Solveeit Logo

Question

Physics Question on Magnetism and matter

A thin circular disk of radius RR is uniformly charged with density σ\sigma > 0 per unit area. The disk rotates about its axis with a uniform angular speed ω\omega. The magnetic moment of the disk is :

A

πR4σω\pi R^{4}\sigma\omega

B

πR42σω\frac{\pi R^{4}}{2}\sigma\omega

C

πR44σω\frac{\pi R^{4}}{4}\sigma\omega

D

2πR4σω2\pi R^{4}\sigma\omega

Answer

πR44σω\frac{\pi R^{4}}{4}\sigma\omega

Explanation

Solution

q2M=MagneticdipolemomentAngularmomentum\frac{q}{2M} = \frac{Magnetic \,dipole \,moment}{Angular \,momentum} \therefore\quad Magnetic dipole moment (M) M=q2M.(MR22)ωM = \frac{q}{2M}. \left(\frac{MR^{2}}{2}\right)\omega =14σ.πR4ω= \frac{1}{4}\sigma.\pi R^{4}\omega