A steel wire of length ' $\ell$ ' has a magnetic moment 'M'. It is then bent into a semicircular arc. The…
A steel wire of length ' $\ell$ ' has a magnetic moment 'M'. It is then bent into a semicircular arc. The new magnetic moment is
$2 \mathrm{M} / \pi$
$\mathrm{M}$
$\mathrm{M} \times \ell$
$\mathrm{M} / \ell$
Solution
When wire is bent in the form of semicircular arc then, $\mathrm{l}=\pi \mathrm{r}$ $\therefore$ The radius of semicircular arc, $\mathrm{r}=1 / \pi$
Distance between two end points of semicircular wire $=2 \mathrm{r}=\frac{21}{\pi}$ $\therefore$ Magnetic moment of semicircular wire $=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l$
But $\mathrm{ml}$ is the magnetic moment of straight wire i.e., $\mathrm{ml}=\mathrm{M}$
$\therefore$ New magnetic moment $=\frac{2}{\pi} \mathrm{M}$
