Educational Galaxy

Electromagnetic Induction — EG's own notes, in simple words

Book: Physics — Part I (CBSE, Class 12)  |  Chapter 6  |  Medium: English

Move a magnet near a coil and a current appears, with no battery at all. That surprise — electromagnetic induction — powers every generator and transformer in the world, and it rests on one idea: a changing magnetic flux creates an emf.

The chapter has two laws to keep apart: Faraday's law gives the size of the induced emf, and Lenz's law gives its direction. Add self-induction, mutual induction and the AC generator, and the chapter is complete.

Diagram (labels on the image): Diagram (labels on the image): a bar magnet sliding into a copper coil wired to a round meter — the meter needle moves as the flux through the coil changes.

A bicycle dynamo lighting a bulb — the same induction idea, worked by the turning wheel.

1. Magnetic flux and Faraday's law

Magnetic flux through a surface is Φ_B = B · A = BA cos θ, the amount of magnetic field passing through it. Its SI unit is the weber (Wb). Flux changes if the field changes, the area changes, or the angle between them changes — and any of those changes induces an emf.

Faraday's law states that the induced emf equals the rate of change of flux: ε = −N dΦ_B/dt for a coil of N turns. The minus sign is Lenz's law, treated below. Note that it is the rate of change that matters, not the size of the flux: a huge steady flux induces nothing.

2. Lenz's law and energy conservation

Lenz's law states that the induced current flows in a direction that opposes the change that produced it. Push a north pole towards a coil and the near end becomes a north pole, repelling the magnet; pull it away and the near end becomes a south pole, attracting it.

Lenz's law is energy conservation in disguise: if the induced current helped the change instead of opposing it, you would get energy for nothing. The extra work you do pushing the magnet against the repulsion is exactly the electrical energy produced.

3. Motional emf

A rod of length l moving at speed v perpendicular to a uniform field B develops an emf ε = Blv between its ends, because the free charges inside are pushed sideways by the magnetic force until the electric field they build up balances it.

If the rod slides on rails forming a closed circuit, the induced current is I = Blv/R and the power delivered is P = B²l²v²/R. A steady external force must keep the rod moving, and the work it does reappears as heat in the resistor — energy is conserved again.

4. Self-induction and mutual induction

Self-induction: a changing current in a coil changes its own flux, inducing a back emf ε = −L dI/dt that opposes the change. The self-inductance L of a solenoid is μ₀n²Al, and its SI unit is the henry (H). An inductor stores energy U = ½LI² in its magnetic field.

Mutual induction: a changing current in one coil induces an emf in a nearby coil, ε₂ = −M dI₁/dt. The mutual inductance M depends on the geometry and on how much flux links the two coils — the principle behind transformers.

5. The AC generator

An AC generator turns mechanical energy into alternating current: a coil rotates in a magnetic field, so its flux changes sinusoidally and the induced emf is ε = NBAω sin ωt. Its peak value is ε₀ = NBAω, reached when the coil's plane is parallel to the field.

Slip rings carry the alternating current out to the circuit — that is what makes it AC rather than DC. Increase the emf by using more turns, a larger area, a stronger field or faster rotation.

Going further — complete NCERT coverage

NCERT section map. 6.1 Introduction, 6.2 The Experiments of Faraday and Henry, 6.3 Magnetic Flux, 6.4 Faraday's Law of Induction, 6.5 Lenz's Law and Conservation of Energy, 6.6 Motional Electromotive Force, 6.7 Inductance (self and mutual), 6.8 AC Generator.

Faraday and Lenz. Flux Φ_B = ∫B·dS (weber). Faraday's law: induced emf ε = −dΦ_B/dt; for a coil of N turns, ε = −N dΦ_B/dt. The minus sign is Lenz's law: the induced current creates a field opposing the change that produced it — push a north pole into a coil and the near end becomes a north pole, resisting the push. Lenz's law is energy conservation in disguise: without the minus sign, a tiny push would generate ever-growing current for free.

Motional emf, inductance, generator. A rod of length l moving at speed v perpendicular to B develops ε = Blv between its ends (the magnetic Lorentz force separates charges until the electric field balances it). Self-inductance: ε = −L dI/dt, with L = μ₀n²Al for a solenoid; energy stored in an inductor = (1/2)LI². Mutual inductance M: emf in coil 2 = −M dI₁/dt. AC generator: a coil rotating in a uniform field gives ε = NBAω sin ωt — sinusoidal emf, maximum when the coil's plane is parallel to the field (flux changing fastest).

Diagram (labels on the image): New PIL schematic: a magnet moving into a coil — the induced emf equals −d(flux)/dt, and Lenz's law says the induced current opposes the change.

