A square loop 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. `(2m_0liL)/(3 pi)`
B. `(mu_0liL)/(2pi)`
C. `(2mu_0Ii)/(3pi)`
D. `(mu_0Ii)/(2pi)`

Answer

C

Explanation

Force on arm AB due to current in conductor XY is
`F_1 = (mu_0)/(4pi) (2 IiL)/((L//2)) = (mu_0Ii)/(pi)`
acting towards XY in the plane of loop .
Force on arm CD due to current in conductor XY is
`F_(2) = (mu_0)/(4pi3)(2liL)/(3(L//2))= (mu_0li)/(3pi)`
acting away form XY in the plane of loop.
`therefore` Net force on the loop `= F_1 -F_2`
`= (mu_0li)/pi[1-1/3] = 2/3 (mu_0Ii)/(pi)`