Physics---Part-1---Class-12

Magnetism and Matter (b) If field lines were entirely confined between two ends of a straight EXAMPLE 5.7 solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must Geomagnetic field frequently asked questions always be zero. For a toroid, this difficulty is absent because it http://www.ngdc.noaa.gov/geomag/ has no ‘ends’. 185 (c) Gauss’s law of magnetism states that the flux of B through any ∫closed surface is always zero s B.∆s = 0 . If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to ∫Gauss’s law of electrostatics, S B.∆s = µ0qm where qm is the (monopole) magnetic charge enclosed by S .] (d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.) (e) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero. 5.4 THE EARTH’S MAGNETISM Earlier we have referred to the magnetic field of the earth. The strength of the earth’s magnetic field varies from place to place on the earth’s surface; its value being of the order of 10–5 T. What causes the earth to have a magnetic field is not clear. Originally the magnetic field was thought of as arising from a giant bar magnet placed approximately along the axis of rotation of the earth and deep in the interior. However, this simplistic picture is certainly not correct. The magnetic field is now thought to arise due to electrical currents produced by convective motion of metallic fluids (consisting mostly of molten iron and nickel) in the outer core of the earth. This is known as the dynamo effect. The magnetic field lines of the earth resemble that of a (hypothetical) magnetic dipole located at the centre of the earth. The axis of the dipole does not coincide with the axis of rotation of the earth but is presently titled by approximately 11.3° with respect to the later. In this way of looking at it, the magnetic poles are located where the magnetic field lines due to the dipole enter or leave the earth. The location of the north magnetic pole is at a latitude of 79.74° N and a longitude of 71.8° W, a place somewhere in north Canada. The magnetic south pole is at 79.74° S, 108.22° E in the Antarctica. The pole near the geographic north pole of the earth is called the north magnetic pole. Likewise, the pole near the geographic south pole is called 2019-20

Physics FIGURE 5.8 The earth as a giant the south magnetic pole. There is some confusion in the magnetic dipole. nomenclature of the poles. If one looks at the magnetic field lines of the earth (Fig. 5.8), one sees that unlike in the case of a bar magnet, the field lines go into the earth at the north magnetic pole (Nm) and come out from the south magnetic pole (Sm). The convention arose because the magnetic north was the direction to which the north pole of a magnetic needle pointed; the north pole of a magnet was so named as it was the north seeking pole. Thus, in reality, the north magnetic pole behaves like the south pole of a bar magnet inside the earth and vice versa. Example 5.8 The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment. Solution From Eq. (5.7), the equatorial magnetic field is, BE = µ0m 4 πr3 We are given t×ha1t0B6 Em~. 0.4 G = 4 × 10–5 T. For r, we take the radius of the earth 6.4 Hence, EXAMPLE 5.8 m = 4 × 10−5 × (6.4 × 106 )3 =4 × 102 × (6.4 × 106)3 (µ0/4π = 10–7) µ0 / 4π = 1.05 × 1023 A m2 This is close to the value 8 × 1022 A m2 quoted in geomagnetic texts. 5.4.1 Magnetic declination and dip Consider a point on the earth’s surface. At such a point, the direction of the longitude circle determines the geographic north-south direction, the line of longitude towards the north pole being the direction of true north. The vertical plane containing the longitude circle and the axis of rotation of the earth is called the geographic meridian. In a similar way, one can define magnetic meridian of a place as the vertical plane which passes through the imaginary line joining the magnetic north and the south poles. This plane would intersect the surface of the earth in a longitude like circle. A magnetic needle, which is free to swing horizontally, would then lie in the magnetic meridian and the north pole of the needle would point towards the magnetic north pole. Since the line joining the magnetic poles is titled with respect to the geographic axis of the earth, the magnetic meridian at a point makes angle with the geographic meridian. FIGURE 5.9 A magnetic needle This, then, is the angle between the true geographic north and free to move in horizontal plane, the north shown by a compass needle. This angle is called the points toward the magnetic magnetic declination or simply declination (Fig. 5.9). The declination is greater at higher latitudes and smaller north-south near the equator. The declination in India is small, it being 186 direction. 2019-20

Magnetism and Matter 0°41′ E at Delhi and 0°58′ W at Mumbai. Thus, at both these places a magnetic needle shows the true north quite accurately. There is one more quantity of interest. If a magnetic needle is perfectly balanced about a horizontal axis so that it can swing in a plane of the magnetic meridian, the needle would make an angle with the horizontal (Fig. 5.10). This is known as the angle of dip (also known as inclination). Thus, dip is the angle that the total magnetic field BE of the earth makes with the surface of the earth. Figure 5.11 shows the magnetic meridian plane at a point P on the surface of the earth. The plane is a section through the earth. The total magnetic field at P can be resolved into a horizontal component HE and a vertical component ZE. The angle that BE makes with HE is the angle of dip, I. FIGURE 5.10 The circle is a FIGURE 5.11 The earth’s section through the earth containing the magnetic magnetic field, BE, its horizontal and vertical components, HE and meridian. The angle between BE and the horizontal component ZE. Also shown are the declination, D and the HE is the angle of dip. inclination or angle of dip, I. In most of the northern hemisphere, the north pole of the dip needle tilts downwards. Likewise in most of the southern hemisphere, the south pole of the dip needle tilts downwards. To describe the magnetic field of the earth at a point on its surface, we need to specify three quantities, viz., the declination D, the angle of dip or the inclination I and the horizontal component of the earth’s field HE. These are known as the element of the earth’s magnetic field. Representing the verticle component by ZE, we have ZE = BE sinI [5.10(a)] HE = BE cosI [5.10(b)] which gives, tan I = Z E [5.10(c)] 187 HE 2019-20

Physics WHAT HAPPENS TO MY COMPASS NEEDLES AT THE POLES? A compass needle consists of a magnetic needle which floats on a pivotal point. When the compass is held level, it points along the direction of the horizontal component of the earth’s magnetic field at the location. Thus, the compass needle would stay along the magnetic meridian of the place. In some places on the earth there are deposits of magnetic minerals which cause the compass needle to deviate from the magnetic meridian. Knowing the magnetic declination at a place allows us to correct the compass to determine the direction of true north. So what happens if we take our compass to the magnetic pole? At the poles, the magnetic field lines are converging or diverging vertically so that the horizontal component is negligible. If the needle is only capable of moving in a horizontal plane, it can point along any direction, rendering it useless as a direction finder. What one needs in such a case is a dip needle which is a compass pivoted to move in a vertical plane containing the magnetic field of the earth. The needle of the compass then shows the angle which the magnetic field makes with the vertical. At the magnetic poles such a needle will point straight down. Example 5.9 In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60°. What is the magnetic field of the earth at this location? Solution It is given that HE = 0.26 G. From Fig. 5.11, we have cos 600 = H E BE EXAMPLE 5.9 BE = HE cos 600 188 = 0.26 = 0.52 G (1/ 2) 2019-20

