Electromagnetism · Chapter 16

Electromagnetic Induction

A changing magnetic flux generates an electric field. This single principle — Faraday's law — is the basis of every electrical generator, transformer, and induction motor ever built.

Prerequisites

Magnetic fields (Ch. 15) \cdot Basic calculus — derivatives and the concept of rate of change

Learning Goals

Calculate magnetic flux through a surface and identify the ways flux can change.
Use Faraday's law to find induced EMF from changing B, area, or angle.
Apply Lenz's law to determine the direction of induced current.
Explain motional EMF from the magnetic force on moving charges.
Connect induction to generators, transformers, and back-EMF in motors.

16.1 Magnetic Flux

Before stating Faraday's law, we need the concept of magnetic fluxthrough a surface. Flux is the total amount of magnetic field threading through an area, accounting for the angle between $\mathbf{B}$ and the surface:

\Phi_B = \int \mathbf{B}\cdot d\mathbf{A} = BA\cos\theta \qquad \text{(uniform field, flat surface)}

(16.1)

The SI unit of magnetic flux is the weber ( $\mathrm{Wb} = \mathrm{T\,m^2}$ ). Flux is maximum when $\mathbf{B}$ is perpendicular to the surface ( $\theta = 0^\circ$ , area faces the field) and zero when $\mathbf{B}$ is parallel to the surface ( $\theta = 90^\circ$ , field skims the area).

16.2 Faraday's Law

Definition 16.1 — Faraday's Law of Electromagnetic Induction

The electromotive force (EMF) induced in a closed loop is equal to the negative rate of change of magnetic flux through the loop:

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

where

N

is the number of turns in a coil. The negative sign encodes Lenz's law.

The key insight is that flux can change in three ways: (1) the magnitude of $\mathbf{B}$ changes, (2) the area of the loop changes, or (3) the angle between $\mathbf{B}$ and the loop changes. Any of these produces an EMF and, if the circuit is closed, a current.

Figure 16.1. Induction simulator. Move the magnet slowly, then quickly: the galvanometer responds to the rate of flux change, not the flux itself. Reverse the magnet direction to test Lenz's law, and increase the number of turns to see why coils amplify induced EMF.

Definition 16.2 — Lenz's Law

The induced current flows in a direction such that the magnetic field it creates opposes the change in flux that caused it.If the flux through a loop is increasing, the induced current creates a field opposing that increase. If flux is decreasing, the induced current tries to maintain it. Lenz's law is a consequence of energy conservation — you must do work to move the magnet.

16.3 Motional EMF

A conductor of length $L$ moving with velocity $\mathbf{v}$ through a magnetic field $\mathbf{B}$ experiences a force $q\mathbf{v}\times\mathbf{B}$ on every charge carrier. This separates charges and creates a potential difference — a motional EMF:

\varepsilon = BLv \qquad \text{(conductor perpendicular to both }\mathbf{B}\text{ and }\mathbf{v}\text{)}

(16.2)

This is the operating principle of the electric generator: a coil of wire rotates in a magnetic field, its area projected onto the field direction varies as $\cos(\omega t)$ , so $\Phi_B = NBA\cos(\omega t)$ , and $\varepsilon = NBA\omega\sin(\omega t)$ . The output is sinusoidal alternating current (AC) — which is why the power grid runs on AC.

Example 16.1 — EMF from a Changing Magnetic Field

A circular loop of radius 10 cm lies in a uniform magnetic field that increases from 0.2 T to 0.8 T in 0.5 s. Find the induced EMF.

Area:

A = \pi r^2 = \pi(0.10)^2 = 3.14\times10^{-2}\,\mathrm{m^2}

Change in flux:

\Delta\Phi = \Delta B\,A = (0.8 - 0.2)(3.14\times10^{-2}) = 1.885\times10^{-2}\,\mathrm{Wb}

EMF:

\varepsilon = -\Delta\Phi/\Delta t = -(1.885\times10^{-2})/0.5 = -0.0377\,\mathrm{V} \approx 37.7\,\mathrm{mV}

Direction:By Lenz's law, the induced current opposes the increasing flux — it flows to create

\mathbf{B}

opposing the increase.

