Electromagnetic Induction

Overview

Electromagnetic induction is the production of an electromotive force (EMF) by changing magnetic flux. This principle is fundamental to generators, transformers, and many modern technologies.

Magnetic Flux

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

For uniform field perpendicular to area:

$$\Phi_B = BA \cos(\theta)$$

Where $\theta$ is angle between $\vec{B}$ and area normal

Unit: Weber (Wb) = T·m² = V·s

Faraday's Law

The induced EMF equals the negative rate of change of magnetic flux:

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

For $N$ loops:

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

Ways to Change Flux

1. Change magnetic field $B$
2. Change area $A$
3. Change angle $\theta$ between $\vec{B}$ and $\vec{A}$
4. Move loop in non-uniform field

Lenz's Law

The induced current creates a magnetic field that opposes the change in flux.

• If $\Phi$ increases → induced $\vec{B}$ opposes the field
• If $\Phi$ decreases → induced $\vec{B}$ reinforces the field

The negative sign in Faraday's law represents Lenz's law.

Motional EMF

A conductor of length $L$ moving with velocity $v$ through field $B$:

$$\varepsilon = BLv \sin(\theta)$$

For perpendicular motion:

$$\varepsilon = BLv$$

Sliding Bar

A bar sliding on rails in a magnetic field:

$$\varepsilon = BLv$$ $$I = \frac{BLv}{R}$$ $$F_{\text{drag}} = BIL = \frac{B^2 L^2 v}{R}$$

Generators

AC Generator

Rotating coil ($N$ turns, area $A$) in magnetic field $B$:

$$\Phi = NBA \cos(\omega t)$$ $$\varepsilon = NBA\omega \sin(\omega t) = \varepsilon_0 \sin(\omega t)$$

Peak EMF:

$$\varepsilon_0 = NBA\omega$$

DC Generator

Uses commutator to convert AC to pulsating DC.

Eddy Currents

Currents induced in bulk conductors by changing magnetic field:

• Cause heating ($I^2R$ losses)
• Create drag forces
• Used in braking systems
• Minimized by laminating cores

Self-Inductance

Inductance

Ratio of flux linkage to current:

$$L = \frac{N\Phi_B}{I}$$

Unit: Henry (H) = Wb/A = V·s/A

Self-Induced EMF

$$\varepsilon_L = -L\frac{dI}{dt}$$

Inductance of Solenoid

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

Where $V = A\ell$ is volume

Energy Stored in Inductor

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

Energy Density

$$u = \frac{B^2}{2\mu_0}$$

Mutual Inductance

EMF induced in one coil by changing current in another:

$$\varepsilon_2 = -M\frac{dI_1}{dt}$$ $$M = k\sqrt{L_1 L_2}$$

Where $k$ is coupling coefficient ($0 \leq k \leq 1$)

RL Circuits

Current Growth (Closing Switch)

$$I(t) = \frac{\varepsilon}{R}(1 - e^{-t/\tau})$$

Current Decay (Opening Switch)

$$I(t) = I_0 e^{-t/\tau}$$

Time Constant

$$\tau = \frac{L}{R}$$

Transformers

Relationship between primary and secondary:

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

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

Power conservation (ideal):

$$P_p = P_s \to V_p I_p = V_s I_s$$

Examples

Example 1: Faraday's Law

A coil (100 turns, area 0.01 m²) in a field that changes from 0.5 T to 0 in 0.1 s.

$$\Delta\Phi = A\Delta B = 0.01 \times (-0.5) = -0.005 \text{ Wb}$$ $$\varepsilon = -N\frac{\Delta\Phi}{\Delta t} = -100 \times \frac{-0.005}{0.1} = 5 \text{ V}$$

Example 2: Motional EMF

A 50 cm rod moves at 10 m/s perpendicular to $B = 0.2$ T.

$$\varepsilon = BLv = 0.2 \times 0.5 \times 10 = 1 \text{ V}$$

Example 3: Generator

A coil (500 turns, 20 cm × 30 cm) rotates at 60 Hz in $B = 0.1$ T.

$$A = 0.2 \times 0.3 = 0.06 \text{ m}^2$$ $$\omega = 2\pi f = 377 \text{ rad/s}$$ $$\varepsilon_0 = NBA\omega = 500 \times 0.1 \times 0.06 \times 377 = 1131 \text{ V}$$

Example 4: Inductor

A solenoid (1000 turns, length 50 cm, radius 2 cm) in a circuit.

$$A = \pi(0.02)^2 = 1.26 \times 10^{-3} \text{ m}^2$$ $$L = \frac{\mu_0 N^2 A}{\ell} = \frac{4\pi \times 10^{-7} \times 10^6 \times 1.26 \times 10^{-3}}{0.5}$$ $$L = 3.16 \times 10^{-3} \text{ H} = 3.16 \text{ mH}$$

Example 5: RL Circuit

An RL circuit ($R = 100$ Ω, $L = 0.5$ H) connected to 10 V. Find current after 5 ms.

$$\tau = \frac{L}{R} = \frac{0.5}{100} = 5 \text{ ms}$$ $$I_{\max} = \frac{\varepsilon}{R} = \frac{10}{100} = 0.1 \text{ A}$$ $$I = 0.1(1 - e^{-1}) = 0.1 \times 0.632 = 63.2 \text{ mA}$$

Example 6: Transformer

A step-down transformer ($N_p = 1000$, $N_s = 50$) has primary at 120 V AC. Secondary load draws 10 A.

$$V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{50}{1000} = 6 \text{ V}$$ $$I_p = I_s \times \frac{N_s}{N_p} = 10 \times \frac{50}{1000} = 0.5 \text{ A}$$

Applications

• Electric generators: Mechanical to electrical energy
• Transformers: Voltage conversion
• Induction motors: AC motors
• Induction heating: Eddy current heating
• Wireless charging: Mutual inductance