A particle of mass `m` and charge `q` is projected into a region having a perpendicular magnetic, field `B`. Find the angle of deviation of the particle as it comes out of the magnetic field if th,, width of the region is (b)
(a) `(2mv)/(Bq) (b) `(mv)/(Bq)` (c) `(mv)/(2 Bq)`

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

A

Explanation

`(a)` The radius of the circular orbit is given by `r=(m upsilon)/(qB).` So the angle of deviation for
`d=(m upsilon)/(2qB)=(r)/(2)` is `sin theta =(d)/(r)=(1)/(2)`
`impliestheta=sin^(-1)((1)/(2))=(pi)/(6)=30^(@)`.