Q. A thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment
A) $M$
B) $\frac{4}{\pi^{2}}M$
C) $\frac{4}{\pi}M$
D) $\frac{\pi}{4}M$
Explanation
(d) Initially for circular coil $L = 2\pi r$ and $M = 1 \times \pi r^{2}$
$= i \times \pi\left( \frac{L}{2\pi} \right)^{2} = \frac{iL^{2}}{4\pi}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Finally for square coil $M^{'} = i \times \left( \frac{L}{4} \right)^{2} = \frac{iL^{2}}{16}\ \ \ \ldots(ii)$
Solving equation (i) and (ii) $M^{'} = \frac{\pi M}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.