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(C) When a magnetic field is perpendicular to the motion of a charged particle,the magnetic force provides the necessary centripetal force.$F_{c} = F_

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is independent of both $R$ and $v$

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(C) When a magnetic field is perpendicular to the motion of a charged particle,the magnetic force provides the necessary centripetal force.$F_{c} = F_{m}$$\frac{m v^{2}}{R} = B q v$From this,the radius of the circular path is given by:$R = \frac{m v}{B q}$The time period $T$ of the circular motion is the time taken to complete one full circumference:$T = \frac{2 \pi R}{v}$Substituting the expression for $R$:$T = \frac{2 \pi}{v} \left( \frac{m v}{B q} \right)$$T = \frac{2 \pi m}{B q}$Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the radius $R$ and the speed $v$.

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

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