac.html are general and valid for any region of space in which a magnetic field EXAMPLE or/and an electric field exist. 6.8 AC GENERATOR The phenomenon of electromagnetic induction has been technologically exploited in many ways. An exceptionally important application is the generation of alternating currents (ac). The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine. In this section, we shall describe the basic principles behind this machine. The Yugoslav inventor Nicola Tesla is credited with the development of the machine. As was pointed out in Section 6.3, one method to induce an emf or current in a loop is through a change in the loop’s orientation or a change in its effective area. As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos q, where q is the angle between A and B. This method of producing a FIGURE 6.13 AC Generator flux change is the principle of operation of a Reprint 2025-26 Electromagnetic Induction simple ac generator. An ac generator converts mechanical energy into electrical energy. The basic elements of an ac generator are shown in Fig. 6.13. It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes. When the coil is rotated with a constant angular speed w, the angle q between the magnetic field vector B and the area vector A of the coil at any instant t is q = wt (assuming q = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. (6.1), the flux at any time t is FB = BA cos q = BA cos wt From Faraday’s law, the induced emf for the rotating coil of N turns is then, dΦB d ε= – N = – NBA (cos ωt ) dt d t Thus, the instantaneous value of the emf is ε= NBA ωsin ωt (6.19) where NBAw is the maximum value of the emf, which occurs when sin wt = ±1. If we denote NBAw as e0, then e = e0 sin wt (6.20) Since the value of the sine fuction varies between +1 and –1, the sign, or polarity of the emf changes with time. Note from Fig. 6.14 that the emf has its extremum value when q = 90° or q = 270°, as the change of flux is greatest at these points. The direction of the current changes periodically and therefore the current is called alternating current (ac). Since w = 2pn, Eq (6.20) can be written as e = e0sin 2p n t (6.21) where n is the frequency of revolution of the generator’s coil. Note that Eq. (6.20) and (6.21) give the instantaneous value of the emf and e varies between +e0 and –e0 periodically. We shall learn how to determine the time-averaged value for the alternating voltage and current in the next chapter. In commercial generators, the mechanical energy required for rotation of the armature is provided by water falling from a height, for example, from dams. These are called hydro-electric generators. Alternatively, water is heated to produce steam using coal or other sources. The steam at high pressure produces the rotation of the armature. These are called thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear power generators. Modern day generators produce electric power as high as 500 MW, i.e., one can light 171 Reprint 2025-26 Physics FIGURE 6.14 An alternating emf is generated by a loop of wire rotating in a magnetic field. up 5 million 100 W bulbs! In most generators, the coils are held stationary and it is the electromagnets which are rotated. The frequency of rotation is 50 Hz in India. In certain countries such as USA, it is 60 Hz. Example 6.10 Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Solution Here n = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing Eq. (6.19) e0 = NBA (2 p n) = 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5 = 0.314 V 6.10 The maximum voltage is 0.314 V. We urge you to explore such alternative possibilities for power generation. EXAMPLE Reprint 2025-26 Electromagnetic Induction SUMMARY 1. The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as, FB = B.A = BA cos q where q is the angle between B and A. 2. Faraday’s laws of induction imply that the emf induced in a coil of N turns is directly related to the rate of change of flux through it, dΦB ε = − N d t Here FB is the flux linked with one turn of the coil. If the circuit is closed, a current I = e/R is set up in it, where R is the resistance of the circuit. 3. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. The negative sign in the expression for Faraday’s law indicates this fact. 4. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its ends is e = Bl v 5. Inductance is the ratio of the flux-linkage to current. It is equal to NF/I. 6. A changing current in a coil (coil 2) can induce an emf in a nearby coil (coil 1). This relation is given by, d I 2 ε1 = − M 12 d t The quantity M12 is called mutual inductance of coil 1 with respect to coil 2. One can similarly define M21. There exists a general equality, M12 = M21 7. When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by, d I ε = − L d t L is the self-inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. 8. The self-inductance of a long solenoid, the core of which consists of a magnetic material of relative permeability mr, is given by L = mr m0 n2 Al where A is the area of cross-section of the solenoid, l its length and n the number of turns per unit length. 9. In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of N turn and area A is rotated at n revolutions per second in a uniform magnetic field B, then the motional emf produced is e = NBA (2pn) sin (2pnt) where we have assumed that at time t = 0 s, the coil is perpendicular to the field. 173 Reprint 2025-26 Physics Quantity Symbol Units Dimensions Equations Magnetic Flux FB Wb (weber) [M L2 T –2 A–1] FB = B i A EMF e V (volt) [M L2 T –3 A–1] e = − d( NΦB )/d t Mutual Inductance M H (henry) [M L2 T –2 A–2] e1 = − M 12 ( dI 2 /d t ) Self Inductance L H (henry) [M L2 T –2 A–2] ε = − L ( d I /d t ) POINTS TO PONDER 1. Electricity and magnetism are intimately related. In the early part of the nineteenth century, the experiments of Oersted, Ampere and others established that moving charges (currents) produce a magnetic field. Somewhat later, around 1830, the experiments of Faraday and Henry demonstrated that a moving magnet can induce electric current. 2. In a closed circuit, electric currents are induced so as to oppose the changing magnetic flux. It is as per the law of conservation of energy. However, in case of an open circuit, an emf is induced across its ends. How is it related to the flux change? 3. The motional emf discussed in Section 6.5 can be argued independently from Faraday’s law using the Lorentz force on moving charges. However, even if the charges are stationary [and the q (v × B) term of the Lorentz force is not operative], an emf is nevertheless induced in the presence of a time-varying magnetic field. Thus, moving charges in static field and static charges in a time-varying field seem to be symmetric situation for Faraday’s law. This gives a tantalising hint on the relevance of the principle of relativity for Faraday’s law. EXERCISES 6.1 Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f ). Reprint 2025-26 Electromagnetic Induction FIGURE 6.15

5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. ⊳ Example 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳

5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x Fig. 5.3(a) Reprint 2025-26 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x →0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is Reprint 2025-26 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg . g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts

5.10 POWER Often it is interesting to know not only the work u Example 5.10 An elevator can carry a done on an object, but also the rate at which maximum load of 1800 kg (elevator + this work is done. We say a person is physically passengers) is moving up with a constant fit if he not only climbs four floors of a building speed of 2 m s–1. The frictional force opposing but climbs them fast. Power is defined as the the motion is 4000 N. Determine the time rate at which work is done or energy is minimum power delivered by the motor to transferred. the elevator in watts as well as in horse The average power of a force is defined as the power. ratio of the work, W, to the total time t taken Answer The downward force on the elevator is W Pav = F = m g + = (1800 × 10) + 4000 = 22000 N t Ff The motor must supply enough power to balanceThe instantaneous power is defined as the this force. Hence,limiting value of the average power as time interval approaches zero, P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ d W 5.11 COLLISIONS P = (5.20) d t In physics we study motion (change in position). The work dW done by a force F for a displacement At the same time, we try to discover physical dr is dW = F.dr. The instantaneous power can quantities, which do not change in a physical also be expressed as process. The laws of momentum and energy conservation are typical examples. In this d r P = F. section we shall apply these laws to a commonly d t encountered phenomena, namely collisions. Several games such as billiards, marbles or = F.v (5.21) carrom involve collisions.We shall study the where v is the instantaneous velocity when the collision of two masses in an idealised form. force is F. Consider two masses m1 and m2. The particle Power, like work and energy, is a scalar m1 is moving with speed v1i , the subscript ‘i’ quantity. Its dimensions are [ML2T–3]. In the SI, implying initial. We can cosider m2 to be at rest. its unit is called a watt (W). The watt is 1 J s–1. No loss of generality is involved in making such The unit of power is named after James Watt, a selection. In this situation the mass m1 one of the innovators of the steam engine in the collides with the stationary mass m2 and this eighteenth century. is depicted in Fig. 5.10. There is another unit of power, namely the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. 100 (watt) × 10 (hour) Fig. 5.10 Collision of mass m1, with a stationary mass m2. = 1000 watt hour The masses m1 and m2 fly-off in different =1 kilowatt hour (kWh) directions. We shall see that there are = 103 (W) × 3600 (s) relationships, which connect the masses, the = 3.6 × 106 J velocities and the angles. Reprint 2025-26 84 PHYSICS 5.11.1 Elastic and Inelastic Collisions The loss in kinetic energy on collision is In all collisions the total linear momentum is 1 2 1 2 conserved; the initial momentum of the system ∆ K = m 1 v1i − (m 1 + m 2 )v f 2 2 is equal to the final momentum of the system. One can argue this as follows. When two objects 2 1 2 1 m 1 2 collide, the mutual impulsive forces acting over = m 1v1i − v1i [using Eq. (5.22)] the collision time ∆t cause a change in their 2 2 m 1 + m 2 respective momenta : 1 2  m 1  ∆p1 = F12 ∆t = 2 m 1v1i 1 − m 1 + m 2  ∆p2 = F21 ∆t where F12 is the force exerted on the first particle 1 m 1m 2 2by the second particle. F21 is likewise the force = v1i 2 m 1 + m 2exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This which is a positive quantity as expected.implies ∆p1 + ∆p2 = 0 Consider next an elastic collision. Using the above nomenclature with θ1 = θ2 = 0, the The above conclusion is true even though the momentum and kinetic energy conservation forces vary in a complex fashion during the equations are collision time ∆t. Since the third law is true at every instant, the total impulse on the first object m1v1i = m1v1f + m2v2f (5.23) is equal and opposite to that on the second. 2 2 2 m 1v1i = m 1v1 f + m 2 v 2 f (5.24) On the other hand, the total kinetic energy of the system is not necessarily conserved. The From Eqs. (5.23) and (5.24) it follows that, impact and deformation during collision may generate heat and sound. Part of the initial kinetic m 1v1i (v 2 f − v1i ) = m 1v1 f (v 2 f − v1 f ) energy is transformed into other forms of energy. A useful way to visualise the deformation during or, v 2 f (v1i − v1 f ) = v12i − v12f collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains = (v1i − v1 f )(v 1i + v 1 f ) its original shape without loss in energy, then the initial kinetic energy is equal to the final Hence, ∴ v 2 f = v1i + v1 f (5.25) kinetic energy but the kinetic energy during the Substituting this in Eq. (5.23), we obtaincollision time ∆t is not constant. Such a collision is called an elastic collision. On the other hand (m 1 − m 2 ) v1 f = v1i (5.26)the deformation may not be relieved and the two m 1 + m 2 bodies could move together after the collision. A 2m 1v1icollision in which the two particles move together and v 2 f = (5.27) m 1 + m 2after the collision is called a completely inelastic collision. The intermediate case where the Thus, the ‘unknowns’ {v1f, v2f} are obtained in deformation is partly relieved and some of the terms of the ‘knowns’ {m1, m2, v1i}. Special cases initial kinetic energy is lost is more common and of our analysis are interesting. is appropriately called an inelastic collision. Case I : If the two masses are equal 5.11.2 Collisions in One Dimension v1f = 0 Consider first a completely inelastic collision v2f = v1i in one dimension. Then, in Fig. 5.10, The first mass comes to rest and pushes off the θ 1 = θ 2 = 0 second mass with its initial speed on collision. Case II : If one mass dominates, e.g. m2 > > m1 m1v1i = (m1+m2)vf (momentum conservation) v1f ~ − v1i v2f ~ 0 m 1 The heavier mass is undisturbed while the v f = v1i (5.22) m 1 + m 2 lighter mass reverses its velocity. Reprint 2025-26 WORK, ENERGY AND POWER 85 ⊳ dimensional, where the initial velocities and the Example 5.11 Slowing down of neutrons: final velocities lie in a plane. In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed 5.11.3 Collisions in Two Dimensions to 103 m s–1 so that it can have a high Fig. 5.10 also depicts the collision of a moving probability of interacting with isotope 235U92 mass m1 with the stationary mass m2. Linear momentum is conserved in such a collision. and causing it to fission. Show that a neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to material making up the light nuclei, usually be the x-y plane. The conservation of the heavy water (D2O) or graphite, is called a z-component of the linear momentum implies moderator. that the entire collision is in the x-y plane. The x- and y-component equations are Answer The initial kinetic energy of the neutron m1v1i = m1v1f cos θ1 + m2v2f cos θ2 (5.28)is 0 = m1v1f sin θ1 − m2v2f sin θ2 (5.29) 1 2 K 1i = m 1v1i 2 One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f, v2f , θ1 and θ2}, andwhile its final kinetic energy from Eq. (5.26) only two equations. If θ 1 = θ 2 = 0, we regain 1 2 1  m 1 − m 2  2 2 Eq. (5.23) for one dimensional collision. K 1 f = m 1v1 f = m 1 v1i 2 2  m 1 + m 2  If, further the collision is elastic, 1 2 1 2 1 2 m1v1i = m1v1 f + m 2v 2 f (5.30) The fractional kinetic energy lost is 2 2 2 2 We obtain an additional equation. That still K 1 f  m 1 − m 2  f 1 = = leaves us one equation short. At least one of K 1i  m 1 + m 2  the four unknowns, say θ1, must be made known while the fractional kinetic energy gained by the for the problem to be solvable. For example, θ1 moderating nuclei K2f /K1i is can be determined by moving a detector in an angular fashion from the x to the y axis. Given f2 = 1 − f1 (elastic collision) {m1, m2, v1i , θ1} we can determine {v1f , v2f, θ2} 4m 1m 2 from Eqs. (5.28)-(5.30). = 2 ⊳ (m1 + m 2 ) Example 5.12 Consider the collision depicted in Fig. 5.10 to be between two One can also verify this result by substituting billiard balls with equal masses m1 = m2.from Eq. (5.27). The first ball is called the cue while the For deuterium m2 = 2m1 and we obtain second ball is called the target. The f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard player wants to ‘sink’ the target neutron’s energy is transferred to deuterium. For ball in a corner pocket, which is at an carbon f1 = 71.6% and f2 = 28.4%. In practice, angle θ2 = 37°. Assume that the collision ishowever, this number is smaller since head-on elastic and that friction and rotational collisions are rare. ⊳ motion are not important. Obtain θ1. If the initial velocities and final velocities of Answer From momentum conservation, since both the bodies are along the same straight line, the masses are equal then it is called a one-dimensional collision, or head-on collision. In the case of small spherical v1i = v 1f + v 2f bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 ⋅ or v 1i 2 = ( v1 f + v 2 f ) ( v1 f + v 2 f ) which is at rest. In general, the collision is two- = v1 f 2 + v 2 f 2 + 2 v1 f .v 2 f Reprint 2025-26 86 PHYSICS 2 2 The matter simplifies greatly if we consider= (5.31) { v1 f + v 2 f + 2v1 f v 2 f cos (θ1 + 37 ° ) } spherical masses with smooth surfaces, and assume that collision takes place only when the Since the collision is elastic and m1 = m2 it follows bodies touch each other. This is what happensfrom conservation of kinetic energy that in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v 2 f 2 (5.32) In our everyday world, collisions take place only when two bodies touch each other. But considerComparing Eqs. (5.31) and (5.32), we get a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and or θ1 + 37° = 90° going away in some direction. Here we have to deal with forces involving action at a distance. Thus, θ1 = 53° Such an event is called scattering. The velocities This proves the following result : when two equal and directions in which the two particles go away masses undergo a glancing elastic collision with depend on their initial velocities as well as the one of them at rest, after the collision, they will type of interaction between them, their masses, move at right angles to each other. ⊳ shapes and sizes. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that d V ( x ) F ( x ) = − d x x f F ( x ) d x or Vi − V f = ∫ x i 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 5. The elastic potential energy of a spring of force constant k and extension x is 1 2 V ( x ) = k x 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. Reprint 2025-26 WORK, ENERGY AND POWER 87 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 5.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 5. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. Reprint 2025-26 88 PHYSICS EXERCISES 5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 5.11 Reprint 2025-26 WORK, ENERGY AND POWER 89 5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this Fig. 5.12 potential must ‘turn back’ when it reaches x = ± 2 m. 5.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not Fig. 5.13 normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 5.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 5.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 5.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? Reprint 2025-26 90 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 kˆ N where ˆ,i ˆ,j kˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ? Fig. 5.14 Reprint 2025-26 WORK, ENERGY AND POWER 91 5.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ? 5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a Fig. 5.15 frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ? 5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m ? 5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ? 5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? 5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Reprint 2025-26 CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a 6.1 Introduction point mass having no size.) We applied the results of our 6.2 Centre of mass study even to the motion of bodies of finite size, assuming 6.3 Motion of centre of mass that motion of such bodies can be described in terms of the

