Physics • Year 12 • Module 6 • Lesson 11
AC Generators — Changing Flux
Build HSC Band 5–6 extended-response technique on deriving and applying the AC generator equations, evaluating design modifications, and constructing flux–emf sketch graphs with full justification.
1. Data + derivation: Snowy Hydro turbine generator (Band 5–6)
8 marks Band 5–6
Context. The Snowy Hydro Scheme’s Tumut 3 Power Station in NSW uses water-driven turbines to rotate large multi-pole generator coils in strong magnetic fields. A simplified model of one such generator has the following parameters: N = 120 turns, B = 1.50 T, coil dimensions 30.0 cm × 20.0 cm, and the turbine rotates at 50 Hz to produce mains-frequency electricity.
Q1. Answer all parts below.
- (a) Calculate the coil area and peak emf of this generator. Show all working including unit conversions. (3 marks)
- (b) Write expressions for the flux Φ(t) and the emf ε(t) as functions of time, starting from the position where the coil is perpendicular to the field. Include numerical values. (2 marks)
- (c) Calculate the instantaneous emf at t = 3.0 ms after the coil passes through the perpendicular position. (2 marks)
- (d) The engineers want to reduce the peak emf by 40% while keeping the rotation frequency constant at 50 Hz. State two different design changes that could each achieve this individually, and explain the physics behind each. (3 marks)
2. Sketch and analyse — flux and emf graphs for two generators (Band 5–6)
7 marks Band 5–6
Scenario. Two generators (X and Y) have identical coil dimensions and field strengths. Generator X rotates at 50 Hz. Generator Y rotates at 100 Hz. Both start with the coil perpendicular to the field at t = 0.
Q2. Answer the following.
- (a) On the axes below, sketch one complete cycle of the emf vs time graph for both Generator X and Generator Y on the same set of axes. Label the peak emf and period for each. (3 marks)
- (b) A student states: “Generator Y produces twice the peak emf of Generator X because it spins twice as fast, and it also has twice the frequency.” Evaluate this claim fully, referring to the equation ε0 = NBAω. (2 marks)
- (c) If Generator Y were modified by halving N (number of turns), predict the new peak emf relative to Generator X and sketch how the resulting emf curve would compare. (2 marks)
Use this set of axes to sketch emf vs time for Generator X (solid line) and Generator Y (dashed line). Label ε0 and T for each.
Written responses (2b and 2c):
Q1(a) — Coil area and peak emf (3 marks)
Area: A = 0.300 m × 0.200 m = 0.0600 m² [1 — correct conversion from cm to m AND correct multiplication].
Angular frequency: ω = 2πf = 2π(50) = 314.2 rad s−1 [1 — correct ω calculation].
Peak emf: ε0 = NBAω = (120)(1.50)(0.0600)(314.2) = 3402 V ≈ 3.40 kV [1 — correct numerical answer with units].
Marking notes: Award all three marks only if unit conversions (cm→m) are correct. If student uses cm², deduct the area mark but carry through for remaining marks if method is correct.
Q1(b) — Flux and emf as functions of time (2 marks)
Flux: Φ(t) = NBA cos(ωt) = (120)(1.50)(0.0600) cos(314t) = 10.8 cos(314t) Wb [1]. Note: NBA = 10.8 Wb is the maximum flux value Φ0.
Emf: ε(t) = NBAω sin(ωt) = 3402 sin(314t) V [1]. (Or equivalently 3.40 × 103 sin(314t) V.)
Note for students: The cos function for flux and sin function for emf arise because at t = 0 the coil is perpendicular to B (maximum flux, zero emf). Accept ε0 sin(2πft) = 3402 sin(100πt) as equivalent.
Q1(c) — Instantaneous emf at t = 3.0 ms (2 marks)
ωt = 314.2 × (3.0 × 10−3) = 0.9425 rad [1 — correct conversion and multiplication].
ε = 3402 sin(0.9425) = 3402 × 0.8090 = 2752 V ≈ 2.75 kV [1 — correct evaluation with units].
Marking notes: Deduct 1 mark if t is not converted to seconds. Accept answers in range 2.70–2.80 kV due to rounding of ω.
Q1(d) — Design changes to reduce peak emf by 40% (3 marks)
A 40% reduction means the new peak emf = 0.60 × 3402 = 2041 V. Since ε0 = NBAω and ω is fixed, any one of N, B, or A must be reduced by 40%.
Change 1 — Reduce N from 120 to 72 turns: Since ε0 is directly proportional to N, reducing N by 40% reduces peak emf by 40%. Physics: fewer turns means less total flux linkage change per revolution, so less emf induced per Faraday’s Law (ε = −NdΦ/dt). [1 for identifying a valid reduction AND explaining physics]
Change 2 — Reduce B from 1.50 T to 0.90 T: Since ε0 ∝ B, reducing B by 40% reduces peak emf by 40%. Physics: a weaker magnetic field means less flux (Φ = NBA cosωt is smaller), so the rate of change of flux is smaller at every instant, inducing a smaller emf. [1 for second valid change with physics explanation]
Accept also: Reduce coil area A by 40% (same reasoning — smaller area intercepts less flux). Any two of N, B, A qualify for full marks. [1 for stating two valid, distinct changes]
Note: Award a maximum of 2 marks if only one change is given, even if fully explained.
Q2(a) — Sketch guidance (3 marks)
Generator X (f = 50 Hz): T = 20 ms; peak emf = ε0(X) = NBAωX. Sine wave starting at zero, crossing zero at 10 ms, reaching positive peak at 5 ms and negative peak at 15 ms, completing one full cycle at 20 ms. [1 — correct T = 20 ms labelled]
Generator Y (f = 100 Hz): ωY = 2 × ωX, so T = 10 ms; peak emf = 2 × ε0(X) (double the amplitude of X). Sine wave completing one full cycle at 10 ms, with double amplitude. [1 — correct T = 10 ms; 1 — correct amplitude double that of X, both labelled]
Marking criteria: Award 1 mark for each of: (i) correct period and shape for X; (ii) correct period (half of X) for Y; (iii) correct amplitude ratio (Y = 2 × X) labelled on axes.
Q2(b) — Evaluate student’s claim (2 marks)
The claim is correct. Since ε0 = NBAω and ωY = 2ωX (identical N, B, A), the peak emf of Y is indeed twice that of X [1]. The frequency of the output equals the mechanical rotation frequency, so Generator Y’s output frequency is also double (100 Hz vs 50 Hz) [1]. Both parts of the claim are correctly stated, so the student’s full claim is valid.
Q2(c) — Modified Generator Y with N halved (2 marks)
Modified Y (call it Y′) has N′ = N/2 and ω′ = ωY = 2ωX. Peak emf of Y′ = (N/2)(B)(A)(2ωX) = NBAωX = ε0(X). So Y′ has the same peak emf as Generator X, but half the period (T = 10 ms, f = 100 Hz) [1]. The sketch of Y′ would be a sine wave with the same amplitude as Generator X but oscillating twice as fast (period 10 ms) [1].
Marking criteria: 1 mark for correct numerical reasoning to the conclusion ε0(Y′) = ε0(X); 1 mark for correctly describing the sketch (same amplitude, half period).