Electromagnetic Induction

Oersted showed in 1820 that an electric current produces a magnetic field. The natural next question was whether a magnetic field can produce an electric current. Michael Faraday in England and Joseph Henry in the United States answered it around 1830, independently: a steady magnetic field does nothing, but a changing magnetic field, or a conductor moving through a magnetic field, drives a current in a closed circuit. The phenomenon is called electromagnetic induction, the emf produced is the induced emf, and the current it drives is the induced current.

Why the board cares: every generator in every power station, every transformer on the grid, the induction cooktop, the metal detector and the wireless charger run on this one idea. In the paper, the chapter shows up as a chain of derivations that all start from the same line, ε = −dΦ_B/dt. If you can write that line and know what Φ_B is in each situation, you can reproduce everything that follows.

The rationalised NCERT chapter has these sections: Faraday and Henry's experiments, magnetic flux, Faraday's law, Lenz's law and conservation of energy, motional emf, inductance (mutual and self) and the AC generator. The older sections on eddy currents and the quantitative energy treatment of the moving arm were removed from the textbook, though the energy argument still makes a good 3-mark question and is included below.

NCERT describes three experiments. You should be able to describe each in two or three lines and state what it proves.

Experiment 6.1: magnet and coil. A coil C₁ is connected to a galvanometer. When the north pole of a bar magnet is pushed towards the coil, the galvanometer deflects. When the magnet is held still, there is no deflection. When the magnet is pulled away, the deflection reverses. Moving the coil instead of the magnet gives the same result, and moving faster gives a larger deflection. Conclusion: relative motion between magnet and coil induces a current, and the current lasts only while the motion lasts.

Experiment 6.2: two coils. The bar magnet is replaced by a second coil C₂ carrying a steady current from a battery. Moving C₂ towards or away from C₁ produces the same deflections as the magnet did. A current-carrying coil behaves like a magnet, so this is the same effect.

Experiment 6.3: no motion at all. Both coils are kept fixed. C₂ is connected to a battery through a tapping key. The galvanometer in C₁ shows a momentary deflection when the key is pressed and a deflection in the opposite direction when it is released. With the key held pressed, there is nothing. Conclusion: what matters is not motion but a change in the magnetic field through C₁. Faraday later found that an iron rod inserted along the axis of the coils makes the deflection much larger.

A classic 2-mark question (NCERT Example 6.1) asks how to get a larger deflection: move the magnet faster, use a stronger magnet, use a coil with more turns or larger area, or insert a soft-iron core. And how to show the induced current without a galvanometer: replace it with a small torch bulb or an LED, which glows briefly.

Magnetic flux through a surface measures how much of the field passes through it. For a plane surface of area A placed in a uniform field B, with θ the angle between B and the normal to the surface,

Φ_B = B · A = BA cos θ

For a general surface split into small elements dA_i, Φ_B = ∑ B_i · dA_i. Flux is a scalar. Its SI unit is the weber (Wb); 1 Wb = 1 T m². The tesla itself is 1 Wb m⁻², which is why the field is sometimes called flux density.

Three cases to keep straight. When the field is perpendicular to the surface (parallel to the normal), θ = 0 and Φ_B = BA, the maximum. When the field lies in the plane of the surface, θ = 90° and Φ_B = 0. When the surface is turned through 180°, the flux changes sign, from +BA to −BA, so the change in flux is 2BA. This last point is the whole trick in NCERT Example 6.3 and in the search-coil type numericals.

For a coil of N turns, each turn links the same flux, so the total flux linkage is NΦ_B. The flux through a coil can change in three ways, and each gives a family of exam questions: B changes with time (a solenoid whose current is changed, a decaying field), the area changes (a loop being pulled out of a field, a loop deformed from a circle into a narrow line), or the angle changes (a coil rotating in a field, which is the generator).

Faraday summarised his experiments in one law. The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. In symbols,

ε = −dΦ_B/dt

The negative sign is Lenz's law, the direction of the emf, which is the subject of the next section. For a closely wound coil of N turns the flux through each turn changes at the same rate, so the emf is N times larger:

ε = −N dΦ_B/dt

Because Φ_B = BA cos θ, the emf can be produced by changing B, or A, or θ, or any combination. The examiner accepts the law in words or symbols but wants both when the question says 'state'.

