An electron moving with a velocity `v` along the positive `x`-axis approaches a circular current carrying loop as shown in the fig. the magnitude of magnetic force on electron at this instant is

Options

A. `(mu_(0))/(2) (eviR^(2))/((x^(2)+R^(2))^(3//2))`
B. `mu_(0) (eviR^(2)x)/((x^(2)+R^(2))^(3//2))`
C. `(mu_(0))/(4pi) (eviR^(2)x)/((x^(2)+R^(2))^(3//2))`
D. `0`

Answer

D

Explanation

`vec(F)=q[vhatj]xxB(-hati)]=0`
Because `B` as well as `v` are along `x` axis.