Physics 9702 · Paper 4

Faraday's Law

Change the flux through a circuit — by moving it, spinning it, or growing the field — and an e.m.f. appears. One law sets its size, and it works in both directions.

The statement (two marks)

E = \dfrac{\Delta(N\Phi)}{\Delta t}

Symbols

$E$ = induced e.m.f. (V)
$\Delta(N\Phi)$ = change in flux linkage (Wb)
$\Delta t$ = time taken for the change (s)

: the induced e.m.f. is proportional to the rate of change of flux linkage. The proportionality claim is M1; completing “rate of change of flux linkage” is A1 — identical wording across three sessions (9702/42/M/J/24 Q7(a)) (9702/42/F/M/25 Q7(a)) (9702/41/O/N/25 Q7(a)).
Why e.m.f. and not just voltage: pushing the conductor does work on the charges, so the loop acts as a source of electrical energy — a battery made of motion.

Worked example

A coil in a growing field (2023 paper)

A 3000-turn coil of area $3.4\times10^{-4}$ m² sits with its plane perpendicular to a field that grows steadily from 0 to 4.0 $\mu\text{T}$ in 0.020 s. Show the induced e.m.f. is $2.0\times10^{-4}$ V (9702/42/O/N/23 Q7(b)(i)).

E.m.f. = rate of change of flux linkage: $E = NA\,\dfrac{\Delta B}{\Delta t}$ .
$E = 3000 \times 3.4\times10^{-4} \times \dfrac{4.0\times10^{-6}}{0.020} = 2.0\times10^{-4}\ \text{V}$ .

Answer

E = 2.0\times10^{-4}\ \text{V}

Worked example

Faraday backwards: an aircraft's wings (2025)

An aircraft with wingspan 68 m flies through the Earth's vertical field component of 38 $\mu\text{T}$ , inducing 0.54 V across its wingtips. Find the flux cut in 15 s, and the aircraft's speed (9702/41/O/N/25 Q7(b)).

Reverse the law: flux cut = e.m.f. × time = $0.54 \times 15 = 8.1\ \text{Wb}$ .
That flux filled an area $A = \Phi/B = 8.1 / 38\times10^{-6} = 2.1\times10^{5}\ \text{m}^2$ .
The wings sweep a rectangle (speed × time) × span: $v = \dfrac{2.1\times10^{5}}{15 \times 68} = 210\ \text{m s}^{-1}$ .

Answer

\Phi = 8.1\ \text{Wb};\quad v \approx 210\ \text{m s}^{-1}

A rod on the move: E = BLv

A rod of length

L

moving sideways at speed

v

sweeps area

Lv

every second, so it cuts flux at rate

BLv

— that is its e.m.f. A 2025 question stretched this: a uniformly accelerating rod has

v = at

, so

E = BLat

— the straight-line E–t graph through the origin was itself the evidence that

E \propto v

, and one point on it gave

B = 4.3\times10^{-5}

T (9702/42/F/M/25 Q7(b)).

PreviousMagnetic Flux & Flux Linkage NextFrom Φ–t Graphs to E–t Graphs

Faraday's Law

Change the flux through a circuit — by moving it, spinning it, or growing the field — and an e.m.f. appears. One law sets its size, and it works in both directions.

The statement (two marks)

E = \dfrac{\Delta(N\Phi)}{\Delta t}

Symbols

$E$ = induced e.m.f. (V)
$\Delta(N\Phi)$ = change in flux linkage (Wb)
$\Delta t$ = time taken for the change (s)

: the induced e.m.f. is proportional to the rate of change of flux linkage. The proportionality claim is M1; completing “rate of change of flux linkage” is A1 — identical wording across three sessions (9702/42/M/J/24 Q7(a)) (9702/42/F/M/25 Q7(a)) (9702/41/O/N/25 Q7(a)).
Why e.m.f. and not just voltage: pushing the conductor does work on the charges, so the loop acts as a source of electrical energy — a battery made of motion.

Worked example

A coil in a growing field (2023 paper)

E.m.f. = rate of change of flux linkage: $E = NA\,\dfrac{\Delta B}{\Delta t}$ .
$E = 3000 \times 3.4\times10^{-4} \times \dfrac{4.0\times10^{-6}}{0.020} = 2.0\times10^{-4}\ \text{V}$ .

Answer

E = 2.0\times10^{-4}\ \text{V}

Worked example

Faraday backwards: an aircraft's wings (2025)

Reverse the law: flux cut = e.m.f. × time = $0.54 \times 15 = 8.1\ \text{Wb}$ .
That flux filled an area $A = \Phi/B = 8.1 / 38\times10^{-6} = 2.1\times10^{5}\ \text{m}^2$ .
The wings sweep a rectangle (speed × time) × span: $v = \dfrac{2.1\times10^{5}}{15 \times 68} = 210\ \text{m s}^{-1}$ .

Answer

\Phi = 8.1\ \text{Wb};\quad v \approx 210\ \text{m s}^{-1}

A rod on the move: E = BLv

A rod of length

L

moving sideways at speed

v

sweeps area

Lv

every second, so it cuts flux at rate

BLv

— that is its e.m.f. A 2025 question stretched this: a uniformly accelerating rod has

v = at

, so

E = BLat

— the straight-line E–t graph through the origin was itself the evidence that

E \propto v

, and one point on it gave

B = 4.3\times10^{-5}

T (9702/42/F/M/25 Q7(b)).