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. `(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

`t=(mv_(0))/(B_(0)q)=(v_(0))/(B_(0) alpha)`
`x/r=(sqrt(3))/2 = sin theta :. Theta= 60^(@)`

`t_(QA)=T/6=(pi)/(3 B_(0) alpha)`
Therefore `x`-coordinate of particle at any time
`tgt(pi)/(3B_(0)alpha)` will be
`x=(sqrt(3))/2 (v_(0))/(B_(0)alpha)+v_(0)(t-(pi)/(3B_(0)alpha)) cos60^(@)`
`=(sqrt(3))/2 (v_(0))/(B_(a) alpha)+(v_(0))/2 (t-(pi)/(3B_(0)alpha))`