A particle of charge per unit mass `alpha` is released from origin with velocity `vec v = v_0 hat i` in a magnetic field
`vec B = -B_0 hat k` for `x le sqrt 3/2 v_0/(B_0 alpha)`
and `vec B = 0` for `x gt sqrt 3/2 v_0/(B_0 alpha)`
The x-coordinate of the particle at time `t(gt pi/(3B_0 alpha))` would be

Options

A. `(sqrt(3))/(2)(v_(0))/(B_(0)alpha)+(sqrt(3))/(2)v_(0)(t-(pi)/(B_(0)alpha))`
B. `(sqrt(3))/(2)(v_(0))/(B_(0)alpha)+v_(0)(t-(pi)/(3B_(0)alpha))`
C. `(sqrt(3))/(2)(v_(0))/(B_(0)alpha)+v_(0)/(2)(t-(pi)/(3B_(0)alpha))`
D. `(sqrt(3))/(2)(v_(0))/(B_(0)alpha)+(v_(0)t)/(2)`

Answer

c

Explanation

`r=(mv_0)/(B_0q)=(v_0)/(B_0alpha), x/r=(sqrt3)/2 =sin theta`
`implies theta=60^@`
`t_(OA)=T/6=pi/(3B_0alpha)`
Therefore, x-coordinate of particle at
any time `tgtpi/(3B_0alpha)` will be
`x=(sqrt3)/2 (v_0)/(B_0alpha)+v_0(t-pi/(3B_0alpha))cos 60^@`
`=(sqrt3)/2 (v_0)/(B_0alpha)+v_0/2(t-pi/(3B_0alpha))`