Q. A thin circular disk of radius $R$ is uniformly charged with density $\sigma > 0$ per unit area. The disk rotates about its axis with a uniform angular speed $\omega.$ The magnetic moment of the disk is
A) $2\pi R^{4}\sigma\omega$
B) $\pi R^{4}\sigma\omega$
C) $\frac{\pi R^{4}}{2}\sigma\omega$
D) $\frac{\pi R^{4}}{4}\sigma\omega$
Explanation
(d) Let us consider the disc to be made up of large number of concentric elementary rings.
Consider one such ring of radius $x$ and thickness $dx.$
Charge on this elementary ring,
$dq = \sigma \times 2\pi xdx$
[image:4_image103.emf]$\ \ \ \ \ \ = \ 2\pi\sigma xdx$
Current associated with this elementary ring,
$dI = \frac{dq}{dt}$
$\ \ \ \ \ = dr \times f = \ \sigma\omega xdx$
Magnetic moment of this elementary ring,
$dM = dI\pi x^{2} = \pi\sigma\omega x^{3}dx$
$\therefore$ Magnetic moment of the entire disc,
$M = \int_{0}^{R}{dM}$
$= \pi\sigma\omega\int_{0}^{R}{x^{3}dx} = \frac{1}{4}\pi R^{4}\sigma\omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.