AQA A-Level Physics Paper 2, June 2025: Question 20

1 mark · Medium difficulty · Multiple Choice

Determine the current required to magnetically support a wire whose length is halved and diameter is doubled compared to an initial wire.

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Question

Question 20 shows a horizontal wire carrying current I perpendicular to horizontal uniform magnetic field lines of flux density B. The diagram indicates that the magnetic force supports the wire's weight. The wire has length L and diameter d. The wire is replaced by a second wire of the same material, with length L/2 and diameter 2d. Four options are given for the required current: A (I/2), B (I), C (2I), and D (4I).

Mark scheme

Show the mark scheme Mark scheme for question 20 showing the correct answer as option D (4I), assessed under AO2.

How to answer it

AQA A-Level Physics • Magnetic Fields • Multiple Choice

Balancing Magnetic Force and Gravitational Force on a Current-Carrying Wire

What this question tests

This question assesses your ability to combine the equation for magnetic force on a straight conductor ( F = BIl ) with the definition of mass in terms of density and geometry ( m = ρV ), and apply proportional reasoning to deduce how changes in wire dimensions affect the equilibrium current required to support its weight.

Question 20 Analysis

Balancing Forces: Magnetic Levitation of a Conductor

✅ Correct Option: D (4I)

The required current to support the wire is 4I.

Mark Scheme: 1 mark for selecting option D [AO2].

💡 Key Physics Concepts

  • Magnetic force on wire: F = B I L sin(θ) . Since the wire is perpendicular to the field, θ = 90° , so F = B I L .
  • Weight of wire: W = m g .
  • Mass & Volume: Mass is density times volume: m = ρ V = ρ A L , where cross-sectional area A = π (d/2)² = (π d²) / 4 .
  • Equilibrium condition: Magnetic force equals weight: B I L = m g .

📐 Step-by-Step Derivation & Calculation

  1. Set up the equilibrium equation for the first wire:
    B × I × L = m × g
  2. Substitute mass in terms of density, diameter, and length:
    m = ρ × V = ρ × (π d² / 4) × L
    Therefore:
    B × I × L = ρ × (π d² / 4) × L × g
  3. Cancel length (L) from both sides:
    Notice that L appears on both sides of the equation and completely cancels out!
    B × I = ρ × (π d² / 4) × g
  4. Express current in terms of variables:
    I = [(ρ π g) / (4 B)] × d²
    Since ρ , g , and B are constant:
    I ∝ d² (the required current is purely proportional to the square of the diameter and completely independent of wire length).
  5. Apply the changes for the second wire:
    • New length = L / 2 (has no effect on required current).
    • New diameter = 2d .
    Inew ∝ (2d)² = 4d² = 4 × I .
    Hence, Inew = 4I (Option D).

🧠 Exam Technique & Scaling Strategy

  • Always cancel common terms first: In scaling/ratio questions, write out the full algebraic equation before substituting any numbers. Identifying that L cancels out saves valuable time and prevents unnecessary arithmetic mistakes.
  • Area scales with the square: Whenever a linear dimension of a cross-section changes (diameter or radius), remember area changes by the factor squared: (2d)² = 4d² .
  • Check Fleming's Left-Hand Rule: While not required for this numerical calculation, verify directions for conceptual clarity: if field is towards the right and magnetic force must be upward to counter weight (downward), current must flow into the plane / along the designated arrow.

❌ Common Misconceptions & Traps

  • The Length Trap (Choosing C, 2I): Students often calculate: "Length is halved (×1/2), diameter is doubled so area is ×4, mass is ×2, so current must be ×2". They forget that halving the length also halves the magnetic force for a given current ( F = BIL ), which cancels the length factor!
  • Linear Area Scaling (Choosing B, I): Forgetting to square the diameter, assuming mass is proportional to d instead of d² . If mass were proportional to d , the factors would be 1/2 × 2 = 1 , incorrectly leading to I .
  • Inverting the Relationship (Choosing A, I/2): Confusing force with current, or multiplying where they should divide.

Topics

Physics · 3.7 Fields and their consequences (A-level only) · 3.4 Mechanics and materials

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.