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.

4.6 AMPERE’S CIRCUITAL LAW There is an alternative and appealing way in which the Biot-Savart law may be expressed. Ampere’s circuital law considers an open surface with a boundary (Fig. 4.12). The surface has current passing through it. We consider the boundary to be made up of a number of small line elements. Consider one such element of length dl. We take the value of the tangential component of the FIGURE 4.12 117magnetic field, Bt, at this element and multiply it by the Reprint 2025-26 Physics length of that element dl [Note: Btdl=B.dl]. All such products are added together. We consider the limit as the lengths of elements get smaller and their number gets larger. The sum then tends to an integral. Ampere’s law states that this integral is equal to µ0 times the total current passing through the surface, i.e., “B.dl = µ0I [4.13(a)] where I is the total current through the surface. The integral is taken over the closed loop coinciding with the boundary C of the surface. The relation above involves a sign-convention, given by the right-hand rule. Let the fingers of the right-hand be curled in the sense the boundary is traversed in the loop integral “B.dl. Then the direction of the thumb gives the sense in which the Andre Ampere (1775 – current I is regarded as positive. 1836) Andre Marie Ampere For several applications, a much simplified version of was a French physicist, mathematician and chemist Eq. [4.13(a)] proves sufficient. We shall assume that, in who founded the science of such cases, it is possible to choose the loop (called electrodynamics. Ampere an amperian loop) such that at each point of the was a child prodigy loop, either who mastered advanced (i) B is tangential to the loop and is a non-zero constant mathematics by the age of B, or 12. Ampere grasped the significance of Oersted’s (ii) B is normal to the loop, or discovery. He carried out a (iii) B vanishes. large series of experiments Now, let L be the length (part) of the loop for which B to explore the relationship is tangential. Let Ie be the current enclosed by the loop. between current electricity Then, Eq. (4.13) reduces to, and magnetism. These investigations culminated BL =µ0Ie [4.13(b)] in 1827 with the When there is a system with a symmetry such as for publication of the a straight infinite current-carrying wire in Fig. 4.13, the ‘Mathematical Theory of Ampere’s law enables an easy evaluation of the magnetic–1836) Electrodynamic Pheno- mena Deduced Solely from field, much the same way Gauss’ law helps in Experiments’. He hypo- determination of the electric field. This is exhibited in the thesised that all magnetic Example 4.8 below. The boundary of the loop chosen is(1775 phenomena are due to a circle and magnetic field is tangential to the circulating electric circumference of the circle. The law gives, for the left hand currents. Ampere was side of Eq. [4.13 (b)], B. 2πr. We find that the magnetic humble and absent- field at a distance r outside the wire is tangential and minded. He once forgot anAMPERE given by invitation to dine with the Emperor Napoleon. He died B × 2πr = µ0 I, of pneumonia at the age of 61. His gravestone bears B = µ0 I/ (2πr) (4.14)ANDRE the epitaph: Tandem Felix The above result for the infinite wire is interesting (Happy at last). from several points of view. (i) It implies that the field at every point on a circle of radius r, (with the wire along the axis), is same in 118 magnitude. In other words, the magnetic field Reprint 2025-26 Moving Charges and Magnetism possesses what is called a cylindrical symmetry. The field that normally can depend on three coordinates depends only on one: r. Whenever there is symmetry, the solutions simplify. (ii) The field direction at any point on this circle is tangential to it. Thus, the lines of constant magnitude of magnetic field form concentric circles. Notice now, in Fig. 4.1(c), the iron filings form concentric circles. These lines called magnetic field lines form closed loops. This is unlike the electrostatic field lines which originate from positive charges and end at negative charges. The expression for the magnetic field of a straight wire provides a theoretical justification to Oersted’s experiments. (iii) Another interesting point to note is that even though the wire is infinite, the field due to it at a non-zero distance is not infinite. It tends to blow up only when we come very close to the wire. The field is directly proportional to the current and inversely proportional to the distance from the (infinitely long) current source. (iv) There exists a simple rule to determine the direction of the magnetic field due to a long wire. This rule, called the right-hand rule*, is: Grasp the wire in your right hand with your extended thumb pointing in the direction of the current. Your fingers will curl around in the direction of the magnetic field. Ampere’s circuital law is not new in content from Biot-Savart law. Both relate the magnetic field and the current, and both express the same physical consequences of a steady electrical current. Ampere’s law is to Biot-Savart law, what Gauss’s law is to Coulomb’s law. Both, Ampere’s and Gauss’s law relate a physical quantity on the periphery or boundary (magnetic or electric field) to another physical quantity, namely, the source, in the interior (current or charge). We also note that Ampere’s circuital law holds for steady currents which do not fluctuate with time. The following example will help us understand what is meant by the term enclosed current. Example 4.7 Figure 4.13 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a. EXAMPLE 4.7 FIGURE 4.13 * Note that there are two distinct right-hand rules: One which gives the direction of B on the axis of current-loop and the other which gives direction of B for a straight conducting wire. Fingers and thumb play different roles in 119 the two. Reprint 2025-26 Physics Solution (a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop, L = 2 π r Ie = Current enclosed by the loop = I The result is the familiar expression for a long straight wire B (2π r) = µ0I µ0 I B = 2 π r [4.15(a)] 1 B ∝ (r > a) r Now the current enclosed Ie is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,  π r 2  Ir 2 I e = I 2 a 2  π  = a I r 2 B (2 π r ) = µ0 2 Using Ampere’s law, a  µ0 I  B = 2 r [4.15(b)]  2πa  B ∝ r (r < a) FIGURE 4.14 Figure (4.14) shows a plot of the magnitude of B with distance r 4.7 from the centre of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section. This example possesses the required symmetry so that Ampere’s EXAMPLE law can be applied readily. It should be noted that while Ampere’s circuital law holds for any loop, it may not always facilitate an evaluation of the magnetic field in every case. For example, for the case of the circular loop discussed in Section 4.5, it cannot be applied to extract the simple expression B = µ0I/2R [Eq. (4.12)] for the field at the centre of the loop. However, there exists a large number of situations of high symmetry where the law 120 can be conveniently applied. We shall use it in the next section to calculate Reprint 2025-26 Moving Charges and Magnetism the magnetic field produced by a commonly used and very useful magnetic system: the solenoid.

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.