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 the, width of the region is `b`, which is very slightly less than `(mv)/(2Bq)`

Explanation

The radius of path ` r=(mv)/(qB)`
For `d lt r,` we have sin `theta = d/r`
`(a) d= ((mv)/(qB)) = r, therefore sin theta= (((mv)/(qB)))/(((mv)/(qB)))=1 or theta = (pi)/(2) ` rad
`(b) d= ((mv)/(2qB)) lt r , sin theta = d/r = ((mv)/(2qB))/((mv)/(qB)) = 1/2 or theta =pi/6 rad`
`(c) d = ((2mv)/(qB))gt r`, the deviation of particle is therefore `theta= pi` rad.