A rectangular loop of sides `27cmxx12cm` carries a current of `12A`. It is placed with its longer side parallel to the long straight conductor `3*0cm` apart and carrying a current of `20A`. Figure. (i) Find the net force on the loop (ii) What will be the net force on the loop if the current in the loop be reversed?

Answer

(i) `3*46xx10^-4N`, directed towards long conductor (ii) `3*46xx10^-4N` directed away from the long conductor

Explanation

Attractive force on side `PQ=(mu_0)/(4pi)(2i_1i_2l)/(r)`,
Repulsive force on the side `RS=(mu_0)/(4pi)(2i_1i_2l)/((r+b))`.
Net attractive force on the loop
`=(mu_0)/(4pi)2i_1i_2[1/r-(1)/(r+b)]l`
When current is reversed, then net repulsive force on the loop `=(mu_0)/(4pi)2i_1i_2[1/r-(1)/(r+b)]l`.