Magnetism and Matter EARTH’S MAGNETIC FIELD It must not be assumed that there is a giant bar magnet deep inside the earth which is causing the earth’s magnetic field. Although there are large deposits of iron inside the earth, it is highly unlikely that a large solid block of iron stretches from the magnetic north pole to the magnetic south pole. The earth’s core is very hot and molten, and the ions of iron and nickel are responsible for earth’s magnetism. This hypothesis seems very probable. Moon, which has no molten core, has no magnetic field, Venus has a slower rate of rotation, and a weaker magnetic field, while Jupiter, which has the fastest rotation rate among planets, has a fairly strong magnetic field. However, the precise mode of these circulating currents and the energy needed to sustain them are not very well understood. These are several open questions which form an important area of continuing research. The variation of the earth’s magnetic field with position is also an interesting area of study. Charged particles emitted by the sun flow towards the earth and beyond, in a stream called the solar wind. Their motion is affected by the earth’s magnetic field, and in turn, they affect the pattern of the earth’s magnetic field. The pattern of magnetic field near the poles is quite different from that in other regions of the earth. The variation of earth’s magnetic field with time is no less fascinating. There are short term variations taking place over centuries and long term variations taking place over a period of a million years. In a span of 240 years from 1580 to 1820 AD, over which records are available, the magnetic declination at London has been found to change by 3.5°, suggesting that the magnetic poles inside the earth change position with time. On the scale of a million years, the earth’s magnetic fields has been found to reverse its direction. Basalt contains iron, and basalt is emitted during volcanic activity. The little iron magnets inside it align themselves parallel to the magnetic field at that place as the basalt cools and solidifies. Geological studies of basalt containing such pieces of magnetised region have provided evidence for the change of direction of earth’s magnetic field, several times in the past. 5.5 MAGNETISATION AND MAGNETIC INTENSITY The earth abounds with a bewildering variety of elements and compounds. In addition, we have been synthesising new alloys, compounds and even elements. One would like to classify the magnetic properties of these substances. In the present section, we define and explain certain terms which will help us to carry out this exercise. We have seen that a circulating electron in an atom has a magnetic moment. In a bulk material, these moments add up vectorially and they can give a net magnetic moment which is non-zero. We define magnetisation M of a sample to be equal to its net magnetic moment per unit volume: M = mnet (5.11) V M is a vector with dimensions L–1 A and is measured in a units of A m–1. 189 Consider a long solenoid of n turns per unit length and carrying a current I. The magnetic field in the interior of the solenoid was shown to be given by 2019-20

Physics B0 = µ0 nI (5.12) If the interior of the solenoid is filled with a material with non-zero magnetisation, the field inside the solenoid will be greater than B0. The net B field in the interior of the solenoid may be expressed as B = B0 + Bm (5.13) where Bm is the field contributed by the material core. It turns out that this additional field Bm is proportional to the magnetisation M of the material and is expressed as Bm = µ0M (5.14) where µ0 is the same constant (permittivity of vacuum) that appears in Biot-Savart’s law. It is convenient to introduce another vector field H, called the magnetic intensity, which is defined by H = B – M (5.15) µ0 where H has the same dimensions as M and is measured in units of A m–1. Thus, the total magnetic field B is written as B = µ0 (H + M) (5.16) We repeat our defining procedure. We have partitioned the contribution to the total magnetic field inside the sample into two parts: one, due to external factors such as the current in the solenoid. This is represented by H. The other is due to the specific nature of the magnetic material, namely M. The latter quantity can be influenced by external factors. This influence is mathematically expressed as M= χH (5.17) where χ , a dimensionless quantity, is appropriately called the magnetic susceptibility. It is a measure of how a magnetic material responds to an external field. Table 5.2 lists χ for some elements. It is small and positive for materials, which are called paramagnetic. It is small and negative for materials, which are termed diamagnetic. In the latter case M and H are opposite in direction. From Eqs. (5.16) and (5.17) we obtain, B = µ0(1 + χ)H (5.18) = µ µ H 0 r = µH (5.19) where µr= 1 + χ, is a dimensionless quantity called the relative magnetic permeability of the substance. It is the analog of the dielectric constant in electrostatics. The magnetic permeability of the substance is µ and it has the same dimensions and units as µ0; µ = µ0µr = µ0 (1+χ). The three quantities χ, µ and µ are interrelated and only one of r 190 them is independent. Given one, the other two may be easily determined. 2019-20

Magnetism and Matter TABLE 5.2 MAGNETIC SUSCEPTIBILITY OF SOME ELEMENTS AT 300 K Diamagnetic substance χ Paramagnetic substance χ Bismuth –1.66 × 10–5 Aluminium 2.3 × 10–5 Copper –9.8 × 10–6 Calcium 1.9 × 10–5 Diamond –2.2 × 10–5 Chromium 2.7 × 10–4 Gold –3.6 × 10–5 Lithium 2.1 × 10–5 Lead –1.7 × 10–5 Magnesium 1.2 × 10–5 Mercury –2.9 × 10–5 Niobium 2.6 × 10–5 Nitrogen (STP) –5.0 × 10–9 Oxygen (STP) 2.1 × 10–6 Silver –2.6 × 10–5 Platinum 2.9 × 10–4 Silicon –4.2 × 10–6 Tungsten 6.8 × 10–5 Example 5.10 A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current Im. Solution (a) The field H is dependent of the material of the core, and is H = nI = 1000 × 2.0 = 2 ×103 A/m. (b) The magnetic field B is given by B = µr µ0 H = 400 × 4π ×10–7 (N/A2) × 2 × 103 (A/m) = 1.0 T (c) Magnetisation is given by M = (8(Bµ×r–µµ1000H5H–Aµ)//0mµH0 )/µ0 = = (µr – 1)H = 399 × H EXAMPLE 5.10 ≅ (d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = µr n0 (I + IM). Using I = 2A, B = 1 T, we get IM = 794 A. 5.6 MAGNETIC PROPERTIES OF MATERIALS 191 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 5.3 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. 2019-20

Physics Diamagnetic TABLE 5.3 Ferromagnetic –1 ≤ χ < 0 Paramagnetic χ >> 1 0 ≤ µr < 1 µr >> 1 0< χ< ε µ >> µ0 µ < µ0 1< µr < 1+ ε µ > µ0 FIGURE 5.12 5.6.1 Diamagnetism Behaviour of magnetic field lines Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field. In other words, near a unlike the way a magnet attracts metals like iron, it would repel a (a) diamagnetic, diamagnetic substance. (b) paramagnetic Figure 5.12(a) shows a bar of diamagnetic material placed in an external substance. magnetic field. The field lines are repelled or expelled and the field inside the material is reduced. In most cases, as is evident from Table 5.2, this reduction is slight, being one part in 105. When placed in a non-uniform magnetic field, the bar will tend to move from high to low field. 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. 5.6.2 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 192 attracted to a magnet. 2019-20

