AQA GCSE Physics Physics Paper 2 (Higher), June 2025: Question 7

12 marks · High Demand difficulty · Extended Answer

Determine the direction of force on a current-carrying conductor, calculate initial acceleration using the motor effect and Newton's second law, and explain the continuous rotation and vertical moment in a simple electric motor.

Practise this question

Question

Figure 7 displays an apparatus demonstrating the motor effect: a copper rod resting on metal rails between the north pole of an upper magnet and the south pole of a lower magnet, connected to a power supply with arrows A (up), B (right), C (down), and D (left). Question 07.1 asks for the direction of rod acceleration from choices A, B, C, or D. Question 07.2 gives rod mass of 5.0 g, length 0.070 m, magnetic flux density 0.30 T, and current 2.0 A, asking to calculate initial acceleration. Figure 8 shows a rectangular coil ABCD in a simple electric motor between north and south magnetic poles with a split-ring commutator and battery contacts. Question 07.3 asks to explain how magnetic fields cause continuous rotation, and 07.4 asks why resultant moment is zero when the coil is vertical.

Mark scheme

Show the mark scheme Mark scheme for Question 7: 07.1 gives answer B (1 mark). 07.2 shows calculation: F = 0.30 × 2.0 × 0.070 = 0.042 N, then 0.042 = 0.005 × a, yielding a = 8.4 m/s² (5 marks). 07.3 awards 4 marks for: magnetic fields interaction produces forces on opposite sides AB and CD; AB and CD move in opposite directions; ends of coil swap contacts every half-revolution; current and forces on each side reverse. 07.4 awards 2 marks for: there is no current in the wire in this position, so no magnetic force acts and resultant moment is zero. Total 12 marks.

How to answer it

Electromagnetism: The Motor Effect & Electric Motors

SPEC CHECK • 4.7.2 & 4.7.3

What this question tests

This question assesses your core understanding of how magnetic fields interact with current-carrying conductors:

  • Using Fleming's Left-Hand Rule in a three-dimensional setup.
  • Linking magnetic force ( F = B × I × l ) with Newton's Second Law ( F = m × a ), including standard unit conversions (grams to kilograms).
  • Explaining how a split-ring commutator enables continuous rotation in a simple DC motor.
  • Explaining why torque/resultant moment drops to zero when the motor coil passes the vertical position.
Question 07.1 • 1 Mark

Direction of Acceleration of the Copper Rod

Determining motion using Fleming's Left-Hand Rule

✅ Correct Answer

Tick box B (towards the right along the rails).

💡 Key Knowledge: Left-Hand Rule

  • First Finger: Magnetic Field (North to South). The North pole is above, South pole is below → points Downwards.
  • seCond Finger: Conventional Current (+ to −). Tracing from the positive supply rail to the negative rail → flows towards the viewer / out of the page along the copper rod.
  • Thumb: Motion / Force → points Right (Direction B).

🧠 Exam Technique

The copper rod is on rails. A force acting upwards (A) or downwards (C) would not cause it to roll along the tracks. Since the current flows perpendicular to the rails and the field is vertical, the resulting force must be along the rails (either B or D). Align your hand carefully with the paper!

[1 mark] for selecting B only.
Question 07.2 • 5 Marks

Calculating Initial Acceleration

Two-step calculation combining magnetic force and acceleration equations

📐 Step-by-Step Calculation

  1. Convert mass to standard SI units (kg):
    m = 5.0 g = 5.0 ÷ 1000 = 0.0050 kg (or 5.0 × 10⁻³ kg)
  2. Calculate the magnetic force acting on the rod:
    Formula: F = B × I × l
    Substitution: F = 0.30 T × 2.0 A × 0.070 m
    Force: F = 0.042 N
  3. Substitute force and mass into Newton's Second Law:
    Formula: F = m × a
    Substitution: 0.042 = 0.0050 × a
  4. Rearrange to solve for acceleration:
    a = 0.042 ÷ 0.0050
  5. Final answer with correct value:
    a = 8.4 m/s²

❌ Common Errors & Traps

  • Unit conversion trap: Leaving mass as 5.0 g gives an acceleration of 0.0084 m/s². You lose the unit conversion mark!
  • Missing the first equation: You must calculate force using F = BIl first to unlock the subsequent method marks.

🧠 Mark Scheme Insight

Even if you made an error converting the mass (e.g. using 5.0 kg or 0.05 kg), marks 3, 4, and 5 can still be awarded as error carried forward (ecf), provided you correctly substituted and rearranged with your values!

Mark Breakdown:
• [1 mark] for correct substitution: F = 0.30 × 2.0 × 0.070
• [1 mark] for calculating force: F = 0.042 (N)
• [1 mark] for equating to mass × acceleration with converted mass: 0.042 = 0.005 × a
• [1 mark] for rearranging: a = 0.042 / 0.005
• [1 mark] for correct answer: 8.4 (m/s²)
Question 07.3 • 4 Marks

Explaining Continuous Rotation in a Motor

How magnetic interactions and the commutator sustain continuous turning

✅ Full Mark Model Answer (4/4)

  1. The interaction between the magnetic field of the permanent magnets and the magnetic field around the wire produces forces on the sides of the coil (AB and CD).
  2. Because the current flows in opposite directions along AB and CD, the forces act in opposite directions (one side moves up, the other moves down), causing it to turn.
  3. Every half-revolution (180°), the split-ring commutator swaps contact from one brush to the other.
  4. This reverses the direction of the current in each side of the coil, so the forces reverse on each side, keeping the coil rotating continuously in the same direction.

❌ Severe Misconception

Do NOT describe a generator! The examiner explicitly instructed: "do not accept answers describing a generator". This question is about an electric motor (electrical energy → kinetic energy), NOT electromagnetic induction or inducing potential difference.

🧠 Exam Tip: The Role of the Commutator

Without the split-ring commutator, the coil would simply flip until vertical and then oscillate to a halt. Always state both: what changes (the current reverses every half-turn) and why that matters (the forces keep pushing the coil in the same rotational direction).

Mark Breakdown:
• [1 mark] Magnetic fields produce forces on the sides AB and CD
• [1 mark] Forces on AB and CD act in opposite directions
• [1 mark] Commutator reverses connection/swaps contacts every half-turn
• [1 mark] Current reverses so forces reverse, sustaining continuous rotation
Question 07.4 • 2 Marks

Zero Moment at the Vertical Position

Why turning effect disappears at 90°

✅ Model Answer

  • When vertical, there is no current in the wire.
  • Therefore, no magnetic force acts on the sides of the coil (so resultant moment is zero).

💡 Key Knowledge: Commutator Gaps

At the exact vertical position, the gaps in the split-ring commutator line up with the brushes/contacts. This momentarily breaks the circuit: Current (I) = 0. Since F = BIl , if I = 0 , then F = 0 . Momentum carries the coil past this dead spot!

🧠 Examiner Insight on Dependent Marks

Mark Point 2 is strictly dependent on Mark Point 1. You cannot score the mark for saying "there is no force / no moment" unless you have clearly explained that there is no current in the wire.

Mark Breakdown:
• [1 mark] There is no current in the wire
• [1 mark] (Dependent on MP1) So there is no (magnetic) force acting

Topics

Physics · P7: Magnetism and Electromagnetism · P5: Forces

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