Electromagnetic Induction

Michael Faraday and, independently, Joseph Henry discovered that an electromotive force (emf) is generated in a coil whenever the magnetic flux linked with it changes. Three simple experiments capture the essential idea. First, when a bar magnet is moved towards or away from a coil connected to a galvanometer, a deflection is seen only while the magnet is moving; no current flows when it is stationary, however strong the field. Second, moving one current-carrying coil towards or away from a second coil produces the same effect — a deflection is seen in the second coil's galvanometer only during relative motion. Third, even without any motion, simply switching the current on or off in a primary coil (or changing its value with a rheostat) induces a momentary current in a nearby secondary coil, because the flux linked with the secondary changes as the primary's field builds up or collapses.

These observations are unified through the idea of magnetic flux, the measure of the number of magnetic field lines passing through a surface. For a plane surface of area A placed in a uniform field B, with the normal to the surface making an angle θ with B, the flux is Φ_B = B·A = BA cosθ. Flux is a scalar quantity; it is maximum when the field is normal to the surface (θ = 0°) and zero when the field lies in the plane of the surface (θ = 90°). For a coil of N tightly wound turns, each of area A, the total flux linkage is NΦ_B, since the same flux effectively threads every turn.

Faraday's law of electromagnetic induction states that the magnitude of the emf induced in a circuit equals the rate of change of magnetic flux linkage through it: ε = -N(dΦ_B/dt). The negative sign is not just a bookkeeping detail — it encodes Lenz's law (covered next) and fixes the direction of the induced current. A larger rate of change of flux, whether from a stronger field, larger area, or faster motion, always produces a larger emf.

Worked example. A circular coil of radius 5 cm has 50 turns and is held with its plane perpendicular to a magnetic field that increases uniformly from 0.2 T to 0.5 T in 3 s. Find the emf induced in the coil.

1. Area of the coil: A = πr² = π × (0.05 m)² = 7.85 × 10^-3 m².
2. Rate of change of field: dB/dt = (0.5 - 0.2)/3 = 0.1 T/s.
3. Induced emf: ε = N × A × (dB/dt) = 50 × 7.85 × 10^-3 × 0.1 ≈ 3.93 × 10^-2 V ≈ 39.3 mV.

Lenz's law states that the direction of an induced current is always such as to oppose the change in magnetic flux that causes it. If the flux through a loop is increasing, the induced current flows in a direction that sets up its own magnetic field opposing that increase; if the flux is decreasing, the induced current flows so as to try to maintain it. This opposition is the physical origin of the minus sign in Faraday's law.

Consider the north pole of a bar magnet being pushed towards a stationary coil. The flux through the coil (in the direction from the magnet towards the coil) is increasing. By Lenz's law, the induced current must oppose this increase, so it flows in the direction that makes the face of the coil nearer the magnet behave like a north pole. Since like poles repel, the coil pushes back against the approaching magnet. To keep pushing the magnet in, an external agent must therefore do positive work against this magnetic repulsion. That mechanical work is exactly what appears as electrical energy in the circuit, which is eventually dissipated as heat (I²R) in the coil's resistance. If the induced current instead assisted the motion (attracted the approaching magnet), the magnet would accelerate on its own, current would grow without any external work being done, and energy would be created from nothing — a violation of the law of conservation of energy. Lenz's law is therefore not an independent postulate; it is required by energy conservation.

Worked example. A coil of 100 turns and resistance 10 Ω is linked by a flux that changes (due to an approaching magnet) from 2 × 10^-5 Wb to 8 × 10^-5 Wb per turn in 0.4 s. Find the induced emf and current, and state the direction rule that applies.

1. Change of flux per turn: ΔΦ = (8 - 2) × 10^-5 = 6 × 10^-5 Wb.
2. Induced emf: ε = N(ΔΦ/Δt) = 100 × (6 × 10^-5/0.4) = 100 × 1.5 × 10^-4 = 1.5 × 10^-2 V = 15 mV.
3. Induced current: I = ε/R = 0.015/10 = 1.5 × 10^-3 A = 1.5 mA.
4. Direction: since flux (from the approaching magnet) is increasing, by Lenz's law the induced current flows so as to make the near face of the coil repel the magnet, opposing its approach.

