Explanation

(b) Suppose, the length of wire is $l.$
When, wire formed in circular loop then radius of loop,
$r = \frac{1}{2\pi}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because l = 2\pi r)$
Magnet dipole, $M_{1} = iA$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = i\pi r^{2} = i\pi \times \left( \frac{1}{2\pi} \right)^{2} = i \times \frac{l^{2}}{4\pi}$
When, wire formed in square loop then the side of loop,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = \frac{1}{4}$
Magnet dipole, $M_{2} = iA$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = i \times a^{2} = i \times \frac{l^{2}}{16}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{M_{1}}{M_{2}} = \frac{4}{\pi}$