Physics • Year 12 • Module 6 • Lesson 11
AC Generators — Changing Flux
Apply your understanding of peak emf calculations, the flux–emf phase relationship, and generator design to real data, graph interpretation, and multi-step scenarios.
1. Calculate peak emf — generator parameter table
Four AC generators (A–D) have the parameters listed in the table. Complete the empty cells. Show your working below the table. 8 marks (2 each)
| Generator | N (turns) | B (T) | A (m²) | ω (rad s−1) | Peak emf ε0 (V) |
|---|---|---|---|---|---|
| A | 100 | 0.40 | 4.0 × 10−3 | 314 | |
| B | 50 | 0.30 | 0.020 | 10 | |
| C | 200 | 0.25 | 100 | 50.0 | |
| D | 0.60 | 8.0 × 10−3 | 50 | 72.0 |
Working space:
2. Interpret emf and flux graphs
The graph below shows the magnetic flux (Φ) and induced emf (ε) for a rotating coil as a function of time. 8 marks
Figure 2.1. Flux and emf for a rotating coil. Period T = 20 ms. Points P, Q, R, S are labelled for reference. Amplitudes are not to the same scale.
2.1 Identify the frequency of rotation of the coil. (1 mark)
2.2 At which labelled point(s) is the coil parallel to the magnetic field? Justify your answer with reference to the graph. (2 marks)
2.3 Describe the phase relationship between the flux and emf curves shown. What does this tell you about when the emf is greatest? (2 marks)
2.4 At point R (t = 10 ms), describe the orientation of the coil relative to the field and state the values of flux and emf at that instant. (3 marks)
3. Predict and justify — modifying a generator
A student has a small AC generator with N = 40 turns, B = 0.20 T, A = 5.0 × 10−3 m², and f = 25 Hz. 6 marks
3.1 Calculate the original peak emf and the output frequency. (2 marks)
3.2 The student wants to double the peak emf without changing N, B, or A. Predict what single change to the generator they could make, calculate the new ω, and state what effect this has on the output frequency. (2 marks)
3.3 A second student claims: “Using a stronger magnet doubles the peak emf but also doubles the output frequency.” Evaluate this claim. (2 marks)
4. Compare AC generator and DC motor contact mechanisms
Complete the table to compare slip rings and split-ring commutators. 8 marks (1 per cell)
| Feature | Slip rings | Split-ring commutator |
|---|---|---|
| Number of segments | ||
| Does it reverse current? | ||
| Output type | ||
| Used in which device |
Q1 — Generator parameter table
Generator A: ε0 = NBAω = (100)(0.40)(4.0 × 10−3)(314) = 50.2 V.
Generator B: ε0 = (50)(0.30)(0.020)(10) = 3.0 V.
Generator C: A = ε0 ÷ (NBω) = 50.0 ÷ (200 × 0.25 × 100) = 50.0 ÷ 5000 = 0.010 m² (= 1.0 × 10−2 m²).
Generator D: N = ε0 ÷ (BAω) = 72.0 ÷ (0.60 × 8.0 × 10−3 × 50) = 72.0 ÷ 0.24 = 300 turns.
Q2.1 — Frequency
T = 20 ms = 0.020 s; f = 1/T = 1/0.020 = 50 Hz.
Q2.2 — Coil parallel to field
The coil is parallel to the magnetic field when the emf is at its maximum magnitude. From the graph, the emf is at its positive peak at t = 5 ms (near point Q) and at its negative peak at t = 15 ms. These are the moments when the coil sides are moving perpendicular to the field, cutting field lines at the greatest rate. At point Q (t = 5 ms) the emf is maximum, so the coil is parallel to B. [1 for identifying t = 5 ms and/or t = 15 ms; 1 for correct justification linking to maximum emf]
Q2.3 — Phase relationship
The flux and emf curves are 90° (one quarter-cycle = 5 ms) out of phase: when flux is at its maximum (t = 0, t = 20 ms), emf is zero; when flux is zero and changing most rapidly (t = 10 ms), emf is at its maximum magnitude. The emf is greatest when the rate of change of flux (dΦ/dt) is largest — that is, when flux is crossing zero. [1 for 90° phase difference; 1 for stating emf is greatest when flux is changing fastest / crossing zero]
Q2.4 — Point R (t = 10 ms)
At t = 10 ms the coil has completed half a rotation from its initial perpendicular position, so it is now perpendicular to the field again, on the opposite side. [1] The flux is at its minimum (most negative): Φ = −NBA. [1] The emf is zero at this instant because the rate of change of flux is zero (flux is at a turning point). [1]
Q3.1 — Original peak emf
ω = 2πf = 2π(25) = 157 rad s−1. ε0 = NBAω = (40)(0.20)(5.0 × 10−3)(157) = 6.28 V. Output frequency = 25 Hz (same as rotation frequency). [1 for ε0; 1 for frequency]
Q3.2 — Doubling peak emf by changing ω
Since ε0 = NBAω, to double ε0 without changing N, B, or A, the student must double ω. New ω = 2 × 157 = 314 rad s−1, corresponding to f = 50 Hz. The output frequency also doubles to 50 Hz because the output frequency equals the mechanical rotation frequency. [1 for identifying ω must double with correct calculation; 1 for stating output frequency also doubles to 50 Hz]
Q3.3 — Evaluate claim about stronger magnet
The claim is partially correct. Doubling B does double the peak emf (since ε0 = NBAω, and B is directly proportional) [1]. However, the claim that output frequency also doubles is incorrect: the output frequency equals the mechanical rotation frequency, which is determined by how fast the coil spins, not by the strength of the magnetic field. Changing B has no effect on frequency. [1]
Q4 — Compare slip rings and commutator
Number of segments: Slip rings: two continuous rings (one per brush contact). Commutator: one ring split into two halves (or more for multi-coil motors).
Does it reverse current? Slip rings: No — the current passes through unchanged. Commutator: Yes — connections swap every half-turn, reversing current in the external circuit.
Output type: Slip rings: alternating current (AC). Commutator: pulsating direct current (DC — always in one direction in the external circuit).
Used in which device: Slip rings: AC generator (alternator). Commutator: DC motor / DC generator.