Magnetism and Matter The individual atoms (or ions or molecules) of a paramagnetic material Magnetic materials, domain, etc.: possess a permanent magnetic dipole moment of their own. On account http://www.nde-ed.org/EducationResources/CommunityCollege/ of the ceaseless random thermal motion of the atoms, no net magnetisation MagParticle/Physics/MagneticMatls.htm 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 5.12(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, as is evident from Table 5.2, 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. Experimentally, one finds that the magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T , M = C B0 [5.20(a)] T or equivalently, using Eqs. (5.12) and (5.17) χ = C µ0 [5.20(b)] T This is known as Curie’s law, after its discoverer Pieree Curie (1859- 1906). The constant C is called Curie’s constant. Thus, 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 Ms, at which point all the dipoles are perfectly aligned with the field. Beyond this, Curie’s law [Eq. (5.20)] is no longer valid. 5.6.3 Ferromagnetism FIGURE 5.13 (a) Randomly Ferromagnetic substances are those which gets strongly magnetised when oriented domains, placed in an external magnetic field. They have strong tendency to move (b) Aligned domains. from a region of weak magnetic field to strong magnetic field, i.e., they get strongly attracted to a magnet. 193 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. 5.13(a). When we apply an external magnetic field B0, the domains orient themselves in the direction of B0 and simultaneously the domain oriented in the direction of B0 grow in size. This existence of domains and their motion in B0 are not speculations. One may observe this under a microscope after sprinkling a liquid suspension of powdered 2019-20

Physics Hysterisis in magnetic materials: ferromagnetic substance of samples. This motion of suspension can be http://hyperphysics.phy-astr.gsu.edu/hbase/solids/hyst.html observed. Figure 5.12(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. It is a phase transition reminding us of the melting of a solid crystal. The temperature of transition from ferromagnetic to paramagnetism is called the Curie temperature Tc. Table 5.4 lists the Curie temperature of certain ferromagnets. The susceptibility above the Curie temperature, i.e., in the paramagnetic phase is described by, χ = T C (T > Tc ) (5.21) − Tc TABLE 5.4 CURIE TEMPERATURE T OF SOME C FERROMAGNETIC MATERIALS Material Tc (K) Cobalt 1394 Iron 1043 Fe2O3 Nickel 893 631 Gadolinium 317 194 EXAMPLE 5.11 Example 5.11 A domain in ferromagnetic iron is in the form of a cube of side length 1µm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27×10–24 A m2. 2019-20

Magnetism and Matter Solution The volume of the cubic domain is V = (10–6 m)3 = 10–18 m3 = 10–12 cm3 Its mass is volume × density = 7.9 g cm–3 × 10–12 cm3= 7.9 × 10–12 g It is given that Avagadro number (6.023 × 1023) of iron atoms have a mass of 55 g. Hence, the number of atoms in the domain is N = 7.9 × 10−12 × 6.023 × 1023 55 = 8.65 × 1010 atoms The maximum possible dipole moment mmax is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus, mmax = (8.65 × 1010) × (9.27 × 10–24) = 8.0 × 10–13 A m2 EXAMPLE 5.11 The consequent magnetisation is Mmax = mmax/Domain volume = 8.0 × 10–13 Am2/10–18 m3 = 8.0 × 105 Am–1 The relation between B and H in ferromagnetic materials is complex. It is often not linear and it depends on the magnetic history of the sample. Figure 5.14 depicts the behaviour of the material as we take it through one cycle of magnetisation. Let the material be unmagnetised initially. We place it in a solenoid and increase the current through the solenoid. The magnetic field B in the material rises and saturates as depicted in the curve Oa. This behaviour represents the alignment and merger of domains until no further enhancement is possible. It is pointless to increase the current (and hence the magnetic intensity H ) beyond this. Next, we decrease H and reduce it to zero. At H = 0, B ≠ 0. This is represented by the curve ab. The value of B at H = 0 is called retentivity or remanence. In Fig. 5.14, BR ~ 1.2 T, where the subscript R denotes retentivity. The domains are not completely randomised even though the external driving field has been removed. Next, the current in the solenoid is reversed and slowly increased. Certain domains are flipped until the net field inside stands nullified. This is represented by the curve bc. The value of FIGURE 5.14 The magnetic H at c is called coercivity. In Fig. 5.14 Hc ~ –90 A m–1. As hysteresis loop is the B-H curve for the reversed current is increased in magnitude, we once again obtain saturation. The curve cd depicts this. The ferromagnetic materials. saturated magnetic field Bs ~ 1.5 T. Next, the current is reduced (curve de) and reversed (curve ea). The cycle repeats itself. Note that the curve Oa does not retrace itself as H is reduced. For a given value of H, B is not unique but depends on previous history of the sample. This phenomenon is called hysterisis. The word hysterisis means lagging behind (and not ‘history’). 5.7 PERMANENT MAGNETS AND ELECTROMAGNETS Substances which at room temperature retain their ferromagnetic property 195 for a long period of time are called permanent magnets. Permanent 2019-20

Physics magnets can be made in a variety of ways. One can hold an iron rod in the north-south direction and hammer it repeatedly. The method is illustrated in Fig. 5.15. The illustration is from a 400 year old book to emphasise that the making of permanent magnets is an old art. One can also hold a steel rod and stroke it with one end of a bar magnet a large number of times, always in the same sense to make a permanent magnet. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current. The FIGURE 5.15 A blacksmith magnetic field of the solenoid magnetises the rod. forging a permanent magnet by The hysteresis curve (Fig. 5.14) allows us to select suitable striking a red-hot rod of iron materials for permanent magnets. The material should have kept in the north-south high retentivity so that the magnet is strong and high coercivity so that the magnetisation is not erased by stray magnetic fields, direction with a hammer. The sketch is recreated from an temperature fluctuations or minor mechanical damage. illustration in De Magnete, a Further, the material should have a high permeability. Steel is work published in 1600 and one-favoured choice. It has a slightly smaller retentivity than authored by William Gilbert, soft iron but this is outweighed by the much smaller coercivity the court physician to Queen of soft iron. Other suitable materials for permanent magnets are alnico, cobalt steel and ticonal. Elizabeth of England. Core of electromagnets are made of ferromagnetic materials which have high permeability and low retentivity. Soft iron is a suitable material for electromagnets. On placing a soft iron rod in a solenoid and passing a current, we increase the magnetism of the solenoid by a thousand fold. When we switch off the solenoid current, the magnetism is effectively switched off since the soft iron core has a low retentivity. The arrangement is shown in Fig. 5.16. India’s Magnetic Field: http://www.iigm.res.in 196 FIGURE 5.16 A soft iron core in solenoid acts as an electromagnet. In certain applications, the material goes through an ac cycle of magnetisation for a long period. This is the case in transformer cores and telephone diaphragms. The hysteresis curve of such materials must be narrow. The energy dissipated and the heating will consequently be small. The material must have a high resistivity to lower eddy current losses. You will study about eddy currents in Chapter 6. Electromagnets are used in electric bells, loudspeakers and telephone diaphragms. Giant electromagnets are used in cranes to lift machinery, and bulk quantities of iron and steel. 2019-20

