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)Ii)/(2 pi)`
B. `(2 mu_(0)IiL)/(3 pi)`
C. `(mu_(0)IiL)/(2 pi)`
D. `(2 mu_(0)Ii)/(3 pi)`

Answer

d

Explanation

The net magnetic force on BC and AD sections of loop will be zero (the force on both section cancel each other because they are acting in opposite direction) as these are perpendicular to the current carrying wire XY.
Magnetic force on section AB will be:
` F_(1) = mu_(0)/(4 pi) (2I iL)/((L//2)) = (mu_(0) I i)/pi`, attractive
Magnetic force on section CD will be :
`F_(2) = mu_(0)/(4 pi) (2I iL)/(3(L//2)) = (mu_(0)I i)/(3 pi)` , repulsive
Net magnetic force , `F_("net") = F_(1) - F_(2)`
`= (mu_(0)I i)/pi [1 - 1/3] = (2 mu_(0)I i)/(3 pi)`, attractive