JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Magnetic Effects of Current

The magnetic dipole moment of a current loop is $m=IA$, so for the square of side $\ell$ we have $m=I{\ell }^2$. Reshaping the same wire into a circle keeps its length fixed, so the perimeter condition $4\ell =2\pi r$ fixes the new radius at $r=\frac{2\ell }{\pi }$. The circle then encloses an area $A=\pi r^2=\pi {\left(\frac{2\ell }{\pi }\right)}^2=\frac{4{\ell }^2}{\pi }$, and carrying the same current its moment becomes $I\cdot \frac{4{\ell }^2}{\pi }=\frac{4I{\ell }^2}{\pi }=\frac{4m}{\pi }$.