Magnetism and Matter MAPPING INDIA’S MAGNETIC FIELD Because of its practical application in prospecting, communication, and navigation, the magnetic field of the earth is mapped by most nations with an accuracy comparable to geographical mapping. In India over a dozen observatories exist, extending from Trivandrum (now Thrivuvananthapuram) in the south to Gulmarg in the north. These observatories work under the aegis of the Indian Institute of Geomagnetism (IIG), in Colaba, Mumbai. The IIG grew out of the Colaba and Alibag observatories and was formally established in 1971. The IIG monitors (via its nation-wide observatories), the geomagnetic fields and fluctuations on land, and under the ocean and in space. Its services are used by the Oil and Natural Gas Corporation Ltd. (ONGC), the National Institute of Oceanography (NIO) and the Indian Space Research Organisation (ISRO). It is a part of the world-wide network which ceaselessly updates the geomagnetic data. Now India has a permanent station called Gangotri. 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 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, B = µ0m (along axis) 2πr3 = – µ0 m (along equator) 4πr3 4. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero φB = ∑ Bi∆S = 0 197 all area elements ∆S 5. The earth’s magnetic field resembles that of a (hypothetical) magnetic dipole located at the centre of the earth. The pole near the geographic north pole of the earth is called the north magnetic pole. Similarly, the pole near the geographic south pole is called the south magnetic pole. This dipole is aligned making a small angle with the rotation axis of the earth. The magnitude of the field on the earth’s surface ≈ 4 × 10–5 T. 2019-20

Physics 6. Three quantities are needed to specify the magnetic field of the earth on its surface – the horizontal component, the magnetic declination, and the magnetic dip. These are known as the elements of the earth’s magnetic field. 7. Consider a material placed in an external magnetic field B0. The magnetic intensity is defined as, H = B0 µ0 The magnetisation M of the material is its dipole moment per unit volume. The magnetic field B in the material is, B = µ0 (H + M) 8. For a linear material M = χ H. So that B = µ H and χ is called the magnetic susceptibility of the material. The three quantities, χ, the relative magnetic permeability µr, and the magnetic permeability µ are related as follows: µ = µ µ 0 r µr = 1+ χ 9. Magnetic materials are broadly classified as: diamagnetic, paramagnetic, and ferromagnetic. For diamagnetic materials χ is negative and small and for paramagnetic materials it is positive and small. Ferromagnetic materials have large χ and are characterised by non-linear relation between B and H. They show the property of hysteresis. 10. Substances, which at room temperature, retain their ferromagnetic property for a long period of time are called permanent magnets. Physical quantity Symbol Nature Dimensions Units Remarks [MLT–2 A–2] Permeability of µ0 Scalar [MT–2 A–1] T m A–1 µ0/4π = 10–7 free space B Vector T (tesla) 104 G (gauss) = 1 T [L–2 A] Magnetic field, m Vector [ML2T–2 A–1] A m2 Magnetic induction, φB Scalar [L–1 A] W (weber) W = T m2 Magnetic flux density [L–1 A] Magnetic moment - - Magnetic flux [MLT–2 A–2] Magnetisation M Vector A m–1 Magnetic moment H Vector A m–1 Magnetic intensity Volume Magnetic field B = µ0 (H + M) strength χ Scalar - M = χH Magnetic susceptibility µ Scalar - B = µ µ H r 0 r Relative magnetic T m A–1 permeability µ Scalar N A–2 µ = µ µ 0 r Magnetic permeability B=µH 198 2019-20

Magnetism and Matter 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. The earth’s magnetic field is not due to a huge bar magnet inside it. The earth’s core is hot and molten. Perhaps convective currents in this core are responsible for the earth’s magnetic field. As to what ‘dynamo’ effect sustains this current, and why the earth’s field reverses polarity every million years or so, we do not know. 5. 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. 6. 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. 7. The phenomenon of magnetic hysteresis is reminiscent of similar behaviour concerning the elastic properties of materials. Strain may not be proportional to stress; here H and B (or M) are not linearly related. The stress-strain curve exhibits hysteresis and area enclosed by it represents the energy dissipated per unit volume. A similar interpretation can be given to the B-H magnetic hysteresis curve. 8. Diamagnetism is universal. It is present in all materials. But it is weak and hard to detect if the substance is para- or ferromagnetic. 9. 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. 199 2019-20

200 Physics EXERCISES 5.1 Answer the following questions regarding earth’s magnetism: (a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field. (b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain? (c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground? (d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole? (e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T–1 located at its centre. Check the order of magnitude of this number in some way. (f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all? 5.2 Answer the following questions: (a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably? (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why? (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents? (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past? (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion? (f ) Interstellar space has an extremely weak magnetic field of the order of 10–12 T. Can such a weak field be of any significant consequence? Explain. [Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.] 5.3 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? 5.4 A short bar magnet of magnetic moment m = 0.32 J T –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? 2019-20

Magnetism and Matter 5.5 A closely wound solenoid of 800 turns and area of cross section 201 5.6 2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which 5.7 the solenoid acts like a bar magnet. What is its associated magnetic 5.8 moment? 5.9 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, 5.10 what is the magnitude of torque on the solenoid when its axis makes 5.11 an angle of 30° with the direction of applied field? 5.12 A bar magnet of magnetic moment 1.5 J T –1 lies aligned with the 5.13 direction of a uniform magnetic field of 0.22 T. 5.14 (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)? 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. (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? A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10–2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s–1. What is the moment of inertia of the coil about its axis of rotation? A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place. At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location. A short bar magnet has a magnetic moment of 0.48 J T –1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.) If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located? 2019-20

Physics 5.15 A short bar magnet of magnetic movement 5.25 × 10–2 J T –1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved. ADDITIONAL EXERCISES 202 5.16 Answer the following questions: (a) Why does a paramagnetic sample display greater magnetisation 5.17 5.18 (for the same magnetising field) when cooled? 5.19 (b) Why is diamagnetism, in contrast, almost independent of temperature? (c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty? (d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields? (e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why? (f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet? Answer the following questions: (a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet. (b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy? (c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement. (d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer? (e) A certain region of space is to be shielded from magnetic fields. Suggest a method. A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.) A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The 2019-20

