Induction

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. Induction a plane of area A placed in a uniform magnetic field B (Fig. . ) can be written as Φ B = B . A = BA cos θ ( . ) where θ is angle between B and A. The notion of the area as a vector has been discussed earlier in Chapter . Equation ( . ) 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. . , then the magnetic flux through the surface is given by B Φ = B A B A i i ... = all i i ∑B A i ( . ) where ‘all’ stands for summation over all the area elements dAi comprising the surface and Bi is the magnetic field at the area element dAi. The SI unit of magnetic flux is weber (Wb) or tesla meter squared (T m2). Magnetic flux is a scalar quantity. Induction EXAMPLE . Solution The angle θ made by the area vector of the coil with the magnetic field is °. From Eq. ( . ), the initial magnetic flux is Φ = BA cos θ – . Wb Final flux, Φmin = The change in flux is brought about in . s. From Eq. ( . ), the magnitude of the induced emf is given by ( – B Φ Φ Δ Δ Δ – . mV . And the magnitude of the current is – mA . I Ω 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 . A circular coil of radius cm, turns and resistance 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 ° in . 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 . × – T. Solution Initial flux through the coil, ΦB (initial) = BA cos θ = . × – × (π × – ) × cos 0º = 3π × – Wb Final flux after the rotation, ΦB (final) = . × – × (π × – ) × cos ° = –3π × – Wb Therefore, estimated value of the induced emf is, N Φ Δ Δ = × (6π × – )/ . = . × – V I = ε/R = . × – 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. EXAMPLE . Induction EXAMPLE . Example . Figure . 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 . Solution (i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux. (ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux. (iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field. Example . (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. . ) 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. EXAMPLE . EXAMPLE . FIGURE . (d) Predict the polarity of the capacitor in the situation described by Fig. . . FIGURE . 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. Induction where we have used dx/dt = –v which is the speed of the conductor PQ. The induced emf Blv is called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit. It is also possible to explain the motional emf expression in Eq. ( . ) by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ. The work done in moving the charge from P to Q is, W = qvBl Since emf is the work done per unit charge, W ε = = Blv This equation gives emf induced across the rod PQ and is identical to Eq. ( . ). 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 ( . ) since v = . 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 , 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 . A metallic rod of m length is rotated with a frequency of rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. . ). A constant and uniform magnetic field of T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring? EXAMPLE . Interactive animation on motional emf: EXAMPLE . FIGURE . 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. ( . ), 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 Bv ε = . Hence, Bv Induction EXAMPLE . Example . A wheel with metallic spokes each . m long is rotated with a speed of rev/min in a plane normal to the horizontal component of earth’s magnetic field HE at a place. If HE = . G at the place, what is the induced emf between the axle and the rim of the wheel? Note that G = – T. Solution Induced emf = ( / ) ω B R2 = ( / ) × 4π × . × – × ( . ) = . × – V The number of spokes is immaterial because the emf’s across the spokes are in parallel. Induction EXAMPLE . When the induced emf is non-zero, the current I is (in magnitude) Bl v I (b) FIGURE . The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude B l v F x b b x b ≤ < ≤ < The Joule heating loss is J P I r B l v x b b x b ≤ < ≤ < One obtains similar expressions for the inward motion from x = 2b to x = . One can appreciate the whole process by examining the sketch of various quantities displayed in Fig. . (b). 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. Induction EXAMPLE . where n2l is the total number of turns of S2. From Eq. ( . ), M21 = μ0n1n2πr 1l ( . ) Using Eq. ( . ) and Eq. ( . ), we get M12 = M21= M (say) ( . ) 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 r2 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 . 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 = πr 1B2 I μ π = M12 I2 Thus, M μ π From Eq. ( . ) M M μ π Note that we calculated M12 from an approximate value of Φ1, assuming the magnetic field B2 to be uniform over the area π r1 . However, we can accept this value because r1 << r2. Now, let us recollect Experiment . in Section . . 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. ( . ), we have N1Φ1 = MI2 For currents varrying with time, ( ( N MI Φ Since induced emf in coil C1 is given by ( – N Φ ε1 = We get, – I M ε1 = 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. . . 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 B N I Φ ∝ B L N I Φ ( . ) 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. ( . ), the induced emf is given by ( B – N Φ ε = – I L ε = ( . ) 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 ( )( )( B N nl n I A Φ μ Induction I Al n2 μ where nl is the total number of turns. Thus, the self-inductance is, L I Β ΝΦ 0n Al μ ( . ) 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 permiability), then, L n Al μ μ ( . ) 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 W I If we ignore the resistive losses and consider only inductive effect, then using Eq. ( . ), W I L I Total amount of work done in establishing the current I is I W W L I I 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. . . 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 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 = 0º at t = ). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. ( . ), 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, – – (cos dt B N NBA Φ Thus, the instantaneous value of the emf is = NBA sin ( . ) where NBAω is the maximum value of the emf, which occurs when sin ωt = ± . If we denote NBAω as ε0, then ε = ε0 sin ωt ( . ) Since the value of the sine fuction varies between + and – , the sign, or polarity of the emf changes with time. Note from Fig. . that the emf has its extremum value when θ = 90º or θ = 270º, 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 ( . ) can be written as ε = ε0sin 2π ν t ( . ) where ν is the frequency of revolution of the generator’s coil. Note that Eq. ( . ) and ( . ) give the instantaneous value of the emf and ε varies between +ε0 and –ε0 periodically. We shall learn how to determine the time-averaged value for the alternating voltage and current in the next chapter. FIGURE . AC Generator EXAMPLE . 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 MW, i.e., one can light up million W bulbs! In most generators, the coils are held stationary and it is the electromagnets which are rotated. The frequency of rotation is Hz in India. In certain countries such as USA, it is Hz. Example . Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a turn coil of area . m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of . T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Solution Here f = . Hz; N = , A = . m2 and B = . T. Employing Eq. ( . ) ε0 = NBA ( π ν) = × . × . × × . × . = . V The maximum voltage is . V. We urge you to explore such alternative possibilities for power generation. FIGURE . An alternating emf is generated by a loop of wire rotating in a magnetic field. Induction SUMMARY . The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as, ΦB = B A i = BA cos θ where θ is the angle between B and A. . 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, B N Φ ε = − 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. . 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. . 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 . Changing magnetic fields can set up current loops in nearby metal (any conductor) bodies. They dissipate electrical energy as heat. Such currents are called eddy currents. . Inductance is the ratio of the flux-linkage to current. It is equal to NΦ/I. Induction having a span of m, if the Earth’s magnetic field at the location has a magnitude of × – T and the dip angle is °.