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Our calculus book, Stewart, has a problem where they claim that for a metal cable (inner radius $r$) encased in insulation (outer radius $R$), the speed of an electrical impulse is given by

$$v = - k \left(\frac{r}{R}\right)^2 \ln \left(\frac{r}{R}\right)$$

where $k$ is a positive constant.

My question What I would like to know is the physical justification for their claim.

My thoughts There claim is somewhat surprising, since for sufficiently high insulation R, with r fixed, the speed of the impulse decreases (by L'Hospital) with more insulation.


EDIT: I received this email after contacting Brooks/Cole, the publisher of the textbook. The response didn't really help unfortunately.

Hi Professor ...,

I just heard back from the author regarding your query: “I can understand why Professor ... thinks this equation is counterintuitive, but it is in fact correct. I have been >trying to track down the source that I used in devising this problem, but unfortunately I >can’t seem to find it right now.” I will certainly let you know if he is able to track >down the source information. I’m sorry I can’t give you a more concrete answer at this >time. Best, ...

[JIRA] (KYTS-1199) Content Feedback from Instructor for ISBN: 0495014281 Essential >Calculus: Early Transcendentals 1st edition.

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@Mark Thanks for the edit! I didn't realize I could use LaTeX or I would have. –  Curious Oct 14 '11 at 16:20
    
Is this a coaxial lead or just some dielectric around some wire. Does that wording, "cable" or "insulation" come from that calculus book? –  Georg Oct 14 '11 at 16:51
    
@Georg That is the exact wording from Stewart, using the terminology "metal cable" and "insulation". –  Curious Oct 15 '11 at 6:49
    
Interesting, if we denote $x=\frac{r}{R}$ then $x^2\ln{x}$ has a minimum: $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$. It's a strange behavior in the physical sense, i think. It would be useful if you will add the text of the original problem. –  Martin Gales Oct 15 '11 at 8:49
1  
Terms like ( ln(r/R) ) remind of formulas of coaxial cables. I'd post this question in some elctronics forum. –  Georg Nov 20 '11 at 12:57

1 Answer 1

I think i came to the origins of this equation. In all likelihood, this equation describes not a speed of an electrical impulse but a direct current power transmitted via a superconducting coaxial cable.

A proof:

Consider a simple transmission DC coaxial cable. To eliminate the energy losses due to Joule heating in the cable, the inner(of radius $r$) and outer(of radius$R$) conductors are made from a superconductor. The inner conductor is insulated by a dielectric material. How much power can be transferred through the cable?

Let the maximum allowable magnetic field induction on the surface of the superconductor be $B_\text{max}$ and the maximum electric field in the insulating interlayer be $E_\text{max}$. Let a current through the cable be $I$.

Then the following holds:

$$B_\text{max}=\frac{\mu_0I}{2\pi r}\Rightarrow I=\frac{2\pi}{\mu_0}rB_\text{max}$$

Since we are dealing with superconductors they keep the potential on the surface(as well as the linear charge density λ) constant.

That means the following holds:

$$E_\text{max}=\frac{\lambda}{2\pi\epsilon_0r}$$

The potential difference between inner and outer conductors:

$$U=\frac{\lambda}{2\pi\epsilon_0}\int_{r}^{R}\frac{dl}{l}=\frac{\lambda}{2\pi\epsilon_0}\ln\frac{R}{r}=rE_\text{max}\ln\frac{R}{r} $$

So, the power transmission:

$$P=UI=\frac{2\pi}{\mu_0}E_\text{max}B_\text{max}r^2\ln\frac{R}{r}$$

This is the same function as in the question, only with different constant $k$

To analyze the result let's introduce the ratio $x=\frac{R}{r}$:

$$P=UI=\frac{2\pi}{\mu_0}E_\text{max}B_\text{max}R^2\left(\frac{\ln x}{x^2}\right)$$

One can see that at $R=\text{const}$, $P$ as a function of $x$ has a maximum. This happens at $x=\sqrt{e}$

So the maximum power transfer in the DC superconducting transmission cable:

$$P_\text{max}=\frac{\pi}{\mu_0e}E_\text{max}B_\text{max}R^2$$

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