Magnetism and Matter 5.20 earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant 5.21 magnetic fields at points 4.0 cm below the cable? 5.22 A compass needle free to turn in a horizontal plane is placed at the 5.23 centre of circular coil of 30 turns and radius 12 cm. The coil is in a 5.24 vertical plane making an angle of 45° with the magnetic meridian. 5.25 When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the earth’s magnetic field at the location. (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field? A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10–31 kg). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.] A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10–23 J T–1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law) A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A? The magnetic moment vectors µ and µ associated with the intrinsic s l spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: µs = –(e/m) S, µ = –(e/2m)l l Which of these relations is in accordance with the result expected classically ? Outline the derivation of the classical result. 203 2019-20

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 204 use of a new born baby?” The phenomenon of electromagnetic induction 2019-20

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 JOSEPH HENRY (1797 – 1878) HENRY The discovery and understanding of electromagnetic Josheph Henry [1797 – induction are based on a long series of experiments carried 1878] American experimental out by Faraday and Henry. We shall now describe some physicist, professor at of these experiments. Princeton University and first director of the Smithsonian Experiment 6.1 Institution. He made important improvements in electro- Figure 6.1 shows a coil C1* connected to a galvanometer magnets by winding coils of G. When the North-pole of a bar magnet is pushed 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 the opposite direction, which indicates reversal of the another. 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 pushed towards the coil, the pointer in In Fig. 6.2 the bar magnet is replaced by a second coil 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 205 conducting material and are prepared using wires which are coated with insulating material. 2019-20

Physics FIGURE 6.2 Current is moved towards the coil C1, the galvanometer shows a induced in coil C1 due to motion deflection. This indicates that electric current is induced in of the current carrying coil C2. 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 relative motion is not an absolute requirement. Figure 6.3 shows two coils C1 and C2 held stationary. Coil C1 is connected to galvanometer G while the second coil C2 is connected to a battery through a tapping key K. Interactive animation on Faraday’s experiments and Lenz’s law: FIGURE 6.3 Experimental set-up for Experiment 6.3. http://micro.magnet.fsu.edu/electromagnet/java/faraday2/ 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 206 familiar with the notion of magnetic flux, Φ B. Magnetic flux is defined in the same way as electric flux is defined in Chapter 1. Magnetic flux through 2019-20

Electromagnetic Induction a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can be written as Φ B = B . A = BA cos θ (6.1) where θ 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 + ... = Bi .dAi (6.2) B 1 12 2 all where ‘all’ stands for summation over all the area elements dAi FIGURE 6.4 A plane of 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 FIGURE 6.5 Magnetic field Bi Experiment 6.1 and moving a current-carrying coil C2 towards at the ith area element. dAi or away from coil C1 in Experiment 6.2, change the magnetic flux associated with coil C1. The change in magnetic flux induces represents area vector of the emf in coil C1. It was this induced emf which caused electric ith area element. current to flow in coil C1 and through the galvanometer. A plausible explanation for the observations of Experiment 6.3 is as follows: When the tapping key K is pressed, the current in 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 207 can be damaged due to the induced emfs (and the resulting currents) when the electromagnet is turned on or off. 2019-20

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 ε = – dΦB (6.3) dt MICHAEL FARADAY (1791–1867) Michael Faraday [1791– The negative sign indicates the direction of ε and hence 1867] Faraday made the direction of current in a closed loop. This will be numerous contributions to discussed in detail in the next section. science, viz., the discovery of electromagnetic In the case of a closely wound coil of N turns, change induction, the laws of of flux associated with each turn, is the same. Therefore, electrolysis, benzene, and the expression for the total induced emf is given by the fact that the plane of polarisation is rotated in an ε = –N dΦB (6.4) electric field. He is also dt credited with the invention of the electric motor, the The induced emf can be increased by increasing the electric generator and the number of turns N of a closed coil. transformer. He is widely regarded as the greatest From Eqs. (6.1) and (6.2), we see that the flux can be experimental scientist of varied by changing any one or more of the terms B, A and the nineteenth century. θ. In Experiments 6.1 and 6.2 in Section 6.2, the flux is changed by varying B. The flux can also be altered by changing the shape of a coil (that is, by shrinking it or stretching it) in a magnetic field, or rotating a coil in a magnetic field such that the angle θ between B and A changes. In these cases too, an emf is induced in the respective coils. EXAMPLE 6.2 EXAMPLE 6.1 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 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. 208 Example 6.2 A square loop of side 10 cm and resistance 0.5 Ω 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. 2019-20

Electromagnetic Induction Solution The angle θ made by the area vector of the coil with the magnetic field is 45°. From Eq. (6.1), the initial magnetic flux is Φ = BA cos θ = 0.1 × 10–2 Wb 2 Final flux, Φmin = 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 ε= ∆ΦB = (Φ – 0) = 10–3 = 1.0 mV ∆t 2 × 0.7 ∆t And the magnitude of the current is I = ε = 10–3 V = 2 mA EXAMPLE 6.2 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 Ω 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, ΦB (initial) = BA cos θ = 3.0 × 10–5 × (π ×10–2) × cos 0° = 3π × 10–7 Wb Final flux after the rotation, ΦB (final) = 3.0 × 10–5 × (π ×10–2) × cos 180° = –3π × 10–7 Wb Therefore, estimated value of the induced emf is, ε = N ∆Φ ∆t = 500 × (6π × 10–7)/0.25 EXAMPLE 6.3 209 = 3.8 × 10–3 V I = ε/R = 1.9 × 10–3 A Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant. 2019-20

Physics FIGURE 6.6 6.5 LENZ’S LAW AND CONSERVATION OF ENERGY Illustration of In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced Lenz’s law. 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: 210 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. Now consider the correct case shown in Fig. 6.6(a). In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, a person has to do work in moving the magnet. Where does the energy spent by the person go? This energy is dissipated by Joule heating produced by the induced current. 2019-20

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 EXAMPLE 6.4 Solution EXAMPLE 6.5 211 (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. 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. (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. 2019-20

Physics FIGURE 6.8 (d) Predict the polarity of the capacitor in the situation described by Fig. 6.9. EXAMPLE 6.5 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. (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 ΦB enclosed by the loop PQRS will be ΦB = Blx FIGURE 6.10 The arm PQ is moved to the left Since x is changing with time, the rate of change side, thus decreasing the area of the of flux ΦB will induce an emf given by: rectangular loop. This movement ε = – dΦB = – d (Blx ) dt dt induces a current I as shown. = –Bl dx = Blv (6.5) dt 212 2019-20

Electromagnetic Induction where we have used dx/dt = –v which is the speed of the conductor PQ. Interactive animation on motional emf: The induced emf Blv is called motional emf. Thus, we are able to produce http://ngsir.netfirms.com/englishhtm/Induction.htm induced emf by moving a conductor instead of varying the magnetic field, http://webphysics.davidson.edu/physlet_resources/bu_semester2/index.html 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, ε=W 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 EXAMPLE 6.6 213 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 axis is present everywhere. What is the emf between the centre and the metallic ring? 2019-20

