Electromagnetic Induction and AC

Master Faraday's law, Lenz's law, self and mutual inductance, and AC circuits for JEE Physics.

Electromagnetic Induction deals with the generation of emf due to changing magnetic flux. This forms the basis of generators, transformers, and AC circuits.

Overview

graph TD
    A[EM Induction] --> B[Faraday's Law]
    A --> C[Inductance]
    A --> D[AC Circuits]
    B --> B1[Induced EMF]
    B --> B2[Lenz's Law]
    C --> C1[Self Inductance]
    C --> C2[Mutual Inductance]
    D --> D1[LCR Circuit]
    D --> D2[Resonance]

Faraday’s Law of Induction

The induced emf in a circuit equals the negative rate of change of magnetic flux:

$$\boxed{\varepsilon = -\frac{d\Phi_B}{dt}}$$

For N turns:

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

Magnetic Flux

$$\Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta$$

Ways to Change Flux

  1. Change magnetic field B
  2. Change area A
  3. Change angle θ
  4. Move loop in non-uniform field

Lenz’s Law

The induced current opposes the change that produces it.

This is the reason for the negative sign in Faraday’s law.

JEE Tip
To find direction: If flux is increasing, induced current creates opposing field. If decreasing, induced current creates field in same direction.

Motional EMF

For a rod moving in magnetic field:

$$\varepsilon = Blv$$

For a rotating rod:

$$\varepsilon = \frac{1}{2}B\omega l^2$$

Eddy Currents

Currents induced in bulk conductors when exposed to changing magnetic fields.

Applications:

  • Electromagnetic braking
  • Induction heating
  • Metal detectors

Reduction: Laminated cores

Self Inductance

$$\Phi = LI$$ $$\varepsilon = -L\frac{dI}{dt}$$

For a solenoid:

$$L = \mu_0 n^2 V = \frac{\mu_0 N^2 A}{l}$$

Energy stored:

$$U = \frac{1}{2}LI^2$$

Mutual Inductance

$$\Phi_{12} = MI_2$$ $$\varepsilon_1 = -M\frac{dI_2}{dt}$$

Coupling coefficient:

$$M = k\sqrt{L_1 L_2}$$

where $0 \leq k \leq 1$

AC Circuits

AC Voltage

$$V = V_0 \sin(\omega t)$$
  • Peak value: $V_0$
  • RMS value: $V_{rms} = \frac{V_0}{\sqrt{2}}$
  • Mean value: 0 (over full cycle)

Pure Resistive Circuit

$$I = \frac{V_0}{R}\sin(\omega t)$$

Current in phase with voltage.

Pure Inductive Circuit

$$I = \frac{V_0}{\omega L}\sin(\omega t - \frac{\pi}{2})$$

Current lags voltage by 90°. Inductive reactance: $X_L = \omega L$

Pure Capacitive Circuit

$$I = \omega C V_0 \sin(\omega t + \frac{\pi}{2})$$

Current leads voltage by 90°. Capacitive reactance: $X_C = \frac{1}{\omega C}$

LCR Series Circuit

Impedance

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Phase Angle

$$\tan\phi = \frac{X_L - X_C}{R}$$

Current

$$I = \frac{V}{Z}$$

Resonance

At resonance: $X_L = X_C$

$$\omega_0 = \frac{1}{\sqrt{LC}}$$ $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$

At resonance:

  • Impedance Z = R (minimum)
  • Current maximum
  • Phase angle φ = 0

Quality Factor

$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

Power in AC Circuits

$$P = V_{rms} I_{rms} \cos\phi$$

where $\cos\phi$ = power factor

CircuitPower FactorPower
Pure R1$I^2R$
Pure L00
Pure C00
LCR at resonance1$I^2R$

Transformer

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$

Step-up: $N_s > N_p$ Step-down: $N_s < N_p$

Efficiency: $\eta = \frac{P_{out}}{P_{in}}$

Practice Problems

  1. A coil has 100 turns. If flux changes from 0.1 Wb to 0.3 Wb in 0.02 s, find induced emf.

  2. An LCR circuit has L = 1 H, C = 1 μF, R = 100 Ω. Find resonance frequency and Q-factor.

  3. A transformer steps down 220 V to 22 V. If secondary has 100 turns, find primary turns.

Quick Check
Why are transformer cores made of laminated sheets?

Further Reading