AQA A-Level Physics Paper 2, June 2023: Question 24

✅ Correct Answer

D — 395 V

💡 Key Knowledge

• Magnetic flux linkage when the coil is tilted at an angle θ to the field is NΦ = BAN cos(ωt) .
• By Faraday's law, induced EMF is ε = -d(NΦ)/dt = BANω sin(ωt) .
• Maximum EMF occurs when the plane of the coil is parallel to the magnetic field lines ( sin(ωt) = 1 ):
  ε_max = BANω
• Angular speed relationship: ω = 2πf .

Step 1: Identify all given quantities with correct SI units
• Number of turns, N = 200
• Radius, r = 0.10 m
• Frequency, f = 50 rev s⁻¹ = 50 Hz
• Magnetic flux density, B = 0.20 T

Step 2: Calculate cross-sectional area (A)
A = πr² = π × (0.10)² = 0.01π m² ≈ 3.1416 × 10⁻² m²

Step 3: Calculate angular frequency (ω)
ω = 2πf = 2 × π × 50 = 100π rad s⁻¹ ≈ 314.16 rad s⁻¹

Step 4: Compute peak EMF (ε_max)
ε_max = B × A × N × ω
ε_max = 0.20 × (0.01π) × 200 × (100π)
ε_max = 0.20 × 200 × 0.01 × 100 × π² = 40 × π²
ε_max = 40 × 9.8696 = 394.78 V ≈ 395 V (to 3 s.f.)

❌ Common Errors & Distractor Traps

• Selecting B (126 V): The most frequent trap! Occurs when forgetting to convert linear frequency f into angular speed ω ( ε = BANf = 40π ≈ 126 V ). Remember, ω = 2πf !
• Selecting A (63 V): Caused by forgetting 2π and also confusing radius with diameter ( r = 0.05 m instead of 0.10 m ), or dividing by extra factors of 2.
• Using πr instead of πr²: Forgetting to square the radius when calculating the cross-sectional area of a circle.

🧠 Exam Technique & Examiner Insights

• Quick sanity check: Because both A and ω contain a factor of π , your working should contain π² ≈ 9.87 ≈ 10 .
• Spotting distractors: Notice that 395 ÷ π ≈ 126 . When you see two options related by a factor of π (approx. 3.14), examiners are explicitly testing whether you remembered ω = 2πf .
• Check the orientation: The axis is perpendicular to the field and in the plane of the coil, which is the standard geometry that produces continuous change in flux linkage and full alternating EMF.