A square loop $\mathrm{ABCD}$ carrying a current $\mathrm{i}$, is placed near and coplanar with a long straight conductor $\mathrm{XY}$ carrying a current $\mathrm{I}$, the net force on the loop will be:

• Correct answer
  A

  $\frac{2{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}}{3\mathrm{\pi }}$
• B

  $\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}}{2\mathrm{\pi }}$
• C

  $\frac{2{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}\mathrm{L}}{3\mathrm{\pi }}$
• D

  $\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}\mathrm{L}}{2\mathrm{\pi }}$

Solution

${\mathrm{F}}_1=\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}\mathrm{L}}{2\mathrm{\pi }\frac{\mathrm{L}}{2}\mathrm{ }}=\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}}{\mathrm{\pi }}\mathrm{ }$

${\text{F}}_{\text{2}}=\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}\mathrm{L}}{2\mathrm{\pi }\frac{3\mathrm{L}}{2}}=\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}}{3\mathrm{\pi }}$



$\therefore {\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\mathrm{F}}_1-{\mathrm{F}}_2$

${\mathrm{F}}_{\mathrm{n}\mathrm{e}\mathrm{t}}=\frac{2}{3}\frac{{\mathrm{\mu }}_0\mathrm{I}\mathrm{i}}{\mathrm{\pi }}$

Hence, the correct answer is option: 'A'.