Electromagnetic Induction — Faraday's Law, Lenz's Law, Inductance

The Core Idea

A changing magnetic flux through a circuit induces an EMF. This single idea — Faraday’s law — is behind generators, transformers, and most of electrical engineering. Lenz’s law tells us the direction: the induced current opposes the change that caused it.

CBSE Class 12 boards give this chapter 6-8 marks. JEE Main tests 1-2 questions, often involving motional EMF or self/mutual inductance.

graph TD
    A[Changing Magnetic Flux?] --> B{How is flux changing?}
    B -->|B changing| C[Time-varying field]
    B -->|Area changing| D[Motional EMF]
    B -->|Angle changing| E[Rotating coil]
    C --> F["EMF = -dΦ/dt"]
    D --> G[EMF = Blv]
    E --> H[EMF = NBAω sin ωt]
    F --> I{Direction?}
    I --> J[Lenz's Law: opposes change]
    G --> I
    H --> I

Essential Formulas

\varepsilon = -\frac{d\Phi_B}{dt} = -N\frac{d\Phi}{dt}

For $N$ turns. The negative sign is Lenz’s law — the induced EMF opposes the change in flux.

For a rod of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$ :

\varepsilon = Blv

\varepsilon = -L\frac{dI}{dt}

For a solenoid: $L = \mu_0 n^2 Al$ where $n$ = turns per unit length, $A$ = area, $l$ = length.

Energy stored: $U = \frac{1}{2}LI^2$

\varepsilon_2 = -M\frac{dI_1}{dt}

$M$ depends on geometry. For two coaxial solenoids: $M = \mu_0 n_1 n_2 A l$

Solved Examples

Example 1 (Easy — CBSE)

A circular coil of 200 turns and area $0.1$ m $^2$ is in a field of 0.2 T. If the field drops to zero in 0.01 s, find the induced EMF.

$\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 200 \times \frac{0.2 \times 0.1}{0.01} = 200 \times 2 = \mathbf{400 \text{ V}}$

Example 2 (Medium — JEE Main)

A rod of length 50 cm moves at 2 m/s in a field of 0.5 T. Find the motional EMF.

$\varepsilon = Blv = 0.5 \times 0.5 \times 2 = \mathbf{0.5 \text{ V}}$

Example 3 (Hard — JEE Main)

A solenoid of 500 turns, length 0.5 m, and area 4 cm $^2$ carries current increasing at 10 A/s. Find the self-induced EMF.

$L = \mu_0 n^2 Al = 4\pi \times 10^{-7} \times (1000)^2 \times 4 \times 10^{-4} \times 0.5 = 2.51 \times 10^{-4}$ H

$\varepsilon = L \frac{dI}{dt} = 2.51 \times 10^{-4} \times 10 = \mathbf{2.51 \text{ mV}}$

Common Mistakes to Avoid

Mistake 1 — Confusing flux with field. Flux $\Phi = BA\cos\theta$ . Even if $B$ is constant, flux changes if area or angle changes.

Mistake 2 — Getting Lenz’s law direction wrong. The induced current creates a field that opposes the change in flux, not the flux itself. If flux is increasing, induced current opposes the increase.

Mistake 3 — Forgetting $N$ for coils with multiple turns. EMF $= N \times d\Phi/dt$ . Each turn contributes.

Mistake 4 — Using motional EMF formula for non-uniform fields. $\varepsilon = Blv$ assumes uniform $B$ . For non-uniform fields, integrate.

Mistake 5 — Wrong units for inductance. $L$ is in henry (H). $1 \text{ H} = 1 \text{ V s/A}$ .

Practice Questions

Q1. A coil has 100 turns. Flux through each turn changes from 0.02 Wb to 0.05 Wb in 0.1 s. Find EMF.

$\varepsilon = N \Delta\Phi/\Delta t = 100 \times 0.03/0.1 = 30$ V.

Q2. A conducting rod of 1 m slides on rails at 5 m/s in a 0.4 T field. Find EMF and current if resistance is 2 $\Omega$ .

$\varepsilon = Blv = 0.4 \times 1 \times 5 = 2$ V. $I = 2/2 = 1$ A.

Q3. An inductor of 0.5 H carries 2 A. Find energy stored.

$U = \frac{1}{2}LI^2 = 0.5 \times 0.5 \times 4 = 1$ J.

Q4. Current in a coil changes from 5 A to 1 A in 0.2 s. If $L = 0.4$ H, find induced EMF.

$\varepsilon = L \times |\Delta I/\Delta t| = 0.4 \times 4/0.2 = 8$ V.

Q5. Two coils have mutual inductance 0.3 H. Current in first changes at 50 A/s. Find EMF in second.

$\varepsilon = M \times dI/dt = 0.3 \times 50 = 15$ V.

Q6. A coil rotates in a magnetic field. EMF is $\varepsilon_0 \sin(\omega t)$ . At what angle is EMF maximum?

$\sin(\omega t) = 1$ when $\omega t = \pi/2$ . The coil is parallel to the field (flux is zero, rate of change of flux is maximum).

FAQs

What is eddy current?

When a bulk conductor moves through a magnetic field (or field changes through it), loops of induced current form within the conductor. These are eddy currents — they cause heating and are used in induction cooktops and electromagnetic braking.

