If a charged particle of charge to mass ratio `(q/m) = alpha` centers in a magnetic field of strength B at a speed `v=(2 alpha d) (B),` then

Options

A. angle subtended by the path of charged particle in magnetic field at the centre of circular path is `2pi`
B. The charge will move on a circular path and then will come out from magnetic field at some distance from the point of indertion
C. The time for which particle will be in the magnetic field is `(2pi)/(alphaB)`
D. Angle subtended by the path of charged particle in magnetic field at the centre of circular path is `pi//2`

Answer

b

Explanation

`r=(mv)/(qB)=v/(Balpha)=((2alphad)(B))/((Balpha))=2d`

i.e., the electron will move out after travelling on a semicircular path of radius `r=2d`.