Explanation

(a) The effective length of magnet $2l = 31.4\ cm = 0.314\ m$
Pole strength $m = 0.5\ Am$
So, the magnetic moment, $M = m \times 2l$
$= (0.5 \times 0.314){Am}^{2}$
$= 0.157\ Am^{2}$
When magnet is bent in the form of semicircle (of diameter d), then length of magnet $= \pi\frac{d}{2}$
$\therefore\ \ 31.4 = \frac{\pi d}{2}$
$\Rightarrow \ \ \ \ \ \ d = \frac{31.4 \times 2}{3.14} = 20\ cm$
$\therefore$ Effective length of magnet
$2l^{'} = d = 20\ cm = 0.2\ cm$
Hence, its magnetic moment will be
$M^{'} = m \times 2l'$
$\ \ \ \ \ \ \ = 0.5 \times 0.2 = 0.1\ Am^{2}$