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?

  • $\begingroup$ $Q\omega R$ has units $\mathrm{A\cdot m}$ which doesn't match the unit of current $\mathrm{A}$ $\endgroup$ – Alfred Centauri May 1 '17 at 21:53
  • $\begingroup$ The relationship between velocity and angular velocity is $\mathbf v = \boldsymbol\omega \times\mathbf r$, not $\mathbf v=\boldsymbol\omega r$. The latter is obviously nonsensical - $\boldsymbol\omega$ points out of the plane and $\mathbf v$ stays within it. $\endgroup$ – Emilio Pisanty May 1 '17 at 22:00
  • $\begingroup$ That said: what do you mean by 'total current in the disk'? The surface current density? The total charge crossing a radius per unit time? If the latter, your book is off by a factor of $2\pi$ (since over one revolution, i.e. time $1/\nu=2\pi/\omega$, you know that charge $Q$ crosses the radius). As to how you got your result - no idea; you should clarify. $\endgroup$ – Emilio Pisanty May 1 '17 at 22:03
  • $\begingroup$ @EmilioPisanty I made the mistake of making the symbols bold as if they were vector quantities, since they usually are. I am dealing here only with scalar magnitudes. From reading your comment, I believe my answer is incorrect since I cannot associate a linear speed $\omega R$ with every point on the disk; I can only associate this with the outermost edge which is actually at radius $R$. The linear speed of other points on the disk will be dependent on their radii. $\endgroup$ – nguzman May 2 '17 at 4:20
  • $\begingroup$ Your expression for $K$ is correct, but it's still not clear what you're defining $I$ to be and how you're calculating it. $\endgroup$ – Emilio Pisanty May 2 '17 at 6:04

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.


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