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 takeen by the particle to complete one revolution is

Options

A. `(2pim)/(qB)`
B. `(2pimq)/B`
C. `(2piqB)/m`
D. `(pimq)/B`

Answer

A

Explanation

For the charged particle, traveling along a circular path, in a transverse magnetic field, magnetic force=C.P.force `qvB=(mv^(2))/r therefore v=(qvb)/m `but `T=(2pi)/omega=(2pir)/v` Using (1) in (2), we get `T=(2pir xx m)/(qBr)=(2pim)/(qB)`