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.

1.13 GAUSS’S LAW As a simple application of the notion of electric flux, let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in Fig. 1.22. The flux through an area element DS is q ∆φ = E i ∆ S = 2 rˆ i ∆S (1.28) 4 πε0 r where we have used Coulomb’s law for the electric field due to a single charge q. The unit vector ˆr is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element DS and ˆr have the same direction. Therefore, FIGURE 1.22 Flux q ∆φ = 2 ∆ S (1.29) through a sphere 4 πε0 r enclosing a point since the magnitude of a unit vector is 1. charge q at its centre. The total flux through the sphere is obtained by adding up flux through all the different area elements: 29 Reprint 2025-26 Physics q φ = Σ 2 ∆ S all ∆S 4 π ε0 r Since each area element of the sphere is at the same distance r from the charge, FIGURE 1.23 Calculation of the q q S φ = Σ ∆ S = flux of uniform electric field all ∆S 4 πε0 r 2 4 πεo r 2 through the surface of a cylinder. Now S, the total area of the sphere, equals 4pr 2. Thus, q 2 q φ = 2 × 4 πr = (1.30) 4 πε0 r ε0 Equation (1.30) is a simple illustration of a general result of electrostatics called Gauss’s law. We state Gauss’s law without proof: Electric flux through a closed surface S = q/e0 (1.31) q = total charge enclosed by S. The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. We can see that explicitly in the simple situation of Fig. 1.23. Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field E. The total flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent the flux through the surfaces 1 and 2 (of circular cross-section) of the cylinder and f3 is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface 3 at every point is perpendicular to E, so by definition of flux, f3 = 0. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E. Therefore, f1 = –E S1, f2 = +E S2 S1 = S2 = S where S is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero. The great significance of Gauss’s law Eq. (1.31), is that it is true in general, and not only for the simple cases we have considered above. Let us note some important points regarding this law: (i) Gauss’s law is true for any closed surface, no matter what its shape or size. (ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface. (iii) In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, 30 represents only the total charge inside S. Reprint 2025-26 Electric Charges and Fields (iv) The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge. (As you go close to the charge, the field grows without any bound.) However, the Gaussian surface can pass through a continuous charge distribution. (v) Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface. (vi) Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law. Example 1.10 The electric field components in Fig. 1.24 are Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. FIGURE 1.24 Solution (a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and DS is ± p/2. Therefore, the flux f = E.DS is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is EL = ax1/2 = aa1/2 (x = a at the left face). The magnitude of electric field at the right face is ER = a x1/2 = a (2a)1/2 (x = 2a at the right face). The corresponding fluxes are fL= EL.DS = ∆ S E L ⋅ nˆ L =EL DS cosq = –EL DS, since q = 180° = –ELa2 EXAMPLE fR= ER.DS = ER DS cosq = ER DS, since q = 0° = ERa2 1.10 Net flux through the cube 31 Reprint 2025-26 Physics = fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2] = aa5/2 2 – 1 ( ) = 800 (0.1)5/2 2 – 1 ( ) 1.10 = 1.05 N m2 C–1 (b) We can use Gauss’s law to find the total charge q inside the cube. We have f = q/e0 or q = fe0. Therefore, EXAMPLE q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C. Example 1.11 An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 ˆi N/C for x > 0 and E = –200 ˆi N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (Fig. 1.25). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? Solution (a) We can see from the figure that on the left face E and DS are parallel. Therefore, the outward flux is ˆi i ∆S fL= E.DS = – 200 = + 200 DS, since ˆi i ∆ S = – DS = + 200 × p (0.05)2 = + 1.57 N m2 C–1 On the right face, E and DS are parallel and therefore fR = E.DS = + 1.57 N m2 C–1. (b) For any point on the side of the cylinder E is perpendicular to DS and hence E.DS = 0. Therefore, the flux out of the side of the cylinder is zero. (c) Net outward flux through the cylinder f = 1.57 + 1.57 + 0 = 3.14 N m2 C–1 FIGURE 1.25 1.11 (d) The net charge within the cylinder can be found by using Gauss’s law which gives q = e0f = 3.14 × 8.854 × 10–12 C EXAMPLE = 2.78 × 10–11 C Reprint 2025-26 Electric Charges and Fields

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.