Year 12 Physics Module 6 ⏱ ~20 min 5 MC · 2 Short Answer Lesson 19 of 21

AC Generators, Changing Flux

You'll be able to calculate peak emf from coil parameters and rotation rate.

A generator converts mechanical input into electrical output. In the ideal model used here, a coil rotates uniformly in a uniform magnetic field, so its flux linkage changes sinusoidally and the induced emf alternates.

Today's hook: In an ideal rotating-coil generator, why is the induced emf zero when flux linkage is greatest, yet greatest when flux linkage passes through zero?
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Orient and retrieve

Recall Faraday’s law, predict the rotating-coil pattern and identify the lesson’s ideal-model assumptions.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A rectangular coil rotates at constant speed in a uniform magnetic field.

  1. At what orientation is the flux through the coil maximum? Minimum?
  2. At what orientation is the rate of change of flux maximum?
  3. 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…

Prerequisite check: Lessons 12–14 established magnetic flux $\Phi=BA\cos\theta$, flux linkage $N\Phi$ and Faraday’s law $\varepsilon=-d(N\Phi)/dt$. This lesson applies those ideas to an ideal coil rotating uniformly in a uniform field. The generator produces an alternating emf; an alternating current flows only when the output circuit is closed through a load.
Learning Intentions
goals

Know, AC Generator Structure

  • AC generator: coil rotating in magnetic field, slip rings, brushes and external load
  • Single-turn flux: $\Phi=BA\cos(\omega t)$; flux linkage: $N\Phi=NBA\cos(\omega t)$
  • Induced emf: $\varepsilon = NBA\omega\sin(\omega t)$

Understand, Flux, Rate of Change, and Emf

  • Maximum flux occurs when the coil plane is perpendicular to $B$ ($\omega t = 0$)
  • Maximum emf occurs when the coil plane is parallel to $B$ ($\omega t = 90°$)
  • 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
Scan these before reading
vocab
AC generator (alternator)A device that converts mechanical input into an alternating emf via electromagnetic induction; a closed load permits alternating current.
Slip ringsContinuous conducting rings that maintain electrical contact with a rotating coil without swapping its external connections.
Peak emf ($\varepsilon_0$)The maximum value of induced emf: $\varepsilon_0 = NBA\omega$.
Angular frequency ($\omega$)The rate of rotation in radians per second: $\omega = 2\pi f = 2\pi/T$.
Misconceptions to fix
✗ Wrong: The emf is maximum when the flux is maximum.
✓ Right: The emf depends on the rate of change of flux, not its value. When flux is at maximum, it is momentarily not changing (the rate is zero), so emf is zero. Emf is maximum when flux is changing most rapidly (at the zero-crossing).
✗ Wrong: AC generators use a split-ring commutator, just like DC motors.
✓ Right: AC generators use slip rings, which maintain continuous contact without swapping the external connections. A split-ring commutator would rectify the naturally alternating emf into a pulsating unidirectional output.
Cross-lesson links: L10 compared machines that use electrical input to produce rotation with machines that use mechanical input to produce electrical output. Lessons 12–14 supplied the flux, Faraday and Lenz foundations used here. L16 extends the model to power-station generators.

Core Content

2

Model the rotating-coil generator

Connect the coil, field, area normal, flux linkage and slip rings in one signed model.

1
How an AC Generator Works
+5 XP

From rotation to alternating voltage

Mechanical input turns a coil relative to a magnetic field. In the ideal two-pole model, one revolution produces one complete emf cycle. The rotating area normal makes the single-turn flux vary as a cosine, so Faraday’s law produces an alternating terminal emf. More turns, a stronger field, a larger coil area or faster rotation increases the peak emf.

AC generator: a coil turns in the uniform field between N and S poles; θ is the angle between B and the area normal; slip rings and brushes carry the alternating emf to the load.NSuniform field, Bθarea normalloadslip rings and brushes → load
Figure 1. The angle $\theta$ is measured between $\vec B$ and the chosen area normal. Continuous slip rings transfer the naturally alternating terminal emf without rectifying it.
Text description
  • A north pole on the left and a south pole on the right make a uniform magnetic field B pointing from left to right.
  • A rectangular coil, drawn in perspective, rotates about a vertical axis between the poles. Its area normal makes an angle θ with B.
  • The two ends of the coil connect to two separate, continuous slip rings on the axle. Brushes press on the rings and carry the naturally alternating emf to the load, without rectifying it.
Flux and emf in an AC generator

$\Phi = BA\cos(\omega t)$  , single-turn flux (Wb)

$N\Phi = NBA\cos(\omega t)$  , flux linkage (Wb-turn)

$\varepsilon = -\dfrac{d(N\Phi)}{dt} = NBA\omega\sin(\omega t)$  , induced emf (V)

$\varepsilon_0 = NBA\omega$  , peak emf (V)

Key insight: Flux and emf are 90° out of phase. When the coil plane is perpendicular to $B$ ($\omega t = 0$), flux is maximum but its rate of change is zero, so emf is zero. When the coil plane is parallel to $B$ ($\omega t = 90°$), flux is zero but changing fastest, so emf magnitude is maximum.