16.4 Inductance and Transformers

A coil opposes changes in current through itself because any change in current changes the flux through its own loops — this is self-inductance. The induced back-EMF is:

\varepsilon = -L dI/dt \qquad (self-inductance)

(16.3)

The inductance $L$ (in henries, $\mathrm{H} = \mathrm{V\,s/A}$ ) depends only on the coil geometry. For a solenoid: $L = \mu_0n^2V$ , where $n$ is turns per meter and $V$ is the volume. Energy stored in an inductor: $U = \frac{1}{2}LI^2$ .

A transformer uses mutual inductance — the flux from one coil threading a second coil — to step voltage up or down. With $N_1$ turns in the primary and $N_2$ in the secondary, and assuming all flux is shared:

\frac{V_2}{V_1} = \frac{N_2}{N_1} \qquad \text{(ideal transformer)}

(16.4)

Power is conserved ( $I_1V_1 = I_2V_2$ ), so stepping voltage up steps current down by the same factor. This is why electrical power is transmitted at high voltage (low current, low resistive loss $I^2R$ ) and stepped down near homes. Without the transformer — without induction — the modern power grid would be impossible.

Definition 16.3 — Common Traps

Flux is not just field strength: it also depends on area and angle.
Constant flux gives no EMF: a large steady flux induces nothing unless it changes.
The minus sign is physical: Lenz's law enforces energy conservation by opposing the change.
Motional EMF needs geometry: $\varepsilon = BLv assumes B, L, and v are mutually perpendicular$
Transformers need changing current: ideal transformers work with AC, not steady DC.

Example 16.2 — Step-Down Transformer

A transformer has 2000 primary turns and 100 secondary turns. The primary is conne $cted to 240 V AC. Find the secondary voltage and the secondary current if the load is 12 \Omega$

Turns ratio:

N_2/N_1 = 100/2000 = 1/20

Secondary voltage:

V_2 = V_1(N_2/N_1) = 240(1/20) = 12\,\mathrm{V}

Secondary current:

I_2 = V_2/R = 12/12 = 1\,\mathrm{A}

Primary current:

I_1 = I_2(N_2/N_1) = 1(1/20) = 0.05\,\mathrm{A}

(power conserved:

240(0.05)=12(1)=12\,\mathrm{W}

✓)

Exercises — 16.1–16.4 Electromagnetic Induction

A circular loop of radius 10 cm sits in a field that increases from 0.2 T to 0.8 T in 0.5 s. Find the induced EMF.

Straightforward

A transformer with 2000 primary and 100 secondary turns connects to 240 V. Find the secondary voltage.

Straightforward

A conducting rod of length 0.4 m moves at 3 m/s perpendicular to a 0.4 T magnetic field. Find the induced EMF.

Intermediate

4.A bar magnet (N-pole leading) is pushed toward a coil, then pulled away. In each case, determine the direction of induced current and the force on the magnet. Explain how this exemplifies energy conservation.

Intermediate

5.Derive the sinusoidal EMF output of an AC generator from first principles using Faraday's law. What determines the peak voltage? How does the motor differ from the generator in principle?

Challenging

Key Takeaways

$Magnetic flux: \Phi_B = BA cos \theta — the total field threading an area$ .
$Faraday's law: \varepsilon = -N d\Phi/dt — any change in flux (from B, area, or angle) induces an EMF$ .
Lenz's law: induced current opposes the flux change that caused it (energy conservation).
$Motional EMF: \varepsilon = BLv for a rod of length L moving at speed v in field B$ .
$Self-inductance: \varepsilon = -L dI/dt — a coil resists changes in its own current$ .
$Transformer: V_{2}/V_{1} = N_{2}/N_{1} — steps voltage up or down; enables the power grid$ .