Electromagnetic Induction: Faraday's and Lenz's Laws

Electromagnetic induction — the generation of an EMF by a changing magnetic flux — is one of the most important phenomena in physics. It is the principle behind generators, transformers, induction hobs, and wireless charging. Michael Faraday and Joseph Henry independently discovered the effect in 1831, and Faraday formulated the law that bears his name.

Faraday’s Law

Faraday’s law states that the magnitude of the induced EMF is equal to the rate of change of flux linkage through the circuit:

ε = −N dΦ/dt = −d(NΦ)/dt

where:

ε is the induced EMF (V)
N is the number of turns
dΦ/dt is the rate of change of magnetic flux (Wb s⁻¹)
The negative sign represents Lenz’s law (see below)

In many exam questions, the change is uniform over a time interval, so:

ε = NΔΦ / Δt

(taking the magnitude and applying Lenz’s law separately for direction).

Worked Example 1

A coil of 200 turns has a flux through it that decreases uniformly from 5.0 × 10⁻³ Wb to 1.0 × 10⁻³ Wb in 0.020 s. What is the induced EMF?

ΔΦ = 5.0 × 10⁻³ − 1.0 × 10⁻³ = 4.0 × 10⁻³ Wb

ε = NΔΦ / Δt = 200 × 4.0 × 10⁻³ / 0.020 = 40 V

Worked Example 2: Inducing an EMF by Changing Area

A square coil of side 0.10 m with 50 turns is in a uniform field of 0.30 T (perpendicular to the coil face). The coil is compressed (squashed) so that its area halves in 0.050 s. Calculate the average induced EMF.

Initial flux linkage: NΦ₁ = NBA = 50 × 0.30 × 0.010 = 0.15 Wb

Final flux linkage: NΦ₂ = 50 × 0.30 × 0.005 = 0.075 Wb

ε = Δ(NΦ) / Δt = (0.15 − 0.075) / 0.050 = 0.075 / 0.050 = 1.5 V

Lenz’s Law

Lenz’s law states that the direction of the induced EMF (and hence the induced current, if the circuit is complete) is such that it opposes the change that produces it.

This is a direct consequence of the conservation of energy. If the induced current aided the change rather than opposing it, the process would feed on itself and energy would be created from nothing.

How to Apply Lenz’s Law Systematically

Identify the change in flux (is it increasing or decreasing?)
The induced current creates a magnetic field that opposes this change
If flux is increasing → induced current creates a field opposing the external field
If flux is decreasing → induced current creates a field in the same direction as the external field
Use the right-hand grip rule to find which direction of current produces the required field

Examples of Lenz’s Law in Action

Magnet moving into a coil: When a north pole of a magnet moves towards a coil, the increasing flux induces a current that creates a magnetic field opposing the approach — the coil acts as a north pole facing the magnet, repelling it. You must do work to push the magnet in, and this work is converted to electrical energy.

Magnet moving away from a coil: When the north pole moves away, the decreasing flux induces a current that creates a south pole facing the retreating magnet, attracting it and opposing its departure.

Falling magnet through a metal tube: A magnet falling through a copper tube induces currents in the tube that create magnetic fields opposing the magnet’s motion. The magnet falls much more slowly than it would in free fall — the gravitational potential energy is converted to electrical energy (and then heat) in the tube.

The EMF of a Straight Conductor

When a straight conductor of length L moves with velocity v perpendicular to a magnetic field B, an EMF is induced:

ε = BLv

This can be derived from Faraday’s law: the conductor sweeps out an area at rate dA/dt = Lv, so the rate of change of flux is dΦ/dt = B × Lv.

Worked Example 3

A metal rod of length 0.50 m moves at 4.0 m s⁻¹ perpendicular to a magnetic field of 0.20 T. What EMF is induced across the rod?

ε = BLv = 0.20 × 0.50 × 4.0 = 0.40 V

Worked Example 4: Aircraft Wing as a Moving Conductor

A Boeing 747 has a wingspan of 64 m and flies at 250 m s⁻¹ through the Earth’s magnetic field. The vertical component of the Earth’s field is 40 μT. Calculate the EMF induced across the wingspan.

