What is the net force on the square coil

Options

A. `25 xx 10^(-7) N` moving towards wire
B. `25 xx 10^(-7)N` moving away from wire
C. `35 xx 10^(-7)N` moving towards wire
D. `35 xx 10^(-7) N` moving away from wire

Answer

A

Explanation

`F_(1) = (mu_(0))/(2pi xx 2 xx 10^(-2)) xx 2 xx 1 xx 15 xx 10^(-2)`
`F_(2) = (mu_(0) xx 2 xx 1 xx 15 xx 10^(-2))/(2pi xx 12 xx 10^(-2))`
now `F_("net") = F_(1) - F_(2)`
`= (4pi xx 10^(-7) xx 2 xx 1 xx 15 xx 10^(-2))/(2pi xx 2 xx 10^(-2)) (1- (1)/(6))`