A particle with charge q and mass m enters a region of uniform magnetic field B that is pointing into the page/screen. The velocity of the particle is perpendicular to B. The particle will move in a circular path of radius r due to the magnetic force. How much work is done by the magnetic force on the particle during a half revolution?

Question

BlackTom AI · Accepted Answer

Magnetic force is always perpendicular to the velocity: F = q(v × B), so its instantaneous power P = F · v = q(v × B) · v = 0. Therefore the magnetic force does no work.

1) Relevant knowledge points and concepts
- Lorentz magnetic force: F = q v × B (vector cross product).
- Work done by a force along a path: W = ∫ F · ds = ∫ F · v dt.
- If the force is everywhere perpendicular to the velocity, the dot product F · v = 0, so no work is done.
- For circular motion under a magnetic field, the magnetic force provides centripetal force but does not change the speed: q v B = m v^2 / r (magnitudes).

2) Step-by-step reasoning
- Write work: W = ∫ F · ds = ∫ F · v dt.
- Substitute F: W = ∫ q(v × B) · v dt.
- Use property of triple product: (v × B) · v = 0 because any vector dotted with its own cross product is zero.
- Thus integrand is zero for all times, so the integral over any portion of the path (including a half revolution) is zero.

3) Intermediate calculation steps
- Instantaneous power: P(t) = F · v = q(v × B) · v = 0.
- Work over half-turn (time t1 to t2): W = ∫_{t1}^{t2} 0 dt = 0.

4) Why the correct answer is correct
- The magnetic force changes only the direction of velocity (causing circular motion) and not its magnitude. Since work equals change in kinetic energy and kinetic energy remains constant, the work done is zero.

5) Why the other options are incorrect
- πr/(qvB): dimensionally inconsistent and arises from mismanipulating variables; not energy.
- "We have to know whether the charge is positive or negative": sign of q affects direction of curvature but not that F ⟂ v, so work remains zero regardless of sign.
- 2πqvbr and qvBπr: both are nonzero expressions (and one has a lowercase b typo); they imply nonzero work (force × distance) but ignore the perpendicularity that makes F · v = 0.
- Any expression giving a nonzero number contradicts the fundamental vector relation (v × B) · v = 0.

6) Conclusion
The work done by the magnetic force during a half revolution is 0, matching the provided correct answer.

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