# Magnetic field of coaxial cylindrical capacitor

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?

• Define your Amperian loop. Oct 27, 2020 at 19:44
• @RobJeffries Thank you! I added a picture.
– Jeff
Oct 27, 2020 at 20:04