A particle of mass m and charge q enters a region of magnetic field (as shown) with speed v at t = 0. There is a region in which the magnetic field is absent as shown. The particle after entering the region collide elastically with a rigid wall. Time t after which the velocity of particle become antiparallel to its initial velocity is -

Options

A. `(m(pi+4))/(2qB)`
B. `(m)/(4qB)(pi+2)`
C. `(m)/(qB)(pi+2)`
D. `(m)/(4qB)(2pi+3)`

Answer

A

Explanation

R=`(mv)/(qB)` ltBrgt `sintheta=(R)/(sqrt(2)xxR)=(1)/(sqrt(2)), theta=45^(@)`
After coming out of B q will collide with the wall

time to exit B
`t_(1)=(pim)/(4qB)`
time to travel in region where B is absent is `t_(2)`

`(R)/(sqrt(2)s)=cos45`
s=R
`t_(2)=(s)/(v)=(R)/(v)=(mv)/(vqB)=(m)/(qB)` ltBrgt total time = `2t_(2)=(2m)/(qB)` ltBrgt time for returing journey in B
`t_(3)=(pim)/(4qB)` ltBrgt total time = `(pim)/(4qB)+(pim)/(4qB)+2q(m)/(qB)`
`(m)/(qB)[(pi)/(2)+2]=(m)/(2qB)[pi+4]`