5.9 THE POTENTIAL ENERGY OF A SPRING The spring force is an example of a variable force which is conservative. Fig. 5.7 shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force Fs is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 5.7(b)] or negative [Fig. 5.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as Fs = − kx The constant k is called the spring constant. Its unit is N m-1. The spring is said to be stiff if k is large and soft if k is small. Fig. 5.7 Illustration of the spring force with a block Suppose that we pull the block outwards as in attached to the free end of the spring. Fig. 5.7(b). If the extension is xm, the work done by (a) The spring force Fs is zero when the the spring force is displacement x from the equilibrium position is zero. (b) For the stretched spring x > 0 xm xm and Fs < 0 (c) For the compressed spring d x x < 0 and Fs > 0.(d) The plot of Fs versus x. Fs d x = −∫kx W s = ∫ 0 0 The area of the shaded triangle represents the work done by the spring force. Due to the k x m2 opposing signs of Fs and x, this work done is = − (5.15) 2 2 negative, W s = −kx m / 2 . This expression may also be obtained by considering the area of the triangle as in The same is true when the spring is Fig. 5.7(d). Note that the work done by the compressed with a displacement xc (< 0). Theexternal pulling force F is positive since it overcomes the spring force. spring force does work Ws = − kx c2 / 2 while the Reprint 2025-26 WORK, ENERGY AND POWER 81 2 and vice versa, however, the total mechanical external force F does work + kxc / 2. If the block energy remains constant. This is graphically is moved from an initial displacement xi to a depicted in Fig. 5.8. final displacement xf , the work done by the spring force Ws is xf k x 2f k x i2 (5.17) k x d x = − Ws = − ∫ 2 2 x i Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to xi ; x i k x i2 k x i2 k x dx = − Ws = − ∫ 2 2 x i = 0 (5.18) Fig. 5.8 Parabolic plots of the potential energy V and The work done by the spring force in a cyclic kinetic energy K of a block attached to a spring obeying Hooke’s law. The two plotsprocess is zero. We have explicitly demonstrated are complementary, one decreasing as the that the spring force (i) is position dependent other increases. The total mechanical only as first stated by Hooke, (Fs = − kx); (ii) energy E = K + V remains constant. does work which only depends on the initial and final positions, e.g. Eq. (5.17). Thus, the spring ⊳ Example 5.8 To simulate car accidents, autoforce is a conservative force. manufacturers study the collisions of moving We define the potential energy V(x) of the spring cars with mounted springs of different springto be zero when block and spring system is in the constants. Consider a typical simulation withequilibrium position. For an extension (or a car of mass 1000 kg moving with a speedcompression) x the above analysis suggests that 18.0 km/h on a smooth road and colliding kx 2 with a horizontally mounted spring of spring V(x) = (5.19) constant 5.25 × 103 N m–1. What is the 2 maximum compression of the spring ?You may easily verify that − dV/dx = −k x, the spring force. If the block of mass m in Fig. 5.7 is extended to xm and released from rest, then its Answer At maximum compression the kinetic total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the where x lies between – xm and + xm, will be given by potential energy of the spring. The kinetic energy of the moving car is 1 2 1 2 1 2 k x m = k x + m v 1 2 2 2 K = mv2 2where we have invoked the conservation of mechanical energy. This suggests that the speed 1 3 and the kinetic energy will be maximum at the = × 10 × 5 × 5 2 equilibrium position, x = 0, i.e., K = 1.25 × 104 J 1 2 1 2 m v m = k x m where we have converted 18 km h–1 to 5 m s–1 [It is 2 2 useful to remember that 36 km h–1 = 10 m s–1]. where vm is the maximum speed. At maximum compression xm, the potential energy V of the spring is equal to the kinetic k or v m = x m energy K of the moving car from the principle of m conservation of mechanical energy. Note that k/m has the dimensions of [T-2] and our equation is dimensionally correct. The 1 2 V = k x m kinetic energy gets converted to potential energy 2 Reprint 2025-26 82 PHYSICS = 1.25 × 104 J We obtain xm = 2.00 m We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction. ⊳ We conclude this section by making a few Fig. 5.9 The forces acting on the car. remarks on conservative forces. (i) Information on time is absent from the above 1 2 ∆K = Kf − Ki = 0 − m v discussions. In the example considered 2 above, we can calculate the compression, but The work done by the net force is not the time over which the compression 1 2 occurs. A solution of Newton’s Second Law W = − kx m −µm g x m 2 for this system is required for temporal information. Equating we have (ii) Not all forces are conservative. Friction, for 1 2 1 2 example, is a non-conservative force. The m v = k x m + µm g x m 2 2 principle of conservation of energy will have Now µmg = 0.5 × 103 × 10 = 5 × 103 N (taking to be modified in this case. This is illustrated g =10.0 m s-2). After rearranging the above in Example 5.9. equation we obtain the following quadratic(iii) The zero of the potential energy is arbitrary. equation in the unknown xm. It is set according to convenience. For the spring force we took V(x) = 0, at x = 0, i.e. the 2 2 k x m + 2µm g x m − m v = 0 unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later chapter we shall see that for the force where we take the positive square root since due to the universal law of gravitation, the zero is best defined at an infinite distance xm is positive. Putting in numerical values we obtain from the gravitational source. However, once the zero of the potential energy is fixed in a xm = 1.35 m given discussion, it must be consistently which, as expected, is less than the result in adhered to throughout the discussion. You Example 5.8. cannot change horses in midstream ! If the two forces on the body consist of a conservative force Fc and a non-conservative⊳ force Fnc , the conservation of mechanical energy Example 5.9 Consider Example 5.8 taking formula will have to be modified. By the WE the coefficient of friction, µ, to be 0.5 and theorem calculate the maximum compression of the spring. (Fc+ Fnc) ∆x = ∆K But Fc ∆x = − ∆V Answer In presence of friction, both the spring Hence, ∆(K + V) = Fnc ∆x force and the frictional force act so as to oppose ∆E = Fnc ∆x the compression of the spring as shown in where E is the total mechanical energy. Over Fig. 5.9. the path this assumes the form We invoke the work-energy theorem, rather Ef − Ei = Wnc than the conservation of mechanical energy. where Wnc is the total work done by the The change in kinetic energy is non-conservative forces over the path. Note that Reprint 2025-26 WORK, ENERGY AND POWER 83 unlike the conservative force, Wnc depends on Our electricity bills carry the energy the particular path i to f. ⊳ consumption in units of kWh. Note that kWh is a unit of energy and not of power.

5.7 A short bar magnet has a magnetic moment of 0.48 J T –1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet. 153 Reprint 2025-26 Physics Chapter Six ELECTROMAGNETIC INDUCTION 6.1 INTRODUCTION Electricity and magnetism were considered separate and unrelated phenomena for a long time. In the early decades of the nineteenth century, experiments on electric current by Oersted, Ampere and a few others established the fact that electricity and magnetism are inter-related. They found that moving electric charges produce magnetic fields. For example, an electric current deflects a magnetic compass needle placed in its vicinity. This naturally raises the questions like: Is the converse effect possible? Can moving magnets produce electric currents? Does the nature permit such a relation between electricity and magnetism? The answer is resounding yes! The experiments of Michael Faraday in England and Joseph Henry in USA, conducted around 1830, demonstrated conclusively that electric currents were induced in closed coils when subjected to changing magnetic fields. In this chapter, we will study the phenomena associated with changing magnetic fields and understand the underlying principles. The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction. When Faraday first made public his discovery that relative motion between a bar magnet and a wire loop produced a small current in the latter, he was asked, “What is the use of it?” His reply was: “What is the 154 use of a new born baby?” The phenomenon of electromagnetic induction Reprint 2025-26 Electromagnetic Induction is not merely of theoretical or academic interest but also of practical utility. Imagine a world where there is no electricity – no electric lights, no trains, no telephones and no personal computers. The pioneering experiments of Faraday and Henry have led directly to the development of modern day generators and transformers. Today’s civilisation owes its progress to a great extent to the discovery of electromagnetic induction.

5.2 Notions of work and kinetic landscape, all are said to be working. In physics, however, energy : The work-energy the word ‘Work’ covers a definite and precise meaning. theorem Somebody who has the capacity to work for 14-16 hours a5.3 Work day is said to have a large stamina or energy. We admire a

5.5 MAGNETIC PROPERTIES OF MATERIALS The discussion in the previous section helps us to classify materials as diamagnetic, paramagnetic or ferromagnetic. In terms of the susceptibility χ, a material is diamagnetic if χ is negative, para- if χ is positive and small, and ferro- if χ is large and positive. A glance at Table 5.2 gives one a better feeling for these materials. Here ε is a small positive number introduced to quantify paramagnetic materials. Next, we describe these materials in some detail. TABLE 5.2 Diamagnetic Paramagnetic Ferromagnetic –1 ≤ χ < 0 0 < χ < ε χ >> 1 0 ≤ µr < 1 1< µr < 1+ ε µr >> 1 µ < µ0 µ > µ0 µ >> µ0 5.5.1 Diamagnetism Diamagnetic substances are those which have tendency to move from FIGURE 5.7 stronger to the weaker part of the external magnetic field. In other words, Behaviour of unlike the way a magnet attracts metals like iron, it would repel a magnetic field lines diamagnetic substance. near a Figure 5.7(a) shows a bar of diamagnetic material placed in an external (a) diamagnetic, magnetic field. The field lines are repelled or expelled and the field inside (b) paramagnetic the material is reduced. In most cases, this reduction is slight, being one substance. part in 105. When placed in a non-uniform magnetic field, the bar will tend to move from high to low field. 147 Reprint 2025-26 Physics The simplest explanation for diamagnetism is as follows. Electrons in an atom orbiting around nucleus possess orbital angular momentum. These orbiting electrons are equivalent to current-carrying loop and thus possess orbital magnetic moment. Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. This happens due to induced current in accordance with Lenz’s law which you will study in Chapter 6. Thus, the substance develops a net magnetic moment in direction opposite to that of the applied field and hence repulsion. Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride. Diamagnetism is present in all the substances. However, the effect is so weak in most cases that it gets shifted by other effects like paramagnetism, ferromagnetism, etc. The most exotic diamagnetic materials are superconductors. These are metals, cooled to very low temperatures which exhibits both perfect conductivity and perfect diamagnetism. Here the field lines are completely expelled! χ = –1 and µr = 0. A superconductor repels a magnet and (by Newton’s third law) is repelled by the magnet. The phenomenon of perfect diamagnetism in superconductors is called the Meissner effect, after the name of its discoverer. Superconducting magnets can be gainfully exploited in variety of situations, for example, for running magnetically levitated superfast trains. 5.5.2 Paramagnetism Paramagnetic substances are those which get weakly magnetised when placed in an external magnetic field. They have tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get weakly attracted to a magnet. The individual atoms (or ions or molecules) of a paramagnetic material possess a permanent magnetic dipole moment of their own. On account of the ceaseless random thermal motion of the atoms, no net magnetisation is seen. In the presence of an external field B0, which is strong enough, and at low temperatures, the individual atomic dipole moment can be made to align and point in the same direction as B0. Figure 5.7(b) shows a bar of paramagnetic material placed in an external field. The field lines gets concentrated inside the material, and the field inside is enhanced. In most cases, this enhancement is slight, being one part in 105. When placed in a non-uniform magnetic field, the bar will tend to move from weak field to strong. Some paramagnetic materials are aluminium, sodium, calcium, oxygen (at STP) and copper chloride. For a paramagnetic material both χ and µr depend not only on the material, but also (in a simple fashion) on the sample temperature. As the field is increased or the temperature is lowered, the magnetisation increases until it reaches the saturation value at which point all the dipoles are perfectly aligned with the field. Reprint 2025-26 Magnetism and Matter 5.5.3 Ferromagnetism Ferromagnetic substances are those which gets strongly magnetised when placed in an external magnetic field. They have strong tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet. The individual atoms (or ions or molecules) in a ferromagnetic material possess a dipole moment as in a paramagnetic material. However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. The explanation of this cooperative effect requires quantum mechanics and is beyond the scope of this textbook. Each domain has a net magnetisation. Typical domain size is 1mm and the domain contains about 1011 atoms. In the first instant, the magnetisation varies randomly from domain to domain and there is no bulk magnetisation. This is shown in Fig. 5.8(a). When we apply an external magnetic field B0, the domains FIGURE 5.8 orient themselves in the direction of B0 and simultaneously the domain (a) Randomly oriented in the direction of B0 grow in size. This existence of domains and oriented domains, their motion in B0 are not speculations. One may observe this under a (b) Aligned domains. microscope after sprinkling a liquid suspension of powdered ferromagnetic substance of samples. This motion of suspension can be observed. Fig. 5.8(b) shows the situation when the domains have aligned and amalgamated to form a single ‘giant’ domain. Thus, in a ferromagnetic material the field lines are highly concentrated. In non-uniform magnetic field, the sample tends to move towards the region of high field. We may wonder as to what happens when the external field is removed. In some ferromagnetic materials the magnetisation persists. Such materials are called hard magnetic materials or hard ferromagnets. Alnico, an alloy of iron, aluminium, nickel, cobalt and copper, is one such material. The naturally occurring lodestone is another. Such materials form permanent magnets to be used among other things as a compass needle. On the other hand, there is a class of ferromagnetic materials in which the magnetisation disappears on removal of the external field. Soft iron is one such material. Appropriately enough, such materials are called soft ferromagnetic materials. There are a number of elements, which are ferromagnetic: iron, cobalt, nickel, gadolinium, etc. The relative magnetic permeability is >1000! The ferromagnetic property depends on temperature. At high enough temperature, a ferromagnet becomes a paramagnet. The domain structure disintegrates with temperature. This disappearance of magnetisation with temperature is gradual. SUMMARY 1. The science of magnetism is old. It has been known since ancient times that magnetic materials tend to point in the north-south direction; like 149 Reprint 2025-26 Physics magnetic poles repel and unlike ones attract; and cutting a bar magnet in two leads to two smaller magnets. Magnetic poles cannot be isolated. 2. When a bar magnet of dipole moment m is placed in a uniform magnetic field B, (a) the force on it is zero, (b) the torque on it is m × B, (c) its potential energy is –m.B, where we choose the zero of energy at the orientation when m is perpendicular to B. 3. Consider a bar magnet of size l and magnetic moment m, at a distance r from its mid-point, where r >>l, the magnetic field B due to this bar is, µ0 m B = 2 π r 3 (along axis) µ0 m = – 3 (along equator) 4 π r 4. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero B B i S 0 all area elements S 5. Consider a material placed in an external magnetic field B0. The magnetic intensity is defined as, B0 H = µ0 The magnetisation M of the material is its dipole moment per unit volume. The magnetic field B in the material is, B = m0 (H + M) 6. For a linear material M = c H. So that B = m H and c is called the magnetic susceptibility of the material. The three quantities, c, the relative magnetic permeability mr, and the magnetic permeability m are related as follows: m = m0 mr mr = 1+ c 7. Magnetic materials are broadly classified as: diamagnetic, paramagnetic, and ferromagnetic. For diamagnetic materials c is negative and small and for paramagnetic materials it is positive and small. Ferromagnetic materials have large c. 8. Substances, which at room temperature, retain their ferromagnetic property for a long period of time are called permanent magnets. Reprint 2025-26 Magnetism and Matter Physical quantity Symbol Nature Dimensions Units Remarks Permeability of µ0 Scalar [MLT–2 A–2] T m A–1 µ0/4π = 10–7 free space Magnetic field, B Vector [MT–2 A–1] T (tesla) 104 G (gauss) = 1 T Magnetic induction, Magnetic flux density Magnetic moment m Vector [L–2 A] A m2 Magnetic flux φB Scalar [ML2T–2 A–1] W (weber) W = T m2 Magnetic momentMagnetisation M Vector [L–1 A] A m–1 Volume Magnetic intensity H Vector [L–1 A] A m–1 B = µ0 (H + M) Magnetic field strength Magnetic χ Scalar - - M = χH susceptibility Relative magnetic µr Scalar - - B = µ0 µr H permeability Magnetic permeability µ Scalar [MLT–2 A–2] T m A–1 µ = µ0 µr N A–2 B = µ H POINTS TO PONDER 1. A satisfactory understanding of magnetic phenomenon in terms of moving charges/currents was arrived at after 1800 AD. But technological exploitation of the directional properties of magnets predates this scientific understanding by two thousand years. Thus, scientific understanding is not a necessary condition for engineering applications. Ideally, science and engineering go hand-in-hand, one leading and assisting the other in tandem. 2. Magnetic monopoles do not exist. If you slice a magnet in half, you get two smaller magnets. On the other hand, isolated positive and negative charges exist. There exists a smallest unit of charge, for example, the electronic charge with value |e| = 1.6 ×10–19 C. All other charges are integral multiples of this smallest unit charge. In other words, charge is quantised. We do not know why magnetic monopoles do not exist or why electric charge is quantised. 3. A consequence of the fact that magnetic monopoles do not exist is that the magnetic field lines are continuous and form closed loops. In contrast, the electrostatic lines of force begin on a positive charge and terminate on the negative charge (or fade out at infinity). 4. A miniscule difference in the value of χ, the magnetic susceptibility, yields radically different behaviour: diamagnetic versus paramagnetic. For diamagnetic materials χ = –10–5 whereas χ = +10–5 for paramagnetic materials. 151 Reprint 2025-26 Physics 5. There exists a perfect diamagnet, namely, a superconductor. This is a metal at very low temperatures. In this case χ = –1, µr = 0, µ = 0. The external magnetic field is totally expelled. Interestingly, this material is also a perfect conductor. However, there exists no classical theory which ties these two properties together. A quantum-mechanical theory by Bardeen, Cooper, and Schrieffer (BCS theory) explains these effects. The BCS theory was proposed in1957 and was eventually recognised by a Nobel Prize in physics in 1970. 6. Diamagnetism is universal. It is present in all materials. But it is weak and hard to detect if the substance is para- or ferromagnetic. 7. We have classified materials as diamagnetic, paramagnetic, and ferromagnetic. However, there exist additional types of magnetic material such as ferrimagnetic, anti-ferromagnetic, spin glass, etc. with properties which are exotic and mysterious. EXERCISES 5.1 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet? 5.2 A short bar magnet of magnetic moment m = 0.32 JT –1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case? 5.3 A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? 5.4 If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field? 5.5 A bar magnet of magnetic moment 1.5 J T –1 lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? 5.6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 –4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. Reprint 2025-26 Magnetism and Matter (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?

