# Resistance of a toroidal conductor

1. Oct 16, 2011

### runnergirl

1. The problem statement, all variables and given/known data
Consider a segment of a toroidal (doughnut-shaped) resistor with a horizontal cross-section (see attachment for the figure). Show that the resistance between the flat ends having a circular cross-section is given by
R = $\frac{\phi_o}{σπ(√b-√a)^2}$

2. Relevant equations
Laplace's equation: $\nabla^2\phi = 0$.
E-field in terms of the potential: E=$-\nabla\phi$
(both for cylindrical coordinates)
I = ∫$J\cdot dA$
J = σE
R = $\frac{\phi}{I}$
3. The attempt at a solution
From Laplace's equation we know that the potential will only vary with the angle

$\phi$ and it will vary linearly : $\phi = k_1\phi+k_2$

due to BC $\phi(0) = 0 = k_2$ and $\phi(\phi_o) = k_1\phi_o = V_o$

the E-field is $\frac{-k_1}{r}\phi-direction$

Using the equation for the current: I = ∫$-σk_1∫\frac{1}{r}drdz$ where the dz portion is the height of the strip and the integrand goes from a to b (the change of radius).

Where I'm having issue is setting up the height of the strip. If you look at the attached picture I've drawn out what I think should be the height, but the integral gets pretty complex and I get some complex numbers when I apply the limits. If someone would please show me my mistake, I would be most grateful.

#### Attached Files:

• ###### problem.jpg
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2. Oct 17, 2011

### rude man

Have you got a figure describing the problem? I can't figure out where the connection to the resistor are made. If I cut the toroid at a c ross-section & stretched it out, would the connections be to the two freed ends? Can't be that simple I guess ...
PLUS - the problem's answer makes no sense. How can resistance be proportional to potential? Are you sure you copied the problem correctly?

3. Oct 17, 2011

### runnergirl

The figure is that pie shaped figure in the beginning that I drew on the attachment. That zoomed in circle is also similar to what they show as well with the dr strip and radius, r.

Using Ohm's Law is how you relate the potential difference between the flat ends and the current flowing through the wedge. So $\phi = IR$ and solve for R.

I included the question word for word out of the book. And on the attachment I included both figures that would given with the problem.

Thanks for helping again.

4. Oct 17, 2011

### rude man

OK, I think I finally figured out what $\phi_o$ is. I got confused since they use
φ for both potential and angle. So $\phi_o$ apparently represents the section of toroid cut out from a whole toroid (in other words, if it were a whole toroid, $\phi_o$ would = 2π. Right or wrong, I will proceed on that assumption.

This would then be, as you say, somewhat similar to our old nemesis, the mag toroid, but much simpler math-wise. From your illustration (BTW some of it's missing in your scan) it looks like you set up your origin at the center of the toroid again, so we'll go with that. The equation for the cross-section circle of radius R is then

(r-r0)^2 + y^2 = R^2 where R = (b-a)/2 and r0 = (b+a)/2.

I would take a differential cross-sectional slice dy just like you did. The conductance of this cross-section, extended over the toroid angle $\phi_o$ is

dG = (σ/$\phi_o$r)(2√(a^2 - y^2)dr
but y^2 = R^2 - (r-r0)^2
so dG = (σ/$\phi_o$r)(2√{a^2 - [R^2 - (r-r0)^2]}dr

where conductance = 1/resistance.

Then, just integrate G = ∫dG from r=a to r=b and finally R = 1/G.

In other words, I would not work with Laplace's equation at all. In fact, I'm not sure how to go about it that way.

5. Oct 18, 2011

### runnergirl

So attached is my solution. I am off by a minus sign, but overall everything else works out nicely. The integral again isn't all that pretty but does simplify nicely. And if anyone does find the negative sign mess up (should be with applying the arcsin or somewhere in there) please let me know.

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