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 dε = Bv dr B ωr dr ∫ ∫ ∫ε = = = 2 00 Note that we have used v = ω r. This gives ε = 1 × 1.0 × 2π × 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 θ 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 π R2 × θ = 1 R 2θ 2π 2 where R is the radius of the circle. Hence, the induced emf is EXAMPLE 6.6 ε =B × d 1 R 2θ  = 1 BR2 dθ = BωR 2 dt  2  2 dt 2 [Note: dθ =ω = 2πν ] dt 214 This expression is identical to the expression obtained by Method I and we get the same value of ε. 2019-20

Electromagnetic Induction Example 6.7 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 The number of spokes is immaterial because the emf’s across the spokes are in parallel. 6.7 ENERGY CONSIDERATION: A QUANTITATIVE STUDY In Section 6.5, we discussed qualitatively that Lenz’s law is consistent with the law of conservation of energy. Now we shall explore this aspect further with a concrete example. Let r be the resistance of movable arm PQ of the rectangular conductor shown in Fig. 6.10. We assume that the remaining arms QR, RS and SP have negligible resistances compared to r. Thus, the overall resistance of the rectangular loop is r and this does not change as PQ is moved. The current I in the loop is, I=ε r = Blv (6.7) r On account of the presence of the magnetic field, there will be a force on the arm PQ. This force I (l × B), is directed outwards in the direction opposite to the velocity of the rod. The magnitude of this force is, F = I l B = B2l 2v r where we have used Eq. (6.7). Note that this force arises due to drift velocity of charges (responsible for current) along the rod and the consequent Lorentz force acting on them. Alternatively, the arm PQ is being pushed with a constant speed v, the power required to do this is, P =Fv = B2l 2v2 (6.8) r The agent that does this work is mechanical. Where does this mechanical energy go? The answer is: it is dissipated as Joule heat, and is given by PJ = I 2r =  Blv 2 r = B2l 2v2 r r 215 which is identical to Eq. (6.8). 2019-20

Physics Thus, mechanical energy which was needed to move the arm PQ is converted into electrical energy (the induced emf) and then to thermal energy. There is an interesting relationship between the charge flow through the circuit and the change in the magnetic flux. From Faraday’s law, we have learnt that the magnitude of the induced emf is, ε = ∆ΦB ∆t However, ε = Ir = ∆Q r ∆t Thus, ∆Q = ∆ΦB r Example 6.8 Refer to Fig. 6.12(a). The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance. (a) FIGURE 6.12 Solution Let us first consider the forward motion from x = 0 to x = 2b The flux ΦB linked with the circuit SPQR is ΦB = B l x 0≤x <b EXAMPLE 6.8 = B l b b ≤ x < 2b The induced emf is, ε = − dΦB dt = −Bl v 0 ≤ x < b 216 = 0 b ≤ x < 2b 2019-20

Electromagnetic Induction When the induced emf is non-zero, the current I is (in magnitude) I = Bl v r (b) FIGURE 6.12 The force required to keep the arm PQ in constant motion is I l B. Its direction is to the left. In magnitude F = B2l 2v 0≤x <b r b ≤ x < 2b =0 The Joule heating loss is PJ = I 2r = B2l 2v2 0≤x <b EXAMPLE 6.8 r b ≤ x < 2b =0 One obtains similar expressions for the inward motion from x = 2b to 217 x = 0. One can appreciate the whole process by examining the sketch of various quantities displayed in Fig. 6.12(b). 2019-20

Physics 6.8 EDDY CURRENTS So far we have studied the electric currents induced in well defined paths in conductors like circular loops. Even when bulk pieces of conductors are subjected to changing magnetic flux, induced currents are produced in them. However, their flow patterns resemble swirling eddies in water. This effect was discovered by physicist Foucault (1819-1868) and these currents are called eddy currents. Consider the apparatus shown in Fig. 6.13. A copper plate is allowed to swing like a simple pendulum between the pole pieces of a strong magnet. It is found that the motion is damped and in a little while the plate comes to a halt in the magnetic field. We can explain this phenomenon on the basis of electromagnetic induction. Magnetic flux associated with the plate keeps on changing as the plate moves in and out of the region between magnetic poles. The flux change induces eddy currents in the plate. Directions of eddy currents are opposite when the plate swings into the region between the poles and when it swings out of the region. If rectangular slots are made in the copper plate as shown FIGURE 6.13 Eddy currents are in Fig. 6.14, area available to the flow of eddy currents is less. generated in the copper plate, Thus, the pendulum plate with holes or slots reduces while entering electromagnetic damping and the plate swings more freely. and leaving the region of Note that magnetic moments of the induced currents (which oppose the motion) depend upon the area enclosed by the magnetic field. currents (recall equation m = I A in Chapter 4). This fact is helpful in reducing eddy currents in the metallic cores of transformers, electric motors and other such devices in which a coil is to be wound over metallic core. Eddy currents are undesirable since they heat up the core and dissipate electrical energy in the form of heat. Eddy currents are minimised by using laminations of metal to make a metal core. The laminations are separated by an insulating material like lacquer. The plane of the laminations must be arranged parallel to the magnetic field, so that they cut across the eddy current paths. This arrangement reduces the strength of the eddy currents. Since the dissipation of electrical energy into heat depends on the square of the strength of electric current, heat loss is substantially reduced. FIGURE 6.14 Cutting slots Eddy currents are used to advantage in certain applications like: in the copper plate reduces the effect of eddy currents. (i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. When the 218 electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth. (ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. 2019-20

Electromagnetic Induction (iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it. (iv) Electric power meters: The shiny metal disc in the electric power meter (analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil. You can observe the rotating shiny disc in the power meter of your house. ELECTROMAGNETIC DAMPING Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and PVC, respectively. Fix them vertically with clamps on retort stands. Take a small cylinderical magnet having diameter slightly smaller than the inner diameter of the pipes and drop it through each pipe in such a way that the magnet does not touch the sides of the pipes during its fall. You will observe that the magnet dropped through the PVC pipe takes the same time to come out of the pipe as it would take when dropped through the same height without the pipe. Note the time it takes to come out of the pipe in each case. You will see that the magnet takes much longer time in the case of aluminium pipe. Why is it so? It is due to the eddy currents that are generated in the aluminium pipe which oppose the change in magnetic flux, i.e., the motion of the magnet. The retarding force due to the eddy currents inhibits the motion of the magnet. Such phenomena are referred to as electromagnetic damping. Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas aluminium is a conductor. 6.9 INDUCTANCE 219 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 2019-20

