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

$\text{ Force on arm }AB\text{ due to current in conductor }XY\text{ is }$

$F_1=\frac{{\mu }_0}{4\pi }\frac{2IiL}{(L/2)}=\frac{{\mu }_{0}Ii}{\pi }$

acting towards X Y in the plane of loop. Force on arm CD due to current in conductor X Y is

$F_2=\frac{{\mu }_0}{4\pi }\frac{2IiL}{3(L/2)}=\frac{{\mu }_{0}Ii}{3\pi }$

acting away from X Y in the plane of loop.
therefore Net force on the loop $=F_1-F_2$

$=\frac{{\mu }_{0}Ii}{\pi }\left[1-\frac{1}{3}\right]=\frac{2}{3}\frac{{\mu }_{0}Ii}{\pi }$