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(B) The magnetic Lorentz force experienced by a charged particle moving in a magnetic field is $F = qvB$. This magnetic force provides the necessary c

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does not depend on $v$ and $R$ both.

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(B) The magnetic Lorentz force experienced by a charged particle moving in a magnetic field is $F = qvB$. This magnetic force provides the necessary centripetal force for circular motion: $qvB = \frac{mv^2}{R}$. Solving for the radius $R$, we get $R = \frac{mv}{qB}$. The period of revolution $T$ is defined as the time taken to complete one full circle, given by $T = \frac{2\pi R}{v}$. Substituting the expression for $R$ into the formula for $T$, we get $T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$. Since $T$ depends only on the mass $m$, charge $q$, and magnetic field $B$, it is independent of both the linear speed $v$ and the radius $R$.

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

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