Two model numericals. NCERT Example 6.2: a square loop of side 10 cm and resistance 0.5 Ω stands vertically in the east-west plane; a field of 0.10 T pointing north-east is reduced to zero in 0.70 s at a steady rate. The normal to the loop points north, so θ = 45°. Initial flux = 0.10 × (0.10)² × cos 45° = 7.07 × 10⁻⁴ Wb. Emf = 7.07 × 10⁻⁴/0.70 = 1.0 × 10⁻³ V, current = 1.0 × 10⁻³/0.5 = 2.0 × 10⁻³ A.

NCERT Example 6.3: a coil of radius 10 cm, 500 turns and resistance 2 Ω has its plane perpendicular to the horizontal component of the earth's field, 3.0 × 10⁻⁵ T, and is rotated about a vertical diameter through 180° in 0.25 s. Flux per turn goes from +BA to −BA, so the change is 2BA and the total change in flux linkage is 2NBA = 2 × 500 × 3.0 × 10⁻⁵ × π × (0.10)² = 9.42 × 10⁻⁴ Wb. Average emf = 9.42 × 10⁻⁴/0.25 = 3.8 × 10⁻³ V, average current = 1.9 × 10⁻³ A.

A useful by-product: the charge that flows through a coil of resistance R when the flux linkage changes by ΔΦ is q = ΔΦ/R, independent of how fast the change happens. This is the principle of the ballistic galvanometer and of the search coil.

Heinrich Lenz gave the rule for the sign in Faraday's law. The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. Note the wording: it opposes the change in flux, not the flux itself.

Apply it in two steps. First decide whether the flux through the loop is increasing or decreasing. Then choose the induced current so that its own field opposes that change: if the flux is increasing, the induced field points against the external field; if decreasing, it points along it. Two standard cases. The north pole of a magnet approaches a coil: flux increases, so the near face of the coil becomes a north pole to push the magnet away, and the current seen from the magnet is anticlockwise. The north pole is withdrawn: flux decreases, the near face becomes a south pole to pull the magnet back, and the current is clockwise as seen from the magnet.

Why Lenz's law is conservation of energy. Suppose the opposite were true and the near face became a south pole when the north pole approached. The magnet would be attracted, gain speed, cause a bigger induced current, be attracted more strongly, and so on, generating kinetic energy and electrical energy from nothing. That is impossible. In the real case, the magnet is repelled and you must do work to push it in; that work is exactly the energy dissipated as heat by the induced current. The direction given by Lenz's law is the only one consistent with conservation of energy.

Two consequences that appear as questions. A magnet dropped through a vertical copper tube falls much more slowly than through a plastic tube, because the induced currents in the tube oppose its motion. A copper ring placed over the core of a solenoid jumps up when the current is switched on, because the sudden increase in flux induces a current whose field is opposite to the solenoid's, so the ring is repelled. The old NCERT section on eddy currents, now deleted, was built on the same law: currents induced in the body of a conductor oppose its motion, giving electromagnetic damping, magnetic braking and induction furnaces.

NCERT Exercise 6.2 uses the same reasoning with a change of area. If an irregularly shaped wire loop in a field pointing into the page is pulled into a circle, its area and hence its flux increase, so the induced current is anticlockwise (producing a field out of the page). If a circular loop is squeezed into a narrow straight line, its flux decreases, so the current is clockwise.

Derivation from Faraday's law. A rectangular conductor PQRS lies in a uniform field B directed into the page. The arm PQ, of length l, is free to slide and is moved to the left with velocity v; RQ and SP are of length x at some instant. The flux through PQRS is Φ_B = Blx. As PQ moves, x decreases, so

ε = −dΦ_B/dt = −Bl dx/dt = Blv

since dx/dt = −v. This emf is called the motional emf. If the circuit has resistance R the induced current is I = Blv/R.

Explanation by the Lorentz force. The rod PQ contains free electrons. Each electron, carried along with velocity v in a field B, experiences a force q(v × B) directed along the rod. The force does work qvBl in moving a charge from one end to the other, and emf is work per unit charge, so ε = Blv again. This is the argument to give when the question says 'explain using the Lorentz force' and it works even for an isolated rod with no closed circuit: charges pile up at the ends until the electric force balances the magnetic force, and a potential difference Blv appears across the rod.