1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional

1.14 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

1.00 nF for two values of R: (i) R = 100 W and (ii) R = 200 W. For the source applied vm = 1 100 V. w0 for this case is = 1.00×106 LC FIGURE 7.14 Variation of im with w for tworad/s. cases: (i) R = 100 W, (ii) R = 200 W, We see that the current amplitude is L = 1.00 mH. maximum at the resonant frequency. Since im = vm / R at resonance, the current amplitude for case (i) is twice to that for case (ii). Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is vm/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. 189 Reprint 2025-26 Physics Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Solution Given R = 200 Ω, C = 15.0 µF = 15.0 × 10 −6 F V = 220 V, ν = 50Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z = R 2 + X C2 = R 2 + (2πνC )− 2 = (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10 −6 F )−2 = (200 Ω )2 + (212.3 Ω )2 = 291.67 Ω Therefore, the current in the circuit is V 220 V I = = = 0.755 A Z 291.5 Ω (b) Since the current is the same throughout the circuit, we have V R = I R = (0.755 A)(200 Ω=) 151V VC = I X C = (0.755 A)(212.3 Ω=) 160.3 V The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem: V R +C = V R2 + VC2 7.6 = 220 V Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor EXAMPLE is equal to the voltage of the source. 7.7 POWER IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinwt applied to a series RLC circuit drives a current in the circuit given by i = im sin(wt + f) where v m − 1  X C − X L  i m = Z and φ = tan  R  190 Therefore, the instantaneous power p supplied by the source is Reprint 2025-26 Alternating Current p = v i = (v m sin ωt ) × [ i m sin(ωt + φ)] v m i m = [ cosφ− cos(2ωt + φ)] (7.29) 2 The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.29). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, v m i m v m i m P = cos φ = cos φ 2 2 2 = V I cos φ [7.30(a)] This can also be written as, P = I 2 Z cos φ [7.30(b)] So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called the power factor. Let us discuss the following cases: CaseCaseCaseCaseCase (i)(i)(i)(i)(i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case φ = 0, cos φ = 1. There is maximum power dissipation. CaseCaseCaseCaseCase (ii)(ii)(ii)(ii)(ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is π/2. Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. CaseCaseCaseCaseCase (iii)(iii)(iii)(iii)(iii) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.30) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. CaseCaseCaseCaseCase (iv)(iv)(iv)(iv)(iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2 R. That is, maximum power is dissipated in a circuit (through R) at resonance. ExampleExampleExampleExampleExample 7.77.77.77.77.7 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. SolutionSolutionSolutionSolutionSolution (a) We know that P = I V cosφ where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission. EEEE (b)Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. EXAMPLEXAMPLEXAMPLEXAMPLEXAMPLE We can improve the power factor (tending to 1) by making Z tend to 7.77.77.77.77.7 us understand, with the help of a phasor diagram (Fig. 7.15) R. Let 191 Reprint 2025-26 Physics FIGURE 7.15 how this can be achieved. Let us resolve I into two components. Ip along the applied voltage V and Iq perpendicular to the applied voltage. Iq as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, 7.7 we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I¢q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I¢q cancel EXAMPLE each other and P is effectively Ip V. Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 mF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor. Solution (a) To find the impedance of the circuit, we first calculate XL and XC. XL = 2 pnL = 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W 1 X C = 2 π νC 1 = 6 = 4 Ω 2 × 3.14 × 50 × 796 × 10− Therefore, Z = R 2 + ( X L − X C )2 = 3 2 + (8 − 4)2 = 5 W X C − X L 7.8 (b) Phase difference, f = tan–1 R −1  4 − 8  EXAMPLE = tan  3 = −53. 1°192 Reprint 2025-26 Alternating Current Since f is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I 2 R im 1  283  = Now, I = 2 2  5 = 40A EXAMPLE Therefore, P = (40 A )2 × 3 Ω= 4800 W (d) Power factor = cos cos –53.1 0.6 7.8 Example 7.9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is 1 1 ω0 = = LC 25.48 × 10 −3 × 796 × 10 −6 = 222.1rad/s ω0 221.1 νr = = Hz = 35.4Hz 2 π 2 × 3.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3 Ω The rms current at resonance is V V  283  1 = = = = 66.7 A Z R  2  3 The power dissipated at resonance is P = I 2 × R = (66.7)2 × 3 = 13.35 kW EXAMPLE You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. 7.9 Example 7.10 At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the EXAMPLE circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. 7.10 193 Reprint 2025-26 Physics 7.8 TRANSFORMERS For many purposes, it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value. This is done with a device called transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.16(a) or on separate limbs of the core as in Fig. 7.16(b). One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. FIGURE 7.16 Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other, (b) two coils on separate limbs of the core. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let f be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it. Then the induced emf or voltage es, in the secondary with Ns turns is dφ εs = − N s (7.31) d t The alternating flux f also induces an emf, called back emf in the primary. This is dφ εp = − N p (7.32) d t But ep = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation es = vs Reprint 2025-26 Alternating Current where vs is the voltage across the secondary. Therefore, Eqs. (7.31) and (7.32) can be written as dφ v s = − N s [7.31(a)] d t dφ v p = − N p [7.32(a)] d t From Eqs. [7.31 (a)] and [7.32 (a)], we have v s N s = (7.33) v p N p Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v, ipvp = isvs (7.34) Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (7.33) and (7.34), we have i p v s N s = = (7.35) i s v p N p Since i and v both oscillate with the same frequency as the ac source, Eq. (7.35) also gives the ratio of the amplitudes or rms values of corresponding quantities. Now, we can see how a transformer affects the voltage and current. We have:  N s   N p  Vs = V p and I s = I p (7.36)  N p   N s  That is, if the secondary coil has a greater number of turns than the primary (Ns > Np), the voltage is stepped up (Vs > Vp). This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (Np/Ns < 1 and Is < Ip). For example, if the primary coil of a transformer has 100 turns and the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A. If the secondary coil has less turns than the primary (Ns < Np), we have a step-down transformer. In this case, Vs < Vp and Is > Ip. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: (i) Flux Leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor 195 Reprint 2025-26 Physics design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I 2R). In high current, low voltage windings, these are minimised by using thick wire. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I 2R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. SUMMARY 1. An alternating voltage v = v m sin ω t applied to a resistor R drives a v m current i = im sinwt in the resistor, i m = . The current is in phase with R the applied voltage. 2. For an alternating current i = im sin wt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a special value of current is used. It is called root mean square (rms) current and is donoted by I: i m I = = 0.707 i m 2 Similarly, the rms voltage is defined by vm V = = 0.707 v m 2 We have P = IV = I 2R 3. An ac voltage v = vm sin wt applied to a pure inductor L, drives a current in the inductor i = im sin (wt – p/2), where im = vm/XL. XL = wL is called inductive reactance. The current in the inductor lags the voltage by p/2. The average power supplied to an inductor over one complete cycle is zero. Reprint 2025-26 Alternating Current 4. An ac voltage v = vm sinwt applied to a capacitor drives a current in the capacitor: i = im sin (wt + p/2). Here, v m 1 i m = , X C = is called capacitive reactance. X C ωC The current through the capacitor is p/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero. 5. For a series RLC circuit driven by voltage v = vm sin wt, the current is given by i = im sin (wt + f) v m where i m = R 2 + ( X C − X L )2 −1 X C − X L and φ = tan R is called the impedance of the circuit. Z = R 2 + ( X C − X L )2 The average power loss over a complete cycle is given by P = V I cosf The term cosf is called the power factor. 6. In a purely inductive or capacitive circuit, cosf = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current. 7. The phase relationship between current and voltage in an ac circuit can be shown conveniently by representing voltage and current by rotating vectors called phasors. A phasor is a vector which rotates about the origin with angular speed w. The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. The analysis of an ac circuit is facilitated by the use of a phasor diagram. 8. A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by  N s  Vs =  N p  V p and the currents are related by  N p  Is = I p  N s  If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer. 197 Reprint 2025-26 Physics Physical quantity Symbol Dimensions Unit Remarks 2 –3 v m rms voltage V [M L T A –1] V V = , v m is the 2 amplitude of the ac voltage. i m rms current I [ A] A I = , im is the amplitude of 2 the ac current. Reactance: Inductive XL [M L2 T –3 A–2] XL = L Capacitive XC [M L2 T –3 A–2] XC = 1/C Impedance Z [M L2 T –3 A–2] Depends on elements present in the circuit. 1 Resonant wr or w0 [T –1] Hz w0  for a frequency LC series RLC circuit ω0 L 1 Quality factor Q Dimensionless Q = = for a series R ω0 C R RLC circuit. Power factor Dimensionless = cosf, f is the phase difference between voltage applied and current in the circuit. POINTS TO PONDER 1. When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is v m = 2V = 2(240) = 340 V 2. The power rating of an element used in ac circuits refers to its average power rating. 3. The power consumed in an ac circuit is never negative. 4. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac 198 currents, as the dc ampere is derived. An ac current changes direction Reprint 2025-26 Alternating Current with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. 5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is V RC = VR2 + VC2 and not VR + VC since VC is p/2 out of phase of VR. 6. Though in a phasor diagram, voltage and current are represented by vectors, these quantities are not really vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element. 8. In a RLC circuit, resonance phenomenon occur when XL = XC or 1 ω0 = . For resonance to occur, the presence of both L and C LC elements in the circuit is a must. With only one of these (L or C ) elements, there is no possibility of voltage cancellation and hence, no resonance is possible. 9. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. 10. In generators and motors, the roles of input and output are reversed. In a motor, electric energy is the input and mechanical energy is the output. In a generator, mechanical energy is the input and electric energy is the output. Both devices simply transform energy from one form to another. 11. A transformer (step-up) changes a low-voltage into a high-voltage. This does not violate the law of conservation of energy. The current is reduced by the same proportion. 199 Reprint 2025-26 Physics EXERCISES 7.1 A 100 W resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? 7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? 7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. 7.4 A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. 7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. 7.6 A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? 7.7 A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? 7.8 Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W. FIGURE 7.17 (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. Reprint 2025-26 Chapter Eight ELECTROMAGNETIC WAVES 8.1 INTRODUCTION In Chapter 4, we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency. Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities. These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time- varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to Reprint 2025-26 Physics the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and light. Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today. In this chapter, we first discuss the need for displacement current and its consequences. Then we present a descriptive account of electromagnetic waves. James Clerk Maxwell The broad spectrum of electromagnetic waves, (1831 – 1879) Born in stretching from g rays (wavelength ~10–12 m) to long Edinburgh, Scotland, radio waves (wavelength ~106 m) is described. was among the greatest physicists of the nineteenth century. He 8.2 DISPLACEMENT CURRENT derived the thermal We have seen in Chapter 4 that an electrical current velocity distribution of produces a magnetic field around it. Maxwell showed molecules in a gas and that for logical consistency, a changing electric field must was among the first to obtain reliable also produce a magnetic field. This effect is of great estimates of molecular importance because it explains the existence of radio parameters from waves, gamma rays and visible light, as well as all other measurable quantities forms of electromagnetic waves. like viscosity, etc. To see how a changing electric field gives rise to Maxwell’s greatest a magnetic field, let us consider the process of acheivement was the unification of the laws of charging of a capacitor and apply Ampere’s circuital electricity and law given by (Chapter 4) magnetism (discovered by Coulomb, Oersted, “B.dl = m0 i (t) (8.1)(1831–1879) Ampere and Faraday) to find magnetic field at a point outside the capacitor. into a consistent set of Figure 8.1(a) shows a parallel plate capacitor C which equations now called Maxwell’s equations. is a part of circuit through which a time-dependent From these he arrived at current i (t) flows . Let us find the magnetic field at a the most important point such as P, in a region outside the parallel plateMAXWELL conclusion that light is capacitor. For this, we consider a plane circular loop of an electromagnetic radius r whose plane is perpendicular to the direction wave. Interestingly, of the current-carrying wire, and which is centred Maxwell did not agree symmetrically with respect to the wire [Fig. 8.1(a)]. FromCLERK with the idea (strongly suggested by the symmetry, the magnetic field is directed along the Faraday’s laws of circumference of the circular loop and is the same in electrolysis) that magnitude at all points on the loop so that if B is theJAMES electricity was magnitude of the field, the left side of Eq. (8.1) is B (2p r). particulate in nature. So we have B (2pr) = m0i (t) (8 .2) 202 Reprint 2025-26 Electromagnetic Waves Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand side is zero and not m0i, since no current passes through the surface of Fig. 8.1(b) and (c). So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig. 8.1(c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field! If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/e0 (see Eq. 2.41). The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux FE through the surface S ? Using Gauss’s law, it is 1 Q Q ΦE = E A = A = (8.3) ε0 A ε0 Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using Eq. (8.3), we have dΦE d  Q  1 d Q = FIGURE 8.1 A d t d t  ε0 = ε0 d t parallel plate capacitor C, as part of This implies that for consistency, a circuit through which a time  dΦE  dependent current i (t) flows, (a) a loop of ε0  d t  = i (8.4) radius r, to determine This is the missing term in Ampere’s circuital law. If we generalise magnetic field at a this law by adding to the total current carried by conductors through point P on the loop; the surface, another term which is e0 times the rate of change of electric (b) a pot-shaped surface passingflux through the same surface, the total has the same value of current i through the interior for all surfaces. If this is done, there is no contradiction in the value of B between the capacitor obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop shown in (a) as itsis non-zero no matter which surface is used for calculating it. B at a rim; (c) a tiffin- point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with inside, as it should be. The current carried by conductors due to flow of the circular loop as charges is called conduction current. The current, given by Eq. (8.4), is a its rim and a flat circular bottom S new term, and is due to changing electric field (or electric displacement, between the capacitor an old term still used sometimes). It is, therefore, called displacement plates. The arrows show uniform electriccurrent or Maxwell’s displacement current. Figure 8.2 shows the electric field between the and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates. The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing 203 Reprint 2025-26 Physics charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ✒0 (d✂E/dt)). So we have dΦE i = i c + i d = i c + ε0 (8.5) d t In explicit terms, this means that outside the capacitor plates, we have only conduction current ic = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., ic = 0, and there is only displacement current, so that id = i. The generalised (and correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. The generalised law is dΦE B idl = µ0 i c + µ0 ε0 (8.6) Ñ∫ dt and is known as Ampere-Maxwell law. In all respects, the displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero since the electric field E does not change with time. In other FIGURE 8.2 (a) The cases, for example, the charging capacitor above, both conduction electric and magnetic and displacement currents may be present in different regions of fields E and B between space. In most of the cases, they both may be present in the same the capacitor plates, at region of space, as there exist no perfectly conducting or perfectly the point M. (b) A cross insulating medium. Most interestingly, there may be large regions sectional view of Fig. (a). of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P). The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field. So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is * They are still not perfectly symmetrical; there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of 204 electric field. Reprint 2025-26 Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Thus, time- dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5). One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section. MAXWELL’S EQUATIONS IN VACUUM 1. “E.dA = Q/✒0 (Gauss’s Law for electricity) 2. “B.dA = 0 (Gauss’s Law for magnetism) –dΦB 3. “E.dl == (Faraday’s Law) d t d ΦE + µ0 ε0 (Ampere – Maxwell Law) 4. “B.dl = µ0 i c d t

