Suppose an infinitely long hollow cylinder of radius $R$ with surface charge density $\sigma$ oriented in the $\mathbf{\hat z}$ direction is spinning with angular velocity $\omega(t)$. The speed changes only very slowly, so in this problem, we ignore displacement current. Therefore, we may approximate the magnetic field on the inside as $$\mathbf B(t)=\mu_0 R\sigma\omega(t)\mathbf{\hat z}.$$ But what is the electric field $\mathbf E$ on the inside of the cylinder? Symmetry tells us that $\mathbf E$ on the inside of the cylinder must have no $\mathbf{\hat z}$ component, and must be invariant with respect the angular direction $\phi$. This problem becomes very easy to solve (using the integral form of Faraday's law) if $\mathbf E$ is entirely in the $\mathbf{\hat\phi}$ direction. But I don't see why it should be. And if it is not, I do not know what to do to find $\mathbf E$.
Is there a justification for why $\mathbf E$ should be in the $\mathbf{\hat\phi}$ direction?
Best Answer
The justification is of course lenz's law. The $\hat\phi$ of the electric field is because it will push the charges in such a direction that will counter the changing flux produced by the changing angular velocity $\omega(t)$. If the angular velocity increases with time, it leads to increase in flux through any loop considered inside the cylinder (outside too). So the induced electric field will tend to reduce the increasing flux by pushing charges such that they reduce the apparent current thereby reducing the flux. For this purpose, induced $\bar E$ HAS to be circumferential. Any other direction will not cancel the changing flux.