Explain why a charged particle moving parallel to a uniform magnetic field experiences no magnetic force, whilst the same particle moving perpendicular to the field experiences a maximum magnetic force.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A proton moves through a region of uniform magnetic field.
Model answer (4 marks)
The magnetic force on a charged particle is given by F = qvB sinθ, where θ is the angle between the velocity vector and the magnetic field. If the particle moves parallel to the field, θ = 0°, so sin0° = 0 and F = 0 – no magnetic force acts. If the particle moves perpendicular to the field, θ = 90°, so sin90° = 1 and F = qvB – the force is at its maximum. The force is always perpendicular to both v and B (right‑hand rule), changing only the direction of motion, not its speed.
Examiner tips
- State the formula F = qvB sinθ and explain the role of θ
- Show the two cases: θ = 0° gives sin0° = 0, θ = 90° gives sin90° = 1
- Mention that the force is perpendicular to v and B and does no work
Common mistakes
- Confusing the direction of the force with the direction of the field
- Using F = qvB for all angles instead of including sinθ
- Failing to note that the force does no work because it is perpendicular to the velocity
Mark scheme (4 marks)
- The magnetic force on a charged particle depends on the component of velocity perpendicular to the magnetic field (F = qvB sinθ).
- When the particle moves parallel to the field, θ = 0°, so sin 0° = 0 and the magnetic force is zero.
- When the particle moves perpendicular to the field, θ = 90°, so sin 90° = 1 and the force is at its maximum value F = qvB.
- The magnetic force acts perpendicular to both the velocity and the field (by the right-hand rule / cross-product nature of the force), so it changes the direction of the particle but does no work on it.
Key terms in this question
magnetic force · uniform magnetic field · perpendicular · parallel
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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