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

• Correct answer
  A

  $\frac{3 M}{\pi}$
• B

  $\frac{4 M}{\pi}$
• C

  $\frac{M}{\pi}$
• D

  $\frac{2 M}{\pi}$

Solution

$M=m \cdot L$
$R \theta=L$ $\frac{R \pi}{3}=L$ $R=\frac{3 L}{\pi}$ $M^{\prime}=m(2 R) \sin 30^{\circ}$ $=m(2) \frac{3 L}{\pi} \times \frac{1}{2}=\frac{3}{\pi} m L=\frac{3 M}{\pi}$

Hence, the correct answer is option: 'A'.