A rigid square loop of side $a$ and carrying current $I_2$ is laying on a horizontal surface near a long current $I_1$ wire in the same plane as shown in figure. The net force on the loop due to the wire will be
JEE MAIN 2019, Diffcult
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${{\text{F}}_3}{\text{& }}{{\text{F}}_4}\,{\text{cancel}}\,{\text{each}}\,{\text{other}}{\text{. }}$

${\text{Force on PQ will be }}{{\text{F}}_1} = {2_B}{\text{Ia}}$

${\text{ = }}{{\text{I}}_2}\frac{{{\mu _0}{I_1}}}{{2\pi {\text{a}}}}{\text{a}}$

$ = \frac{{{\mu _0}I}}{{2\pi {\text{a}}}}{\text{a}}$

$ = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi }}$

Force on $RS$ will be $F_{2}=I_{2} B_{2} a$

${=\mathrm{I}_{2} \frac{\mu_{0} \mathrm{I}_{1}}{2 \pi 2 \mathrm{a}} \mathrm{a}}$

${=\frac{\mu_{0} I_{1} I_{2}}{4 \pi}}$

Net force $=\mathrm{F}_{1}-\mathrm{F}_{2}=\frac{\mu_{0} I_1 \mathrm{I}_{2}}{4 \pi}$ repulsion

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