Question

A magnetised wire of magnetic moment $M$ and length $l$ is bent in the form of one-fourth of a circle of radius $r$. Then its new magnetic moment is.

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

Answer 1

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



After, bending the wire into one-fourth of a circle of radius $r$. Then it's new magnetic moment$M^{{}^{\prime }}$ becomes,

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

Where, $\theta$ is angle subtended at the center.

Here, $\theta =\frac{\pi }{2}={90}^{\circ }$

$\Rightarrow M^{{}^{\prime }}=\frac{2M\sin \left(\frac{{90}^{\circ }}{2}\right)}{\frac{\pi }{2}}$

$=\frac{4M}{\pi }\times \sin {45}^{\circ }=\frac{4M}{\pi }\times \frac{1}{\sqrt{2}}$

$\therefore M^{{}^{\prime }}=\frac{2\sqrt{2}M}{\pi }$

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