1.4 DIMENSIONS OF PHYSICAL QUANTITIES formula of the given physical quantity. For example, the dimensional formula of the volume The nature of a physical quantity is described is [M° L3 T°], and that of speed or velocity is by its dimensions. All the physical quantities [M° L T-1]. Similarly, [M° L T–2] is the dimensional represented by derived units can be expressed formula of acceleration and [M L–3 T°] that of in terms of some combination of seven mass density. fundamental or base quantities. We shall call An equation obtained by equating a physical these base quantities as the seven dimensions quantity with its dimensional formula is called of the physical world, which are denoted with the dimensional equation of the physical square brackets [ ]. Thus, length has the quantity. Thus, the dimensional equations are dimension [L], mass [M], time [T], electric current the equations, which represent the dimensions [A], thermodynamic temperature [K], luminous of a physical quantity in terms of the base intensity [cd], and amount of substance [mol]. quantities. For example, the dimensional The dimensions of a physical quantity are the equations of volume [V], speed [v], force [F ] and powers (or exponents) to which the base mass density [ρ] may be expressed as quantities are raised to represent that [V] = [M0 L3 T0] quantity. Note that using the square brackets [v] = [M0 L T–1] [ ] round a quantity means that we are dealing [F] = [M L T–2] with ‘the dimensions of’ the quantity. [ρ] = [M L–3 T0] In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] The dimensional equation can be obtained and [T]. For example, the volume occupied by from the equation representing the relations between the physical quantities. Thean object is expressed as the product of length, dimensional formulae of a large number andbreadth and height, or three lengths. Hence the wide variety of physical quantities, derived fromdimensions of volume are [L] × [L] × [L] = [L]3 = [L3]. the equations representing the relationships As the volume is independent of mass and time, among other physical quantities and expressed it is said to possess zero dimension in mass [M°], in terms of base quantities are given in zero dimension in time [T°] and three Appendix 9 for your guidance and ready dimensions in length. reference. Similarly, force, as the product of mass and acceleration, can be expressed as 1.6 DIMENSIONAL ANALYSIS AND ITS Force = mass × acceleration APPLICATIONS = mass × (length)/(time)2 The recognition of concepts of dimensions, which The dimensions of force are [M] [L]/[T]2 = guide the description of physical behaviour is [M L T–2]. Thus, the force has one dimension in of basic importance as only those physical Reprint 2025-26 8 PHYSICS quantities can be added or subtracted which such as angle as the ratio (length/length), have the same dimensions. A thorough refractive index as the ratio (speed of light in understanding of dimensional analysis helps us vacuum/speed of light in medium) etc., has no in deducing certain relations among different dimensions. physical quantities and checking the derivation, Now we can test the dimensional consistency accuracy and dimensional consistency or or homogeneity of the equation homogeneity of various mathematical 2expressions. When magnitudes of two or more x = x 0 + v 0 t + (1/2 ) a t physical quantities are multiplied, their units for the distance x travelled by a particle or body should be treated in the same manner as in time t which starts from the position x0 with ordinary algebraic symbols. We can cancel an initial velocity v0 at time t = 0 and has uniformidentical units in the numerator and acceleration a along the direction of motion. denominator. The same is true for dimensions The dimensions of each term may be written as of a physical quantity. Similarly, physical [x] = [L] quantities represented by symbols on both sides of a mathematical equation must have the same [x0 ] = [L] dimensions. [v0 t] = [L T–1] [T] = [L] [(1/2) a t2] = [L T–2] [T2]1.6.1 Checking the Dimensional Consistency of Equations = [L] As each term on the right hand side of this The magnitudes of physical quantities may be equation has the same dimension, namely that added together or subtracted from one another of length, which is same as the dimension of only if they have the same dimensions. In other left hand side of the equation, hence this words, we can add or subtract similar physical equation is a dimensionally correct equation. quantities. Thus, velocity cannot be added to It may be noted that a test of consistency of force, or an electric current cannot be subtracted dimensions tells us no more and no less than a from the thermodynamic temperature. This test of consistency of units, but has the simple principle called the principle of advantage that we need not commit ourselves homogeneity of dimensions in an equation is to a particular choice of units, and we need not extremely useful in checking the correctness of worry about conversions among multiples and an equation. If the dimensions of all the terms sub-multiples of the units. It may be borne in are not same, the equation is wrong. Hence, if mind that if an equation fails this consistency we derive an expression for the length (or test, it is proved wrong, but if it passes, it is distance) of an object, regardless of the symbols not proved right. Thus, a dimensionally correct appearing in the original mathematical relation, equation need not be actually an exact when all the individual dimensions are (correct) equation, but a dimensionally wrong simplified, the remaining dimension must be (incorrect) or inconsistent equation must be that of length. Similarly, if we derive an equation wrong. of speed, the dimensions on both the sides of ⊳equation, when simplified, must be of length/ Example 1.3 Let us consider an equation time, or [L T–1]. Dimensions are customarily used as a 1 2 m v = m g h preliminary test of the consistency of an 2 equation, when there is some doubt about the where m is the mass of the body, v its correctness of the equation. However, the velocity, g is the acceleration due to dimensional consistency does not guarantee gravity and h is the height. Check correct equations. It is uof dimensionless quantities or functions. The correct. arguments of special functions, such as the trigonometric, logarithmic and exponential Answer The dimensions of LHS are functions must be dimensionless. A pure [M] [L T–1 ]2 = [M] [ L2 T–2] number, ratio of similar physical quantities, = [M L2 T–2] Reprint 2025-26 UNITS AND MEASUREMENT 9 The dimensions of RHS are string, that oscillates under the action of [M][L T–2] [L] = [M][L2 T–2] the force of gravity. Suppose that the period = [M L2 T–2] of oscillation of the simple pendulum depends on its length (l), mass of the bobThe dimensions of LHS and RHS are the same and hence the equation is dimensionally correct. ⊳ (m) and acceleration due to gravity (g). Derive the expression for its time period ⊳ using method of dimensions. Example 1.4 The SI unit of energy is J = kg m2 s–2; that of speed v is m s–1 and Answer The dependence of time period T on of acceleration a is m s–2. Which of the the quantities l, g and m as a product may be formulae for kinetic energy (K) given below written as : can you rule out on the basis of T = k lx gy mz dimensional arguments (m stands for the mass of the body) : where k is dimensionless constant and x, y (a) K = m2 v3 and z are the exponents. (b) K = (1/2)mv2 By considering dimensions on both sides, we (c) K = ma have (d) K = (3/16)mv2 o o 1 1 x 1 –2 y 1 z [L M T ]=[L ] [L T ] [M ] (e) K = (1/2)mv2 + ma = Lx+y T–2y Mz Answer Every correct formula or equation must On equating the dimensions on both sides, have the same dimensions on both sides of the we have equation. Also, only quantities with the same x + y = 0; –2y = 1; and z = 0 physical dimensions can be added or 1 1 subtracted. The dimensions of the quantity on So that x = , y = – , z = 0 2 2the right side are [M2 L3 T–3] for (a); [M L2 T–2] for (b) and (d); [M L T–2] for (c). The quantity on the Then, T = k l½ g–½ right side of (e) has no proper dimensions since two quantities of different dimensions have been l or, T = kadded. Since the kinetic energy K has the g dimensions of [M L2 T–2], formulas (a), (c) and (e) Note that value of constant k can not be obtainedare ruled out. Note that dimensional arguments by the method of dimensions. Here it does notcannot tell which of the two, (b) or (d), is the matter if some number multiplies the right sidecorrect formula. For this, one must turn to the of this formula, because that does not affect itsactual definition of kinetic energy (see dimensions.Chapter 5). The correct formula for kinetic energy is given by (b). ⊳ l Actually, k = 2π so that T = 2π ⊳ 1.6.2 Deducing Relation among the g Physical Quantities The method of dimensions can sometimes be Dimensional analysis is very useful in deducing used to deduce relation among the physical relations among the interdependent physical quantities. For this we should know the quantities. However, dimensionless constants dependence of the physical quantity on other cannot be obtained by this method. The method quantities (upto three physical quantities or of dimensions can only test the dimensional linearly independent variables) and consider it validity, but not the exact relationship between as a product type of the dependence. Let us take physical quantities in any equation. It does not an example. distinguish between the physical quantities having same dimensions.⊳ Example 1.5 Consider a simple A number of exercises at the end of this pendulum, having a bob attached to a chapter will help you develop skill in dimensional analysis. Reprint 2025-26 10 PHYSICS SUMMARY 1. Physics is a quantitative science, based on measurement of physical quantities. Certain physical quantities have been chosen as fundamental or base quantities (such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity). 2. Each base quantity is defined in terms of a certain basic, arbitrarily chosen but properly standardised reference standard called unit (such as metre, kilogram, second, ampere, kelvin, mole and candela). The units for the fundamental or base quantities are called fundamental or base units. 3. Other physical quantities, derived from the base quantities, can be expressed as a combination of the base units and are called derived units. A complete set of units, both fundamental and derived, is called a system of units. 4. The International System of Units (SI) based on seven base units is at present internationally accepted unit system and is widely used throughout the world. 5. The SI units are used in all physical measurements, for both the base quantities and the derived quantities obtained from them. Certain derived units are expressed by means of SI units with special names (such as joule, newton, watt, etc). 6. The SI units have well defined and internationally accepted unit symbols (such as m for metre, kg for kilogram, s for second, A for ampere, N for newton etc.). 7. Physical measurements are usually expressed for small and large quantities in scientific notation, with powers of 10. Scientific notation and the prefixes are used to simplify measurement notation and numerical computation, giving indication to the precision of the numbers. 8. Certain general rules and guidelines must be followed for using notations for physical quantities and standard symbols for SI units, some other units and SI prefixes for expressing properly the physical quantities and measurements. 9. In computing any physical quantity, the units for derived quantities involved in the relationship(s) are treated as though they were algebraic quantities till the desired units are obtained. 10. In measured and computed quantities proper significant figures only should be retained. Rules for determining the number of significant figures, carrying out arithmetic operations with them, and ‘rounding off ‘ the u 11. The dimensions of base quantities and combination of these dimensions describe the nature of physical quantities. Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among the physical quantities, etc. A dimensionally consistent equation need not be actually an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong. EXERCISES Note : In stating numerical answers, take care of significant figures. 1.1 Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3. 1.2 Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1. Reprint 2025-26 UNITS AND MEASUREMENT 11 1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ 2 in terms of the new units. 1.4 Explain this statement clearly : “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary : (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light. 1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ? 1.6 Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ? 1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ? 1.8 Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ? 1.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement. 1.10 State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2 1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. 1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? 1.13 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m 0 m = 2 1/2 . 1 − v ( ) Guess where to put the missing c. Reprint 2025-26 12 PHYSICS 1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ? 1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ? 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you). 1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m. Reprint 2025-26 CHAPTER TWO MOTION IN A STRAIGHT LINE 2.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 2.1 Introduction into and out of our lungs and blood flows in arteries and 2.2 Instantaneous velocity and veins. We see leaves falling from trees and water flowing speed down a dam. Automobiles and planes carry people from one 2.3 Acceleration place to the other. The earth rotates once every twenty-four 2.4 Kinematic equations for hours and revolves round the sun once in a year. The sun uniformly accelerated motion itself is in motion in the Milky Way, which is again moving

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1

1.19 A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

1.6 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1, q2, q3, ..., qn in vacuum. What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition. To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force on one charge, say q1, due to two other charges q2, q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by Eq. (1.3) even though other charges are present. FIGURE 1.5 A system of 1 q1q 2 (a) three charges Thus, F12 = 2 ˆr12 (b) multiple charges. 11 4 πε0 r12 Reprint 2025-26 Physics In the same way, the force on q1 due to q3, denoted by F13, is given by 1 q1q 3 F13 = 2 rˆ13 4 πε0 r13 which again is the Coulomb force on q1 due to q3, even though other charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 q1q 2 1 q1q 3 F1 = F12 + F13 = rˆ12 + rˆ13 (1.4) 4 π ε0 r122 4 πε0 r132 The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.5(b). The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n: i.e., 1  q1q 2 q1q 3 q1q n  F1 = F12 + F13 + ...+ F1n =  2 rˆ12 + 2 rˆ13 + ... + 2 1rˆ n  4 πε0  r12 r13 r1n  q1 n q i = ˆr1i 2 (1.5) 4πε0 =∑i 2 r1 i The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle. Example 1.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6? FIGURE 1.6 1.5 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O EXAMPLE 12 from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO. Reprint 2025-26 Electric Charges and Fields Thus, 3 Qq Force F1 on Q due to charge q at A = 2 along AO 4 πε0 l 3 Qq 2 along BOForce F2 on Q due to charge q at B = 4 πε0 l 3 Qq 2 along COForce F3 on Q due to charge q at C = 4 πε0 l 3 Qq 2 along OA, by theThe resultant of forces F2 and F3 is 4 πε0 l 3 Qq parallelogram law. Therefore, the total force on Q = 2 ( rˆ − rˆ ) 4 πε0 l = 0, where ˆr is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. EXAMPLE Consider what would happen if the system was rotated through 60° about O. 1.5 Example 1.6 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge? FIGURE 1.7 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.7. By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F 1ˆr where 1ˆr is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the q 2 same magnitude F = 4 π ε0 l 2 EXAMPLE The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a unit vector along AC. 1.6 13 Reprint 2025-26 Physics Similarly the total force on charge –q at C is F3 = 3 F ˆn , where ˆn is the unit vector along the direction bisecting the ÐBCA. It is interesting to see that the sum of the forces on the three charges 1.6 is zero, i.e., F1 + F2 + F3 = 0 The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof EXAMPLE is left to you as an exercise. 1.7 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where OP = r, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r is given as 1 Q 1 Q rˆ rˆ = E ( r ) = (1.6) 4 πε0 r 2 4 πε0 r 2 where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6) specifies the value of the electric field for each value of the position vector r. The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the force F exerted by a charge Q on a charge q, as 1 Qq F = 2 rˆ (1.7) 4 πε0 r Note that the charge q also exerts an equal and opposite force on the charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus, F(r) = q E(r) (1.8) Equation (1.8) defines the SI unit of electric field as N/C*. Some important remarks may be made here: (i) From Eq. (1.8), we can infer that if q is unity, the electric field due to FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the field (a) due to a electric field due to a charge Q at a point in space may be defined charge Q, (b) due to a as the force that a unit positive charge would experience if placed charge –Q. 14 * An alternate unit V/m will be introduced in the next chapter. Reprint 2025-26 Electric Charges and Fields at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F  E = lim (1.9) q → 0  q  A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.14), the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet. (ii) Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. Thus, the electric field E due to Q is also dependent on the space coordinate r. For different positions of the charge q all over the space, we get different values of electric field E. The field exists at every point in three-dimensional space. (iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector, at each point, points radially inwards. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field E will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry. 1.7.1 Electric field due to a system of charges Consider a system of charges q1, q2, ..., qn with position vectors r1, r2, ..., rn relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by 15position vector r. Reprint 2025-26 Physics Electric field E1 at r due to q1 at r1 is given by 1 q1 E1 = 2 ˆr1P 4 πε0 r1 P where ˆr1P is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P. In the same manner, electric field E2 at r due to q2 at r2 is 1 q 2 E2 = 2 ˆr2P 4 πε0 r2 P where ˆr2P is a unit vector in the direction from q2 to P FIGURE 1.9 Electric field at a point and r2P is the distance between q2 and P. Similar due to a system of charges is the expressions hold good for fields E3, E4, ..., En due to vector sum of the electric fields at charges q3, q4, ..., qn. the point due to individual charges. By the superposition principle, the electric field E at r due to the system of charges is (as shown in Fig. 1.9) E(r) = E1 (r) + E2 (r) + … + En(r) 1 q1 1 q 2 1 q n = 2 rˆ1P + 2 rˆ2 P + ... + 2 rˆn P 4 πε0 r1P 4 πε0 r2 P 4 πε0 rn P 1 n q i E(r) = 2 ˆri P (1.10) 4 πε0 =∑i 1 ri P E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. 1.7.2 Physical significance of electric field You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. (1.5)]. Why then introduce this intermediate quantity called the electric field? For electrostatics, the concept of electric field is convenient, but not really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity. The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time- dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another 16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot Reprint 2025-26 Electric Charges and Fields arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time- dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday and is now among the central concepts in physics. Example 1.7 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’. FIGURE 1.10 Solution In Fig. 1.10(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae = eE/me where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a 2 h 2 h m e distance h is given by t e = = a e e E For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg, E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m, te = 2.9 × 10–9s In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is ap = eE/mp EXAMPLE where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of fall for the proton is 1.7 17 Reprint 2025-26 Physics 2 h 2 h m p –7 t p = = = 1. 3 × 10 s a p e E Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field: e E a p = m p (1 . 6 × 10 − 19 C) × (2. 0 × 10 4 N C −1 ) = −27 1 .67 × 10 kg 12 –2 1.7 = 1 .9 × 10 m s which is enormous compared to the value of g (9.8 m s–2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored EXAMPLE in this example. Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11. FIGURE 1.11 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude (9 × 10 9 Nm 2 C -2 ) × (10 −8 C) E1A = 2 = 3.6 × 104 N C–1 (0.05m) 1.8 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 N C–1 EXAMPLE EA is directed toward the right.18 Reprint 2025-26 Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E1B = 2 = 3.6 × 104 N C–1 (0.05 m) The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E 2B = 2 = 4 × 103 N C–1 (0.15 m) The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 N C–1 EB is directed towards the left. The magnitude of each electric field vector at point C, due to charge q1 and q2 is (9 × 10 9 Nm 2 C –2 ) × (10 − 8 C) E1C = E 2C = 2 = 9 × 103 N C–1 (0.10 m) The directions in which these two vectors point are indicated in Fig. 1.11. The resultant of these two vectors is π π EXAMPLE E C = E1c cos + E 2 c cos = 9 × 103 N C–1 3 3 1.8 EC points towards the right.

