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

Answer

A

Explanation

`F_(AB) = (mu_(o)liL)/(2pi [(L)/(2)])` towards XY
`F_(CD) = (mu_(o)liL)/(2 pi [(3L)/(2)])` away from XY
`F_("net") = F_(AB) - F_(CD)`
`= (2)/(3) (mu_(o)Ii)/(pi)`