Consider an essentially flat conductive disk of radius $R$ spinning at angular speed $\omega$ with surface charge density $\sigma$ and total charge $Q$. I believe that based on the definition of current density, the surface current density should be $$ K = \sigma v=\frac{Q}{\pi R^2}v=\frac{Q}{\pi R^2}{\omega}r$$ The total current in the $I$ disk is then given by $$I=Q{\omega}R$$ However, my textbook gives $$I=Q{\omega}$$ Which formulation is correct in this instance?
[Physics] Electric Current in Spinning Disk
angular velocityelectric-currentelectromagnetism
Best Answer
Well, given a disk with a uniform surface charge density ${\sigma}$ rotating at angular speed ${\omega}$, its linear speed at a point {r} from the center is ${v=\omega r}$, so its surface current is
$$K=\sigma\omega r$$ I think you're right.