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))/(4) (eviR^(2)x)/((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. none of these

Answer

D

Explanation

Direction of electron velocity is along magnetic field lines. Hence F = 0