A straight bar magnet has a magnetic moment M . It is first cut...

Let the initial length of the bar magnet be $L$ and its pole strength be $m$. The initial magnetic moment is $M = m L$.

When the magnet is cut transversely into two equal halves, the length of each half becomes $\dfrac{L}{2}$. The pole strength $m$ remains unchanged.

When one of these halves is bent into a semicircle of radius $R$, the arc length is equal to the length of the piece:
$\pi R = \dfrac{L}{2} \implies R = \dfrac{L}{2 \pi}$

The new effective magnetic length is the shortest distance between the poles, which is the diameter of the semicircle:
$d = 2 R = 2 \left( \dfrac{L}{2 \pi} \right) = \dfrac{L}{\pi}$

The new magnetic moment $M'$ is the product of the pole strength and the new effective length:
$M' = m \times d = m \left( \dfrac{L}{\pi} \right) = \dfrac{M}{\pi}$

Answer: $\dfrac{M}{\pi}$