A square loop of ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be

Options

A. `(mu_(0) li)/(2pi)`
B. `(2mu_(0) liL)/(3pi)`
C. `(mu_(0) liL)/(2pi)`
D. `(2mu_(0) li)/(3pi)`

Answer

D

Explanation

`F = F_(1) - F_(2) = (mu_(0) I)/(2pi ((L)/(2))) "".(iL) - (mu_(0) I)/(2pi ((3L)/(2))) ""(iL) = (2mu_(0) iI)/(3pi)`