Question

A straight magnetised wire of magnetic moment 10 A m${}^2$ is bent into a semi circle. The decrease in the magnetic moment is :

• A. $10\left(\frac{\pi -2}{\pi }\right)A{m}^2$
• B. $\frac{20}{\pi }A{m}^2$
• C. $\frac{5}{\pi }A{m}^2$
• D. $5\pi A{m}^2$

Answer 1

The correct option is A $10\left(\frac{\pi -2}{\pi }\right)A{m}^2$

$M=10A{m}^2=m\times 2l$
$M^{\prime }=m\times d$ and $2l=\frac{\pi d}{2}$
$\therefore d=\frac{2\left(2l\right)}{\pi }$
$M^{\prime }=\frac{m\times \left(2l\right)\times 2}{\pi }=\frac{2M}{\pi }$
Decrease in magnetic moment
$M-M^{\prime }=M-\frac{2M}{\pi }$
$=10\left[\frac{\pi -2}{\pi }\right]A{m}^2$