Question

A square loop $\text{ABCD}$ of side $L$ carrying a current $i_1$ is placed near and co-planar with a long straight conductor $\text{XY}$ carrying a current $i_2$ as shown in the figure. Find the net force on the loop is-

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

Answer 1

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

$F_{AB}=\frac{{\mu }_0{i}_1{i}_{2}L}{2\pi \left(\frac{L}{2}\right)}=\frac{{\mu }_0{i}_1{i}_2}{\pi }\left(\text{towards left}\right)......\left(1\right)$

$F_{CD}=\frac{{\mu }_0{i}_1{i}_{2}L}{2\pi \left(\frac{3L}{2}\right)}=\frac{{\mu }_0{i}_1{i}_2}{3\pi }\left(\text{towards right}\right)......\left(12\right)$

$F_{BC}$ and $F_{DA}$ cancel each other

$\left|\begin{array}{c}{F}_{net}\end{array}\right|=F_{AB}-F_{CD}=\frac{{\mu }_0{i}_1{i}_2}{\pi }(1-\frac{1}{3})$

$\left|\begin{array}{c}{F}_{net}\end{array}\right|=\frac{2{\mu }_0{i}_1{i}_2}{3\pi }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.