(adsbygoogle = window.adsbygoogle || []).push({}); 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 = [itex] \frac{\phi_o}{σπ(√b-√a)^2} [/itex]

2. Relevant equations

Laplace's equation: [itex]\nabla^2\phi = 0[/itex].

E-field in terms of the potential: E=[itex]-\nabla\phi[/itex]

(both for cylindrical coordinates)

I = ∫[itex]J\cdot dA[/itex]

J = σE

R = [itex]\frac{\phi}{I}[/itex]

3. The attempt at a solution

From Laplace's equation we know that the potential will only vary with the angle

[itex]\phi[/itex] and it will vary linearly : [itex]\phi = k_1\phi+k_2[/itex]

due to BC [itex]\phi(0) = 0 = k_2[/itex] and [itex]\phi(\phi_o) = k_1\phi_o = V_o[/itex]

the E-field is [itex]\frac{-k_1}{r}\phi-direction[/itex]

Using the equation for the current: I = ∫[itex]-σk_1∫\frac{1}{r}drdz[/itex] 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.

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# Resistance of a toroidal conductor

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