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. `(2mu_0 Ii)/(3pi)`
B. `(mu_0 Ii)/(2pi)`
C. `(2mu_0 IiL)/(3pi)`
D. `(mu_0IiL)/(2pi)`

Answer

A

Explanation

Force on wire BE will be cancel out by the force on the wire AD.
` F_(AB) = (mu_0 Ii(L) )/(2pi (L//2) ) = (mu_0 Ii)/(pi) ` (attractive)
`F_(CD) = (mu_0 Ii (L) )/(2pi (3L // 2) ) = (mu_o Ii)/(3pi ) ` (repulsive)
` therefore F_("net") = (mu_0 Ii)/(pi) (1 - 1/3) = (2mu_0Ii)/(3pi) `