Resistance of Quarter-Ring Conductor: Did I Get the Answer Right?

[mmiguel1]

I just had this problem on my Electromagnetics final. I want to know if I got this right and I can't find the problem with google.

Homework Statement

Consider a conducting quarter-ring.
It can be envisioned as one piece of a hollow cylinder that has been cut into fourths down its length.
The original hollow cylinder would have had inner radius a, outer radius b, and length h.
In cylindrical coordinates the geometry of this quarter ring is: $$r \in [a,b]$$; $$\phi \in [0,\pi/2]$$; and $$z \in [0,h]$$;
The material of interest has conductivity $$\sigma$$
What is the resistance from the $$\phi=0$$ end to the $$\phi=\pi/2$$ end?

Homework Equations

For a conductor with linear geometry down it's length:
$$R = \frac{L}{\sigma A}$$
Where L = length and A = cross-sectional area;
$$G = \frac{1}{R}$$

The Attempt at a Solution

I considered a slice of the conductor corresponding to a differential angle $$d \phi$$.
The slice is described by:
$$r \in [a,b]$$; $$\phi \in [\phi_0,\phi_0+d\phi]$$; and $$z \in [0,h]$$;
Let the resistance of this slice be dR. Since the total conductor is a series combination of these slices, and since series resistances add,
$$R = \int dR$$

Within the slice of interest, dR, consider a sub-slice located at radius r with a < r < b.
The sub-slice is described by:
$$r \in [r_0,r_0+dr]$$; $$\phi \in [\phi_0,\phi_0+d\phi]$$; and $$z \in [0,h]$$;
Let us call the resistance of this sub-slice dR'. Our original slice, dR, is a parallel combination of many dR' sub-slices. It is therefore more convenient to calculate the conductances for this part of the problem. Define $$dG = \frac{1}{dR}$$ and $$dG' = \frac{1}{dR'}$$. Since parallel conductances add, we have
$$dG = \int dG'$$
As the differential values $$d\phi$$ and $$dr$$ tend to zero, the subslice of interest is linearized along its length.
We can therefore say
$$dG' = \frac{\sigma A}{L} = \frac{\sigma h dr}{r d \phi}$$
Where $$A = h dr$$ and $$L = r d \phi$$
$$dG = \int_a^b \frac{\sigma h}{d \phi} \frac{1}{r} dr$$
$$dG = \frac{\sigma h}{d \phi} \ln \left (\frac{b}{a} \right )$$
$$dR = \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }$$
$$R = \int_0^{\frac{\pi}{2}} \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }$$
$$R = \frac{\pi}{2 \sigma h \ln \left ( \frac{b}{a} \right ) }$$

I was debating this question with my friends after the final. Each of us got different answers. I think my method is correct but I would like some verification.
Thanks!

 

[Redbelly98]

Your derivation looks fine, and your result is certainly true for the case a≈b as well as a=0.

Here is something to check with your friends: If a=0, then the two ends will be in contact, giving a resistance of 0. Your formula correctly reproduces that, since you are dividing by the ln term which approaches infinity as a→0.

Do any of your friends' answers give R=0 when a=0?

 

[mmiguel1]

I honestly can't really remember. I am glad that you think my derivation makes sense. I just wanted to make sure I didn't violate any rules in math since I have never really dealt with reciprocals of integrals before. Thanks Redbelly98!