Question

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. $4\pi {R^4}\sigma \omega $
B. $2\pi {R^4}\sigma \omega $
C. $\dfrac{{\pi {R^4}}}{2}\sigma \omega $
D $\dfrac{{\pi {R^4}}}{4}\sigma \omega $

Answer 1

Hint: When charge is at rest it produces only electric fields but when charge is under motion it produces both electric and magnetic fields. So when a charged disc rotates current is produced due to it as charge is in motion and due to that current, magnetic moment is also produced.
Formula used:
$\dfrac{M}{L} = \dfrac{q}{{2m}}$

Complete answer:
Due to the magnetic effect produced because of current flowing in the rotating disc, a magnetic field is produced which in turn creates the magnetic moment.
When an electron is revolving around the nucleus then that also produces the magnetic moment. Electrons under motion can be considered as a current flow and due to that current flowing, magnetic moment is produced.
The magnetic moment produced will be
$\dfrac{M}{L} = \dfrac{q}{{2m}}$
Where M is the magnetic moment and q is the charge and ‘L’ is the angular momentum of rotation and ‘m’ the mass of the body which is rotating.
Now that result will be valid for all kinds of uniformly charged rotating bodies.
Hence for the uniformly charged disc,
$\eqalign{
  & \dfrac{M}{L} = \dfrac{q}{{2m}} \cr
  & \Rightarrow M = \dfrac{{qL}}{{2m}} \cr
  & \Rightarrow M = \dfrac{{q\left( {I\omega } \right)}}{{2m}} \cr
  & \Rightarrow M = \dfrac{{q\left( {\dfrac{{m{R^2}}}{2}\omega } \right)}}{{2m}} \cr
  & \Rightarrow M = \dfrac{{\sigma A\left( {\dfrac{{m{R^2}}}{2}\omega } \right)}}{{2m}} \cr
  & \Rightarrow M = \dfrac{{\sigma \left( {\pi {R^2}} \right)\left( {\dfrac{{m{R^2}}}{2}\omega } \right)}}{{2m}} \cr
  & \Rightarrow M = \dfrac{{\pi {R^4}}}{4}\sigma \omega \cr} $
Angular momentum of the rotating body will be $L = I\omega $. Moment of inertia of disc will be $I = \dfrac{{m{R^2}}}{2}$. Total charge distributed on the uniform disc will be $q = \sigma A = \sigma \pi {R^2}$. Where R is the radius and $\omega $is the angular velocity of rotation and $\sigma $ is the charge density.

Hence the final answer would be option D.

Note:
This can also be done by the method of integration i.e taking a small ring element and integrating that element for the entire disc by considering the charge on that ring element in terms of area and surface charge density, but that method will be long and difficult one when compared to this one. The given method is only valid when charge distribution is uniform.