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)`

Explanation

`sin theta = (d)/(R ) = (d)/(mv//Bq) = (Bqd)/(mv)`
`theta = sin^(-1)((Bqd)/(mv))`
(a) `d ~= (mv)/(Bq), theta = sin^(-1) (1) = (pi)/(6)`
(b) `d ~= (mv)/(2 Bq), theta = sin^(-1)(1//2) = (pi)/(6)`
( c) Here `d gt R (=(mv)/(Bq))`, the above formula is not applicable.
In this case, the particle enters and emerges from same side.
The particle moves on semi-circular path