Question

A charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is :

• A. $\frac{2\pi qB}{m}$
• B. $\frac{2\pi m}{qB}$
• C. $\frac{2\pi mq}{B}$
• D. $\frac{2\pi q^2}{B}$

Answer 1

The correct option is B $\frac{2\pi m}{qB}$

Equating the centripetal force and magnetic force, we get $\frac{m{v}^2}{r}=qvB\sin {90}^{\circ }$
$\Rightarrow v=\frac{qBr}{m}$
Thus time period of revolution $T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$.