question_answer

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 :     AIEEE  Solved  Paper (Held On 11 May  2011)

A) \[\pi {{R}^{4}}\sigma \omega \]

B) \[\frac{\pi {{R}^{4}}}{2}\sigma \omega \]

C) \[\frac{\pi {{R}^{4}}}{4}\sigma \omega \]

D) \[2\pi {{R}^{4}}\sigma \omega \]

Correct Answer: C

Solution :

                                \[\frac{q}{2M}=\frac{Magnetic\,dipole\,moment}{Angular\,momentum}\]                 \[\therefore \]Magnetic dipole moment (M)                 \[M=\frac{q}{2M}.\left( \frac{M{{R}^{2}}}{2} \right).\omega \]                 \[=\frac{1}{4}\sigma .\pi {{R}^{4}}\omega .\]