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.
$$\frac{{2\pi {q^2}B}}{m}$$
B.
$$\frac{{2\pi mq}}{B}$$
C.
$$\frac{{2\pi m}}{{qB}}$$
D.
$$\frac{{2\pi qB}}{m}$$
Answer :
$$\frac{{2\pi m}}{{qB}}$$
Solution :
Equating magnetic force to centripetal force,
$$\frac{{m{v^2}}}{r} = qvB\sin {90^ \circ }$$
Time to complete one revolution,
$$T = \frac{{2\pi r}}{v} = \frac{{2\pi m}}{{qB}}$$