5.5 MAGNETIC PROPERTIES OF MATERIALS The discussion in the previous section helps us to classify materials as diamagnetic, paramagnetic or ferromagnetic. In terms of the susceptibility χ, a material is diamagnetic if χ is negative, para- if χ is positive and small, and ferro- if χ is large and positive. A glance at Table 5.2 gives one a better feeling for these materials. Here ε is a small positive number introduced to quantify paramagnetic materials. Next, we describe these materials in some detail. TABLE 5.2 Diamagnetic Paramagnetic Ferromagnetic –1 ≤ χ < 0 0 < χ < ε χ >> 1 0 ≤ µr < 1 1< µr < 1+ ε µr >> 1 µ < µ0 µ > µ0 µ >> µ0 5.5.1 Diamagnetism Diamagnetic substances are those which have tendency to move from FIGURE 5.7 stronger to the weaker part of the external magnetic field. In other words, Behaviour of unlike the way a magnet attracts metals like iron, it would repel a magnetic field lines diamagnetic substance. near a Figure 5.7(a) shows a bar of diamagnetic material placed in an external (a) diamagnetic, magnetic field. The field lines are repelled or expelled and the field inside (b) paramagnetic the material is reduced. In most cases, this reduction is slight, being one substance. part in 105. When placed in a non-uniform magnetic field, the bar will tend to move from high to low field. 147 Reprint 2025-26 Physics The simplest explanation for diamagnetism is as follows. Electrons in an atom orbiting around nucleus possess orbital angular momentum. These orbiting electrons are equivalent to current-carrying loop and thus possess orbital magnetic moment. Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. This happens due to induced current in accordance with Lenz’s law which you will study in Chapter 6. Thus, the substance develops a net magnetic moment in direction opposite to that of the applied field and hence repulsion. Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride. Diamagnetism is present in all the substances. However, the effect is so weak in most cases that it gets shifted by other effects like paramagnetism, ferromagnetism, etc. The most exotic diamagnetic materials are superconductors. These are metals, cooled to very low temperatures which exhibits both perfect conductivity and perfect diamagnetism. Here the field lines are completely expelled! χ = –1 and µr = 0. A superconductor repels a magnet and (by Newton’s third law) is repelled by the magnet. The phenomenon of perfect diamagnetism in superconductors is called the Meissner effect, after the name of its discoverer. Superconducting magnets can be gainfully exploited in variety of situations, for example, for running magnetically levitated superfast trains. 5.5.2 Paramagnetism Paramagnetic substances are those which get weakly magnetised when placed in an external magnetic field. They have tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get weakly attracted to a magnet. The individual atoms (or ions or molecules) of a paramagnetic material possess a permanent magnetic dipole moment of their own. On account of the ceaseless random thermal motion of the atoms, no net magnetisation is seen. In the presence of an external field B0, which is strong enough, and at low temperatures, the individual atomic dipole moment can be made to align and point in the same direction as B0. Figure 5.7(b) shows a bar of paramagnetic material placed in an external field. The field lines gets concentrated inside the material, and the field inside is enhanced. In most cases, this enhancement is slight, being one part in 105. When placed in a non-uniform magnetic field, the bar will tend to move from weak field to strong. Some paramagnetic materials are aluminium, sodium, calcium, oxygen (at STP) and copper chloride. For a paramagnetic material both χ and µr depend not only on the material, but also (in a simple fashion) on the sample temperature. As the field is increased or the temperature is lowered, the magnetisation increases until it reaches the saturation value at which point all the dipoles are perfectly aligned with the field. Reprint 2025-26 Magnetism and Matter 5.5.3 Ferromagnetism Ferromagnetic substances are those which gets strongly magnetised when placed in an external magnetic field. They have strong tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet. The individual atoms (or ions or molecules) in a ferromagnetic material possess a dipole moment as in a paramagnetic material. However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. The explanation of this cooperative effect requires quantum mechanics and is beyond the scope of this textbook. Each domain has a net magnetisation. Typical domain size is 1mm and the domain contains about 1011 atoms. In the first instant, the magnetisation varies randomly from domain to domain and there is no bulk magnetisation. This is shown in Fig. 5.8(a). When we apply an external magnetic field B0, the domains FIGURE 5.8 orient themselves in the direction of B0 and simultaneously the domain (a) Randomly oriented in the direction of B0 grow in size. This existence of domains and oriented domains, their motion in B0 are not speculations. One may observe this under a (b) Aligned domains. microscope after sprinkling a liquid suspension of powdered ferromagnetic substance of samples. This motion of suspension can be observed. Fig. 5.8(b) shows the situation when the domains have aligned and amalgamated to form a single ‘giant’ domain. Thus, in a ferromagnetic material the field lines are highly concentrated. In non-uniform magnetic field, the sample tends to move towards the region of high field. We may wonder as to what happens when the external field is removed. In some ferromagnetic materials the magnetisation persists. Such materials are called hard magnetic materials or hard ferromagnets. Alnico, an alloy of iron, aluminium, nickel, cobalt and copper, is one such material. The naturally occurring lodestone is another. Such materials form permanent magnets to be used among other things as a compass needle. On the other hand, there is a class of ferromagnetic materials in which the magnetisation disappears on removal of the external field. Soft iron is one such material. Appropriately enough, such materials are called soft ferromagnetic materials. There are a number of elements, which are ferromagnetic: iron, cobalt, nickel, gadolinium, etc. The relative magnetic permeability is >1000! The ferromagnetic property depends on temperature. At high enough temperature, a ferromagnet becomes a paramagnet. The domain structure disintegrates with temperature. This disappearance of magnetisation with temperature is gradual. SUMMARY 1. The science of magnetism is old. It has been known since ancient times that magnetic materials tend to point in the north-south direction; like 149 Reprint 2025-26 Physics magnetic poles repel and unlike ones attract; and cutting a bar magnet in two leads to two smaller magnets. Magnetic poles cannot be isolated. 