AQA A-Level Physics Paper 2, June 2025: Question 22

Induced EMF Across a Moving Conductor

AQA A-Level Physics • Magnetic Fields & Induction • Multiple Choice

What this question tests

This question assesses your ability to calculate the motional electromotive force (emf) induced across a 3D rectangular conductor moving through a uniform magnetic field. Key areas evaluated include:

Applying the cutting-flux formula: ε = Bvl (or ε = B L v ).
Correctly identifying mutually perpendicular vectors: magnetic field (B), velocity (v), and effective length (L).
Handling unit conversions accurately from centimetres ( cm ) and milliteslas ( mT ) to standard SI units, then converting the result to microvolts ( μV ).

Question 22 Breakdown

Identifying the Correct Dimension and Calculating EMF

✅ Correct Answer

D | 6.25 μV

Awarded for correctly identifying the 5.0 cm dimension as the perpendicular cutting length and executing standard SI unit conversions.

💡 Key Knowledge

Induced EMF formula: ε = B L v for a straight conductor moving at right angles to a uniform magnetic field.
Orthogonality Rule: The vectors for magnetic flux density ( B ), velocity ( v ), and length across which the emf is established ( L ) must all be mutually perpendicular (at 90° to one another).
Free electrons inside the conductor experience a magnetic force F = Bqv that separates charge along the third axis.

📐 Step-by-Step Calculation

Determine the orientation of each vector:
- Magnetic Field (B): Vertical direction (y-axis) → height dimension of 10.0 cm is parallel to B.
- Velocity (v): Horizontal to the right (x-axis) → length dimension of 30.0 cm is parallel to v.
- Induced EMF / Conductor Length (L): Must be mutually perpendicular to both B and v (z-axis, into/out of the page) → L = 5.0 cm .
Convert all values to standard SI units:
- B = 2.5 mT = 2.5 × 10⁻³ T
- v = 5.0 cm s⁻¹ = 0.050 m s⁻¹ = 5.0 × 10⁻² m s⁻¹
- L = 5.0 cm = 0.050 m = 5.0 × 10⁻² m
Substitute into the induced emf formula:
ε = B × L × v

ε = (2.5 × 10⁻³ T) × (0.050 m) × (0.050 m s⁻¹)

ε = 6.25 × 10⁻⁶ V
Convert to microvolts (μV):
ε = 6.25 × 10⁻⁶ V = 6.25 μV → Option D

🧠 Exam Technique: The 3D Orthogonal Grid

Whenever a 3D block moves through a field, assign axes to avoid guessing which dimension is L :

x-axis (horizontal right): Motion ( v = 5.0 cm s⁻¹ , block length = 30 cm)
y-axis (vertical): Field lines ( B , block height = 10 cm)
z-axis (depth): Charge accumulation & induced emf ( L = 5.0 cm )

Since F = -e(v × B) , electrons are driven along the third remaining direction: the 5.0 cm depth!

❌ Common Errors & Distractor Traps

Choosing Option C (12.5 μV): Occurs if you select L = 10.0 cm (the vertical dimension). The vertical dimension is parallel to B , so no emf can develop across it!
Choosing Option B (37.5 μV): Occurs if you select L = 30.0 cm (the dimension along velocity). Charge cannot separate parallel to velocity in this geometry.
Choosing Option A (150 μV): Occurs if you multiply all three dimensions or use face area ( 30 × 10 or 30 × 5 ) incorrectly.
Prefix Slip-ups: Forgetting that mT is 10⁻³ or that cm must be converted to m before calculating.

Mark Scheme Breakdown: 1 mark total (Assessment Objective AO2 – Application of knowledge in unfamiliar context). Full credit is awarded solely for selecting D.