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

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

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