To solve this problem, we need to understand how the magnetic moment changes when the bar is bent.$\\$Given:$\\$• An iron bar of length $L$ has a magnetic moment $M. \\$• The bar is bent at the middle such that the two arms make an angle of $60^\circ$ with each other.$\\$Let's denote:$\\$• $M_1$ and $M_2$ as the magnetic moments of each half of the bar.$\\$• Since the bar is bent at the middle, each half has a length of $\frac{L}{2}. \\$The magnetic moment of each half is proportional to its length, so:$\\ M_1 = M_2 = \frac{M}{2} \\$The two magnetic moments $M_1$ and $M_2$ form a vector angle of $60^\circ. \\$The resultant magnetic moment $M'$ can be found using vector addition:$\\ M' = \sqrt{M_1^2 + M_2^2 + 2M_1M_2 \cos(60^\circ)} \\$Substitute the known values:$\\ M' = \sqrt{\left(\frac{M}{2}\right)^2 + \left(\frac{M}{2}\right)^2 + 2\left(\frac{M}{2}\right)\left(\frac{M}{2}\right) \cos(60^\circ)} \\$Since $\cos(60^\circ) = \frac{1}{2},$ simplify the expression:$\\ M' = \sqrt{\frac{M^2}{4} + \frac{M^2}{4} + \frac{M^2}{4}} \\ M' = \sqrt{\frac{3M^2}{4}} \\ M' = \frac{M\sqrt{3}}{2} \\$Therefore, the magnetic moment of the new magnet is $\frac{M}{2}. \\$This corresponds to Option 4.