1.10 ELECTRIC DIPOLE An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space. By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole. The total charge of the electric dipole is obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and –q are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r >> 2a), the fields due to q and –q nearly cancel out. The electric field due to a dipole therefore falls off, at large distance, faster than like 1/r 2 (the dependence on r of the field due to a single charge q). These qualitative ideas are borne out by the explicit calculation as follows: 1.10.1 The field of an electric dipole The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. The results are simple for the following two cases: (i) when the point is on the dipole axis, and (ii) when it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre. The electric field at any general point P is obtained by adding the electric fields E–q due to the charge –q and E+q due to the charge q, by the parallelogram law of vectors. (i) For points on the axis Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Fig. 1.17(a). Then q E − q = − 2 pˆ [1.13(a)] 4 πε0 (r + a ) where ˆp is the unit vector along the dipole axis (from –q to q). Also q E + q = pˆ [1.13(b)] 23 4 π ε0 (r − a )2 Reprint 2025-26 Physics The total field at P is q  1 1  pˆ − E = E + q + E − q =   (r + a )2 4 π ε0  (r − a )2  q 4 a r = ˆp (1.14) 4 π εo ( r 2 − a 2 )2 For r >> a 4 q a E = 3 pˆ (r >> a) (1.15) 4 πε0 r (ii) For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by q 1 E + q = 2 2 [1.16(a)] 4 πε0 r + a q 1 E – q = 2 2 [1.16(b)] 4 πε0 r + a FIGURE 1.17 Electric field of a dipole and are equal. at (a) a point on the axis, (b) a point The directions of E+q and E–q are as shown in on the equatorial plane of the dipole. Fig. 1.17(b). Clearly, the components normal to the dipole p is the dipole moment vector of axis cancel away. The components along the dipole axis magnitude p = q × 2a and add up. The total electric field is opposite to ˆp. We have directed from –q to q. E = – (E +q + E –q ) cosq ˆp 2 q a = − pˆ (1.17) 4 π εo (r 2 + a 2 )3 / 2 At large distances (r >> a), this reduces to 2 q a E = − pˆ (r >> a ) (1.18) 4 π εo r 3 From Eqs. (1.15) and (1.18), it is clear that the dipole field at large distances does not involve q and a separately; it depends on the product qa. This suggests the definition of dipole moment. The dipole moment vector p of an electric dipole is defined by p = q × 2a ˆp (1.19) that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q. In terms of p, the electric field of a dipole at large distances takes simple forms: At a point on the dipole axis 2 p E = 3 (r >> a) (1.20) 4 πεor At a point on the equatorial plane p 3 (r >> a) (1.21) 24 E = −4 πεor Reprint 2025-26 Electric Charges and Fields Notice the important point that the dipole field at large distances falls off not as 1/r 2 but as1/r 3. Further, the magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and the dipole moment p. We can think of the limit when the dipole size 2a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a point dipole, Eqs. (1.20) and (1.21) are exact, true for any r. 1.10.2 Physical significance of dipoles In most molecules, the centres of positive charges and of negative charges* lie at the same place. Therefore, their dipole moment is zero. CO2 and CH4 are of this type of molecules. However, they develop a dipole moment when an electric field is applied. But in some molecules, the centres of negative charges and of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H2O, is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field. Example 1.9 Two charges ±10 mC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. 1.18(a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.18(b). EXAMPLE FIGURE 1.18 1.9 * Centre of a collection of positive point charges is defined much the same way ∑ q i ri as the centre of mass: rcm = i . ∑ q i 25 i Reprint 2025-26 Physics Solution (a) Field at P due to charge +10 mC 10 −5 C 1 = − 12 2 −1 −2 × 2 −4 2 4 π (8.854 × 10 C N m ) (15 − 0.25) × 10 m = 4.13 × 106 N C–1 along BP Field at P due to charge –10 mC 10 –5 C 1 = −12 2 −1 −2 × 2 − 4 2 4 π (8.854 × 10 C N m ) (15 + 0.25) × 10 m = 3.86 × 106 N C–1 along PA The resultant electric field at P due to the two charges at A and B is = 2.7 × 105 N C–1 along BP. In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ± q, 2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude 2 p E = 3 (r/a >> 1) 4 πε0 r where p = 2a q is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from –q to q). Here, p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m Therefore, 2 × 5 × 10 − 8 C m 1 E = −12 2 −1 −2 × 3 −6 3 = 2.6 × 105 N C–1 4 π (8.854 × 10 C N m ) (15) × 10 m along the dipole moment direction AB, which is close to the result obtained earlier. (b) Field at Q due to charge + 10 mC at B 10 −5 C 1 = −12 2 −1 − 2 × 2 2 −4 2 4 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 m = 3.99 × 106 N C–1 along BQ Field at Q due to charge –10 mC at A 10 −5 C 1 = − 12 2 −1 − 2 × 2 2 −4 2 4 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 m = 3.99 × 106 N C–1 along QA. Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is 0.25 6 –1 = 2 × 2 2 × 3.99 × 10 N C along BA 1.9 15 + (0.25) = 1.33 × 105 N C–1 along BA. As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to EXAMPLE 26 the axis of the dipole: Reprint 2025-26 Electric Charges and Fields p E = 3 (r/a >> 1) 4 π 0ε r 5 × 10 −8 Cm 1 = −12 2 –1 –2 × 3 −6 3 4 π (8.854 × 10 C N m ) (15) × 10 m = 1.33 × 105 N C–1. The direction of electric field in this case is opposite to the direction EXAMPLE of the dipole moment vector. Again, the result agrees with that obtained before. 1.9

1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

1.11 DIPOLE IN A UNIFORM EXTERNAL FIELD Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in Fig. 1.19. (By permanent dipole, we mean that p exists irrespective of E; it has not been induced by E.) There is a force qE on q and a force –qE on –q. The net force on the dipole is zero, since E is uniform. However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole. When the net force is zero, the torque (couple) is independent of the origin. Its magnitude equals the magnitude of FIGURE 1.19 Dipole in a each force multiplied by the arm of the couple (perpendicular uniform electric field. distance between the two antiparallel forces). Magnitude of torque = q E × 2 a sinq = 2 q a E sinq Its direction is normal to the plane of the paper, coming out of it. The magnitude of p × E is also p E sinq and its direction is normal to the paper, coming out of it. Thus, t = p × E (1.22) This torque will tend to align the dipole with the field E. When p is aligned with E, the torque is zero. What happens if the field is not uniform? In that case, the net force will evidently be non-zero. In addition there will, in general, be a torque on the system as before. The general case is involved, so let us consider the simpler situations when p is parallel to E or antiparallel to E. In either case, the net torque is zero, but there is a net force on the dipole if E is not uniform. Figure 1.20 is self-explanatory. It is easily seen that when p is parallel to E, the dipole has a net force in the direction of increasing field. When p is antiparallel to E, the net force on the dipole is in the direction of decreasing field. In general, the force depends on the orientation of p with respect to E. This brings us to a common observation in frictional electricity. A comb run through dry hair attracts pieces of FIGURE 1.20 Electric force on a paper. The comb, as we know, acquires charge through dipole: (a) E parallel to p, (b) E friction. But the paper is not charged. What then explains antiparallel to p. the attractive force? Taking the clue from the preceding 27 Reprint 2025-26 Physics discussion, the charged comb ‘polarises’ the piece of paper, i.e., induces a net dipole moment in the direction of field. Further, the electric field due to the comb is not uniform. This non-uniformity of the field makes a dipole to experience a net force on it. In this situation, it is easily seen that the paper should move in the direction of the comb! 1.12 CONTINUOUS CHARGE DISTRIBUTION We have so far dealt with charge configurations involving discrete charges q1, q2, ..., qn. One reason why we restricted to discrete charges is that the mathematical treatment is simpler and does not involve calculus. For many purposes, however, it is impractical to work in terms of discrete charges and we need to work with continuous charge distributions. For example, on the surface of a charged conductor, it is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents. It is more feasible to consider an area element DS (Fig. 1.21) on the surface of the conductor (which is very small on the macroscopic scale but big enough to include a very large number of electrons) and specify the charge DQ on that element. We then define a surface charge density s at the area element by ∆Q σ = (1.23) ∆S We can do this at different points on the conductor and thus arrive at a continuous function s, called the surface charge density. The surface charge density s so defined ignores the quantisation of charge and the discontinuity in charge distribution at the microscopic level*. s represents macroscopic surface charge density, which in a sense, is a smoothed out average of the microscopic charge density over an area element DS which, as said before, is large microscopically but small macroscopically. The units for s are C/m2. FIGURE 1.21 Similar considerations apply for a line charge distribution and a volume Definition of linear, charge distribution. The linear charge density l of a wire is defined by surface and volume ∆ Q charge densities. λ = (1.24) In each case, the ∆ l element (Dl, DS, DV) where Dl is a small line element of wire on the macroscopic scale that, chosen is small on however, includes a large number of microscopic charged constituents, the macroscopic and DQ is the charge contained in that line element. The units for l are scale but contains C/m. The volume charge density (sometimes simply called charge density) a very large number is defined in a similar manner: of microscopic constituents. ∆ Q ρ= (1.25) ∆ V where DQ is the charge included in the macroscopically small volume element DV that includes a large number of microscopic charged constituents. The units for r are C/m3. The notion of continuous charge distribution is similar to that we adopt for continuous mass distribution in mechanics. When we refer to 28 * At the microscopic level, charge distribution is discontinuous, because they are discrete charges separated by intervening space where there is no charge. Reprint 2025-26 Electric Charges and Fields the density of a liquid, we are referring to its macroscopic density. We regard it as a continuous fluid and ignore its discrete molecular constitution. The field due to a continuous charge distribution can be obtained in much the same way as for a system of discrete charges, Eq. (1.10). Suppose a continuous charge distribution in space has a charge density r. Choose any convenient origin O and let the position vector of any point in the charge distribution be r. The charge density r may vary from point to point, i.e., it is a function of r. Divide the charge distribution into small volume elements of size DV. The charge in a volume element DV is rDV. Now, consider any general point P (inside or outside the distribution) with position vector R (Fig. 1.21). Electric field due to the charge rDV is given by Coulomb’s law: 1 ρ ∆ V ∆ E = 2 rˆ' (1.26) 4 πε0 r' where r¢ is the distance between the charge element and P, and ˆr¢ is a unit vector in the direction from the charge element to P. By the superposition principle, the total electric field due to the charge distribution is obtained by summing over electric fields due to different volume elements: 1 ρ ∆V rˆ ' E ≅ Σ (1.27) all ∆V r' 2 4 πε0 Note that r, r¢, ˆ′r all can vary from point to point. In a strict mathematical method, we should let DV®0 and the sum then becomes an integral; but we omit that discussion here, for simplicity. In short, using Coulomb’s law and the superposition principle, electric field can be determined for any charge distribution, discrete or continuous or part discrete and part continuous.

