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

Explanation

The magnetic force `F` is `_|_^(ar)` to `v` as well as `B`, since `v` is `_|_^(ar)` to `B`, charged particle will below circular path inside magnetic field.

`AB` is circular arc, sketch two tangent lines at `A` and `B`, if we make perpendicular lines on tangent lines passing through `A` and `B`, these perpendiculars will meet at centre `O` of circle.
Here, `theta`: angle of deviation.
In `Delta OBC`
`sin theta = (d)/(R ) = (d)/(mv//Bq) = (Bqd)/(mv)`
`theta = sin^(-1) ((Bqd)/(mv))`
Time-period `T = (2 pim)/(Bq)`
Time to travel circular arc of angle `theta`
`t = (2 pim)/(Bq).(theta)/(2 pi) = (m theta)/(Bq), theta`: in radian
Deflection `_|_^(ar)` to initial motion dierction:
`AC = R - R cos theta = R(1 - cos theta)`
`= R[1 - (1 - 2 sin^(2) theta//2)] = 2R sin^(2) (theta//2)`
If `theta` is small, `sin theta ~= theta = d//R`
`AC = 2R((theta)/(2))^(2) = 2R((d)/(2R))^(2) = (d^(2))/(2R)`
`sin theta = (Bqd)/(mv)`
`(sin theta) le 1` i.e. `d le R (= (mv)/(Bq))`
This formula is valid if`d le R (= (mv)/(Bq))`
If `d gt R( = (mv)/(Bq))`, the particle enters and emerges from same side.