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.

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

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.