Cylindrical capacitor with 2 dielectrics

I want to calculate $\vec{E}$ and $\vec{H}$ inside this capacitor:

So this would be a cylindrical capacitor but with two different dielectrics in it (they both occupy the same amount of space, excuse me if the drawing I made is not symmetrical). The dielectrics are not ideal, so there is current going through them. We could state then that each material has a $\epsilon$ and a $\sigma$. Let's say that these quantities are $\sigma_1$, $\epsilon_1$, $\sigma_2$ and $\epsilon_2$.

The potential difference in the plates is $V_o$ so the electric field will be the same in each material.

$$E = E_1 = E_2 = \frac{V_o}{d}$$

However, different current densities will flow in each material. They can be easily calculated using Ohm's law.

The problem arises when I try to find the magnetic field $\vec{H}$. There is no symmetry in this problem, so I can't use Ampére and take a circulation around the center of the capacitor because $H$ will not be constant for a given radius around the central axis. Also, the fact that different currents are flowing in each material also confuses me. I don't know how to calculate $H$ for a system like this.

Is there a way to find $\vec{H}$ inside this capacitor? Or maybe some software in which I could simulate this?

• Just handwaving here, happy if it actually helps you. Did you try using $\nabla \times H$ = $J_f + \frac{\partial D}{\partial t}$? since you know $D$ and the free current density $J_f$ is the current density produced due to the voltage source alone(exclude magnetization). This immediately tells you that since the currents are longitudinal, the magnetic field $has$ to be circumferential. Oct 28, 2016 at 15:46
• @PrasadMani Hi there, I know that it has to be circumferential but I wanted to know its value (as a function of the conductivities, the permittivities and $V_o$) , not only its direction. Oct 28, 2016 at 15:48
• Which is why you have to calculate the RHS of the above maxwell's equation; again, handwaving, sorry! Oct 28, 2016 at 15:49
• What is your question? Why do you need magnetic fields here? Where (inside or outside the capacitor) would you like to know the magnetic fields, if any? Oct 28, 2016 at 17:03
• @freecharly It is an exercise our teacher gave us, and he told us to calculate $\vec{E}$ and $\vec{H}$ inside the capacitor. Oct 28, 2016 at 17:06