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Q.
Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius $R$ with constant speed $v.$ The time period of the motion
NTA AbhyasNTA Abhyas 2022
Solution:
When magnetic field is perpendicular to motion of charged particle, then
Centripetal force = magnetic force
$ie, \, \, \frac{m v^{2}}{R}=Bqv$
or $R=\frac{m v}{B q}$
Further, time period of the motion
$T=\frac{2 \pi R}{v}=\frac{2 \pi \left(\frac{m v}{B q}\right)}{v}$
or $T=\frac{2 \pi m}{B q}$
It is independent of both $R \, \, and \, v.$