Q. A charge $Q$ is uniformly distributed over the surface of non-conducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure
A) [image:4_image80.emf]
B) [image:4_image81.emf]
C) [image:4_image82.emf]
D) [image:4_image83.emf]
Explanation
(a) $dB = \frac{\mu_{0}(dq)}{2r}\left( \frac{\omega}{2\pi} \right)$
$B = \int_{}^{}{dB} = \frac{\mu_{0}\omega}{4\pi}.\frac{Q}{\pi R^{2}}2\pi\int_{0}^{R}\frac{rdr}{r}$
$B = \frac{\mu_{0}\omega Q}{2\pi R^{2}}.R$
$B = \frac{\mu_{0}\omega Q}{2\pi R}$
$B \propto \frac{1}{R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.