Question

A straight magnetised wire of magnetic moment $10A{m}^2$ 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 $10\left(\frac{\pi -2}{\pi }\right)A{m}^2$

We know that the length of a circular path is $2\pi r$ where r is its radius.

So the length(L) of a semi circle is $L=\pi r.$

If the wire is bent into a semi circle, then the straight distance(L') between the poles is the diameter i.e. $L^{\prime }=2r.$

Thus $L^{\prime }=\frac{2L}{\pi }$

The magnetic moment is directly proportional to the straight distance between the poles.

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

$\frac{M^{\prime }-M}{M}=\frac{2-\pi }{\pi }$

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

But the change in moment$=M-M^{\prime }=M\times \frac{\pi -2}{\pi }=10\times \frac{\pi -2}{\pi }A{m}^2$