Question

A bar magnet of length $\ell$ and magnetic dipole moment $M$ is bent in the form of an arc, as shown in the figure. The new magnetic dipole moment will be -

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

Answer 1

The correct option is C $\frac{3M}{\pi }$



Given, $\theta ={60}^{\circ }=\frac{\pi }{3}\text{rad}$

The new magnetic length of the dipole will be, $l^{\prime }=r$

Now, $M=m\cdot l$ and $M^{\prime }=m\cdot l^{\prime }$

$\therefore \frac{M}{M^{\prime }}=\frac{l}{l^{\prime }}\dots \left(1\right)$

Now, $\text{angle in radians}=\frac{\text{arc length}}{\text{radius}}$
$\therefore \frac{\pi }{3}=\frac{l}{r}\Rightarrow l=\frac{\pi r}{3}$

In the equilateral triangle, marked by dotted lines, $l^{\prime }=r$

$\therefore \frac{l}{l^{\prime }}=\frac{\pi }{3}\dots \left(2\right)$

From and , we get,
$\frac{M}{M^{\prime }}=\frac{\pi }{3}$

$\therefore M^{\prime }=\frac{3M}{\pi }$

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

Alternative method:

Using the short cut formula about bending of a bar magnet,$M^{{}^{\prime }}=\frac{2M\sin \left(\frac{\theta }{2}\right)}{\theta }$

$=\frac{2M\sin {30}^{\circ }}{\frac{\pi }{3}}=\frac{3M}{\pi }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.