Consider a straight conducting rod PQ of length l resting on two parallel horizontal rails that are connected at one end through a resistor R, the whole arrangement lying in a uniform magnetic field B directed into the plane. If the rod is moved with velocity v perpendicular to its own length and to B, the area of the circuit enclosed by the rails increases (or decreases), so the flux linked with the circuit changes with time, and by Faraday's law an emf is induced: as the rod sweeps out area (l × v) per unit time, dΦ_B/dt = Blv, giving ε = Blv.

The same result follows directly from the force on the free charge carriers inside the moving rod, without invoking flux at all. Each charge q inside the rod moves with the rod's velocity v through the field B and experiences a magnetic (Lorentz) force F = qv × B. This force pushes positive charges to one end of the rod and negative charges to the other, until the resulting electrostatic field inside the rod balances the magnetic force at equilibrium. The work done per unit charge in moving from one end of the rod to the other against this separation is the motional emf, ε = Blv, exactly matching the flux-rule result. This equivalence is an important consistency check: Faraday's law and the direct force-on-charges (motional emf) picture always agree.

If the rod-and-rail circuit has a total resistance R, a current I = ε/R = Blv/R flows in the closed loop. By the same physics in reverse, once current flows in the rod, the field exerts a force F = BIl on the rod, and by Lenz's law this force always opposes the rod's motion — it acts as a magnetic "brake".

Worked example. A rod of length 1 m slides on frictionless rails with a constant speed of 2 m/s, perpendicular to a uniform field B = 0.5 T. The rails are connected through a resistor R = 2 Ω. Find (a) the induced emf, (b) the current, and (c) the retarding force on the rod.

1. Motional emf: ε = Blv = 0.5 × 1 × 2 = 1 V.
2. Current: I = ε/R = 1/2 = 0.5 A.
3. Retarding force: F = BIl = 0.5 × 0.5 × 1 = 0.25 N, directed opposite to the rod's velocity.

The rod-on-rails system of the previous section provides a clean quantitative check of energy conservation. To keep the rod moving at constant velocity v against the retarding magnetic force F = BIl, an external agent must supply mechanical power P_mech = Fv. This must equal the electrical power delivered to the resistor, P_elec = I²R (assuming the rod itself has negligible resistance). Substituting I = Blv/R shows that both expressions reduce to the same quantity, B²l²v²/R, confirming that mechanical work done against the induced force is converted, without loss, into electrical energy, which is then dissipated as heat. No energy is created or destroyed — induction is simply a mechanism for converting one form of energy into another.

Worked example. A rod of length 0.5 m moves at a constant 4 m/s through a field B = 0.4 T, with the circuit resistance R = 4 Ω. Verify that mechanical and electrical power match.

1. Emf: ε = Blv = 0.4 × 0.5 × 4 = 0.8 V.
2. Current: I = ε/R = 0.8/4 = 0.2 A.
3. Retarding force: F = BIl = 0.4 × 0.2 × 0.5 = 0.04 N.
4. Mechanical power: P_mech = Fv = 0.04 × 4 = 0.16 W.
5. Electrical power: P_elec = I²R = (0.2)² × 4 = 0.16 W. The two match exactly, as expected.

The same induction mechanism operates inside any bulk piece of conducting metal (not just thin wires) whenever the flux through it changes — for instance, a solid metal plate swinging through a magnetic field, or the iron core of a transformer carrying an alternating flux. Closed loops of current called eddy currents are set up within the body of the conductor, circulating in whirlpool-like paths. By Lenz's law these eddy currents oppose the change producing them, and since the conductor has finite resistance, they dissipate energy as heat (I²R losses), which can substantially reduce the efficiency of electrical machines.

Eddy currents are put to deliberate use in several devices: electromagnetic braking in trains and roller-coasters (a strong field induces eddy currents in a moving metal wheel or rail, and the resulting opposing force provides smooth, contactless braking), induction furnaces (eddy currents generated by a high-frequency field heat a metal sample enough to melt it), and the damping mechanisms of galvanometers and induction-type energy meters. Where eddy currents are undesirable — particularly in transformer cores and the armatures of motors and generators — their magnitude is reduced by laminating the core: building it from thin sheets of iron insulated from one another rather than one solid block. This breaks up the large-area conducting loops available to the eddy currents, greatly increasing their effective resistance and cutting down the associated heat loss, while barely affecting the desired magnetic behaviour of the core.

