question_answer

A charge \[(+Q)\] is uniformly distributed on a circular disc of radius R. The disc rotates with constant angular speed \[\omega \] about its own axis. The magnetic field at the centre of the disc is [Assume that the disc is in the XY - plane and rotates in anticlockwise direction]

A) \[\frac{{{\mu }_{0}}Q\omega }{2\pi R}\]

B) \[\frac{{{\mu }_{0}}Q\omega }{\pi R}\]

C) \[\frac{{{\mu }_{0}}Q\omega }{4\pi R}\]

D) \[\frac{2{{\mu }_{0}}Q\omega }{\pi R}\]  

Correct Answer: A

Solution :

\[dq=\left( \frac{Q}{\pi {{R}^{2}}} \right)\left( 2\pi rdr \right)\]

\[i=\frac{dq}{T}=\frac{dq}{\left( 2gp/\omega  \right)}=\frac{\omega \left( dq \right)}{2\pi }\]

\[B=\int\limits_{0}^{R}{\frac{{{\mu }_{0}}i}{2r}=\frac{{{\mu }_{0}}}{2}\int\limits_{0}^{R}{\frac{\omega \left( \frac{Q}{\pi {{R}^{2}}} \right)2\pi rdr}{\left( 2\pi  \right)r}}}\]

\[=\frac{{{\mu }_{0}}Q\omega }{2\pi R}\]