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

Explanation

Mass of the particle `= m` .
`charge = q`,
`width = d`,
(a ) If `d = mV/qb`
then `d` is equal the radius `q` is the angle between the radius and tangent which is equal to `(pi)/(2)`
(b ) if `d = distance travelled` ,
`(1)/(2)` of radius along x- distance
`d = V_(0) r`
i, e, `i = (d) /(Vx)` .....(1)
`v_(y) = u_(y) + r_(y) l`
From (i) putting the value of `i`,
`tan sin theta = (1)/(2)`
`rArr sin theta = tan ^(-1) ((1)/(2))`
` = 26.4 = 30 ^@ = (pi) / (6)`
(c ) The angle between the initial direction and final direction of velocity is `pi` .