If the rod moves at an angle θ to the field, only the perpendicular component counts, ε = Blv sin θ. A rod moving parallel to the field has no emf. NCERT Exercise 6.7 is a wire 10 m long, lying east-west, falling at 5.0 m s⁻¹ through the horizontal component of the earth's field, 0.30 × 10⁻⁴ T. The emf is 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻³ V. With v downward and B_H northward, v × B points east, so positive charge accumulates at the eastern end: the eastern end is at higher potential and the current in the wire, if the circuit were closed, would flow from west to east through it.

Rod rotating about one end. A metallic rod of length l rotates with angular speed ω about an axis through one end, perpendicular to a uniform field B. A small element of length dr at distance r from the axis moves with speed v = ωr, so it contributes dε = Bωr dr. Integrating from 0 to l: ε = Bω ∫ r dr = ½Bωl². Equivalently, the rod sweeps area ½l²θ, so dΦ/dt = B · ½l²ω. NCERT Exercise 6.5: l = 1.0 m, ω = 400 rad s⁻¹, B = 0.5 T gives ε = ½ × 0.5 × 400 × 1 = 100 V. For a wheel with many spokes, every spoke gives the same emf and they are in parallel, so the emf between axle and rim does not depend on the number of spokes.

Energy consideration. With the arm PQ carrying current I = Blv/R, the field exerts a force F = BIl = B²l²v/R on it, directed against its motion (Lenz's law). To keep the arm moving at constant v an external agent must push with this force, delivering power P = Fv = B²l²v²/R. The Joule heat in the circuit is I²R = (Blv/R)² R = B²l²v²/R, the same. Mechanical work in equals electrical energy out.

An electric current produces a magnetic field, so the flux through a coil is proportional to the current that produces it, whether that current is in the coil itself or in a neighbouring coil. For a coil of N turns, the flux linkage NΦ_B ∝ I, and the constant of proportionality is called inductance. It depends only on the geometry of the coils and the medium, not on the current, just as capacitance depends on geometry and not on charge. Inductance is a scalar; its SI unit is the henry (H), and 1 H = 1 Wb A⁻¹ = 1 V s A⁻¹. Its dimensions are [M L² T⁻² A⁻²].

Mutual inductance is the flux linkage in one coil per unit current in a neighbouring coil: N₁Φ₁ = M₁₂I₂. Equivalently, it is the emf induced in one coil when the current in the other changes at 1 A s⁻¹: ε₁ = −M₁₂ dI₂/dt. The SI unit is the henry.

Derivation for two long coaxial solenoids (the standard 5-mark). Solenoid S₁ of radius r₁ and n₁ turns per unit length sits inside solenoid S₂ of radius r₂ and n₂ turns per unit length; both have length l. Pass current I₂ through S₂. The field inside it is B₂ = μ₀n₂I₂, and this field passes through all n₁l turns of S₁, each of area πr₁². So N₁Φ₁ = (n₁l)(μ₀n₂I₂)(πr₁²) = M₁₂I₂, giving M₁₂ = μ₀n₁n₂πr₁²l. Now pass I₁ through S₁ instead. Its field μ₀n₁I₁ exists only inside radius r₁, so the flux through each turn of S₂ is μ₀n₁I₁πr₁² (not πr₂²), and N₂Φ₂ = (n₂l)(μ₀n₁I₁)(πr₁²) = M₂₁I₁. Hence M₂₁ = M₁₂ = M = μ₀n₁n₂πr₁²l. With a core of relative permeability μ_r, M = μ_rμ₀n₁n₂πr₁²l.

M grows with the number of turns of each coil, with the area of the inner coil, and with a magnetic core; it falls when the coils are moved apart or their axes are turned away from each other, because less of one coil's flux threads the other. NCERT Example 6.7 does the same calculation for two concentric coplanar coils with r₁ ≪ r₂: the large coil's field at its centre, μ₀I₂/2r₂, is nearly uniform over the small coil, so M = μ₀πr₁²/2r₂.

NCERT Exercise 6.9: M = 1.5 H and the current in one coil goes from 0 to 20 A in 0.5 s. Change in flux linkage with the other coil = MΔI = 1.5 × 20 = 30 Wb. The induced emf during the change is 30/0.5 = 60 V.

A changing current in a coil changes the flux through the coil itself, so an emf is induced in the same coil. This is self-induction. Self-inductance L is the flux linkage per unit current, NΦ_B = LI, or equivalently the emf induced in the coil when the current through it changes at 1 A s⁻¹: ε = −L dI/dt. SI unit: henry. Its dimensional formula is [M L² T⁻² A⁻²].

