Electromagnetic Induction PDF

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 Oers...

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 2024-25  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. 6.2 THE EXPERIMENTS OF FARADAY AND HENRY JOSEPH HENRY (1797 – 1878) The discovery and understanding of electromagnetic induction are based on a long series of experiments carried Josheph Henry [1797 – out by Faraday and Henry. We shall now describe some 1878] American experimental of these experiments. physicist, professor at Princeton University and first Experiment 6.1 director of the Smithsonian Institution. He made important Figure 6.1 shows a coil C1* connected to a galvanometer improvements in electro- G. When the North-pole of a bar magnet is pushed magnets by winding coils of insulated wire around iron towards the coil, the pointer in the galvanometer deflects, pole pieces and invented an indicating the presence of electric current in the coil. The electromagnetic motor and a deflection lasts as long as the bar magnet is in motion. new, efficient telegraph. He The galvanometer does not show any deflection when the discoverd self-induction and magnet is held stationary. When the magnet is pulled investigated how currents in away from the coil, the galvanometer shows deflection in one circuit induce currents in another. the opposite direction, which indicates reversal of the current’s direction. Moreover, when the South-pole of the bar magnet is moved towards or away from the coil, the deflections in the galvanometer are opposite to that observed with the North-pole for similar movements. Further, the deflection (and hence current) is found to be larger when the magnet is pushed towards or pulled away from the coil faster. Instead, when the bar magnet is held fixed and the coil C1 is moved towards or away from the magnet, the same effects are observed. It shows that it is the relative motion between the magnet and the coil that is responsible for generation (induction) of electric current in the coil. Experiment 6.2 FIGURE 6.1 When the bar magnet is In Fig. 6.2 the bar magnet is replaced by a second coil pushed towards the coil, the pointer in C2 connected to a battery. The steady current in the the galvanometer G deflects. coil C2 produces a steady magnetic field. As coil C2 is * Wherever the term ‘coil’ or ‘loop’ is used, it is assumed that they are made up of conducting material and are prepared using wires which are coated with insulating material. 155 2024-25  Physics moved towards the coil C 1, the galvanometer shows a deflection. This indicates that electric current is induced in coil C1. When C2 is moved away, the galvanometer shows a deflection again, but this time in the opposite direction. The deflection lasts as long as coil C2 is in motion. When the coil C2 is held fixed and C1 is moved, the same effects are observed. Again, it is the relative motion between the coils that induces the electric current. Experiment 6.3 The above two experiments involved relative motion between a magnet and a coil and between two coils, respectively. Through another experiment, Faraday showed that this FIGURE 6.2 Current is relative motion is not an absolute requirement. Figure 6.3 induced in coil C1 due to motion shows two coils C1 and C2 held stationary. Coil C1 is connected of the current carrying coil C2. to galvanometer G while the second coil C2 is connected to a battery through a tapping key K. FIGURE 6.3 Experimental set-up for Experiment 6.3. It is observed that the galvanometer shows a momentary deflection when the tapping key K is pressed. The pointer in the galvanometer returns to zero immediately. If the key is held pressed continuously, there is no deflection in the galvanometer. When the key is released, a momentory deflection is observed again, but in the opposite direction. It is also observed that the deflection increases dramatically when an iron rod is inserted into the coils along their axis. 6.3 MAGNETIC FLUX Faraday’s great insight lay in discovering a simple mathematical relation to explain the series of experiments he carried out on electromagnetic induction. However, before we state and appreciate his laws, we must get familiar with the notion of magnetic flux, F B. Magnetic flux is defined in 156 the same way as electric flux is defined in Chapter 1. Magnetic flux through 2024-25  Electromagnetic Induction a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can be written as F B = B. A = BA cos q (6.1) where q is angle between B and A. The notion of the area as a vector has been discussed earlier in Chapter 1. Equation (6.1) can be extended to curved surfaces and nonuniform fields. If the magnetic field has different magnitudes and directions at various parts of a surface as shown in Fig. 6.5, then the magnetic flux through the surface is given by Φ = B. dA + B. dA +... = B 1 1 2 2 ∑ Bi. dA i all (6.2) FIGURE 6.4 A plane of where ‘all’ stands for summation over all the area elements dAi surface area A placed in a comprising the surface and Bi is the magnetic field at the area element uniform magnetic field B. dAi. The SI unit of magnetic flux is weber (Wb) or tesla meter squared (T m2). Magnetic flux is a scalar quantity. 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 i th area element. dAi as follows: When the tapping key K is pressed, the current in represents area vector of the i th 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 2024-25  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) dt 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. MICHAEL FARADAY (1791–1867) 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 science, viz., the discovery ε = –N (6.4) dt 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 polarisation is rotated in an From Eqs. (6.1) and (6.2), we see that the flux can be electric field. He is also varied by changing any one or more of the terms B, A and 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 A 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 small torch light. The relative motion between the two coils will cause EXAMPLE 6.1 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 for his innovative skills. EXAMPLE 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 the magnitudes of induced emf and current during this time-interval. 158 2024-25  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 ∆ΦB (Φ – 0) 10 –3 ε= = = = 1.0 mV ∆t ∆t 2 × 0.7 And the magnitude of the current is ε EXAMPLE 6.2 10 –3 V I = = = 2 mA R 0.5Ω Note that the earth’s magnetic field also produces a flux through the 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. 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 EXAMPLE 6.3 I = e/R = 1.9 × 10–3 A Note that the magnitudes of e and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant. 159 2024-25  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. 2024-25  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. EXAMPLE 6.4 (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. Note that there are no induced current as long as the loops are 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. EXAMPLE 6.5 (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 constant during the passage out of the field region? The field is normal to the loops. 161 2024-25  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. EXAMPLE 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 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 ) rectangular loop. This movement dt dt induces a current I as shown. dx 162 = – Bl = Blv (6.5) dt 2024-25  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 EXAMPLE 6.6 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 axis is present everywhere. What is the emf between the centre and the metallic ring? 163 2024-25  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 ωR2 ε =∫ dε = ∫ Bv dr = ∫ B ωr dr = 0 0 2 Note that we have used v = w r. This gives 1 e = × 1.0 × 2 π × 50 × (12 ) 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 θ 1 2 π R2 × R θ = 2π 2 where R is the radius of the circle. Hence, the induced emf is d 1 2  1 2 dθ Bω R 2 e =B × R θ BR = dt  2  = EXAMPLE 6.6 2 dt 2 dθ [Note: = ω = 2πν ] dt This expression is identical to the expression obtained by Method I and we get the same value of e. 164 2024-25  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 EXAMPLE 6.7 = (1/2) × 4π × 0.4 × 10–4 × (0.5)2 = 6.28 × 10–5 V The number of spokes is immaterial because the emf’s across the 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 dI ∝ dt dt 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 of turns of coils S1 and S2, respectively. 165 2024-25  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 = M12 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, ( ) (µ n I ) N1Φ1 = (n1l ) πr12 0 2 2 = µ0n1n 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 M12 = µ0n1n2πr 12l (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 ( ) (µ n I ) N 2Φ2 = (n 2l ) πr12 0 1 1 where n2l is the total number of turns of S2. From Eq. (6.10), M21 = µ0n1n2πr 12 l (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. 2024-25  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