Magnetic field of coaxial cylindrical capacitor

Question

I have a coaxial cylindrical capacitor as shown, with inner radius a and outer radius b.

The potential difference across both cylinders is V. I need the magnetic field everywhere when the inner cylinder rotates with constant angular speed $\omega$. I'm kinda lost here and I don't know if what I'm doing makes sense. I used Gauss' law to find the electric field and the information about the potential difference to find $\lambda = \frac{2 \pi \epsilon_0 V}{ln(b/a)}$. Then, with Ampère's law, using the Amperian loop shown below,

$\oint \vec{B} d\vec{l} = BL = \mu_0 I$. I thought that if $s > a$, there's no current, so $B=0$. But if $s < a$, $I = \lambda \omega a L$ (because $v = \omega a$) and then $B = \frac{2 \pi \mu_0 \epsilon_0 V \omega a}{ln(b/a)}$, pointing up or down along the axis of the cylinder, depending on the direction of the angular velocity. Is this correct? If not, what should I do?

Answer 1

I don't think you're correct in inferring where the magnetic field will be zero. It will be zero only in the region inside the capacitor plates and will diminish as we come out of the capacitor vertically. As for the regions inside the inner plate and outside the capacitor magnetic field will be zero. This is because due to rotation of the inner plate, an opposite current will be generated in the outer plate. I think you can proceed from here.