Question

A square loop $ABCD$, carrying a current $I_1$, 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

• A. $\frac{{\mu }_0{I}_1{I}_2}{2\pi }$
• B. $\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi }$
• C. $\frac{2{\mu }_0{I}_1{I}_{2}L}{2\pi }$
• D. $\frac{2{\mu }_0{I}_1{I}_2}{3\pi }$

Answer 1

The correct option is D $\frac{2{\mu }_0{I}_1{I}_2}{3\pi }$

Force on arm $AB$ due to current in conductor $XY$ is
$F_1=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_{2}L}{\left(L/2\right)}=\frac{{\mu }_0{I}_1{I}_2}{\pi }$ acting towards the wire in the plane of loop.
Force on arm $CD$ due to current in conductor $XY$ is
$F_2=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_{2}L}{\left(3L/2\right)}=\frac{{\mu }_0{I}_1{I}_2}{3\pi }$ away from the wire the plane of loop.
$\therefore$ Net force on the loop $=F_1-F_2$
$=\frac{{\mu }_0{I}_1{I}_2}{\pi }[1-\frac{1}{3}]=\frac{2}{3}\frac{{\mu }_0{I}_1{I}_2}{\pi }$.