AC Generators — Changing Flux
On 28 October 1831, Michael Faraday at the Royal Institution, London, rotated a copper disc between the poles of a horseshoe magnet and detected a continuous direct current on his galvanometer — the world's first dynamo. He recorded in his diary: "Very distinct evidence of an induced current." Every power station turbine ever built — from Niagara Falls (1895) to Snowy 2.0 — is the direct successor of that copper disc spinning between two magnets.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A rectangular coil rotates at constant speed in a uniform magnetic field.
- At what orientation is the flux through the coil maximum? Minimum?
- At what orientation is the rate of change of flux maximum?
- Will the induced emf be a steady DC or an alternating AC voltage?
Warm-up — Faraday's law tells us that the induced emf depends on…
Know — AC Generator Structure
- AC generator: coil rotating in magnetic field, slip rings, brushes
- Flux through coil: $\Phi = NBA\cos(\omega t)$
- Induced emf: $\varepsilon = NBA\omega\sin(\omega t)$
Understand — Flux, Rate of Change, and Emf
- Maximum flux occurs when the coil is perpendicular to $B$ ($\omega t = 0$)
- Maximum emf occurs when flux is changing fastest ($\omega t = 90°$)
- Flux and emf are 90° out of phase
Can Do — Calculate and Sketch
- Calculate peak emf from coil parameters and rotation rate
- Sketch flux vs time and emf vs time for a rotating coil
- Explain why slip rings (not a commutator) are used in AC generators
Core Content
From rotation to alternating voltage
Watch a turbine at a power station: steam spins a shaft at 50 revolutions per second; the shaft rotates a coil inside a strong magnetic field; and at the output terminals, a voltage oscillates back and forth at exactly 50 Hz — the AC electricity that reaches every home in Australia. The coil's rotation continuously changes how much of the magnetic field passes through it, and that changing flux is what drives the alternating voltage. The more turns on the coil, the larger the field, and the faster the rotation, the greater the peak emf.
$\Phi = NBA\cos(\omega t)$ — flux through the coil (Wb)
$\varepsilon = -\dfrac{d\Phi}{dt} = NBA\omega\sin(\omega t)$ — induced emf (V)
$\varepsilon_0 = NBA\omega$ — peak emf (V)
Key insight: Flux and emf are 90° out of phase. When the coil is perpendicular to $B$ ($\omega t = 0$), flux is maximum but its rate of change is zero — so emf is zero. When the coil is parallel to $B$ ($\omega t = 90°$), flux is zero but changing fastest — so emf is maximum.
The slip rings maintain continuous electrical contact with the rotating coil. Unlike a commutator, they do not reverse the current — they simply pass the alternating voltage to the external circuit. This is why the output is AC.
A student says: "An AC generator uses a split-ring commutator just like a DC motor." Is this correct? Explain the difference and why each device needs its particular contact mechanism.
AC generator: $\Phi = NBA\cos(\omega t)$ (Wb); $\varepsilon = NBA\omega\sin(\omega t)$ (V); peak emf $\varepsilon_0 = NBA\omega$. Slip rings (not commutator) maintain contact without reversing current → output is AC. Flux and emf are 90° out of phase.
Pause — copy the highlighted generator formulas and slip-ring note into your book before moving on.
The emf in an AC generator is maximum when the coil is…
Understanding why one is maximum when the other is zero
We just saw that the generator produces $\varepsilon = NBA\omega\sin(\omega t)$ while flux follows $\cos(\omega t)$. That raises a question: why are they 90° out of phase — why is emf zero when flux is maximum? This card answers it → emf = $-d\Phi/dt$; at a flux peak the rate of change is zero, so emf is zero.
Faraday's law tells us $\varepsilon = -d\Phi/dt$. Since emf is the derivative of flux, whenever flux is at a maximum or minimum (its turning point), the rate of change is zero — and so is the emf. Whenever flux is passing through zero with maximum slope, the emf is at its peak.
Think of a sine curve and its derivative (cosine). The derivative is maximum where the original function has its steepest slope — at the zero crossings. The derivative is zero where the original function peaks. This is exactly the relationship between flux ($\cos$) and emf ($\sin$).
Four key positions
- Coil ⊥ B ($\omega t = 0°$): flux = maximum ($NBA$), emf = 0
- Coil ∥ B ($\omega t = 90°$): flux = 0, emf = maximum ($NBA\omega$)
- Coil ⊥ B again ($\omega t = 180°$): flux = minimum ($-NBA$), emf = 0
- Coil ∥ B again ($\omega t = 270°$): flux = 0, emf = minimum ($-NBA\omega$)
$\Phi(t) = NBA\cos(\omega t)$ — cosine wave
$\varepsilon(t) = NBA\omega\sin(\omega t)$ — sine wave (90° ahead of flux)
$\varepsilon = -d\Phi/dt$: emf is the rate of change of flux. At a flux maximum/minimum $d\Phi/dt = 0$ → emf = 0. At a flux zero-crossing (steepest slope) → emf is maximum. Flux: cosine. Emf: sine. They are 90° out of phase.
Add the highlighted phase relationship rule to your notes before the check below.
When the flux through the coil is at its maximum value, the emf is also at its maximum.
The induced emf in an AC generator follows a sinusoidal pattern over time.
Slip rings in an AC generator reverse the current every half turn.
Using the simulator — when you double the angular speed $\omega$ of the coil, the peak emf…
Calculate peak emf and instantaneous emf
We just saw the 90° phase relationship between flux and emf. That raises a question: how do we actually calculate peak emf and the emf at a specific instant in time? This card answers it → use $\varepsilon_0 = NBA\omega$ with $\omega = 2\pi f$, then $\varepsilon(t) = \varepsilon_0\sin(\omega t)$.
