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

Answer

D

Explanation

`F=q[v(+hati)]xxB(-hati)]=0`
Because B as well as v is are along axis of circular CN.