Any coil carrying a current I sets up a magnetic flux linked with itself, and if I changes, that self-linked flux changes too, inducing a back-emf in the very same coil — this is self-induction. Since flux is proportional to current for a fixed geometry, NΦ_B = LI, where the constant of proportionality L is called the self-inductance of the coil. Combining this with Faraday's law gives the self-induced emf ε = -L(dI/dt): the minus sign shows that this back-emf always opposes the change in current, which is why an inductor resists sudden changes of current much as inertia resists sudden changes of velocity. For a long solenoid of N turns, length l, and cross-sectional area A (with n = N/l turns per unit length), the self-inductance works out to L = μ₀n²Al = μ₀N²A/l, depending only on the solenoid's geometry and the permeability of the core medium, not on the current flowing through it.

Establishing a current in an inductor requires work to be done against the back-emf, and this work is stored as energy in the magnetic field: U = ½LI². This is the magnetic analogue of the energy ½CV² stored in a charged capacitor's electric field.

When two coils are placed close to each other, a changing current in one (the primary) sets up a changing flux that links the second (the secondary), inducing an emf in it even though the two coils are not electrically connected — this is mutual induction. The induced emf in the secondary is ε₂ = -M(dI₁/dt), where M is the mutual inductance of the pair, a geometric quantity depending on the size, shape, number of turns, relative orientation and separation of the two coils, and the medium between them. Remarkably, the mutual inductance is the same whichever coil is treated as primary: M₁₂ = M₂₁ = M (the reciprocity theorem).

Worked example (self-inductance). A solenoid of length 0.5 m and 1000 turns has a cross-sectional area of 4 × 10^-4 m². Find its self-inductance, and the energy stored when it carries a current of 2 A.

1. Turns per unit length: n = N/l = 1000/0.5 = 2000 turns/m.
2. L = μ₀n²Al = (4π × 10^-7) × (2000)² × (4 × 10^-4) × 0.5 ≈ 1.005 × 10^-3 H ≈ 1.0 mH.
3. Energy stored: U = ½LI² = 0.5 × 1.005 × 10^-3 × (2)² ≈ 2.01 × 10^-3 J ≈ 2.0 mJ.

Worked example (mutual inductance). Two coils have a mutual inductance M = 0.2 H. If the current in the primary changes from 0 to 5 A in 0.1 s, find the emf induced in the secondary.

1. dI₁/dt = 5/0.1 = 50 A/s.
2. ε₂ = M(dI₁/dt) = 0.2 × 50 = 10 V.

An AC generator (alternator) applies electromagnetic induction to generate electricity on a large scale. Its essential parts are an armature (a coil of many turns of insulated wire wound on a soft-iron core), strong field magnets between whose poles the armature rotates, slip rings fixed to the armature shaft, and brushes that press against the slip rings to connect the rotating coil to the external circuit. An external mechanical agent — falling water in a hydroelectric plant, steam in a thermal plant, or wind in a turbine — supplies the energy needed to rotate the armature against the magnetic forces resisting its motion (by Lenz's law), and this mechanical energy is converted into electrical energy.

Suppose a coil of N turns and area A rotates with constant angular speed ω in a uniform magnetic field B, about an axis perpendicular to B and lying in the plane of the coil. Taking t = 0 to be the instant when the coil's plane is perpendicular to B (so the flux linked is maximum at that instant), the flux linked at time t is Φ_B = BA cos(ωt). Faraday's law then gives the instantaneous induced emf as ε(t) = -N(dΦ_B/dt) = NBAω sin(ωt) = ε₀ sin(ωt), where ε₀ = NBAω is the peak (maximum) emf. This emf alternates sinusoidally with time: it is zero when the coil plane is perpendicular to B (flux maximum, but momentarily not changing) and maximum when the coil plane is parallel to B (flux zero, but changing fastest). The angular speed of rotation is related to the frequency of the generated AC by ω = 2πf, where f is the number of complete rotations per second (Hz), typically 50 Hz for the Indian power grid.

Worked example. A generator coil has N = 50 turns, area A = 0.15 m², and rotates in a field B = 0.05 T at a frequency of 60 rotations per second. Find the angular speed and the peak emf generated.

1. Angular speed: ω = 2πf = 2π × 60 ≈ 377 rad/s.
2. Peak emf: ε₀ = NBAω = 50 × 0.05 × 0.15 × 377 ≈ 141.4 V.