Derivation for a long solenoid. Let the solenoid have n turns per unit length, cross-sectional area A and length l, carrying current I. The field inside is B = μ₀nI. The total number of turns is N = nl, so the flux linkage is NΦ_B = (nl)(μ₀nI)(A) = μ₀n²AlI. Comparing with NΦ_B = LI, L = μ₀n²Al. If the solenoid is filled with a material of relative permeability μ_r (a soft-iron core), L = μ_rμ₀n²Al, which can be hundreds of times larger. Since n = N/l, the same result reads L = μ₀N²A/l: doubling the turns in the same length gives four times the inductance.

Back emf and the inertia of electricity. By Lenz's law the self-induced emf opposes any change in current. When the current is switched on it opposes the growth; when the current is switched off it opposes the decay, which is why a large spark jumps at a switch that breaks a circuit containing a big inductor. Self-inductance plays the same role for current that mass plays for velocity, so it is called the electrical inertia of the circuit. NCERT Exercise 6.8: the current in a circuit falls from 5.0 A to 0 in 0.1 s and the average induced emf is 200 V, so L = ε/(dI/dt) = 200/(5.0/0.1) = 4 H.

Energy stored in an inductor. To build up a current against the back emf, the source must do work. At an instant when the current is I, the rate of work is |ε| I = LI dI/dt, so dW = LI dI. Integrating from 0 to I: W = ½LI². This work is stored as magnetic energy U = ½LI², in the field of the inductor, just as ½CV² is stored in the electric field of a capacitor.

Substituting L = μ₀n²Al and I = B/μ₀n gives U = ½ μ₀n²Al × B²/μ₀²n² = B²Al/2μ₀. Dividing by the volume Al, the magnetic energy density is u = B²/2μ₀, which is the exact analogue of ½ε₀E² for an electric field.

Finally, when two coils are near each other and both currents change, the emf in coil 1 has both contributions: ε₁ = −L₁ dI₁/dt − M₁₂ dI₂/dt.

The AC generator converts mechanical energy into electrical energy. Principle: when a coil rotates in a uniform magnetic field, the flux through it changes continuously and, by Faraday's law, an alternating emf is induced.

Construction. An armature, a coil of N turns and area A, is mounted on a shaft (the rotor) between the poles of a permanent magnet or an electromagnet (the field magnet). The two ends of the coil are joined to two slip rings that rotate with the shaft, and two stationary carbon brushes press against the rings to carry the current to the external circuit. The rotor is turned by a turbine driven by falling water (hydroelectric), by steam from burning coal (thermal) or from a nuclear reactor.

Derivation. Let the coil rotate with constant angular speed ω. At time t the angle between the field and the normal to the coil is θ = ωt (taking θ = 0 at t = 0). Flux through the coil: Φ_B = BA cos ωt. By Faraday's law, for N turns,

ε = −N dΦ_B/dt = −NBA d(cos ωt)/dt = NBAω sin ωt

So ε = ε₀ sin ωt with ε₀ = NBAω, the maximum emf. Since ω = 2πν, ε = ε₀ sin 2πνt, and ε₀ = 2πνNBA. The emf is maximum when ωt = 90° or 270°, that is when the plane of the coil is parallel to the field: flux is zero there but changing fastest. It is zero when ωt = 0 or 180°, the plane of the coil perpendicular to the field, where flux is maximum but momentarily not changing. The graph of ε against t is a sine curve of period T = 1/ν; the emf changes sign every half rotation, which is why the output is alternating. Its average over a full cycle is zero.

In India the frequency is 50 Hz; in the USA it is 60 Hz. NCERT Example 6.8 (Kamla on the stationary bicycle): N = 100, A = 0.10 m², B = 0.01 T, half a revolution per second, so ω = π rad s⁻¹ and ε₀ = 100 × 0.01 × 0.10 × π = 0.314 V. NCERT Exercise 6.6 asks for the same quantities with N = 20, r = 8.0 cm, ω = 50 rad s⁻¹, B = 3.0 × 10⁻² T and R = 10 Ω: ε₀ = 20 × 0.03 × π(0.08)² × 50 = 0.603 V, average emf over a cycle = 0, I₀ = 0.0603 A, average power = ½ε₀I₀ = 0.018 W, and the power comes from the external agent that keeps the coil turning against the magnetic torque.