A square loop `ABCD`, carrying a current `I_(2)` is placed near and coplanar with a long straight conductor `XY`, carrying a current `I_(1)` as shown in Figure. The net force on the loop will be

Options

A. `(mu_(0) I_(1) I_(2))/(2pi)`
B. `(2mu_(0) I_(1) I_(2))/(3pi)`
C. `(mu_(0) I_(1) I_(2)L)/(2pi)`
D. `(2mu_(0) I_(1) I_(2)L)/(3pi)`

Answer

B

Explanation

Force on side `AB` due to current in long wire
`F_(1) = (mu_(0))/(4pi) (2 I_(1) I_(2) L)/(41pi (L//2)) = (mu_(0) I_(1) I_(2))/(pi)` (attractive).
Force on side `CD`
`F_(1) = (mu_(0))/(4pi) (2 I_(1) I_(2) L)/((3L//2)) = (mu_(0) I_(1) I_(2))/(3pi)` (repulsive)
`:.` Net force on the loop `= F_(1) - F_(2)`
`= (mu_(0) I_(1) I_(2))/(pi) [1 - (1)/(3)] = (2)/(3) (mu_(0) I_(1) I_(2))/(pi)` (attractive)