Question

A bar magnet of length $l$ and magnetic dipole moment $M$ is bent to form an arc which subtends an angle of ${120}^{\circ }$ at centre. The new magnetic dipole moment will be

• A. $\frac{3M}{2\pi }$
• B. $\frac{3\sqrt{3}M}{2\pi }$
• C. $\frac{3M}{\pi }$
• D. $\frac{2M}{\pi }$

Answer 1

The correct option is B $\frac{3\sqrt{3}M}{2\pi }$




Initially $ml=M$

$m$ $\to$ pole strength
$M$ $\to$ magnetic moment


here in $\Delta ABC,AC=CB=r$

$d=rcos{30}^{\circ }+rcos{30}^{\circ }=r\sqrt{3}$

And, $Arclength,l=r\theta =r\left(\frac{2\pi }{3}\right)$

$l=\frac{2\pi r}{3}\Rightarrow r=\frac{3l}{2\pi }$

So, $d=\frac{3l\sqrt{3}}{2\pi }$
So, $M^{\prime }=md=m.\frac{3l\sqrt{3}}{2\pi }=\frac{3\sqrt{3}M}{2\pi }$

Hence, option $\left(B\right)$ is the correct answer.

Alternate solution:

Short Trick: To solve such problems we can directly use the expression given below,

$M^{\prime }=\frac{2M\left(\sin \theta /2\right)}{\theta }$