A bar magnet of magnetic moment  ‘M’ is bent in the form of an arc which makes angle ${60}^0$  . The percentage change in magnetic moment is

If the bar magnet is  bent  in the form of an arc  making an angle $\theta$  at the centre then its new magnetic moment is
                                 $$\begin{gathered} M^1=\frac{2M\sin \frac{\theta }{2}}{\theta \left(rad\right)} \\ \theta =60°=\frac{{\displaystyle {\pi }^C}}{{\displaystyle 3}} \\ {M}^1=\frac{{\displaystyle 2M\sin 30}}{{\displaystyle \pi }}\times 3 \\ {M}^1=\frac{{\displaystyle 2M\times 3}}{{\displaystyle 2\left(\pi \right)}}\times 3=\frac{{\displaystyle 3M}}{{\displaystyle \pi }} \end{gathered}$$ 

Percentage change in magnetic moment is
                            $$\begin{gathered} \frac{\Delta M}{M}\times 100=\frac{M^1-M}{M}\times 100 \\ \begin{array}{l}=\left(\frac{\frac{3M}{\pi }-M}{M}\right)\times 100\\ =\frac{3M-\pi M}{M\pi }\times 100\end{array} \\ \begin{array}{l}=\frac{M(3-\pi )}{M\pi }\times 100\\ =\left(\frac{3-3.14}{3.14}\right)\times 100\end{array} \end{gathered}$$
  
                             $$\begin{gathered} = - 4.5\% \\ =Decreases by 4.5\% \end{gathered}$$