A charged particle `(q, m)` enters perpendicular in a uniform magnetic field `B` and comes out field as shown. The angle of deviation `theta` time taken by particle to cross magnetic field will be

Options

A. `sin^(-1). (Bqd)/(mv), (m theta)/(Bq)`
B. `sin^(-1) .(Bqv)/(md), (m theta)/(Bq)`
C. `cos^(-1) .(Bqd)/(mv), (m theta)/(Bq)`
D. `cos^(-1). (Bqv)/(md), (m theta)/(Bq)`

Answer

A

Explanation

`sin theta = (d)/( R) = (d)/(mv//Bq) = (Bqd)/(mv)`
`theta = sin^(-1) ((Bqd)/(mv))`
`t = (2pi m)/(Bq).(theta)/(2pi) = (m theta)/(Bq)`