IB Physics HLTopic 4 — InductionPaper 1 & 2ε = ε0 sin ωt~17 min read
AC Generators
This is where the whole Induction chapter pays off. Take everything — induced e.m.f., flux linkage, Faraday, Lenz — and simply spin a coil steadily in a magnetic field. As it turns, the flux linkage rises and falls, so the induced e.m.f. swings positive, back to zero, negative, and round again. That smooth back-and-forth voltage is alternating current, and the machine that produces it — sitting in every power station on the planet — is the AC generator.
📘 What you need to know
An AC generator (alternator) turns mechanical energy into alternating electrical energy
A coil is spun in a uniform magnetic field, connected to the circuit by slip rings and brushes
As the coil rotates, the flux linkage changes, inducing an alternating e.m.f. (Faraday’s law)
Flux linkage for a rotating coil is NΦ = BAN cos ωt
The induced e.m.f. is ε = ε0 sin ωt, where the peak is ε0 = BANω
The e.m.f. is 90° out of phase with the flux linkage (e.m.f. is a sine, flux a cosine)
e.m.f. is maximum when the coil is parallel to the field (cutting fastest), zero when perpendicular
ω = 2πf is the angular speed, so frequency and angular speed are linked
Spinning the coil faster increases both the peak e.m.f. and the frequency
How the generator works
A rectangular coil is forced to spin in a uniform magnetic field. Its ends connect to the external circuit through two slip rings pressed by carbon brushes — a sliding contact that lets the coil turn freely without the wires tangling.
As the coil spins, its sides cut through the field lines, so the flux linkage changes
By Faraday’s law, this changing flux linkage induces an e.m.f.
Because the coil’s motion reverses relative to the field twice per turn, the e.m.f. alternates in direction
A centre-reading meter shows the pointer swinging both ways — the signature of AC
A coil spins between the poles; slip rings and brushes carry the alternating e.m.f. to the circuit. The centre-reading meter’s pointer swings both ways as the current reverses each half-turn.
Why the e.m.f. alternates
The key is that the coil’s flux linkage follows a cosine as it turns, and Faraday’s law makes the e.m.f. the rate of change of that — which is a sine. The two are a quarter-cycle apart, so the crucial moments swap over:
Coil orientation
Flux linkage
Cutting field lines
Induced e.m.f.
Perpendicular to field (face-on)
Maximum
Not cutting (moving along lines)
Zero
Parallel to field (edge-on)
Zero
Cutting fastest
Maximum
This is the same “max flux, zero e.m.f.” twist from Faraday’s law, just spinning continuously. When the coil is face-on, it’s momentarily not cutting any lines — its sides are sliding along the field — so the e.m.f. dips to zero even though the flux is greatest. A quarter-turn later, edge-on, the sides slice straight across the lines and the e.m.f. peaks. Picture the coil’s sides and ask “am I cutting lines right now?” — that instantly tells you the e.m.f.
The equations
Starting from the rotating-coil flux linkage and using angular speed ω = 2πf:
Flux linkage of a rotating coilNΦ = BAN cos ωt
Faraday’s law says the e.m.f. is the rate of change of this — and differentiating a cosine gives a sine:
Induced e.m.f. of a rotating coilε = ε0 sin ωt where ε0 = BANωε0 = peak e.m.f. (V) • ω = angular speed (rad s−1) • ω = 2πf
A quick warning the examiners love: the equation ε = ε0 sin ωt is not in your data booklet — you must recognise and recall it. And since ωt is in radians, put your calculator in radian mode before you take that sine, or every instantaneous value will come out wrong.
The generator output is a clean sine wave: ε = ε0 sin ωt. It peaks when the coil is edge-on (cutting fastest) and is zero when face-on.
WE 1
A generator coil of 200 turns and cross-sectional area 8.0 × 10−3 m2 spins at 50 Hz in a uniform field of flux density 0.12 T. Calculate the peak e.m.f. produced.
Step 1 — angular speed from the frequencyω = 2πf = 2π × 50 = 314 rad s⁻¹Step 2 — peak e.m.f.ε₀ = BANω = (0.12)(8.0 × 10⁻³)(200)(314)ε₀ = 60 VRemember ε₀ is the peak value — the top of the sine wave — not the average. And convert the frequency to angular speed with ω = 2πf before you substitute; forgetting the 2π is the classic slip here.
