alleryA proton of mass m, charge e, speed v enters a magnetic field B perpendicular to its velocity. The radius of its circular path is
AeBv/m
Bmv/(eB)
CmB/(ev)
Dev/(mB)
Answer & Solution
Correct answer: B. mv/(eB)
1. Magnetic force provides centripetal force: evB = mv²/r.
2. Cancelling one v from both sides: eB = mv/r.
3. Solving for r: r = mv/(eB).
4. The radius scales with momentum p = mv, so faster protons curve along larger arcs.
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.3_
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