The magnetic moment of an iron bar is $M$. It is now bent in such a way that it forms an arc section of a…
- $\frac{3 M}{\pi}$
- $\frac{4 M}{\pi}$
- $\frac{M}{\pi}$
- $\frac{2 M}{\pi}$
Solution

$R \theta=L$ $\frac{R \pi}{3}=L$ $R=\frac{3 L}{\pi}$ $M^{\prime}=m(2 R) \sin 30^{\circ}$ $=m(2) \frac{3 L}{\pi} \times \frac{1}{2}=\frac{3}{\pi} m L=\frac{3 M}{\pi}$
Asked in: NEET 2024 (Re-NEET)