# EM Induction in non-uniformly conducting ohmic toroid, thought experiment

Assume that there's a conducting toroid with radius of revolution $$R$$ and an own radius of $$D$$ so that the cross section of the toroid is given by $$\pi D^2$$

Assume that there's a circular region of linearly increasing (or decreasing) magnetic field in the center of the toroid, of radius B. Region is marked with red in the picture. The red region has a total cross section of $$A$$ and is uniformly "pierced" by a magnetic field $$B$$ that's going inside of the page. The magnetic field is also increasing linearly with time.

Due to symmetry of the problem, Faraday's law should give an electric field given by expression:

$$\textbf{E(r)} = \frac{KA}{2r\pi} \textbf{e}_\phi$$

where $$K$$ is the rate of increase of magnetic induction. The expression is valid anywhere outside of red area.

So, in case of a conducting toroid, lets say of conductance $$\kappa$$, a current density will be established in the toroid, given by the expression:

$$\textbf{J(r)} = \kappa\textbf{E(r)} = \kappa\frac{KA}{2r\pi} \textbf{e}_\phi$$

If I'm correct in my thinking, there shouldn't be any surface charge build up on the toroid since the necessary electric field is already established by induction.

However, things seem to change as soon as there's an ohmic discontunity in the toroid. For example, assume that there are 2 conductivities now. $$\kappa_1$$ and $$\kappa_2$$. Lets assume that $$x$$ is the portion of the circumference occupied by $$\kappa_1$$ and $$(1-x)$$ by $$\kappa_2$$

Now, Faraday's law when applied along with Ohm's law and the condition that current density has to be continuous in both portions of the toroid, gives the following:

$$E_1(r)=\frac{J(r)}{\kappa_1}$$ $$E_2(r)=\frac{J(r)}{\kappa_2}$$

$$J(r)(\frac{x}{\kappa_1} + \frac{1-x}{\kappa_2}) = KA$$

Which gives:

$$J(r) = \frac{KA}{(\frac{x}{\kappa_1} + \frac{1-x}{\kappa_2})}$$

Or expressed via fields $$E_1$$ and $$E_2$$

$$E_1(r) = \frac{KA}{x + (1-x)\frac{\kappa_1}{\kappa_2}}$$

$$E_2(r) = \frac{KA}{x\frac{\kappa_2}{\kappa_1} + (1-x)}$$

Which would mean that there's a discontinuity in the electrical field along the boundary between 2 conductances. More specifically, there are 2 discontinuities. One when entering the region of lower conductance, there's a slab of surface charge given by $$+\sigma$$ and another when leaving the region of lower conductance, given by $$-\sigma$$ where $$\sigma(r) = \epsilon_0|E_1(r)-E_2(r)|$$

So, at this point, assuming that my previous reasoning is correct, my question is, how do you go about calculating the surface charge that gets influenced on the rest of the toroid, due to $$+\sigma$$ and $$-\sigma$$?

I know how I would deal with a situation where I have a conductor and a charge or a distribution of charge outside the conductor. But here, there seems to be a slab of charge directly inside the conductor. And that leaves me clueless. Can the problem be reformulated as an electrostatics problem?

Edit:

I think it doesn't have to be a toroid for the essence of the question to remain the same. The cross section can also be rectangular. Toroid is just neatly symmetrical and has no sharp boundaries.

• I think I've figured it out. I'll try to do a writeup of the solution when I get the time. Basically, in the 2 conductance case, there will be 2 electric fields. The induced, non-conservative one and a conservative one that gets created as the result of different conductuncaes. Subtracting away the non-conservative one and the current and one is left with a toroidal shell that has proscribed potentials at places where cond. discont. Potential should change uniformly along the polar angle. Then laplace's equation can be solved for the entire space with a bond. cond. given by that solution. – user70746 Dec 23 '18 at 9:32