ε = BLv = 40 × 10⁻⁶ × 64 × 250 = 0.64 V

This is a real (though small) EMF. It cannot drive a useful current because there is no complete circuit, but it demonstrates that electromagnetic induction is happening everywhere around us.

The AC Generator (Alternator)

A coil rotating in a uniform magnetic field experiences a continuously changing flux linkage:

NΦ = NBA cos(ωt)

By Faraday’s law, the induced EMF is:

ε = NBAω sin(ωt)

This gives a sinusoidal EMF with:

Peak EMF: ε₀ = NBAω
Frequency: f = ω / (2π)

The EMF is maximum when the coil is parallel to the field (flux linkage is zero but changing most rapidly) and zero when the coil is perpendicular to the field (flux linkage is maximum but momentarily not changing).

How to Increase the Peak EMF

Looking at ε₀ = NBAω, you can increase the peak EMF by:

Increasing N (more turns)
Increasing B (stronger magnet)
Increasing A (larger coil)
Increasing ω (faster rotation)

Each of these is directly proportional to ε₀.

Worked Example 5: Generator Calculation

A generator has a coil of 500 turns, area 0.080 m², rotating at 50 revolutions per second in a field of 0.30 T. Calculate the peak EMF.

ω = 2π × 50 = 314 rad s⁻¹

ε₀ = NBAω = 500 × 0.30 × 0.080 × 314 = 3770 V ≈ 3.8 kV

Worked Example 6: Generator at a Specific Time

Using the generator above, what is the EMF at t = 0.003 s?

ε = 3770 × sin(314 × 0.003) = 3770 × sin(0.942) = 3770 × 0.809 = 3050 V ≈ 3.1 kV

Eddy Currents

When a conducting material (not just a wire coil) moves through a magnetic field, or is exposed to a changing field, currents are induced in the bulk of the material. These are called eddy currents.

By Lenz’s law, eddy currents always flow in a direction that opposes the change causing them. This creates:

A braking force (used in electromagnetic brakes on trains and roller coasters)
Heating (used in induction cookers)
Energy losses in transformer cores (undesirable — minimised by using laminated cores)

How Induction Cooking Works

An induction hob has a coil beneath the cooking surface that carries a high-frequency alternating current (~25 kHz). This creates a rapidly changing magnetic field. When a ferromagnetic pan is placed on the hob, large eddy currents are induced in the base of the pan, heating it directly. The hob surface stays relatively cool because it is ceramic (non-conducting). Aluminium or copper pans do not work on induction hobs because they are not ferromagnetic and the eddy currents are smaller.

Back-EMF in Motors

When a motor runs, its rotating coil cuts through the magnetic field, inducing an EMF that opposes the supply voltage. This is called the back-EMF.

As the motor speeds up, the back-EMF increases, reducing the net voltage across the coil and therefore reducing the current. This is why motors draw a large current when starting (back-EMF is zero) and a smaller current at full speed.

The relationship is: V_supply = back-EMF + IR_coil

At full speed: back-EMF ≈ V_supply, so the current is small.

Worked Example 7: Back-EMF

A DC motor draws 8.0 A when stalled (not rotating) from a 24 V supply. At full speed, it draws 2.0 A. What is the back-EMF at full speed?

When stalled: R = V/I = 24/8.0 = 3.0 Ω

At full speed: 24 = back-EMF + 2.0 × 3.0 = back-EMF + 6.0

Back-EMF = 18 V

The motor converts electrical power into mechanical power. The power delivered to the back-EMF (which becomes mechanical power) is: P = back-EMF × I = 18 × 2.0 = 36 W. The power wasted as heat is: P = I²R = 2.0² × 3.0 = 12 W. Total: 36 + 12 = 48 W = VI = 24 × 2.0 ✔

Real-World Applications

Regenerative braking: In electric and hybrid vehicles, when the driver brakes, the motor is used as a generator. The kinetic energy of the vehicle is converted to electrical energy (by electromagnetic induction) and stored in the battery. This extends the vehicle’s range. The same principle is used on electric trains.

