A bar magnet of length l and magnetic dipole moment M is bent in...

Given, (\theta=60^{\circ}=\frac{\pi}{3} \mathrm{rad}) The new magnetic length of the dipole will be, (I^{\prime}=r) Now, (M=m \cdot I) and (M^{\prime}=m \cdot I^{\prime}) (\therefore \frac{\mathrm{M}}{\mathrm{M}^{\prime}}=\frac{$\mathrm{I}$}{$\mathrm{I}$^{\prime}} \ldots \quad (1)) Now, angle in radians (=\frac{\text{ arc length }}{\text{ radius }}) (\therefore \frac{\pi}{3}=\frac{1}{r} \Rightarrow 1=\frac{\pi r}{3}) In the equilateral triangle, marked by dotted lines, (I^{\prime}=r) (\therefore \frac{1}{I^{\prime}}=\frac{\pi}{3} \ldots \quad (2)) From and, we get, $$\begin{aligned} & \frac{M}{M^{\prime}}=\frac{\pi}{3} \\ & \therefore M^{\prime}=\frac{3 M}{\pi}\end{aligned}$$) Hence, $(C)$ is$the$correct answer. Alternative method: Using the short cut formula about bending of a bar magnet, (M=\frac{2 M \sin \left(\frac{\theta}{2}\right)}{\theta}) (=\frac{2 M \sin 30^{\circ}}{\frac{\pi}{3}}=\frac{3 M}{\pi}) Hence, (C) is the correct answer.$