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A particle of charge q and mass m moves on a circular path of radius r in a plane normal to a uniform magnetic field B. The time taken by this particle to complete one revolution is ______.

Asked in GSEB Board July 2016 · Circular motion: radius, period and energy

• (1) (2π q²B)/m
• (2) (2π mq)/B
• (3) (2π qB)/m
• (4) (2π m)/(Bq)

Step-by-step solution

Why the other options are wrong

• (1) Nothing in the derivation squares the charge. A unit check settles it: (qB)/m has the unit s⁻¹, so (q²B)/m has the unit C s⁻¹ — a current, not a time.
• (2) This puts q in the numerator, so a larger charge would take longer to go round. In fact a larger charge feels a larger force qvB, turns on a tighter circle and finishes sooner, so q belongs in the denominator.
• (3) (qB)/m is the angular frequency ω, so this option is 2πω, measured in s⁻¹. The period is (2π)/ω — this has the ratio upside down.

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