AQA A-Level Physics Paper 2, June 2023: Question 23
1 mark · Easy difficulty · Multiple Choice
Calculate the induced emf between the wing tips of an aircraft moving horizontally through a vertical magnetic field.
Practise this questionQuestion
Question text
23 The distance between the wing tips of a metal aircraft is 30 m.
The aircraft flies horizontally at a steady speed of 100 m s−1.
The aircraft passes through a vertical magnetic field of flux density 2.0 × 10−7 T.
What is the emf induced between its wing tips?
[1 mark]
A 0.2 μV
B 20 μV
C 300 μV
D 600 μV
Mark scheme
Show the mark scheme
23 D 600 μV
How to answer it
Electromagnetic Induction: Wingtip EMF
This question assesses your understanding of Faraday's Law of Electromagnetic Induction applied to a straight conductor cutting through magnetic field lines (motional EMF). Specifically, it tests your ability to apply the formula ε = Bvl and convert the calculated electromotive force into metric microvolts ( μV ).
Induced EMF in a Moving Aircraft
Multiple Choice (1 Mark)
✅ Correct Answer
Option D: 600 μV
💡 Key Knowledge
- Conductor cutting flux: When a conductor of length l moves with velocity v perpendicular to a magnetic field B, it cuts magnetic flux lines at a rate of:
ΔΦ/Δt = B(ΔA/Δt) = B(l · Δx)/Δt = Bvl - Motional EMF equation:
ε = B v l - Unit Prefixes:
1 μV = 10⁻⁶ V | 1 V = 10⁶ μV
📐 Step-by-Step Calculation
- Identify given values and standard SI units:
Length of conductor (wingspan), l = 30 m
Speed of motion, v = 100 m s⁻¹
Vertical magnetic flux density, B = 2.0 × 10⁻⁷ T - Select and apply the formula:
ε = B × v × l
ε = (2.0 × 10⁻⁷ T) × (100 m s⁻¹) × (30 m) - Calculate the value in Volts (V):
ε = 2.0 × 10⁻⁷ × 3000 = 6.0 × 10⁻⁴ V (or 0.0006 V ) - Convert Volts to microvolts (μV):
ε = (6.0 × 10⁻⁴ V) / (10⁻⁶ V μV⁻¹) = 6.0 × 10² μV = 600 μV
This matches Option D.
🧠 Exam Technique
- Geometric check: Ensure the motion and field are perpendicular. Here, the plane flies horizontally while the field is vertical, so they are at 90° (maximum cutting rate). Only the vertical component of the Earth's field induces an EMF across horizontal wings.
- Quick power of ten arithmetic:
2 × 10⁻⁷ × 10² × 3 × 10¹ = 6 × 10⁻⁴
Since 10⁻⁴ = 100 × 10⁻⁶ , you instantly get 600 μV without risking calculator input errors.
❌ Common Distractors & Traps
- Option A (0.2 μV): Occurs from dividing or mixing up ratios, e.g. 2.0 × 10⁻⁷ / 100 .
- Option B (20 μV): Arises from neglecting the wingspan l = 30 m and misplacing orders of magnitude ( 2.0 × 10⁻⁷ × 100 = 20 μV ).
- Option C (300 μV): Arises from forgetting the factor of 2.0 in the flux density ( 100 × 30 × 10⁻⁷ = 300 μV ).
- Micro vs Milli: Don't confuse μ (10⁻⁶) with m (10⁻³) ; 6.0 × 10⁻⁴ V = 0.6 mV = 600 μV .
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 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.