A particle of charge -q and mass m enters a uniform magnetic field `vecB` (perpendicular to paper inward) at P with a velocity `v_0` at an angle `alpha` and leaves the field at Q with velocity v at angle `beta` as shown in fig.

Options

A. `alpha`=`beta`
B. `v=v_0`
C. `PQ=(2mv_0sin alpha)/(Bq)`
D. The particle remains in field for time `t=(2m(pi-alpha))/(Bq)`

Answer

(a,b,c,d)

Explanation

Option (b) is obvious
`r=(mv_0)/(qB)`
`PQ=2r sin alpha=2(mv_0)/(qB) sin alpha`
`alpha=beta`
Time taken `=T=(2pir)/v`
`T=(2pim)/(qB)`
For t time, `t=T/(2pi)(2pi-2alpha)=(2m)/(qB)(pi-alpha)`