Question

A curved magnet which occupies $\frac{1}{6}$ th of a circle of magnetic moment $\frac{30}{\pi }$ A m${}^2$ is made straight. Find final magnetic moment.

• A. $10A{m}^2$
• B. $20A{m}^2$
• C. $30A{m}^2$
• D. $40A{m}^2$

Answer 1

The correct option is A $10A{m}^2$

$M=\frac{30}{\pi }A{m}^2=m\times r$
$\frac{1^{th}}{6}$ of circumference of a circle
$=\frac{1}{6}\left(2\pi r\right)$
Angle subtended by $\frac{1^{th}}{6}$ of a circle $\left(\theta \right)$
$=\frac{1}{6}\left({360}^{\circ }\right)={60}^{\circ }$ or $\frac{\pi }{3}$
$\therefore$ New magnetic moment M${}^1$
$=m\times 2l$
$=m\times r\pi$
$=(m\times r)\times \frac{\pi }{3}$
$=\frac{30}{\pi }\times \frac{\pi }{3}=10A{m}^2$