Why is the negative sign in Faraday’s law important?

The negative sign represents Lenz’s law — the induced EMF opposes the change in flux. Without it, we’d violate conservation of energy.

What is the difference between self-inductance and mutual inductance?

Self-inductance ( $L$ ) is a coil’s EMF opposing changes in its own current. Mutual inductance ( $M$ ) is one coil inducing EMF in another coil due to changing current.

How does a transformer work?

AC in the primary coil creates a changing magnetic flux, which induces an EMF in the secondary coil (mutual inductance). $V_s/V_p = N_s/N_p$ . Ideal transformers conserve power: $V_p I_p = V_s I_s$ .

Advanced Concepts

Eddy currents — why and where they matter

When a bulk conductor experiences a changing magnetic field, loops of induced current form within the conductor. These are eddy currents. They obey Lenz’s law — they oppose the change that caused them.

Harmful effects: Heating in transformer cores, power loss in rotating machinery. Solution: use laminated cores (thin insulated sheets) to break the current loops and reduce eddy currents.

Useful applications:

Electromagnetic braking in trains and roller coasters — eddy currents in a metal disc moving through a magnetic field create a retarding force. No friction, no wear.
Induction cooktop — high-frequency AC generates eddy currents in the base of a steel utensil, heating it directly. Aluminium vessels do not work (too low resistivity for efficient heating at that frequency).
Metal detectors — eddy currents in a hidden metal object alter the impedance of the detector coil.

AC generator — converting rotation to EMF

A coil of $N$ turns and area $A$ rotating with angular velocity $\omega$ in a uniform field $B$ :

\varepsilon = NBA\omega \sin(\omega t) = \varepsilon_0 \sin(\omega t)

where $\varepsilon_0 = NBA\omega$ is the peak EMF.

The EMF is maximum when the coil is parallel to the field (flux is zero but rate of change is maximum). EMF is zero when the coil is perpendicular to the field (flux is maximum but rate of change is zero).

This is counterintuitive. Remember: EMF depends on the RATE of change of flux, not the flux itself. Flux is maximum when the coil face is perpendicular to $B$ , but at that instant, the flux is momentarily not changing (like the top of a sine wave).

LR and LC circuits — energy storage and oscillation

LR circuit (growth of current): When a battery is connected to an inductor and resistor in series:

I(t) = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)

Time constant $\tau = L/R$ . After one time constant, current reaches 63% of its maximum value.

LC circuit (oscillation): An inductor and capacitor exchange energy — magnetic field energy in $L$ converts to electric field energy in $C$ and back. This is electromagnetic oscillation.

\omega = \frac{1}{\sqrt{LC}}, \quad f = \frac{1}{2\pi\sqrt{LC}}

This is the same resonance frequency we meet in AC circuits.

Transformer — efficiency and losses

An ideal transformer conserves power: $P_{in} = P_{out}$ . Real transformers have losses:

Loss type	Cause	Remedy
Copper loss ( $I^2R$ )	Resistance of windings	Use thick copper wires
Iron loss (eddy currents)	Eddy currents in core	Use laminated core
Hysteresis loss	Repeated magnetisation cycle	Use soft iron core
Flux leakage	Not all flux links both coils	Tight winding, core design

\eta = \frac{P_{out}}{P_{in}} \times 100\%

Good power transformers reach 95–99% efficiency.

CBSE board exams frequently ask: “What are the energy losses in a transformer and how are they minimised?” This is a 5-mark question. List all four losses with causes and remedies.

Additional Solved Examples

Example 4 (JEE Main): A circular coil of 50 turns and radius 0.1 m rotates at 100 rev/s in a field of 0.1 T. Find the peak EMF.

$\varepsilon_0 = NBA\omega = 50 \times 0.1 \times \pi(0.1)^2 \times 2\pi(100)$

$= 50 \times 0.1 \times 0.0314 \times 628.3 = \mathbf{98.7 \text{ V}}$

Example 5 (CBSE): A step-down transformer converts 220 V to 22 V. If the secondary has 50 turns, how many turns does the primary have?

$N_p/N_s = V_p/V_s$ . $N_p = 50 \times 220/22 = \mathbf{500}$ turns.

Additional Practice Questions

Q7. Why is the core of a transformer laminated?

Lamination breaks the large eddy current loops into smaller ones, drastically reducing eddy current losses. Each lamination is insulated from the next by a thin layer of varnish or oxide.

Q8. An LC circuit has $L = 0.01$ H and $C = 0.01$ F. Find the oscillation frequency.

$f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.01 \times 0.01}} = \frac{1}{2\pi \times 0.01} = \frac{1}{0.0628} \approx 15.9$ Hz.

Q9. A transformer has 1000 primary turns and 100 secondary turns. Primary current is 2 A. Find secondary current (ideal).

$I_s = I_p \times N_p/N_s = 2 \times 1000/100 = 20$ A. Step-down in voltage means step-up in current.

Q10. The time constant of an LR circuit is 0.1 s with $R = 10$ $\Omega$ . Find $L$ .

$\tau = L/R$ . $L = \tau \times R = 0.1 \times 10 = 1$ H.