Guitar pickups: A permanent magnet sits inside a coil beneath each guitar string. When the steel string vibrates, it changes the magnetic flux through the coil, inducing a small alternating EMF. This signal is amplified and sent to a loudspeaker. Different pickup designs (single-coil vs humbucker) produce different tonal characteristics.

Electromagnetic braking on roller coasters: At the end of a ride, permanent magnets are mounted on the track and copper or aluminium fins extend from the train. As the train passes through, eddy currents are induced in the fins, creating a strong braking force. No friction or moving parts are involved, making it highly reliable.

Common Exam Mistakes

Confusing flux and rate of change of flux: The induced EMF depends on how FAST the flux is changing, not on the flux itself. A large constant flux induces zero EMF.
Applying Lenz’s law incorrectly: The induced current opposes the CHANGE, not the flux itself. If flux is decreasing, the induced current tries to maintain it (same direction as external field).
Forgetting that ε = BLv applies only to a straight conductor: For a coil, use ε = NBAω sin(ωt) or ε = NΔΦ/Δt.
Mixing up peak and instantaneous EMF: ε₀ = NBAω is the peak. The instantaneous EMF is ε₀ sin(ωt).

Key Points

Faraday’s law: ε = −N dΦ/dt — the induced EMF equals the rate of change of flux linkage
Lenz’s law: the induced EMF opposes the change producing it (conservation of energy)
AC generators produce ε = NBAω sin(ωt) with peak EMF ε₀ = NBAω
EMF is maximum when the coil is parallel to the field (flux changing fastest)
For a straight conductor: ε = BLv
Eddy currents are induced in bulk conductors and always oppose the change
Back-EMF in motors opposes the supply voltage and limits the current at full speed

A-Level Deep Dive: Electromagnetic Induction

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields addresses electromagnetic induction as the production of an emf when the magnetic flux linkage through a circuit changes with time, encompassing Faraday's law (induced emf proportional to the rate of change of flux linkage), Lenz's law (the induced current opposes the change producing it), and the application of these laws to a moving straight conductor ( $\varepsilon = BLv$ ), rotating coils, transformers and eddy-current effects (refer to the official specification document for exact wording). Although introduced in Topic 7, electromagnetic induction reappears synoptically in Topic 11 (capacitor discharge through inductive loads — comparison of time constants), Topic 12 (alternating currents and the moving-coil generator) and is exam-tested on 9PH0 Paper 2 (Topics 6–8 + core) and the synoptic Paper 3. The Edexcel formula booklet provides $\varepsilon = -N\frac{d\Phi}{dt}$ and $\Phi = BA\cos\theta$ — the physical interpretation of the negative sign (Lenz's law) is not in the booklet and must be argued from energy conservation.

Worked example with full mark scheme

Question (8 marks):

A flat rectangular coil of $N = 200$ turns, length $L = 8.0\,\text{cm}$ and width $W = 5.0\,\text{cm}$ rotates at a steady angular speed $\omega = 50\pi\,\text{rad s}^{-1}$ about an axis perpendicular to a uniform magnetic field of flux density $B = 0.040\,\text{T}$ .

(a) Show that the flux linkage through the coil at time $t$ (taking the coil's normal parallel to the field at $t = 0$ ) is given by $N\Phi = NBA\cos(\omega t)$ and evaluate $NBA$ . (3)

(b) Hence derive an expression for the induced emf $\varepsilon(t)$ and calculate the peak emf. (3)

(c) Explain, with reference to Lenz's law, the sign of the induced emf at the instant $t = T/4$ (where $T$ is the rotational period). (2)

Solution with mark scheme:

(a) Step 1 — flux linkage at angle $\theta$ . The magnetic flux through one turn is $\Phi = BA\cos\theta$ , where $\theta$ is the angle between the field and the coil's normal. With $\theta = \omega t$ and $N$ turns:

$N\Phi = NBA\cos(\omega t)$

M1 — quoting $\Phi = BA\cos\theta$ and identifying $\theta = \omega t$ .