1.17 A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) FIGURE 1.31

7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = R E ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radiusof a large number of concentric spherical shells and ρ is the density. On the other hand thewith the smallest one at the centre and the largest one at its surface. A point outside the 4π 3 ρr and mass of the sphere Mr of radius r isearth is obviously outside all the shells. Thus, 3 Reprint 2025-26 GRAVITATION 133 hence its distance from the centre of the earth is (RE + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m M E Eq. (7.5) : = 3 r (7.10) R E If the mass m is situated on the surface of GM E m F (h ) = earth, then r = RE and the gravitational force on ( R E + h )2 (7.13) it is, from Eq. (7.10) The acceleration experienced by the point M E m F = G 2 (7.11) mass is F (h )/ m ≡ g (h ) and we get R E The acceleration experienced by the mass F (h ) GM E . g (h ) = = (7.14)m, which is usually denoted by the symbol g is m ( R E + h )2 related to F by Newton’s 2nd law by relation This is clearly less than the value of g on the F = mg. Thus GM E . g = surface of earth : GM F For h << R E , we can E R E2 g = = 2 (7.12) m R E expand the RHS of Eq. (7.14) : E Acceleration g is readily measurable. RE is a g (h ) = 2 GM 2 = g (1 + h / R E )−2known quantity. The measurement of G by R E (1 + h / R E ) Cavendish’s experiment (or otherwise), combined h << 1 , using binomial expression,with knowledge of g and RE enables one to For R E estimate ME from Eq. (7.12). This is the reason  2h why there is a popular statement regarding g (h ) ≅ g 1 − . (7.15)Cavendish : “Cavendish weighed the earth”.  RE  7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a factor (1 − 2h / R E ).Consider a point mass m at a height h above the Now, consider a point mass m at a depthsurface of the earth as shown in Fig. 7.8(a). The d below the surface of the earth (Fig. 7.8(b)),radius of the earth is denoted by RE . Since this so that its distance from the centre of thepoint is outside the earth, earth is ( R E − d ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If Ms is the mass of the smaller sphere, then, Ms/ME = ( RE – d)3 / RE3 ( 7.16) Since mass of a sphere is proportional to be Fig. 7.8 (a) g at a height h above the surface of the cube of its radius. earth. Reprint 2025-26 134 PHYSICS close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a Ms ME point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W12 Fig. 7.8 (b) g at a depth d. In this case only the smaller W12 = Force × displacement = mg (h2 – h1) (7.20) sphere of radius (RE–d) contributes to g. Thus the force on the point mass is If we associate a potential energy W(h) at a point at a height h above the surface such that F (d) = G Ms m / (RE – d ) 2 (7.17) W(h) = mgh + Wo (7.21) Substituting for Ms from above , we get (where Wo = constant) ; F (d) = G ME m ( RE – d ) / RE 3 (7.18) then it is clear that and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22) a depth d, The work done in moving the particle is just the difference of potential energy between its F ( d ) final and initial positions.Observe that the g(d) = is m constant Wo cancels out in Eq. (7.22). Setting h = 0 in the last equation, we get W ( h = 0 ) = Wo. F (d ) GM E g ( d ) = = 3 ( R E − d ) . h = 0 means points on the surface of the earth. m R E Thus, Wo is the potential energy on the surface of the earth. R E − d = g = g (1 − d / R E ) (7.19) If we consider points at arbitrary distance R E from the surface of the earth, the result just Thus, as we go down below earth’s surface, derived is not valid since the assumption that the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no (1 − d / R E ). The remarkable thing about longer valid. However, from our discussion we know that a point outside the earth, the force of acceleration due to earth’s gravity is that it is gravitation on a particle directed towards the maximum on its surface decreasing whether you centre of the earth is go up or down. G ME m F = 2 (7.23)7.7 GRAVITATIONAL POTENTIAL ENERGY r where ME = mass of earth, m = mass of theWe had discussed earlier the notion of potential particle and r its distance from the centre of theenergy as being the energy stored in the body at earth. If we now calculate the work done inits given position. If the position of the particle lifting a particle from r = r1 to r = r2 (r2 > r1) alongchanges on account of forces acting on it, then a vertical path, we get instead of Eq. (7.20) the change in its potential energy is just the amount of work done on the body by the force. r2 G M m W12 2 d rAs we had discussed earlier, forces for which the =∫r1 r work done is independent of the path are the conservative forces.  1 1  = − G M E m − (7.24) The force of gravity is a conservative force  r2 r1  and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate body arising out of this force, called the a potential energy W(r) at a distance r, such that gravitational potential energy. Consider points Reprint 2025-26 GRAVITATION 135 G M E m W (r ) =− + W1 , (7.25) r valid for r > R , so that once again W12 = W(r2) – W(r1). Setting r = infinity in the last equation, we get W ( r = infinity ) = W1 . Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets W1 equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces Fig. 7.9 on it due to the earth and it is proportional to the mass of the particle. The gravitational The gravitational potential at the centre of potential due to the gravitational force of the the square r = 2 l/2 is ( )earth is defined as the potential energy of a particle of unit mass at that point. From the G m U (r ) = − 4 2 . ⊳earlier discussion, we learn that the gravitational l potential energy associated with two particles of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED distance r is given by If a stone is thrown by hand, we see it falls back Gm 1m 2 to the earth. Of course using machines we can V = – (if we choose V = 0 as r →∞) r shoot an object with much greater speeds and It should be noted that an isolated system of with greater and greater initial speed, the object particles will have the total potential energy that scales higher and higher heights. A natural equals the sum of energies (given by the above query that arises in our mind is the following: equation) for all possible pairs of its constituent ‘can we throw an object with such high initial particles. This is an example of the application speeds that it does not fall back to the earth?’ of the superposition principle. The principle of conservation of energy helps us to answer this question. Suppose the object ⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was a system of four particles placed at the Vf. The energy of an object is the sum of potential vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that the potential at the centre of the square. gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is Answer Consider four masses each of mass m at the corners of a square of side l; See Fig. 7.9. 2 mV fWe have four mass pairs at distance l and two E ( ∞=) W1 + (7.26) 2 diagonal pairs at distance 2 l If the object was thrown initially with a speed Hence, Vi from a point at a distance (h+RE) from the G m 2 G m 2 centre of the earth (RE = radius of the earth), its W (r ) = − 4 − 2 l 2 l energy initially was 2 2 1 2 GmM E 2 G m  1  G m E (h + R E ) = mVi – + W1 (7.27) 5.41 = − 2 (h + R E ) 2 +  = − l  l  2  Reprint 2025-26 136 PHYSICS By the principle of energy conservation ⊳ Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres 2 2 of equal radii R, but mass M and 4 M have mV f mVi GmM E a centre to centre separation 6 R, as shown – = (7.28) 2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed. The R.H.S. is a positive quantity with a A projectile of mass m is projected from the minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly Thus, an object can reach infinity as long as Vi towards the centre of the second sphere. is such that Obtain an expression for the minimum speed v of the projectile so that it reaches mVi 2 GmM E the surface of the second sphere. – ≥ 0 (7.29) 2 (h + R E ) The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) Fig. 7.10corresponds to 1 2 GmM E Answer The projectile is acted upon by two m = (7.30) mutually opposing gravitational forces of the two ( Vi )min 2 h + R E spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces If the object is thrown from the surface of cancel each other exactly. If ON = r, we have the earth, h = 0, and we get G M m 4 G M m = 2GM E r 2 (6 R −r )2 (Vi )min = (7.31) (6R – r)2 = 4r2 R E 6R – r = ±2r r = 2R or – 6R. 2 The neutral point r = – 6R does not concern Using the relation g = GM E / R E , we get us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed (Vi )min = 2 gR E (7.32) which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface Using the value of g and RE, numerically of M is (Vi)min≈11.2 km/s. This is called the escape 1 2 G M m 4 G M mspeed, sometimes loosely called the escape E i = m v − − . velocity. 2 R 5 R Equation (7.32) applies equally well to an At the neutral point N, the speed approaches object thrown from the surface of the moon with zero. The mechanical energy at N is purely g replaced by the acceleration due to Moon’s potential. gravity on its surface and rE replaced by the G M m 4 G M m − EN = − .radius of the moon. Both are smaller than their 2 R 4 R values on earth and the escape speed for the From the principle of conservation of moon turns out to be 2.3 km/s, about five times mechanical energy smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the 1 2 GM 4GM GM GMsurface of the moon having velocities larger than v − − = − − this will escape the gravitational pull of the 2 R 5 R 2R R moon. or Reprint 2025-26 GRAVITATION 137 + h) with speed V. Its 2 2 G M  4 1  traverses a distance 2π(RE v = − time period T therefore is R  5 2  2π( R E + h ) 2π( R E + h )3 / 2 T = = (7.37)  3 G M 1/2 V ⊳ G M E v =  5 R  on substitution of value of V from Eq. (7.35). A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the 2 T = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)heavier sphere 4 M. The calculation of this speed is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (7.38). Earth satellites are objects which revolve around Hence, for such satellites, T is To, where the earth. Their motion is very similar to the motion of planets around the Sun and hence T 0 = 2π R E / g (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near 6.4 × 10 6 T 0 = 2π s circular orbit with a time period of approximately 9.8

7.4 AC VOLTAGE APPLIED TO AN INDUCTOR Figure 7.5 shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v = vm sinw t. Using the Kirchhoff’s loop rule, ε ()t = 0 , and since there ∑ is no resistor in the circuit, d i v − L = 0 (7.10) d t where the second term is the self-induced Faraday FIGURE 7.5 An ac source emf in the inductor; and L is the self-inductance of connected to an inductor. * Though voltage and current in ac circuit are represented by phasors – rotating vectors, they are not vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know. 181 Reprint 2025-26 Physics the inductor. The negative sign follows from Lenz’s law (Chapter 6). Combining Eqs. (7.1) and (7.10), we have d i v v m = = sin ωt (7.11) d t L L Equation (7.11) implies that the equation for i(t), the current as a function of time, must be such that its slope di/dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by vm/L. To obtain the current, we integrate di/dt with respect to circuits: time: di v m series d t = sin(ωt )d t ∫ d t L ∫ RLC and get, and v m i = − cos( ω t ) + constant C ω L L, The integration constant has the dimension of current and is time- R, independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. containing, Using  π  circuits − cos(ωt ) = sin ω t − , we have  2  ac of  π  i = i m sin  ωt − 2  diagrams v m (7.12) where i m = is the amplitude of the current. The quantity w L is ω L analogous to the resistance and is called inductive reactance, denoted Phasor by XL: on XL = w L (7.13) The amplitude of the current is, then v m animation i m = (7.14) X L The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (W). The inductive reactance limits the current in a Interactive http://www.animations.physics.unsw.edu.au//jw/AC.html purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A comparison of Eqs. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by p/2 or one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t1. The current phasor I is p/2 behind the voltage phasor V. When rotated with frequency w counter- clockwise, they generate the voltage and current given by Eqs. (7.1) and 182 (7.12), respectively and as shown in Fig. 7.6(b). Reprint 2025-26 Alternating Current FIGURE 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5. (b) Graph of v and i versus wt. We see that the current reaches its maximum value later than the  T π/ 2 voltage by one-fourth of a period = . You have seen that an  4 ω  inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is  π  p L = i v = i m sin ω t − ×v m sin (ωt )  2  = −i m vm cos (ωt ) sin (ωt ) i m v m = − sin ( 2ωt ) 2 So, the average power over a complete cycle is i m v m PL = − sin ( 2ωt ) 2 i m v m = − sin ( 2ωt ) = 0, 2 since the average of sin (2wt) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Example 7.2 A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. Solution The inductive reactance, X L = 2 πνL = 2 × 3 .14 × 50 × 25 × 10 –3 Ω = 7.85W The rms current in the circuit is EXAMPLE V 220 V I = = = 28 A 7.2 7.85 Ω X L 183 Reprint 2025-26 Physics 7.5 AC VOLTAGE APPLIED TO A CAPACITOR Figure 7.7 shows an ac source e generating ac voltage v = vm sin wt connected to a capacitor only, a purely capacitive ac circuit. When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. That is, a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, as in Fig. 7.7, it limits or regulates the current, but FIGURE 7.7 An ac source does not completely prevent the flow of charge. The connected to a capacitor. capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is q v = (7.15) C From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal, q v m sin ωt = C d q To find the current, we use the relation i = dt d i = (vm C sin ωt ) = ωC v m cos(ωt ) d t  π  Using the relation, cos(ωt ) = sin  ω t + 2  , we have  π  (7.16) i = i m sin  ωt + 2  where the amplitude of the oscillating current is im = wCvm. We can rewrite it as vm i m = (1/ωC ) Comparing it to im= vm/R for a purely resistive circuit, we find that (1/wC) plays the role of resistance. It is called capacitive reactance and is denoted by Xc, Xc= 1/wC (7.17) so that the amplitude of the current is vm i m = (7.18)184 X C Reprint 2025-26 Alternating Current The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. A comparison of Eq. (7.16) with the FIGURE 7.8 (a) A Phasor diagram for the circuit equation of source voltage, Eq. (7.1) shows that in Fig. 7.7. (b) Graph of v and i versus ωt. the current is π/2 ahead of voltage. Figure 7.8(a) shows the phasor diagram at an instant t1. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.8(b) shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is pc = i v = im cos(ωt)vm sin(ωt) = imvm cos(ωt) sin(ωt) i m vm = sin(2ωt ) (7.19) 2 So, as in the case of an inductor, the average power i m v m i m v m PC = sin(2ωt ) = sin(2ωt ) = 0 2 2 since <sin (2ωt)> = 0 over a complete cycle. Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2. Example 7.3 A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Solution When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. EXAMPLE Reducing C will increase reactance and the lamp will shine less brightly than before. 7.3 Example 7.4 A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Solution The capacitive reactance is 1 1 = = 212 Ω X C = EXAMPLE 2 π (50Hz)(15.0 × 10 −6 F ) 2 π νC The rms current is 7.4 185 Reprint 2025-26 Physics V 220 V I = = = 1.04 A X C 212 Ω The peak current is m = 2 I = (1.41)(1.04 A ) = 1.47 A 7.4 i This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by p/2. If the frequency is doubled, the capacitive reactance is halved and EXAMPLE consequently, the current is doubled. Example 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9. FIGURE 7.9 The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. 7.5 SolutionmagnetizesAsthetheironiron increasingrod is inserted,the magneticthe magneticfieldfieldinsideinsideit. theHence,coil the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage EXAMPLE across the bulb. Therefore, the glow of the light bulb decreases. 7.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Figure 7.10 shows a series LCR circuit connected to an ac source e. As usual, we take the voltage of the source to be v = vm sin wt. If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule: d i q L + i R + = v (7.20) d t C We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this problem by two methods. First, we use the technique of FIGURE 7.10 A series LCR circuit phasors and in the second method, we solve connected to an ac source. Eq. (7.20) analytically to obtain the time– 186 dependence of i. Reprint 2025-26 Alternating Current 7.6.1 Phasor-diagram solution From the circuit shown in Fig. 7.10, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be i = im sin(wt+f) (7.21) where fis the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous sections, we shall construct a phasor diagram for the present case. Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that VR is parallel to I, VC is p/2 behind I and VL is p/2 ahead of I. VL, VR, VC and I are shown in Fig. 7.11(a) with apppropriate phase- relations. The length of these phasors or the amplitude of VR, VC and VL are: vRm = im R, vCm = im XC, vLm = im XL (7.22) The voltage Equation (7.20) for the circuit can be written as vL + vR + vC = v (7.23) The phasor relation whose vertical component gives the above equation is FIGURE 7.11 (a) Relation between the phasors VL, VR, VC, and I, (b) Relation VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC) This relation is represented in Fig. 7.11(b). Since for the circuit in Fig. 7.10. VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude ½vCm – vLm½. Since V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC + VL), the pythagorean theorem gives: 2 2 2 vm = v Rm + (v Cm − v Lm ) Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above equation, we have v m2 = (i m R )2 + (i m X C − i m X L )2 = i m2  R 2 + ( X C − X L )2  v m or, i m = 2 2 [7.25(a)] R + ( X C − X L ) By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit: v m i m = [7.25(b)] Z where Z = R 2 + ( X C − X L )2 (7.26) 187 Reprint 2025-26 Physics Since phasor I is always parallel to phasor VR, the phase angle f is the angle between VR and V and can be determined from Fig. 7.12: vCm − v Lm tan φ = v Rm Using Eq. (7.22), we have X C − X L tan φ = (7.27) R Equations (7.26) and (7.27) are graphically shown in Fig. (7.12). FIGURE 7.12 Impedance This is called Impedance diagram which is a right-triangle with diagram. Z as its hypotenuse. Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified. If XC > XL, f is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC < XL, f is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage. Figure 7.13 shows the phasor diagram and variation of v and i with wt for the case XC > XL. Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t1, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution. This is not a general FIGURE 7.13 (a) Phasor diagram of V and I. solution. Additionally, we do have a (b) Graphs of v and i versus w t for a series LCR transient solution which exists even for circuit where XC > XL. v = 0. The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. 7.6.2 Resonance An interesting characteristic of the series RLC circuit is the phenomenon of resonance. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the system’s natural frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large. A familiar example of this phenomenon is a child on a swing. The swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the Reprint 2025-26 Alternating Current rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging, the amplitude of the swinging will be large. For an RLC circuit driven with voltage of amplitude vm and frequency w, we found that the current amplitude is given by v m v m i m = = 2 2 Z R + ( X C − X L ) with Xc = 1/wC and XL = wL. So if w is varied, then at a particular frequency Z = R 2 + 0 2 = R . Thisw0, Xc = XL, and the impedance is minimum ( ) frequency is called the resonant frequency: 1 X c = X L or = ω0 L ω0 C 1 or ω0 = (7.28) LC At resonant frequency, the current amplitude is maximum; im = vm/R. Figure 7.16 shows the variation of im with w in a RLC series circuit with L = 1.00 mH, C =

7.3 REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS In the previous section, we learnt that the current through a resistor is in phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a combination of these circuit elements. In order to show phase relationship between voltage and current in an ac circuit, we use the notion of phasors. The analysis of an ac circuit is facilitated by the use of a phasor diagram. A phasor* is a vector which rotates about the origin with angular speed w, as shown in Fig. 7.4. The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The magnitudes of phasors V and I represent the amplitudes or the peak values vm and im of these oscillating quantities. Figure 7.4(a) shows the FIGURE 7.4 (a) A phasor diagram for the voltage and current phasors and their circuit in Fig 7.1. (b) Graph of v and i versus wt.relationship at time t1 for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1. The projection of voltage and current phasors on vertical axis, i.e., vm sinwt and im sinwt, respectively represent the value of voltage and current at that instant. As they rotate with frequency w, curves in Fig. 7.4(b) are generated. From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero.

