AQA A-Level Physics Paper 2, June 2025: Question 21

1 mark · Medium difficulty · Multiple Choice

Determine the mean induced emf as a rotating coil moves through pi radians from a position parallel to a magnetic field.

Practise this question

Question

Multiple choice question showing a rectangular coil rotating about an axis in a uniform horizontal magnetic field. The plane of the coil is currently horizontal. Given values: cross-sectional area of coil = 20 × 10⁻³ m², number of turns = 500, magnetic flux density = 50 mT, frequency of rotation = 50 Hz. Four options are given for the mean induced emf as it rotates through π rad: A (0), B (0.50), C (100), D (157).

Mark scheme

Show the mark scheme Mark scheme row for question 21 indicating correct answer A with value 0, testing assessment objective AO2.

How to answer it

Mean Induced EMF in a Rotating Coil

📌 What this question tests

This question assesses your understanding of Faraday's Law of Electromagnetic Induction, magnetic flux linkage, and how induced electromotive force (emf) varies over time for a rotating coil in a uniform magnetic field. Specifically, it tests your ability to distinguish between instantaneous peak emf, average emf, and net change in magnetic flux linkage over a defined angular displacement.

Question 21 • Multiple Choice (1 Mark)

Analysis & Solution

AQA A-Level Physics — Magnetic Fields & Induction

✅ Correct Answer

Option A (0 V)

Awarded for identifying that the net change in magnetic flux linkage over a rotation of π radians from the horizontal position is zero, resulting in a mean induced emf of zero.

💡 Key Knowledge

  • Magnetic Flux (Φ): Defined as Φ = B A cos θ, where θ is the angle between the normal to the coil's area and the magnetic field lines.
  • Horizontal Coil: The plane of the coil is parallel to the field lines. Therefore, the normal is perpendicular to the field (θ = 90°), giving an initial flux of Φ₁ = 0.
  • Faraday's Law: The magnitude of the mean induced emf is given by:
    Mean ε = |Δ(NΦ) / Δt|
  • Rotation by π rad (180°): The coil turns upside down, but its plane is once again parallel to the magnetic field (θ = 270°), meaning Φ₂ = 0.

📐 Step-by-Step Calculation & Proof

  1. Determine Initial Magnetic Flux Linkage (NΦ₁):
    The coil is in the horizontal plane (parallel to the field). No field lines pass through the area of the coil.
    Φ₁ = B × A × cos(90°) = 0 Wb
    NΦ₁ = 0 Wb·turns
  2. Determine Final Magnetic Flux Linkage (NΦ₂):
    After rotating through π rad (180°), the coil is again horizontal (parallel to the field lines).
    Φ₂ = B × A × cos(270°) = 0 Wb
    NΦ₂ = 0 Wb·turns
  3. Calculate the Change in Flux Linkage (ΔNΦ):
    Δ(NΦ) = NΦ₂ - NΦ₁ = 0 - 0 = 0 Wb·turns
  4. Apply Faraday's Law for Mean EMF:
    Mean ε = - Δ(NΦ) / Δt = 0 / Δt = 0 V

🧠 Exam Technique & Alternative Perspective

Graphical / Calculus Approach:

  • Because the coil starts parallel to the field, the cutting of flux is at a maximum at t = 0, meaning instantaneous emf starts at a peak: ε(t) = ε₀ cos(ωt) .
  • As the coil rotates through π rad (half a complete period, T/2):
    • During the first quarter turn (0 to π/2), emf is positive.
    • During the second quarter turn (π/2 to π), the emf reverses sign and is negative.
  • By symmetry, the positive area under the emf–time graph exactly equals the negative area. The mean induced emf over this entire interval is therefore identically 0 V.

❌ Common Traps & Distractors

  • Distractor D (157 V): This is the peak emf ( ε₀ = BANω = 50×10⁻³ × 20×10⁻³ × 500 × 2π × 50 ≈ 157 V ). Students who immediately reach for the formula sheet often calculate this without noticing the word "mean".
  • Distractor C (100 V): This is the mean rectified emf ( 2ε₀ / π ≈ 100 V ). Students calculate the average magnitude without accounting for the change in polarity over a half-cycle.
  • Assuming max flux at start: If the coil started perpendicular to the field, Δ(NΦ) would be 2BAN, giving a non-zero average emf over a half turn. Always inspect the starting orientation!

Topics

Physics · 3.7 Fields and their consequences (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.