Physics 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. 6.9.1 Mutual inductance Consider Fig. 6.15 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. 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 = M12I 2 (6.9) 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, ( ) ( )( )N1Φ1 = n1l πr12 µ0n2I2 = µ0n1n2πr12l I2 (6.10) where n1l is the total number of turns in solenoid S1. Thus, from Eq. (6.9) and Eq. (6.10), M12 = µ0n1n2πr12l (6.11) 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 (6.12) FIGURE 6.15 Two long co-axial M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1. solenoids of same The flux due to the current I1 in S1 can be assumed to 220 length l. be confined solely inside S1 since the solenoids are very long. Thus, flux linkage with solenoid S2 is ( ) ( )( )N2Φ2 = n2l πr12 µ0n1I1 2019-20

Electromagnetic Induction where n2l is the total number of turns of S2. From Eq. (6.12), (6.13) M21 = µ0n1n2πr 21l Using Eq. (6.11) and Eq. (6.12), we get M12 = M21= M (say) (6.14) 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 equality M12=M21 is very useful in such situations. 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 r12 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 6.9 Two concentric circular coils, one of small radius r1 and the other of large radius r2, such that r1 << r2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Solution Let a current I2 flow through the outer circular coil. The field at the centre of the coil is B2 = µ0I2 / 2r2. Since the other co-axially placed coil has a very small radius, B2 may be considered constant over its cross-sectional area. Hence, Φ1 = πr12B2 = µ0 πr12 I2 2r2 = M12 I2 Thus, M12 = µ0 πr12 2r2 From Eq. (6.14) M12 = M 21 = µ0 π r12 EXAMPLE 6.9 2r2 Note that we calculated M12 from an approximate value of Φ1, assuming the magnetic field B2 to be uniform over the area π r12. However, we can accept this value because r1 << r2. 221 2019-20

Physics Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil C1 wherever there was any change in current through coil C2. Let Φ1 be the flux through coil C1 (say of N1 turns) when current in coil C2 is I2. Then, from Eq. (6.9), we have N1Φ1 = MI2 For currents varrying with time, d (N1Φ1 ) = d (MI2 ) dt dt Since induced emf in coil C1 is given by ε1 = – d (N1Φ1) dt We get, ε1 = –M dI 2 dt It shows that varying current in a coil can induce emf in a neighbouring coil. The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils. 6.9.2 Self-inductance In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as NΦB ∝ I NΦB = L I (6.15) where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.15), the induced emf is given by ε = – d (NΦB ) dt ε = –L dI (6.16) dt Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross- sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ0 n I (neglecting edge effects, as before). The total flux linked with the solenoid is 222 NΦB = (nl )(µ0n I )(A) 2019-20

Electromagnetic Induction = µ0n2Al I where nl is the total number of turns. Thus, the self-inductance is, L = ΝΦΒ I = µ0n 2 Al (6.17) If we fill the inside of the solenoid with a material of relative permeability µr (for example soft iron, which has a high value of relative permeability), then, L = µr µ0 n 2 Al (6.18) The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε ) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is dW = ε I dt If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.16), dW = L I dI dt dt Total amount of work done in establishing the current I is I W = ∫ dW = ∫ L I dI 0 Thus, the energy required to build up the current I is, W = 1 LI 2 (6.19) 2 This expression reminds us of mv 2/2 for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit). Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. Equation (6.9) would be modified into N1 Φ1 = M11 I1 + M12 I 2 where M11 represents inductance due to the same coil. Therefore, using Faraday’s law, ε1 = −M11 dI1 − M12 dI2 223 dt dt 2019-20

Physics M11 is the self-inductance and is written as L1. Therefore, ε1 = −L1 dI1 − M12 dI 2 dt dt Example 6.10 (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? Solution (a) From Eq. (6.19), the magnetic energy is Interactive animation on ac generator: UB = 1 LI 2 http://micro.magnet.fsu.edu/electromag/java/generator/ac.html 2 EXAMPLE 6.10 = 1 L  B 2 (since B = µ0nI,for a solenoid) 2  µ0n  = 1 ( µ0n 2 Al )  B 2 [from Eq. (6.17)] 2  µ0n  = 1 B2 Al 2µ0 (b) The magnetic energy per unit volume is, uB = UB (where V is volume that contains flux) V = UB Al = B2 (6.20) 2µ0 We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor (refer to Chapter 2, Eq. 2.77), uΕ = 1 ε0E 2 (2.77) 2 In both the cases energy is proportional to the square of the field strength. Equations (6.20) and (2.77) have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist. 224 6.10 AC GENERATOR The phenomenon of electromagnetic induction has been technologically exploited in many ways. An exceptionally important application is the generation of alternating currents (ac). The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine. In this section, we shall describe the basic principles behind this machine. The Yugoslav inventor Nicola Tesla is credited with the development of the machine. As was pointed out in Section 6.3, one method to induce an emf 2019-20

Electromagnetic Induction or current in a loop is through a change in the loop’s orientation or a change in its effective area. As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos θ, where θ is the angle between A and B. This method of producing a flux change is the principle of operation of a simple ac generator. An ac generator converts mechanical energy into electrical energy. The basic elements of an ac generator are shown in Fig. 6.16. It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. The ends of the coil are connected to an external circuit by means FIGURE 6.16 AC Generator of slip rings and brushes. When the coil is rotated with a constant angular speed ω, the angle θ between the magnetic field vector B and the area vector A of the coil at any instant t is θ = ωt (assuming θ = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. (6.1), the flux at any time t is ΦB = BA cos θ = BA cos ωt From Faraday’s law, the induced emf for the rotating coil of N turns is then, ε = – N dΦB = – NBA d (cos ω t ) dt dt Thus, the instantaneous value of the emf is ε = NBA ω sin ωt (6.21) where NBAω is the maximum value of the emf, which occurs when sin ωt = ±1. If we denote NBAω as ε0, then ε = ε0 sin ωt (6.22) Since the value of the sine fuction varies between +1 and –1, the sign, or polarity of the emf changes with time. Note from Fig. 6.17 that the emf has its extremum value when θ = 90° or θ = 270°, as the change of flux is greatest at these points. The direction of the current changes periodically and therefore the current is called alternating current (ac). Since ω = 2πν, Eq (6.22) can be written as ε = ε0sin 2π ν t (6.23) where ν is the frequency of revolution of the generator’s coil. Note that Eq. (6.22) and (6.23) give the instantaneous value of the emf and ε varies between +ε0 and –ε0 periodically. We shall learn how to 225 determine the time-averaged value for the alternating voltage and current in the next chapter. 2019-20

Physics FIGURE 6.17 An alternating emf is generated by a loop of wire rotating in a magnetic field. In commercial generators, the mechanical energy required for rotation of the armature is provided by water falling from a height, for example, from dams. These are called hydro-electric generators. Alternatively, water is heated to produce steam using coal or other sources. The steam at high pressure produces the rotation of the armature. These are called thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear power generators. Modern day generators produce electric power as high as 500 MW, i.e., one can light up 5 million 100 W bulbs! In most generators, the coils are held stationary and it is the electromagnets which are rotated. The frequency of rotation is 50 Hz in India. In certain countries such as USA, it is 60 Hz. 226 EXAMPLE 6.11 Example 6.11 Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Solution Here ν = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing Eq. (6.21) ε0 = NBA (2 π ν) = 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5 = 0.314 V The maximum voltage is 0.314 V. We urge you to explore such alternative possibilities for power generation. 2019-20