Chart / Table — at a glance

Quantity	Symbol	SI unit
Magnetic flux	Φ_B	weber (Wb)
Induced emf	ε	volt (V)
Self-inductance	L	henry (H)
Mutual inductance	M	henry (H)
Motional emf	Blv	volt (V)

Practice set — try it yourself

1. A coil of 100 turns and area 0.01 m² sits in a field that changes from 0.2 T to 0.5 T in 0.1 s. Find the induced emf. Hint: Use ε = N × ΔΦ/Δt, with ΔΦ = A × ΔB.
2. A north pole is pushed towards a coil. What pole appears at the near end, and why? Hint: Apply Lenz's law: the induced effect opposes the approach.
3. A 0.5 m rod moves at 4 m/s perpendicular to a 0.3 T field. Find the motional emf. Hint: Use ε = Blv.
4. Why are transformer cores laminated? Hint: Think about eddy currents in a solid block of metal.
5. An AC generator's coil rotates twice as fast. What happens to its peak emf? Hint: Recall ε₀ = NBAω.

Questions with Answers — full answers

1. State Faraday's law of electromagnetic induction. Answer: Answer: The emf induced in a circuit equals the negative rate of change of magnetic flux through it: ε = −N dΦ_B/dt, where N is the number of turns.
2. State Lenz's law and show that it follows from conservation of energy. Answer: Answer: The induced current opposes the change producing it. If it aided the change, the magnet would be pulled in and gain energy with no work input, violating energy conservation; the work done against the opposition supplies the electrical energy.
3. Derive the motional emf for a rod moving perpendicular to a uniform magnetic field. Answer: Answer: Charges in the rod feel magnetic force qvB along the rod, building up charge at the ends until electric force qE balances it: qE = qvB, so E = vB and the emf across length l is ε = El = Blv.
4. Define self-inductance and give its SI unit. What is the energy stored in an inductor? Answer: Answer: Self-inductance L is the flux linkage per unit current, L = NΦ/I, and the back emf is ε = −L dI/dt. Its SI unit is the henry (H). Energy stored is U = ½LI².
5. Explain the working of an AC generator and give the expression for its emf. Answer: Answer: A coil rotating with angular speed ω in field B has flux Φ = NBA cos ωt, so ε = −dΦ/dt = NBAω sin ωt, alternating sinusoidally. Slip rings pass this alternating emf to the external circuit.

Extension Questions — go further

1. A coil of 100 turns and area 0.02 m² sits in a field that changes from 0.4 T to 0 in 0.1 s (field perpendicular to the coil). Find the average induced emf. Answer: ΔΦ per turn = 0.4 × 0.02 = 0.008 Wb. ε = N ΔΦ/Δt = 100 × 0.008/0.1 = 8 V.
2. A 0.5 m rod moves at 4 m/s perpendicular to a 0.8 T field. Find the motional emf. Answer: ε = Blv = 0.8 × 0.5 × 4 = 1.6 V.
3. Why does Lenz's law have to be true? Answer with energy conservation. Answer: If the induced current aided the change instead of opposing it, the growing current would strengthen the change, inducing still more current — energy would grow without any work input. The minus sign prevents this free-energy loop.
4. A solenoid has n = 1000 turns/m, A = 10 cm², length 0.5 m. Find its self-inductance. Answer: L = μ₀n²Al = 4π × 10⁻⁷ × 10⁶ × 10 × 10⁻⁴ × 0.5 = 2π × 10⁻⁴ ≈ 6.28 × 10⁻⁴ H.
5. An AC generator coil (N = 200, A = 0.05 m²) rotates at 50 rev/s in B = 0.3 T. Find the maximum emf. Answer: ω = 2π × 50 = 100π rad/s. ε_max = NBAω = 200 × 0.3 × 0.05 × 100π = 300π ≈ 942 V.

Board Exam Questions — board-pattern practice

These questions are written by Educational Galaxy in board-exam style for practice. They are not claimed to be real past-year questions.

1. A square loop of side 0.2 m is pulled out of a 0.5 T field at 2 m/s, with resistance 10 Ω. Find the induced current while it is leaving the field. Answer: Answer: Motional emf ε = Blv = 0.5 × 0.2 × 2 = 0.2 V. Current I = ε/R = 0.02 A, in a direction opposing the loss of flux.
2. A coil of inductance 0.5 H carries a current falling from 4 A to 0 in 0.2 s. Find the induced emf. Answer: Answer: ε = L|ΔI/Δt| = 0.5 × 4/0.2 = 10 V, opposing the fall in current.
3. The flux through a coil of 200 turns changes from 0.01 Wb to 0.03 Wb in 0.05 s. Find the average induced emf. Answer: Answer: ε = NΔΦ/Δt = 200 × 0.02/0.05 = 80 V.
4. Why does a generator need more mechanical work when the external circuit draws more current? Answer: Answer: Larger current means a larger opposing magnetic force on the coil (Lenz's law), so more mechanical work is needed to keep it turning — energy is conserved.
5. A magnet is dropped through a copper ring. Describe what happens to its fall and why. Answer: Answer: As the magnet approaches, the changing flux induces a current whose field repels it, slowing the fall. As it leaves, the induced current attracts it, slowing it again. The magnet falls more slowly than in free fall.