7.4 The gravitational (1564-1642) who recognised the fact that all bodies, constant irrespective of their masses, are accelerated towards the earth7.5 Acceleration due to gravity of the earth with a constant acceleration. It is said that he made a public demonstration of this fact. To find the truth, he certainly7.6 Acceleration due to gravity below and above did experiments with bodies rolling down inclined planes and the surface of earth arrived at a value of the acceleration due to gravity which is 7.7 Gravitational potential close to the more accurate value obtained later. energy A seemingly unrelated phenomenon, observation of stars, 7.8 Escape speed planets and their motion has been the subject of attention 7.9 Earth satellites in many countries since the earliest of times. Observations 7.10 Energy of an orbiting since early times recognised stars which appeared in the satellite sky with positions unchanged year after year. The more Summary interesting objects are the planets which seem to have regular Points to ponder motions against the background of stars. The earliest Exercises recorded model for planetary motions proposed by Ptolemy about 2000 years ago was a ‘geocentric’ model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. The only motion that was thought to be possible for celestial objects was motion in a circle. Complicated schemes of motion were put forward by Ptolemy in order to describe the observed motion of the planets. The planets were described as moving in circles with the centre of the circles themselves moving in larger circles. Similar theories were also advanced by Indian astronomers some 400 years later. However a more elegant model in which the Sun was the centre around which the planets revolved – the ‘heliocentric’ model – was already mentioned by Aryabhatta (5th century A.D.) in his treatise. A thousand years later, a Polish monk named Nicolas Copernicus (1473-1543) Reprint 2025-26 128 PHYSICS proposed a definitive model in which the planets of the ellipse (Fig. 7.1a). This law was a moved in circles around a fixed central sun. His deviation from the Copernican model which theory was discredited by the church, but allowed only circular orbits. The ellipse, of notable amongst its supporters was Galileo who which the circle is a special case, is a closed had to face prosecution from the state for his curve which can be drawn very simply as beliefs. follows. It was around the same time as Galileo, a Select two points F1 and F2. Take a length nobleman called Tycho Brahe (1546-1601) of a string and fix its ends at F1 and F2 by hailing from Denmark, spent his entire lifetime pins. With the tip of a pencil stretch the string recording observations of the planets with the taut and then draw a curve by moving the naked eye. His compiled data were analysed pencil keeping the string taut throughout.(Fig. later by his assistant Johannes Kepler (1571- 7.1(b)) The closed curve you get is called an 1640). He could extract from the data three ellipse. Clearly for any point T on the ellipse, elegant laws that now go by the name of Kepler’s the sum of the distances from F1 and F2 is alaws. These laws were known to Newton and constant. F1, F2 are called the focii. Join theenabled him to make a great scientific leap in points F1 and F2 and extend the line toproposing his universal law of gravitation. intersect the ellipse at points P and A as shown 7.2 KEPLER’S LAWS in Fig. 7.1(b). The midpoint of the line PA is The three laws of Kepler can be stated as the centre of the ellipse O and the length PO = follows: AO is called the semi-major axis of the ellipse. 1. Law of orbits : All planets move in elliptical For a circle, the two focii merge into one and orbits with the Sun situated at one of the foci the semi-major axis becomes the radius of the circle. B 2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 7.2). This law comes from the observations that planets appear to move 2b P S S' A slower when they are farther from the sun than when they are nearer. C 2a Fig. 7.1(a) An ellipse traced out by a planet around the sun. The closest point is P and the farthest point is A, P is called the perihelion and A the aphelion. The semimajor axis is half the distance AP. Fig. 7.2 The planet P moves around the sun in an elliptical orbit. The shaded area is the area Fig. 7.1(b) Drawing an ellipse. A string has its ends ∆A swept out in a small interval of time ∆t. fixed at F1 and F2. The tip of a pencil holds the string taut and is moved around. Reprint 2025-26 GRAVITATION 129 3. Law of periods : The square of the time period the law of areas. Gravitation is a central force of revolution of a planet is proportional to the and hence the law of areas follows. cube of the semi-major axis of the ellipse traced ⊳ Example 7.1 Let the speed of the planetout by the planet. at the perihelion P in Fig. 7.1(a) be vP and Table 7.1 gives the approximate time periods the Sun-planet distance SP be rP. Relate of revolution of eight* planets around the sun {rP, vP} to the corresponding quantities at along with values of their semi-major axes. the aphelion {rA, vA}. Will the planet take Table 7.1 Data from measurement of equal times to traverse BAC and CPB ? planetary motions given below confirm Kepler’s Law of Periods Answer The magnitude of the angular (a ≡ Semi-major axis in units of 1010 m. momentum at P is Lp = mp rp vp, since inspection T ≡ Time period of revolution of the planet tells us that rp and v p are mutually in years(y). perpendicular. Similarly, LA = mp rA vA. From Q ≡ The quotient ( T2/a3 ) in units of -34 angular momentum conservation 10 y2 m-3.) mp rp vp = mp rA vA Planet a T Q v p r A = Mercury 5.79 0.24 2.95 or v A r p ⊳ Venus 10.8 0.615 3.00 Since rA > rp, vp > vA . Earth 15.0 1 2.96 Mars 22.8 1.88 2.98 The area SBAC bounded by the ellipse and Jupiter 77.8 11.9 3.01 the radius vectors SB and SC is larger than SBPC Saturn 143 29.5 2.98 in Fig. 7.1. From Kepler’s second law, equal areas Uranus 287 84 2.98 are swept in equal times. Hence the planet will Neptune 450 165 2.99 take a longer time to traverse BAC than CPB. 7.3 UNIVERSAL LAW OF GRAVITATION Legend has it that observing an apple falling from The law of areas can be understood as a a tree, Newton was inspired to arrive at anconsequence of conservation of angular universal law of gravitation that led to anmomentum whch is valid for any central explanation of terrestrial gravitation as well asforce . A central force is such that the force of Kepler’s laws. Newton’s reasoning was thaton the planet is along the vector joining the the moon revolving in an orbit of radius Rm wasSun and the planet. Let the Sun be at the subject to a centripetal acceleration due toorigin and let the position and momentum earth’s gravity of magnitudeof the planet be denoted by r and p respectively. Then the area swept out by the 2 2 V 4π R m =planet of mass m in time interval ∆t is (Fig. a m = T 2 (7.3) R m 7.2) ∆A given by ∆A = ½ (r × v∆t) (7.1) where V is the speed of the moon related to the Hence m / T . The ∆A /∆t =½ (r × p)/m, (since v = p/m) time period T by the relation V = 2πR time period T is about 27.3 days and Rm was = L / (2 m) (7.2) already known then to be about 3.84 × 108m. If where v is the velocity, L is the angular momentum equal to ( r × p). For a central we substitute these numbers in Eq. (7.3), we get a value of am much smaller than the value offorce, which is directed along r, L is a constant acceleration due to gravity g on the surface ofas the planet goes around. Hence, ∆A /∆t is a the earth, arising also due to earth’s gravitational constant according to the last equation. This is attraction. Reprint 2025-26 130 PHYSICS This clearly shows that the force due to The gravitational force is attractive, i.e., the earth’s gravity decreases with distance. If one force F is along – r. The force on point mass m1 assumes that the gravitational force due to the due to m2 is of course – F by Newton’s third law. earth decreases in proportion to the inverse Thus, the gravitational force F12 on the body 1square of the distance from the centre of the due to 2 and F21 on the body 2 due to 1 are related earth, we will have am α R−;m2 g α R−E 2 and we get as F12 = – F21. 2 Before we can apply Eq. (7.5) to objects under g R m = 2 3600 (7.4) consideration, we have to be careful since the a m R E law refers to point masses whereas we deal with in agreement with a value of g 9.8 m s-2 and extended objects which have finite size. If we have the value of am from Eq. (7.3). These observations a collection of point masses, the force on any led Newton to propose the following Universal Law one of them is the vector sum of the gravitational of Gravitation : forces exerted by the other point masses as Every body in the universe attracts every other shown in Fig 7.4. body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short). Stated Mathematically, Newton’s gravitation law reads : The force F on a point mass m2 due to another point mass m1 has the magnitude m 1 m 2 | F | = G 2 (7.5) r Equation (7.5) can be expressed in vector form as m 1 m 2 ɵ m 1 m 2 ɵ – r r F = G = – G ) Fig. 7.4 Gravitational force on point mass m1 is the r 2 ( r 2 vector sum of the gravitational forces exerted m1 m 2 ɵ by m2, m3 and m4. = – G 3 r r The total force on m1 is where G is the universal gravitational constant, Gm 2 m 1 ɵ Gm 3 m 1 ɵ Gm 4 m 1 ɵ r 41 r 21 + r 31 + 1 = ɵr is the unit vector from m1 to m2 and r = r2 – r1 F r212 r412 r312 as shown in Fig. 7.3. ⊳ Example 7.2 Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass 2m placed at the centroid G of the triangle? (b) What is the force if the mass at the O vertex A is doubled ? Take AG = BG = CG = 1 m (see Fig. 7.5) Answer (a) The angle between GC and the positive x-axis is 30° and so is the angle between Fig. 7.3 Gravitational force on m1 due to m2 is along GB and the negative x-axis. The individual forces r where the vector r is (r2– r1). in vector notation are Reprint 2025-26 GRAVITATION 131 cases, a simple law results when you do that : (1) The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell. Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components Fig. 7.5 Three equal masses are placed at the three prependicular to this line cancel out when vertices of the ∆ ABC. A mass 2m is placed summing over all regions of the shell leaving at the centroid G. only a resultant force along the line joining the point to the centre. The magnitude of Gm ( 2m ) ˆ FGA = j this force works out to be as stated above. 1 (2) The force of attraction due to a hollow Gm ( 2m ) FGB = − ˆi cos 30ο − ˆj sin 30ο ( ) sphericalpoint massshellsituatedof uniforminside density,it is zero.on a 1 Qualitatively, we can again understand this Gm 2m ( ) FGC = + ˆi cos 30ο − ˆj sin 30ο result. Various regions of the spherical shell 1 ( ) attract the point mass inside it in various From the principle of superposition and the law directions. These forces cancel each other of vector addition, the resultant gravitational completely. force FR on (2m) is FR = FGA + FGB + FGC 7.4 THE GRAVITATIONAL CONSTANT FR = 2Gm 2 ˆj + 2Gm 2 (− ˆi cos 30ο− ˆj sin 30ο ) The value of the gravitational constant G entering the Universal law of gravitation can be + 2Gm 2 ( ˆi cos 30ο − ˆj sin 30ο ) = 0 determined experimentally and this was first done Alternatively, one expects on the basis of by English scientist Henry Cavendish in 1798. symmetry that the resultant force ought to be The apparatus used by him is schematically zero. shown in Fig.7.6 (b) Now if the mass at vertex A is doubled then ⊳ For the gravitational force between an extended Fig. 7.6 Schematic drawing of Cavendish’s object (like the earth) and a point mass, Eq. (7.5) is not experiment. S1 and S2 are large spheres directly applicable. Each point mass in the extended which are kept on either side (shown object will exert a force on the given point mass and shades) of the masses at A and B. When the big spheres are taken to the other side these force will not all be in the same direction. We of the masses (shown by dotted circles), have to add up these forces vectorially for all the point the bar AB rotates a little since the torque masses in the extended object to get the total force. reverses direction. The angle of rotation can This is easily done using calculus. For two special be measured experimentally. Reprint 2025-26 132 PHYSICS The bar AB has two small lead spheres all the shells exert a gravitational force at the attached at its ends. The bar is suspended from point outside just as if their masses are a rigid support by a fine wire. Two large lead concentrated at their common centre according spheres are brought close to the small ones but to the result stated in section 7.3. The total mass on opposite sides as shown. The big spheres of all the shells combined is just the mass of the attract the nearby small ones by equal and earth. Hence, at a point outside the earth, the opposite force as shown. There is no net force gravitational force is just as if its entire mass of on the bar but only a torque which is clearly the earth is concentrated at its centre. equal to F times the length of the bar,where F is For a point inside the earth, the situation the force of attraction between a big sphere and is different. This is illustrated in Fig. 7.7. its neighbouring small sphere. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque . If θis the angle of twist of the suspended wire, the restoring torque is proportional to θ, equal to τθ. Where τ is the restoring couple per unit angle of twist. τ can be measured independently e.g. by applying a known torque and measuring the angle of twist. The gravitational force between the spherical Mrballs is the same as if their masses are concentrated at their centres. Thus if d is the separation between the centres of the big and Fig. 7.7 The mass m is in a mine located at a depth its neighbouring small ball, M and m their d below the surface of the Earth of mass masses, the gravitational force between the big ME and radius RE. We treat the Earth to be sphere and its neighouring small ball is. spherically symmetric. Mm Again consider the earth to be made up of F = G 2 (7.6) d concentric shells as before and a point mass m situated at a distance r from the centre. The If L is the length of the bar AB , then the torque arising out of F is F multiplied by L. At point P lies outside the sphere of radius r. For the shells of radius greater than r, the point Pequilibrium, this is equal to the restoring torque lies inside. Hence according to result stated inand hence the last section, they exert no gravitational force Mm G 2 L = τθ (7.7) on mass m kept at P. The shells with radius ≤r d make up a sphere of radius r for which the point Observation of θ thus enables one to P lies on the surface. This smaller sphere calculate G from this equation. therefore exerts a force on a mass m at P as if Since Cavendish’s experiment, the its mass Mr is concentrated at the centre. Thus measurement of G has been refined and the the force on the mass m at P has a magnitude currently accepted value is Gm ( M r ) G = 6.67×10-11 N m2/kg2 (7.8) F = 2 (7.9) r We assume that the entire earth is of uniform

7.3 Universal law of above fall towards the earth and there are many other such gravitation phenomena. Historically it was the Italian Physicist Galileo

