A magnet of length L and Moment M is cut into two halves perpendicular to its axis. One piece A is bent into a semicircle of radius R and is joined to the other piece at the poles as shown. The moment of the resulting magnet is,

$\text{Magnetic Moment is proportional to Magnetic Length}$

$\begin{array}{l}\therefore M_B \text{}=M/2\\ \text{For A, Magnetic length} =2R \text{}\\ \text{where} \pi R=\frac{L}{2}\text{}\\ \Rightarrow R= \frac{L}{2\pi } \text{} \\ \text{}\Rightarrow \text{Magnetic length} =2R = L/\pi \\ \text{}\because \text{Magnetic Moment is proportional to Magnetic Length} \\ \text{}\therefore M_A \text{} = \frac{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.}{L }\times M \text{} = \frac{M}{\pi }\end{array}$

$Hence, M^1=M_A+M_B= \frac{M}{\pi }+\frac{M}{2}=\frac{M(2+\pi )}{2\pi }$