A rectangular coil with 100 turns and dimensions 8.0 cm × 5.0 cm rotates at 50 Hz in a uniform magnetic field of 0.40 T.
- Calculate the peak emf generated.
- Calculate the emf 2.0 ms after the coil passes through the position perpendicular to the field.
- Given: $N = 100$, $A = (0.080)(0.050) = 4.0 \times 10^{-3}$ m², $f = 50$ Hz, $B = 0.40$ T.
- Find: $\varepsilon_0 = NBA\omega$.
- Solve: $\omega = 2\pi f = 2\pi(50) = 314$ rad/s.
- Solve: $\varepsilon_0 = (100)(0.40)(4.0 \times 10^{-3})(314) = 50.2$ V.
- Context: At $t = 0$ the coil is perpendicular to $B$, so $\varepsilon = 0$ and the emf follows $\varepsilon = \varepsilon_0\sin(\omega t)$.
- Solve: $\omega t = (314)(2.0 \times 10^{-3}) = 0.628$ rad.
- Solve: $\varepsilon = 50.2\sin(0.628) = 50.2 \times 0.588 = 29.5$ V.
Swap your working with a partner. Check that they: (1) converted cm to m for area, (2) used $\omega = 2\pi f$ (not $\omega = f$), (3) used $\sin(\omega t)$ — not $\cos$ — for emf starting from the perpendicular position, and (4) converted 2.0 ms to seconds.
Step order: $\omega = 2\pi f$ → $\varepsilon_0 = NBA\omega$ → $\varepsilon(t) = \varepsilon_0\sin(\omega t)$. Worked result: $\omega = 314$ rad/s, $\varepsilon_0 \approx 50$ V, $\varepsilon(2\,\text{ms}) \approx 30$ V. Units check: area in m², time in s, angle in radians.
Pause — write the highlighted step order and worked result into your book before moving on.
A coil with 50 turns, area 0.020 m², rotates at 10 rad/s in $B = 0.30$ T. The peak emf is…
Explore how each parameter affects peak emf
- Set $n = 20$, $B = 0.50$ T, $\omega = 4.0$ rad/s. Record the peak emf. Now double $\omega$ to 8.0 rad/s. By what factor does the peak emf change? Verify with $\varepsilon_0 = NBA\omega$.
- Return to $\omega = 4.0$ rad/s. Double $n$ to 40. What happens to peak emf? Explain.
- Sketch the flux vs time and emf vs time curves on the same axes. Mark where each is maximum and zero.
Explain a fundamental design difference
Compare slip rings and a split-ring commutator. For each, describe: (a) what it physically does to the circuit connections, (b) what type of output voltage it produces, and (c) one real device that uses it.
Three of these statements about AC generators are correct. Pick the odd one out.
- An AC generator produces alternating voltage by rotating a coil in a magnetic field.
- Flux varies as $\Phi = NBA\cos(\omega t)$; emf varies as $\varepsilon = NBA\omega\sin(\omega t)$.
- Flux and emf are 90° out of phase — max flux → zero emf; zero flux → max emf.
- Slip rings (not a commutator) pass the alternating voltage to the external circuit.
- Peak emf: $\varepsilon_0 = NBA\omega$. Doubling any single parameter doubles $\varepsilon_0$.
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(2 marks) 1. A coil with 80 turns and area $6.0 \times 10^{-3}$ m² rotates at 25 Hz in a magnetic field of 0.50 T. Calculate the peak emf generated and state the angular frequency.
1 mark: correct $\omega = 2\pi f$ · 1 mark: correct $\varepsilon_0 = NBA\omega$ with substitution
AnalyseBand 5(4 marks) 2. A rotating coil produces emf $\varepsilon = \varepsilon_0\sin(\omega t)$. (a) Explain why the emf is zero when the magnetic flux through the coil is at its maximum. (b) Explain why an AC generator uses slip rings rather than a split-ring commutator, and state what type of output each produces.
2 marks: (a) emf depends on rate of change of flux — at maximum, rate of change is zero (1 mark); linked to Faraday's law (1 mark). 2 marks: (b) slip rings → AC (1 mark); commutator reverses → pulsating DC (1 mark).
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (2 marks): $\omega = 2\pi \times 25 = 157$ rad/s (1 mark). $\varepsilon_0 = NBA\omega = (80)(0.50)(6.0 \times 10^{-3})(157) = 37.7$ V $\approx 38$ V (1 mark).
Q2 (4 marks): (a) Faraday's law states that induced emf equals the rate of change of flux ($\varepsilon = -d\Phi/dt$). When flux is at its maximum, it is momentarily not changing — the rate of change is zero — so no emf is induced (2 marks). (b) Slip rings maintain continuous contact with the rotating coil without reversing the current, so the output is alternating current (AC). A split-ring commutator reverses the circuit connections every half turn, converting the output to pulsating direct current (DC). An AC generator needs slip rings to preserve the alternating nature of the output (2 marks).
Five timed questions on AC generators. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt the start you were asked about Faraday's 1831 copper disc dynamo at the Royal Institution: why did the galvanometer read zero at maximum flux and deflect most at zero flux?
The key insight is that emf depends on the rate of change of flux, not its value. When the disc face was perpendicular to B, flux was maximum but momentarily not changing — so emf was zero. When the disc rotated through the position of zero flux, flux was changing at its fastest rate — so emf peaked. This 90° phase difference between flux and emf is the defining feature of Faraday's dynamo and every AC generator since.
If you doubled the rotation frequency: peak emf doubles (because $\varepsilon_0 = NBA\omega$ and $\omega = 2\pi f$) and the output frequency also doubles — so the sine wave oscillates twice as fast.