The slip rings maintain continuous electrical contact with the rotating coil. Unlike a split-ring commutator, they do not swap the coil’s external connections each half-turn; they pass the naturally alternating terminal emf to the external circuit. With a closed load, the current also alternates.

Energy and load

With an open circuit the generator can have a terminal emf but negligible output current. When a load closes the circuit, current transfers electrical energy to the load. By Lenz’s law, that current produces a magnetic torque opposing the driving rotation, so the external source must do mechanical work. Ideally this becomes electrical energy; real generators also lose energy through resistance, friction and other mechanisms.

Stop and check

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: single-turn flux $\Phi=BA\cos(\omega t)$ in Wb; flux linkage $N\Phi=NBA\cos(\omega t)$ in Wb-turn; $\varepsilon=-d(N\Phi)/dt$. Slip rings maintain contact without reversing the alternating output.

Pause, copy the highlighted generator formulas and slip-ring note into your book before moving on.

The emf magnitude in an ideal AC generator is maximum when the coil plane is…

3

Represent flux linkage and emf

Use four coil positions and paired graphs to connect flux-linkage value, gradient and induced emf.

Beyond the syllabus. A simple AC generator's parts, flux change and emf reasoning are core. Routine use of the $\varepsilon = NBA\omega\sin(\omega t)$ waveform equation, phase calculations and peak-emf arithmetic are extension unless an exam question supplies the relationship through Faraday's law.
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Flux and Emf, The 90° Phase Relationship
+5 XP

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 linkage follows $\cos(\omega t)$. That raises a question: why are they 90° out of phase, why is emf zero when flux linkage is maximum? This card answers it → emf = $-d(N\Phi)/dt$; at a linkage peak the rate of change is zero, so emf is zero.

Faraday's law tells us $\varepsilon=-d(N\Phi)/dt$. Since emf is the negative derivative of flux linkage, whenever linkage is at a maximum or minimum, its rate of change and the emf are zero. At a zero crossing the slope magnitude, and therefore emf magnitude, is greatest.

Key insight

Think of a cosine curve and its derivative. The derivative has greatest magnitude where the original function has its steepest slope, at the zero crossings. It is zero where the original function peaks. Faraday’s minus sign then sets the emf direction relative to the chosen area normal and terminal polarity.

Four key positions

  • Coil plane ⊥ B ($\omega t = 0°$): flux = maximum ($NBA$), emf = 0
  • Coil plane ∥ B ($\omega t = 90°$): flux = 0, emf = maximum ($NBA\omega$)
  • Coil plane ⊥ B again ($\omega t = 180°$): flux = minimum ($-NBA$), emf = 0
  • Coil plane ∥ B again ($\omega t = 270°$): flux = 0, emf = minimum ($-NBA\omega$)
Phase relationship summary

$N\Phi(t) = NBA\cos(\omega t)$  , flux-linkage cosine wave

$\varepsilon(t) = NBA\omega\sin(\omega t)$  , sine wave, one quarter-cycle behind linkage for this convention

Over one turn, flux linkage follows a cosine and emf follows a sine: the emf is zero at the flux-linkage peaks and largest where the flux linkage crosses zero.NΦε+NBA−NBA+NBAω−NBAω0°90°180°270°360°ε = 0 where NΦ peaks;ε is largest where NΦ crosses zero
Figure 2. With $t=0$ chosen when the area normal is parallel to $\vec B$, linkage follows cosine and emf follows sine. The emf is zero at linkage extrema and has greatest magnitude at linkage zero-crossings.
Text description
  • Two graphs share an angle axis from 0° to 360°. The top graph is flux linkage NΦ (solid line); the bottom graph is the induced emf ε (dashed line).
  • NΦ follows a cosine: +NBA at 0°, zero at 90°, −NBA at 180°, zero at 270° and +NBA at 360°.
  • ε follows a sine: zero at 0°, +NBAω at 90°, zero at 180°, −NBAω at 270° and zero at 360°.
  • The emf is zero where the flux linkage is at a peak, and largest in size where the flux linkage crosses zero: a 90° phase difference.

$\varepsilon=-d(N\Phi)/dt$: emf is the signed rate of change of flux linkage. At a linkage maximum or minimum the derivative is zero; at a zero crossing its magnitude is greatest.

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.

4

Explore the ideal generator

Use the canonical simulator, or the static graph above, to test how angular speed changes amplitude and frequency.

Optional interactive

The simulator visualises the ideal rotating-coil model. If it is unavailable, use Figure 2 and $\varepsilon_0=NBA\omega$: doubling $\omega$ doubles both the peak emf and the number of cycles per second.

Interactive Tool, AC Generator Simulator Open fullscreen ↗

Using the simulator, when you double the angular speed $\omega$ of the coil, the peak emf…

5

Calculate peak and instantaneous emf

Convert area and frequency, calculate angular speed, then apply the signed sine model with stated initial orientation.