Step 2 — evaluate $NBA$ . Area $A = L \times W = 0.080 \times 0.050 = 4.0 \times 10^{-3}\,\text{m}^2$ .

$NBA = 200 \times 0.040 \times 4.0 \times 10^{-3} = 0.032\,\text{Wb}$

M1 — substitution with all SI units (cm $\to$ m). A1 — $NBA = 0.032\,\text{Wb}$ (or $3.2 \times 10^{-2}\,\text{Wb}$ ).

(b) Step 1 — apply Faraday's law.

$\varepsilon = -\frac{d(N\Phi)}{dt} = -\frac{d}{dt}\left[NBA\cos(\omega t)\right] = NBA\omega\sin(\omega t)$

M1 — correct differentiation; sign handled correctly (the two minus signs cancel, leaving a positive sine).

Step 2 — peak emf.

$\varepsilon_0 = NBA\omega = 0.032 \times 50\pi = 1.6\pi \approx 5.03\,\text{V}$

M1 — substitution. A1 — $\varepsilon_0 \approx 5.0\,\text{V}$ (2 s.f.).

(c) At $t = T/4$ , $\omega t = \pi/2$ , so the coil's normal is perpendicular to $B$ and the flux linkage is zero, but its rate of change is maximum. The induced current must, by Lenz's law, flow in a sense that creates a magnetic moment opposing the change in external flux — i.e. attempting to restore the previous (decreasing) flux through the coil.

M1 — identifies $d\Phi/dt$ maximum at this instant. A1 — sign of induced emf justified by Lenz's law: induced current opposes the change in flux; quoting energy conservation ("if the induced current aided the change, the coil would self-accelerate, violating energy conservation") earns the A1 unambiguously.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A bar magnet of mass $m = 0.050\,\text{kg}$ is dropped from rest down the vertical axis of a long copper tube. After a short transient, the magnet falls at constant terminal velocity $v_T = 0.12\,\text{m s}^{-1}$ .

(a) State, with a brief justification, the magnitude of the average eddy-current dissipation per unit time once terminal velocity is reached. Take $g = 9.81\,\text{m s}^{-2}$ . (3)

(b) Explain, using Faraday's law and Lenz's law, why a terminal velocity exists. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1) — at terminal velocity, net force is zero, so the magnetic retarding force equals the weight $mg$ .
M1 (AO2.1) — power dissipated equals $F_{\text{mag}} \cdot v_T = mg v_T$ by energy conservation (loss of GPE per unit time $=$ ohmic heating in the tube).
A1 (AO1.1) — $P = mgv_T = 0.050 \times 9.81 \times 0.12 \approx 0.059\,\text{W}$ .

(b)

M1 (AO1.2) — as the magnet falls, the flux through any horizontal cross-section of the tube changes $\Rightarrow$ Faraday's law gives an induced emf, hence eddy currents in the tube wall.
M1 (AO2.1) — by Lenz's law, the induced currents flow in a sense that opposes the change in flux, i.e. they create a magnetic field that retards the falling magnet.
A1 (AO3.1b) — as $v$ increases, $d\Phi/dt$ increases, so the induced emf, current and retarding force all increase; equilibrium is reached when this retarding force equals $mg$ , defining $v_T$ .

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 1, with one mark of AO1.2 (application). Topic 7 induction questions on 9PH0 Paper 2 typically use roughly this AO blend: procedural physics (Faraday/Lenz application) framed by an energy-conservation argument.

Synoptic links

Connects to:

Magnetic flux and flux linkage (Topic 7 earlier sub-strand): $\Phi = BA\cos\theta$ and $N\Phi$ are the inputs to Faraday's law. Without confident use of the cosine geometry, the rate-of-change calculation collapses. Every induction problem starts by writing $\Phi(t)$ explicitly.
Transformers (Topic 7, immediately after this lesson): the transformer equation $V_s/V_p = N_s/N_p$ is a direct consequence of applying Faraday's law to two coupled coils sharing the same time-varying flux. The "ideal transformer" assumption is precisely that all the primary flux links the secondary.