A wire of length (L) meters, carrying a current (I) ampere is bent in the form of a circle. The magnitude of its magnetic moment, in SI units, is

$\frac{IL^2}{4\pi}$

The correct answer is Option (3) → $\frac{IL^2}{4\pi}$

Given:

Length of wire, $L$

Current, $I$

When the wire is bent into a circle, radius of circle:

$r = \frac{L}{2\pi}$

Area of circle:

$A = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}$

Magnetic moment: $M = I \cdot A$

$M = I \cdot \frac{L^2}{4\pi} = \frac{I L^2}{4\pi}$

∴ Magnitude of magnetic moment of the circular loop = $\frac{I L^2}{4\pi}$ A·m²