Explain why the radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field increases when the particle is accelerated through a greater potential difference before entering the field.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A charged particle of mass m and charge q is accelerated from rest through a potential difference V before entering a region of uniform magnetic field of flux density B, directed perpendicular to the particle's velocity.
Model answer (4 marks)
The electric field does work qV, giving the particle kinetic energy ½mv², so a larger V gives a larger speed v on entering the field.
The magnetic force provides the centripetal force: qvB = mv²/r, giving r = mv/(qB).
With m, q and B fixed, r is directly proportional to v; therefore a greater v means a larger radius.
Because the magnetic force is perpendicular to the velocity it does no work, so the speed remains constant during the circular motion, keeping the radius fixed.
The magnetic force provides the centripetal force: qvB = mv²/r, giving r = mv/(qB).
With m, q and B fixed, r is directly proportional to v; therefore a greater v means a larger radius.
Because the magnetic force is perpendicular to the velocity it does no work, so the speed remains constant during the circular motion, keeping the radius fixed.
Examiner tips
- Show the work‑energy relation qV = ½mv² to link V to v. Write the centripetal force equation qvB = mv²/r and solve for r. State that m, q, B are unchanged, so r ∝ v. Mention that the magnetic force does no work, keeping v constant in the field.
Common mistakes
- Confusing the electric field with the magnetic field; forgetting that the magnetic force does no work. Using the wrong formula for radius (e.g., r = mv/(qB) but mis‑applying it). Failing to explain why the speed remains constant once in the magnetic field.
Mark scheme (4 marks)
- The work done by the electric field (qV) equals the kinetic energy gained, so a greater potential difference gives the particle a greater speed on entering the field.
- The magnetic force provides the centripetal force, giving the relationship qvB = mv²/r, which simplifies to r = mv/(qB).
- Since r = mv/(qB) and both m, q, and B are unchanged, a greater speed v directly produces a greater radius r.
- The magnetic force does no work on the particle (force always perpendicular to velocity), so the speed — and hence kinetic energy — remains constant throughout the circular motion, confirming the path is a circle of fixed radius.
Key terms in this question
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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