A particle of charge per unit mass `alpha` is released from origin with a velocity `vecv=v_(0)hati` in a magnetic field
`vec(B)=-B_(0)hatk` for `xle(sqrt(3))/2 (v_(0))/(B_(0)alpha)`
and `vec(B)=0` for `xgt(sqrt(3))/2 (v_(0))/(B_(0)alpha)`
The `x`-coordinate of the particle at time `t((pi)/(3B_(0)alpha))` would be

Options

A. `(sqrt3)/(2) (v_(o))/(B_(o)alpha) + (sqrt3)/(2)v_(o) (t - (pi)/(B_(o)alpha))`
B. `(sqrt3)/(2) (v_(o))/(B_(o)alpha) + v_(o) (t - (pi)/(3B_(o)alpha))`
C. `(sqrt3)/(2) (v_(o))/(B_(o)alpha) + (v_(o))/(2) (t - (pi)/(3B_(o)alpha))`
D. `(sqrt3)/(2) (v_(o))/(B_(o)alpha) + (v_(o)t)/(2)`

Answer

C

Explanation

`r = (mv_(o))/(B_(o)q) = (v_(o))/(B_(o)alpha)`
`(x)/(r ) = (sqrt3)/(2) = sin theta :. theta = 60^(@)`
`t_(oA) = (T)/(6) = (pi)/(3B_(o)alpha)` Therefore `x -` coordinate of particle at any time `t gt (pi)/(3B_(o)alpha)` will be
`x = (sqrt3)/(2) (v_(o))/(B_(o)alpha) + v_(o) (t - (pi)/(3B_(o)alpha)) cos 60^(@) So x = ((sqrt3)/(2) (v_(o))/(B_(o)alpha)) + (v_(o))/(2) (t - (pi)/(3B_(o)alpha))`