alleryThe period of revolution of a charged particle moving in a uniform magnetic field perpendicular to its velocity is
A2πqB/m
BqB/(2πm)
Cm/(2πqB)
D2πm/(qB)
Answer & Solution
Correct answer: D. 2πm/(qB)
1. Magnetic force provides centripetal force: qvB = mv²/r.
2. Orbital speed v = qBr/m, so angular velocity ω = v/r = qB/m.
3. Period T = 2π/ω = 2πm/(qB).
4. Notice T is independent of v and r, so the cyclotron frequency stays constant as the particle gains energy.
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.3_
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