A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of an arc as shown in the figure. The new magnetic dipole moment will be:

Original Magnetic Moment: The magnetic dipole moment ($M$) is defined as the product of the pole strength ($m$) and the magnetic length ($l$). $M = m \times l$

Bent Magnet Geometry: When the magnet is bent into an arc of radius $r$ subtending an angle $\theta$ (in radians) at the center, the length of the arc remains equal to the original length $l$. $l = r\theta \implies r = \frac{l}{\theta}$

New Magnetic Moment ($M'$): The new magnetic moment is the product of the pole strength ($m$) and the straight-line distance (chord) between the two poles. The chord length is given by $2r \sin(\theta/2)$. $M' = m \times 2r \sin(\theta/2)$

Substitution: Substituting $r = l/\theta$: $M' = m \times 2 \left( \frac{l}{\theta} \right) \sin(\theta/2) = \frac{2(ml)}{\theta} \sin(\theta/2) = M \frac{2 \sin(\theta/2)}{\theta}$

Finding the Answer: To match the option $\frac{3M}{\pi}$, we must determine the angle $\theta$. Comparing terms: $\frac{2 \sin(\theta/2)}{\theta} = \frac{3}{\pi}$ If we assume the arc subtends $60^\circ$ (or $\pi/3$ radians): $M' = M \frac{2 \sin(\pi/6)}{\pi/3} = M \frac{2(1/2)}{\pi/3} = \frac{3M}{\pi}$ Thus, for a $60^\circ$ arc, the new magnetic moment is $3M/\pi$.