2. When a bar magnet of dipole moment m is placed in a uniform magnetic field B, (a) the force on it is zero, (b) the torque on it is m × B, (c) its potential energy is –m.B, where we choose the zero of energy at the orientation when m is perpendicular to B. 3. Consider a bar magnet of size l and magnetic moment m, at a distance r from its mid-point, where r >>l, the magnetic field B due to this bar is, µ0 m B = 2 π r 3 (along axis) µ0 m = – 3 (along equator) 4 π r 4. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero B B i S 0 all area elements S 5. Consider a material placed in an external magnetic field B0. The magnetic intensity is defined as, B0 H = µ0 The magnetisation M of the material is its dipole moment per unit volume. The magnetic field B in the material is, B = m0 (H + M) 6. For a linear material M = c H. So that B = m H and c is called the magnetic susceptibility of the material. The three quantities, c, the relative magnetic permeability mr, and the magnetic permeability m are related as follows: m = m0 mr mr = 1+ c 7. Magnetic materials are broadly classified as: diamagnetic, paramagnetic, and ferromagnetic. For diamagnetic materials c is negative and small and for paramagnetic materials it is positive and small. Ferromagnetic materials have large c. 8. Substances, which at room temperature, retain their ferromagnetic property for a long period of time are called permanent magnets. Reprint 2025-26 Magnetism and Matter Physical quantity Symbol Nature Dimensions Units Remarks Permeability of µ0 Scalar [MLT–2 A–2] T m A–1 µ0/4π = 10–7 free space Magnetic field, B Vector [MT–2 A–1] T (tesla) 104 G (gauss) = 1 T Magnetic induction, Magnetic flux density Magnetic moment m Vector [L–2 A] A m2 Magnetic flux φB Scalar [ML2T–2 A–1] W (weber) W = T m2 Magnetic momentMagnetisation M Vector [L–1 A] A m–1 Volume Magnetic intensity H Vector [L–1 A] A m–1 B = µ0 (H + M) Magnetic field strength Magnetic χ Scalar - - M = χH susceptibility Relative magnetic µr Scalar - - B = µ0 µr H permeability Magnetic permeability µ Scalar [MLT–2 A–2] T m A–1 µ = µ0 µr N A–2 B = µ H POINTS TO PONDER 1. A satisfactory understanding of magnetic phenomenon in terms of moving charges/currents was arrived at after 1800 AD. But technological exploitation of the directional properties of magnets predates this scientific understanding by two thousand years. Thus, scientific understanding is not a necessary condition for engineering applications. Ideally, science and engineering go hand-in-hand, one leading and assisting the other in tandem. 2. Magnetic monopoles do not exist. If you slice a magnet in half, you get two smaller magnets. On the other hand, isolated positive and negative charges exist. There exists a smallest unit of charge, for example, the electronic charge with value |e| = 1.6 ×10–19 C. All other charges are integral multiples of this smallest unit charge. In other words, charge is quantised. We do not know why magnetic monopoles do not exist or why electric charge is quantised. 3. A consequence of the fact that magnetic monopoles do not exist is that the magnetic field lines are continuous and form closed loops. In contrast, the electrostatic lines of force begin on a positive charge and terminate on the negative charge (or fade out at infinity). 4. A miniscule difference in the value of χ, the magnetic susceptibility, yields radically different behaviour: diamagnetic versus paramagnetic. For diamagnetic materials χ = –10–5 whereas χ = +10–5 for paramagnetic materials. 151 Reprint 2025-26 Physics 5. There exists a perfect diamagnet, namely, a superconductor. This is a metal at very low temperatures. In this case χ = –1, µr = 0, µ = 0. The external magnetic field is totally expelled. Interestingly, this material is also a perfect conductor. However, there exists no classical theory which ties these two properties together. A quantum-mechanical theory by Bardeen, Cooper, and Schrieffer (BCS theory) explains these effects. The BCS theory was proposed in1957 and was eventually recognised by a Nobel Prize in physics in 1970. 6. Diamagnetism is universal. It is present in all materials. But it is weak and hard to detect if the substance is para- or ferromagnetic. 7. We have classified materials as diamagnetic, paramagnetic, and ferromagnetic. However, there exist additional types of magnetic material such as ferrimagnetic, anti-ferromagnetic, spin glass, etc. with properties which are exotic and mysterious. EXERCISES 5.1 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet? 5.2 A short bar magnet of magnetic moment m = 0.32 JT –1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case? 5.3 A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? 5.4 If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field? 5.5 A bar magnet of magnetic moment 1.5 J T –1 lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? 5.6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 –4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. Reprint 2025-26 Magnetism and Matter (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?

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.

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.