3
Worked Example, AC Generator Emf
+5 XP

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 $\varepsilon(t) = \varepsilon_0\sin(\omega t)$.

Problem

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.

  1. Calculate the peak emf generated.
  2. Calculate the emf 2.0 ms after the coil passes through the position perpendicular to the field.
Step 1, Peak emf
  1. Given: $N = 100$, $A = (0.080)(0.050) = 4.0 \times 10^{-3}$ m², $f = 50$ Hz, $B = 0.40$ T.
  2. Find: $\varepsilon_0 = NBA\omega$.
  3. Solve: $\omega = 2\pi f = 2\pi(50) = 314$ rad/s.
  4. Solve: $\varepsilon_0 = (100)(0.40)(4.0 \times 10^{-3})(314) = 50.2$ V.
Step 2, Instantaneous emf at t = 2.0 ms
  1. Context: At $t = 0$ the coil is perpendicular to $B$, so $\varepsilon = 0$ and the emf follows $\varepsilon = \varepsilon_0\sin(\omega t)$.
  2. Solve: $\omega t = (314)(2.0 \times 10^{-3}) = 0.628$ rad.
  3. Solve: $\varepsilon = 50.2\sin(0.628) = 50.2 \times 0.588 = 29.5$ V.
Peer check

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…

6

Apply, evaluate and consolidate

Test parameter changes, compare slip rings with a commutator and state the limits of the ideal sinusoidal model.

Reference convention and model

The displayed sine/cosine phase assumes an $N$-turn coil of constant area rotating uniformly in a uniform field, with $t=0$ chosen when the area normal is parallel to $\vec B$. Reversing the chosen area normal or swapping output terminals reverses the emf sign but not its amplitude or phase separation from linkage. Real generators can use many coils, poles and shaped fields, so their waveform need not be a perfect sine wave.

Activity 1, Simulator Investigation
ApplyBand 3

Explore how each parameter affects peak emf

  1. Record the simulator’s displayed coil area $A$. Set $N = 20$, $B = 0.50$ T and $\omega = 4.0$ rad/s. Record the peak emf. Now double $\omega$ to 8.0 rad/s. By what factor does the peak emf change? Verify with $\varepsilon_0 = NBA\omega$.
  2. Return to $\omega = 4.0$ rad/s. Double $N$ to 40. What happens to peak emf? Explain.
  3. Sketch the flux vs time and emf vs time curves on the same axes. Mark where each is maximum and zero.
Activity 2, Commutator vs Slip Rings
UnderstandBand 4

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.

Synthesis, connect the ideas
  • An AC generator produces alternating voltage by rotating a coil in a magnetic field.
  • Flux linkage varies as $N\Phi=NBA\cos(\omega t)$; 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 $\varepsilon_0$.
Analyse6 marks

A coil is rotated at a steady rate in a uniform magnetic field. Analyse how the emf generated varies during one complete revolution, with reference to the flux through the coil. (6 marks)

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Unpack the verb

Analyse is this dot point's own verb. The relationship to establish is between two things that are easy to confuse: the flux and its rate of change.

Plan: state Faraday's Law as a rate, describe the flux through the revolution, then read the emf off as the rate of change and locate its extremes.

Framing the whole answer around "rate of change" from the first line is the defence against the misconception this lesson names.
1 of 6
Compare5 marks

Compare the construction and output of a simple AC generator with those of a simple DC generator. (5 marks)

1
Unpack the verb

Compare needs both machines in every point. This dot point covers both generators, so the comparison is within its scope rather than across it.

Plan: establish everything the two share, isolate the single component that differs, then show what that one difference does to the output.

Almost everything about these machines is identical, so leading with the shared parts makes the one difference stand out instead of getting buried.
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Check and apply

Complete a shuffled concept check, then show the calculation and reasoning independently.

Quick recall, AC generators
+5 XP

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.

Short Answer, 6 marks
+5 XP

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 coil’s flux linkage 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).

Show all answers

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 is the negative rate of change of flux linkage, $\varepsilon=-d(N\Phi)/dt$. When linkage is at its maximum, it is momentarily not changing, so its gradient and the emf are zero (2 marks). (b) Slip rings maintain continuous contact without swapping the coil’s external connections, so the naturally alternating emf reaches the load. A split-ring commutator swaps the connections every half-turn and rectifies the output to pulsating DC (2 marks).

8

Retrieve and reflect

Use spaced retrieval, revisit the opening generator question and choose the next lesson.

Check what actually stuck
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quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
How did your thinking change?

At the start you were asked why an ideal rotating-coil generator has zero emf at maximum flux linkage and maximum emf magnitude when linkage passes through zero.

The key insight is that emf depends on the time gradient of flux linkage, not its value. At a linkage maximum or minimum the graph is momentarily flat, so $\varepsilon=-d(N\Phi)/dt=0$. At a zero-crossing its gradient magnitude is greatest, so the emf magnitude peaks. The sign is fixed by the chosen area normal and terminal polarity.

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.

← Previous: Power Transmission and Distribution Next: AC Induction Motors and Generators →

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