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

8.3 ELECTROMAGNETIC WAVES 8.3.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves. The proof of this basic result is beyond the scope of this book, but we can accept it on the basis of rough, qualitative reasoning. Consider a charge oscillating with some frequency. (An oscillating charge is an example of accelerating charge.) This produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. From the preceding discussion, it might appear easy to test the prediction that light is an electromagnetic wave. We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light, say, yellow light. But, alas, that is not possible. The frequency of yellow light is about 6 × 1014 Hz, while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. This is why the experimental demonstration of electromagnetic 205 Reprint 2025-26 Physics wave had to come in the low frequency region (the radio wave region), as in the Hertz’s experiment (1887). Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. Two important achievements in this connection deserve mention. Seven years after Hertz, Jagdish Chandra Bose, working at Calcutta (now Kolkata), succeeded in producing and observing electromagnetic waves of much shorter 8.1 wavelength (25 mm to 5 mm). His experiment, like that of Hertz’s, was confined to the laboratory. At around the same time, Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting EXAMPLE electromagnetic waves over distances of many kilometres. Heinrich Rudolf Hertz Marconi’s experiment marks the beginning of the field of (1857 – 1894) German communication using electromagnetic waves. physicist who was the first to broadcast and 8.3.2 Nature of electromagnetic wavesHEINRICH receive radio waves. He It can be shown from Maxwell’s equations that electric produced electro- and magnetic fields in an electromagnetic wave are magnetic waves, sent them through space, and perpendicular to each other, and to the direction of measured their wave- propagation. It appears reasonable, say from ourRUDOLF length and speed. He discussion of the displacement current. Consider showed that the nature Fig. 8.2. The electric field inside the plates of the capacitor of their vibration, is directed perpendicular to the plates. The magnetic reflection and refraction field this gives rise to via the displacement current is was the same as that ofHERTZ along the perimeter of a circle parallel to the capacitor light and heat waves, plates. So B and E are perpendicular in this case. This establishing their identity for the first time. is a general feature. He also pioneered In Fig. 8.3, we show a typical example of a plane research on discharge of electromagnetic wave propagating along the z direction electricity through gases, (the fields are shown as a function of the z coordinate, at and discovered the(1857–1894) a given time t). The electric field Ex is along the x-axis, photoelectric effect. and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation. We can write Ex and By as follows: Ex= E0 sin (kz–wt) [8.7(a)] By= B0 sin (kz–wt) [8.7(b)] Here k is related to the wave length FIGURE 8.3 A linearly polarised electromagnetic wave, l of the wave by the usual propagating in the z-direction with the oscillating electric field E equation along the x-direction and the oscillating magnetic field B along the y-direction. 2 π k = (8.8) 206 λ Reprint 2025-26 Electromagnetic Waves and ω is the angular frequency. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the wave. The speed of propagation of the wave is (ω/k). Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one finds that ω = ck, where, c = 1/ µ0ε0 [8.9(a)] The relation ω = ck is the standard one for waves (see for example, Section 14.4 of class XI Physics textbook). This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as  2π  2 πν = c  λ  or νλ = c [8.9(b)] It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as B0 = (E0/c) (8.10) We here make remarks on some features of electromagnetic waves. They are self-sustaining oscillations of electric and magnetic fields in free space, or vacuum. They differ from all the other waves we have studied so far, in respect that no material medium is involved in the vibrations of the electric and magnetic fields. But what if a material medium is actually there? We know that light, an electromagnetic wave, does propagate through glass, for example. We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability µ (these describe the factors by which the total fields differ from the external fields). These replace ε0 and µ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes, 1 v = µε (8.11) Thus, the velocity of light depends on electric and magnetic properties of the medium. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. It has been shown by experiments on electromagnetic waves of different wavelengths that this velocity is the same (independent of wavelength) to within a few metres per second, out of a value of 3×108 m/s. The constancy of the velocity of em waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length. The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth. 207 Reprint 2025-26 Physics Example 8.1 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ˆj V/m. What is B at this point? Solution Using Eq. (8.10), the magnitude of B is E B = c 6.3 V/m –8 = 8 = 2.1 × 10 T 3 × 10 m/s 8.1 To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, (+ ˆj ) × (+ ˆk ) = ˆi , B is along the z-direction. EXAMPLE Thus, B = 2.1 × 10–8 ˆk T Example 8.2 The magnetic field in a plane electromagnetic wave is given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Comparing the given equation with   x t   By=B0 Sin 2p +  λ T  spectrum  2π We get, λ = 3 m = 1.26 cm, 0.5 × 10 1 11 and = ν= 1.5 × 10 /2 π = 23.9 GHz T ( ) 8.2 (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/mElectromagnetic http://www.fnal.gov/pub/inquiring/more/light http://imagine.gsfc.nasa.gov/docs/science/ The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as EXAMPLE Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m 8.4 ELECTROMAGNETIC SPECTRUM At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves. The existence of ultraviolet and infrared waves was barely established. By the end of the nineteenth century, X-rays and gamma rays had also been discovered. We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.4). There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected. We briefly describe these different types of electromagnetic waves, in 208 order of decreasing wavelengths. Reprint 2025-26 Electromagnetic Waves FIGURE 8.4 The electromagnetic spectrum, with common names for various part of it. The various regions do not have sharply defined boundaries. 8.4.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. They are generally in the frequency range from 500 kHz to about 1000 MHz. The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. Higher frequencies upto 54 MHz are used for short wave bands. TV waves range from 54 MHz to 890 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. How these waves are transmitted and received is described in Chapter 15. 8.4.2 Microwaves Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes). Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Radar also provides the basis for the speed guns used to time fast balls, tennis- serves, and automobiles. Microwave ovens are an interesting domestic application of these waves. In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of 209the molecules. This raises the temperature of any food containing water. Reprint 2025-26 Physics 8.4.3 Infrared waves Infrared waves are produced by hot bodies and molecules. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum. Infrared waves are sometimes referred to as heat waves. This is because water molecules present in most materials readily absorb infrared waves (many other molecules, for example, CO2, NH3, also absorb infrared waves). After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. Infrared lamps are used in physical therapy. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and re- radiated as infrared (longer wavelength) radiations. This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Infrared detectors are used in Earth satellites, both for military purposes and to observe growth of crops. Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets, video recorders and hi-fi systems. 8.4.4 Visible rays It is the most familiar form of electromagnetic waves. It is the part of the spectrum that is detected by the human eye. It runs from about 4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. Visible light emitted or reflected from objects around us provides us information about the world. Our eyes are sensitive to this range of wavelengths. Different animals are sensitive to different range of wavelengths. For example, snakes can detect infrared waves, and the ‘visible’ range of many insects extends well into the utraviolet. 8.4.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0.6 nm). Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. But fortunately, most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km. UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laser- assisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers. Ozone layer in the atmosphere plays a protective role, and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter 210 of international concern. Reprint 2025-26 Electromagnetic Waves 8.4.6 X-rays Beyond the UV region of the electromagnetic spectrum lies the X-ray region. We are familiar with X-rays because of its medical applications. It covers wavelengths from about 10–8 m (10 nm) down to 10–13 m (10–4 nm). One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary or over exposure. 8.4.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. Table 8.1 summarises different types of electromagnetic waves, their production and detections. As mentioned earlier, the demarcation between different regions is not sharp and there are overlaps. TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Type Wavelength range Production Detection Radio > 0.1 m Rapid acceleration and Receiver’s aerials decelerations of electrons in aerials Microwave 0.1m to 1 mm Klystron valve or Point contact diodes magnetron valve Infra-red 1mm to 700 nm Vibration of atoms Thermopiles and molecules Bolometer, Infrared photographic film Light 700 nm to 400 nm Electrons in atoms emit The eye light when they move from Photocells one energy level to a Photographic film lower energy level Ultraviolet 400 nm to 1nm Inner shell electrons in Photocells atoms moving from one Photographic film energy level to a lower level X-rays 1nm to 10–3 nm X-ray tubes or inner shell Photographic film electrons Geiger tubes Ionisation chamber Gamma rays <10–3 nm Radioactive decay of the -do- nucleus 211 Reprint 2025-26 Physics SUMMARY 1. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current, called displacement current, to remove this inconsistency. This displacement current is due to time-varying electric field and is given by dΦΕ di = ε0 dt and acts as a source of magnetic field in exactly the same way as conduction current. 2. An accelerating charge produces electromagnetic waves. An electric charge oscillating harmonically with frequency n, produces electromagnetic waves of the same frequency n. An electric dipole is a basic source of electromagnetic waves. 3. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 4. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave. For a wave of frequency n, wavelength l, propagating along z-direction, we have E = Ex (t) = E0 sin (kz – w t )   z     z t   = E0 sin 2 π  λ − νt  = E 0 sin 2 π  λ − T   B = By(t) = B0 sin (kz – w t)   z     z t   = B 0 sin 2 π  λ − νt  = B 0 sin  2 π  λ − T   They are related by E0/B0 = c. 5. The speed c of electromagnetic wave in vacuum is related to m0 and e0 (the free space permeability and permittivity constants) as follows: c = 1/ µ0 ε0 . The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. The speed of light, or of electromagnetic waves in a material medium is given by v = 1/ µε where m is the permeability of the medium and e its permittivity. 6. The spectrum of electromagnetic waves stretches, in principle, over an infinite range of wavelengths. Different regions are known by different names; g-rays, X-rays, ultraviolet rays, visible rays, infrared rays, microwaves and radio waves in order of increasing wavelength from 10–2 Å or 10–12 m to 106 m. They interact with matter via their electric and magnetic fields which set in oscillation charges present in all matter. The detailed interaction and so the mechanism of absorption, scattering, etc., depend on the wavelength of the electromagnetic wave, and the nature of the atoms and molecules 212 in the medium. Reprint 2025-26 Electromagnetic Waves POINTS TO PONDER 1. The basic difference between various types of electromagnetic waves lies in their wavelengths or frequencies since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter. 2. Accelerated charged particles radiate electromagnetic waves. The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. Thus, gamma radiation, having wavelength of 10–14 m to 10–15 m, typically originate from an atomic nucleus. X-rays are emitted from heavy atoms. Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. Visible radiation emitted by atoms is, however, much longer in wavelength than atomic size. 3. Infrared waves, with frequencies lower than those of visible light, vibrate not only the electrons, but entire atoms or molecules of a substance. This vibration increases the internal energy and consequently, the temperature of the substance. This is why infrared waves are often called heat waves. 4. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. It is because humans have evolved with visions most sensitive to the strongest wavelengths from the sun. EXERCISES

8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. FIGURE 8.5

27.3 days which is also roughly equal to the Which is approximately 85 minutes. rotational period of the moon about its own axis. ⊳ Example 7.5 The planet Mars has twoSince, 1957, advances in technology have enabled moons, phobos and delmos. (i) phobos hasmany countries including India to launch artificial a period 7 hours, 39 minutes and an orbitalearth satellites for practical use in fields like radius of 9.4 ×103 km. Calculate the masstelecommunication, geophysics and meteorology. of mars. (ii) Assume that earth and mars We will consider a satellite in a circular orbit move in circular orbits around the sun,of a distance (RE + h) from the centre of the earth, with the martian orbit being 1.52 timeswhere RE = radius of the earth. If m is the mass the orbital radius of the earth. What isof the satellite and V its speed, the centripetal the length of the martian year in days ?force required for this orbit is mV 2 Answer (i) We employ Eq. (7.38) with the sun’s F(centripetal) = (7.33) ( R E + h ) mass replaced by the martian mass Mm directed towards the centre. This centripetal force 2 4 π 2 3 T = Ris provided by the gravitational force, which is GM m G m M E 4 π 2 R 3 F(gravitation) = 2 (7.34) Mm = 2 ( R E + h ) G T where ME is the mass of the earth. 2 3 18 Equating R.H.S of Eqs. (7.33) and (7.34) and 4 × ( 3.14 ) × ( 9.4 ) × 10 = -11 2cancelling out m, we get 6.67 × 10 × ( 459 × 60 ) 2 G M E 2 3 18 V = (7.35) 4 × ( 3.14 ) × ( 9.4 ) × 10 ( R E + h ) M m = 2 -5 6.67 × ( 4.59 × 6 ) × 10 Thus V decreases as h increases. From = 6.48 × 1023 kg. equation (7.35),the speed V for h = 0 is (ii) Once again Kepler’s third law comes to our V 2 (h = 0) = GM / R E = gR E (7.36) aid, where we have used the relation T M2 R MS3 2 2 = 3 g = GM / R E . In every orbit, the satellite T E R ES Reprint 2025-26 138 PHYSICS where RMS is the mars -sun distance and RES is − 13  1 2   1    d the earth-sun distance. = 10    ( 24 × 60 × 60 ) 2  ( 1 / 1000 ) 3 km 3  ∴ TM = (1.52)3/2 × 365 = 1.33 ×10–14 d2 km–3 = 684 days Using Eq. (7.38) and the given value of k, We note that the orbits of all planets except the time period of the moon is Mercury and Mars are very close to being 2 T = (1.33 × 10-14)(3.84 × 105)3 circular. For example, the ratio of the semi- T = 27.3 d ⊳ minor to semi-major axis for our Earth is, Note that Eq. (7.38) also holds for elliptical b/a = 0.99986. ⊳ orbits if we replace (RE+h) by the semi-major axis ⊳ of the ellipse. The earth will then be at one of Example 7.6 Weighing the Earth : You the foci of this ellipse. are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the moon R 7.10 ENERGY OF AN ORBITING SATELLITE = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the Using Eq. (7.35), the kinetic energy of the satellite mass of the Earth ME in two different ways. in a circular orbit with speed v is 1 m v 2Answer From Eq. (7.12) we have K i E = 2 g R E2 M E = Gm M E G = , (7.40) 2( R E + h ) 6 2 Considering gravitational potential energy at 9.81 × ( 6.37 × 10 ) = -11 infinity to be zero, the potential energy at distance 6.67 × 10 (Re+h) from the centre of the earth is = 5.97× 1024 kg. The moon is a satellite of the Earth. From G m M E P .E = − (7.41)the derivation of Kepler’s third law [see Eq. ( R E + h ) (7.38)] The K.E is positive whereas the P.E is 2 4 π2R 3 negative. However, in magnitude the K.E. is half T = G M E the P.E, so that the total E is E 4 π2R 3 E = K .E + P .E = − G m M ME = G T 2 2( R E + h ) (7.42) 4 × 3.14 × 3.14 × ( 3.84 ) 3 × 10 24 The total energy of an circularly orbiting = -11 2 satellite is thus negative, with the potential 6.67 × 10 × ( 27.3 × 24 × 60 × 60 ) energy being negative but twice is magnitude of = 6.02 × 1024 kg the positive kinetic energy. Both methods yield almost the same answer, When the orbit of a satellite becomes the difference between them being less than 1%. elliptic, both the K.E. and P.E. vary from point ⊳ to point. The total energy which remains constant is negative as in the circular orbit case. ⊳ Example 7.7 Express the constant k of Eq. This is what we expect, since as we have (7.38) in days and kilometres. Given discussed before if the total energy is positive or k = 10–13 s2 m–3. The moon is at a distance zero, the object escapes to infinity. Satellites of 3.84 × 105 km from the earth. Obtain its are always at finite distance from the earth and time-period of revolution in days. hence their energies cannot be positive or zero. Answer Given k = 10–13 s2 m–3 Reprint 2025-26 GRAVITATION 139 The change in the total energy is⊳ Example 7.8 A 400 kg satellite is in a circular ∆E = Ef – Ei orbit of radius 2RE about the Earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in  E E = G M E m =  the kinetic and potential energies ?  G M2  m R 8 R E  R E  8 Answer Initially, g m R E = 9.81 × 400 × 6. 37 × 106 = 3.13 × 10 9 J ∆ E = G M E m 8 8 E i = − 4 R E The kinetic energy is reduced and it mimics While finally ∆E, namely, ∆K = Kf – Ki = – 3.13 × 109 J. The change in potential energy is twice the G M E m E f = − change in the total energy, namely 8 R E ∆V = Vf – Vi = – 6.25 × 109 J ⊳ SUMMARY 1. Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude m 1m 2 F = G 2 r where G is the universal gravitational constant, which has the value 6.672 ×10–11 N m2 kg–2. 2. If we have to find the resultant gravitational force acting on the particle m due to a number of masses M1, M2, ….Mn etc. we use the principle of superposition. Let F1, F2, ….Fn be the individual forces due to M1, M2, ….Mn, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force FR is then found by vector addition n FR = F1 + F2 + ……+ Fn = ∑ Fi i = 1 where the symbol ‘Σ’ stands for summation. 3. Kepler’s laws of planetary motion state that (a) All planets move in elliptical orbits with the Sun at one of the focal points (b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved. (c) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related by 2  4 π 2  3 T =   R  G M s  where Ms is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major axis, a. 4. The acceleration due to gravity. (a) at a height h above the earth’s surface G M E g ( h ) = 2 ( R E + h ) G M E  2 h  ≈ 2 1 − for h << RE R E  R E  Reprint 2025-26 140 PHYSICS  2 h  G M E g (h ) = g ( 0 ) 1 − where g ( 0 ) = 2  R E  R E (b) at depth d below the earth’s surface is d   1 − g (d ) = G M2 E 1 − d  = g ( 0 ) R E  R E   R E  5. The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is given by G m1 m 2 V = − r where V is taken to be zero at r → ∞. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition. 6. If an isolated system consists of a particle of mass m moving with a speed v in the vicinity of a massive body of mass M, the total mechanical energy of the particle is given by 1 G M m E = m v 2− 2 r That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion. 7. If m moves in a circular orbit of radius a about M, where M >> m, the total energy of the system is G M m E = − 2a with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are G M m K = 2a G M m V = − a 8. The escape speed from the surface of the earth is 2 G M E ve = = 2 gR E R E and has a value of 11.2 km s–1. 9. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere. 10. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle. Reprint 2025-26 GRAVITATION 141 POINTS TO PONDER 1. In considering motion of an object under the gravitational influence of another object the following quantities are conserved: (a) Angular momentum (b) Total mechanical energy Linear momentum is not conserved 2. Angular momentum conservation leads to Kepler’s second law. However, it is not special to the inverse square law of gravitation. It holds for any central force. 3. In Kepler’s third law (see Eq. (7.1) and T2 = KS R3. The constant KS is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)]. 4. An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in “free fall” towards the Earth. 5. The gravitational potential energy associated with two particles separated by a distance r is given by G m 1 m 2 V = – + constant r The constant can be given any value. The simplest choice is to take it to be zero. With this choice G m 1 m 2 V = – r This choice implies that V → 0 as r → ∞. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant. 6. The total mechanical energy of an object is the sum of its kinetic energy (which is always positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative. 7. The commonly encountered expression m g h for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above. 8. Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central. 9. The gravitational force on a particle inside a spherical shell is zero. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible. EXERCISES 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ? Reprint 2025-26 142 PHYSICS 7.2 Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth. 7.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ? 7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun. 7.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly. 7.6 Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence. 7.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? 7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. 7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem. 7.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0. Fig. 7.11 7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. 7.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m). 7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km. 7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ? 7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? Reprint 2025-26 GRAVITATION 143 7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ? 7.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G). 7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ? Reprint 2025-26