A particle of charge `Q` moving with kinetic energy `K` enters a zone of uniform magnetic field `B` along normal to `A_(1)B_(1)` as shown satisfying the condition `dgt(2K)/(QvB)` where `v` is the velocity of the particle .What will be tha angle of deflection in crossing the field at the centra of the curve followed by the particle in the field, if `K=QvBb` ? (`b`=radius of the circle in which the particle rotates)

Options

A. `pi//4`
B. `pi//3`
C. `pi//6`
D. `pi//2`

Answer

D

Explanation

`QvB=(mv^(2))/(r )`
`r=(mv^(2))/(vQB)=(2K)/(vQB)`

Since, `dlt(2K)/(QvB)`, particle will not complete the
semicircle
`sintheta=(d)/(r )=(dvQB)/(2K), K=(QvBd)/(2sintheta)`
`2sintheta=1` for `K=QvBd`
`implies theta=(pi)/(2)`