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The magnetic moment of an iron bar is M. It is now bent in such a way that it forms an arc section of a circle subtending an angle of $60^\circ$ at the centre. The magnetic moment of this arc section is
A
$\dfrac{3M}{\pi}$
B
$\dfrac{4M}{\pi}$
C
$\dfrac{M}{\pi}$
D
$\dfrac{2M}{\pi}$
Detailed Solution
From $R\theta=L$ (arc length equals original bar length), $R=\dfrac{L}{\theta}=\dfrac{3L}{\pi}$ (for $\theta=\pi/3$). The new magnetic moment (effective dipole length is the chord $2R\sin30^\circ$): $M'=m(2R)\sin30^\circ=m\times2\times\dfrac{3L}{\pi}\times\dfrac{1}{2}=\dfrac{3mL}{\pi}=\dfrac{3M}{\pi}$.