Electromagnetic Induction MIGRATION OF BIRDS The migratory pattern of birds is one of the mysteries in the field of biology, and indeed all of science. For example, every winter birds from Siberia fly unerringly to water spots in the Indian subcontinent. There has been a suggestion that electromagnetic induction may provide a clue to these migratory patterns. The earth’s magnetic field has existed throughout evolutionary history. It would be of great benefit to migratory birds to use this field to determine the direction. As far as we know birds contain no ferromagnetic material. So electromagnetic induction seems to be the only reasonable mechanism to determine direction. Consider the optimal case where the magnetic field B, the velocity of the bird v, and two relevant points of its anatomy separated by a distance l, all three are mutually perpendicular. From the formula for motional emf, Eq. (6.5), ε = Blv Taking B = 4 × 10–5 T, l = 2 cm wide, and v = 10 m/s, we obtain ε = 4 × 10–5 × 2 × 10–2 × 10 V = 8 × 10–6 V = 8 µV This extremely small potential difference suggests that our hypothesis is of doubtful validity. Certain kinds of fish are able to detect small potential differences. However, in these fish, special cells have been identified which detect small voltage differences. In birds no such cells have been identified. Thus, the migration patterns of birds continues to remain a mystery. SUMMARY 1. The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as, ΦB = B.A = BA cos θ where θ is the angle between B and A. 2. Faraday’s laws of induction imply that the emf induced in a coil of N turns is directly related to the rate of change of flux through it, ε = −N dΦB dt Here ΦΒ is the flux linked with one turn of the coil. If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit. 3. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. The negative sign in the expression for Faraday’s law indicates this fact. 4. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf ) across its ends is ε = Bl v 5. Changing magnetic fields can set up current loops in nearby metal 227 (any conductor) bodies. They dissipate electrical energy as heat. Such currents are eddy currents. 6. Inductance is the ratio of the flux-linkage to current. It is equal to NΦ/I. 2019-20

Physics 7. A changing current in a coil (coil 2) can induce an emf in a nearby coil (coil 1). This relation is given by, ε1 = −M12 dI 2 dt The quantity M12 is called mutual inductance of coil 1 with respect to coil 2. One can similarly define M21. There exists a general equality, M12 = M21 8. When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by, ε = −L dI dt L is the self-inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. 9. The self-inductance of a long solenoid, the core of which consists of a magnetic material of permeability µr, is given by L = µr µ0 n2 A l where A is the area of cross-section of the solenoid, l its length and n the number of turns per unit length. 10. In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of N turn and area A is rotated at ν revolutions per second in a uniform magnetic field B, then the motional emf produced is ε = NBA ( 2πν) sin (2πνt) where we have assumed that at time t = 0 s, the coil is perpendicular to the field. Quantity Symbol Units Dimensions Equations Magnetic Flux ΦB Wb (weber) [M L2 T –2 A–1] ΦB = B i A EMF ε V (volt) [M L2 T –3 A–1] Mutual Inductance [M L2 T –2 A–2] ε = − d(NΦB )/ dt M H (henry) [M L2 T –2 A–2] H (henry) ε = −M12 (dI2 /dt ) 1 Self Inductance L ε = −L (dI /dt ) 228 POINTS TO PONDER 1. Electricity and magnetism are intimately related. In the early part of the nineteenth century, the experiments of Oersted, Ampere and others established that moving charges (currents) produce a magnetic field. Somewhat later, around 1830, the experiments of Faraday and Henry demonstrated that a moving magnet can induce electric current. 2. In a closed circuit, electric currents are induced so as to oppose the changing magnetic flux. It is as per the law of conservation of energy. However, in case of an open circuit, an emf is induced across its ends. How is it related to the flux change? 3. The motional emf discussed in Section 6.5 can be argued independently from Faraday’s law using the Lorentz force on moving charges. However, 2019-20

Electromagnetic Induction even if the charges are stationary [and the q (v × B) term of the Lorentz force is not operative], an emf is nevertheless induced in the presence of a time-varying magnetic field. Thus, moving charges in static field and static charges in a time-varying field seem to be symmetric situation for Faraday’s law. This gives a tantalising hint on the relevance of the principle of relativity for Faraday’s law. 4. The motion of a copper plate is damped when it is allowed to oscillate between the magnetic pole-pieces. How is the damping force, produced by the eddy currents? EXERCISES 6.1 Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ). FIGURE 6.18 229 2019-20

Physics 6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire. 6.3 FIGURE 6.19 6.4 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 6.5 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is 6.6 the induced emf in the loop while the current is changing? 6.7 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T 6.8 directed normal to the loop. What is the emf developed across the 6.9 cut if the velocity of the loop is 1 cm s–1 in a direction normal to the 6.10 (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 230 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 2019-20

Electromagnetic Induction having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°. ADDITIONAL EXERCISES 6.11 Suppose the loop in Exercise 6.4 is stationary but the current 6.12 feeding the electromagnet that produces the magnetic field is 6.13 gradually reduced so that the field decreases from its initial value 6.14 of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power? A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10 –3 T cm–1 along the negative x-direction (that is it increases by 10 – 3 T cm –1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10 –3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet. Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown. Give the polarity and magnitude of the induced emf. FIGURE 6.20 231 (b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do 2019-20

Physics 6.15 experience magnetic force due to the motion of the rod. Explain. 6.16 (d) What is the retarding force on the rod when K is closed? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s–1) when K is closed? How much power is required when K is open? (f ) How much power is dissipated as heat in the closed circuit? What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21. (b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance. FIGURE 6.21 6.17 A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by, B = – B0 k (r ≤ a; a < R) = 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off ? 232 FIGURE 6.22 2019-20

Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. 2019-20

Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American NICOLA TESLA (1856 – 1943) scientist, inventor and genius. He conceived the idea of the rotating magnetic field, which is the basis of practically all alternating current machinery, and which helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, apply Kirchhoff’s loop rule ∑ ε(t) = 0 (refer to Section the polyphase system of ac power, and the high 3.13), to the circuit shown in Fig. 7.1 to get frequency induction coil vm sin ω t = i R (the Tesla coil) used in radio and television sets and i = vm sin ω t R other electronic equipment. or The SI unit of magnetic field Since R is a constant, we can write this equation as is named in his honour. i = im sin ω t (7.2) where the current amplitude im is given by im = vm (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the FIGURE 7.2 In a pure same time. Clearly, the voltage and current are in phase with resistor, the voltage and each other. current are in phase. The minima, zero and maxima We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current is zero. The fact that the average current is zero, however, does 234 2019-20