Spinning the coil faster
Turn up the rotation speed and ω rises. Look at the two equations: ω appears in the peak (ε0 = BANω) and in the frequency (ω = 2πf). So faster spinning does two things at once:
The peak e.m.f. increases (taller waves) — because ε0 ∝ ω
The frequency increases (waves closer together, shorter period) — because f ∝ ω
Spin faster (bigger ω)
ε0 = BANω
Taller peaks
and f = ω/2π
Higher frequency
Double the angular speed and the output (red) is twice as tall and packed twice as close together — both the peak e.m.f. and the frequency double.
WE 2
An AC generator produces a peak e.m.f. ε at frequency f. The rotational speed of the coil is then doubled. State the new peak e.m.f. and the new frequency, with reasoning.
Step 1 — link angular speed and frequencyω = 2πf, so f ∝ ω
Doubling the speed doubles ω, so the frequency doubles.
Step 2 — effect on the peak e.m.f.ε₀ = BANω, so ε₀ ∝ ω
Doubling ω doubles the peak e.m.f. too.
new peak e.m.f. = 2ε, new frequency = 2fBoth double because ω sits in both relationships. A very common exam mistake is to change only one — but spinning faster raises the peaks and squeezes the waves closer together at the same time.
WE 3
A coil of 120 turns and area 5.0 × 10−3 m2 spins at 25 Hz in a field of 0.30 T. (a) Calculate the peak e.m.f. (b) Calculate the instantaneous e.m.f. 5.0 ms after it passes through zero (going positive).
(a) Step 1 — angular speed, then peak e.m.f.ω = 2πf = 2π × 25 = 157 rad s⁻¹ε₀ = BANω = (0.30)(5.0 × 10⁻³)(120)(157)ε₀ = 28 V(b) Step 2 — use ε = ε₀ sin ωt (radian mode!)ωt = 157 × (5.0 × 10⁻³) = 0.785 radε = 28 × sin(0.785) = 28 × 0.707ε = 20 V0.785 rad is π/4, or an eighth of a cycle, so the e.m.f. has climbed to sin 45° = 0.707 of its peak. If your calculator gave a tiny number, it was in degrees — switch to radians and try again.
⚡ Working an AC-generator question
Angular speed first:ω = 2πf (frequency → rad s−1).
Peak e.m.f.:ε0 = BANω — the top of the sine wave.
Instantaneous e.m.f.:ε = ε0 sin ωt, calculator in radians.
Faster spin? Both ε0 and f scale with ω — change both.
Where’s the peak? Coil edge-on (parallel to field, cutting fastest).
Convert units: area in m2, B in T.
💡 Top tips
ε = ε0 sin ωt is not in the data booklet — learn it.
Always start with ω = 2πf to get angular speed.
For instantaneous values, put your calculator in radian mode.
e.m.f. peaks when the coil is edge-on (parallel to field), zero when face-on.
Faster spinning increases both the peak e.m.f. and the frequency.
The e.m.f. (sine) is 90° out of phase with the flux linkage (cosine).
⚠ Common mistakes
Forgetting ω = 2πf and using f directly in ε0 = BANω
Leaving the calculator in degree mode for ε = ε0 sin ωt
Saying the e.m.f. is maximum when the coil is face-on — it’s zero there
Changing only the frequency (or only the peak) when the coil spins faster — both change
Confusing peak e.m.f. ε0 with the instantaneous or average value
Mixing up the flux-linkage cosine with the e.m.f. sine — they’re 90° apart
Quick recap: An AC generator spins a coil in a magnetic field, using slip rings and brushes to feed the alternating output to the circuit. The changing flux linkage NΦ = BAN cos ωt induces an e.m.f. ε = ε0 sin ωt with peak ε0 = BANω. The e.m.f. is a sine, 90° out of phase with the cosine flux linkage — maximum when the coil is edge-on (cutting fastest) and zero when face-on. Since ω = 2πf, spinning faster raises both the peak e.m.f. and the frequency.
And that completes the Induction chapter — from a single moving rod all the way to the generators that light up cities. You can now find an induced e.m.f., calculate flux and flux linkage, apply Faraday’s and Lenz’s laws, and describe how alternating current is made. Everything after this — transformers, the grid, AC circuits — is built on exactly these five ideas. Well done getting through the whole set.
AC generators spinning your head?
Book a free meeting and we’ll work through ε = ε0 sin ωt, the peak-e.m.f. formula, and why faster spinning changes everything.
