Suppose we have a long coaxial cable like in the following figure:

Coaxial cable I want to compute the self-inductance $L$ of the cable. As I understand, given a coil with $N$ closely winded turns, through which a current $I$ goes through, then $$N \Phi = L I,$$

where $\Phi$ is the flux of the induced magnetic field (by the current) through the surface $S$ bounded (approximately) by one turn of the coil.

I also understand that we can imagine a litle cable is winded around the edges of the blue rectangle in the figure carrying a uniform constant corrent $i$, and we can do this multiple (N) times around the whole cylinder, creating a coil. If the length $l$ is sufficiently big, the magnetic field has a $\phi$-direction (azimuthal?).

Now, up to here I have no problem, but after computing $\Phi = \frac{\mu_0 I l}{2\pi} \ln(b/a)$, the book says:

$$L = \frac{\Phi}{I} = \frac{\mu_0 l}{2\pi} \ln(b/a)$$

and apparently this is the same result I find everywhere. Now, my doubt is: shouldn't it be $N\Phi = \frac{L I}{N}$ so that

$$L = \frac{\mu_0 N^2 l}{2\pi} \ln(b/a)?$$

On the right hand side, I divided by $N$ because $I$ in this case is the total current through each ring (in opposite directions for the inner and outer ring) and thus little cable carries a current $i = I/N$.

Am I doing something wrong or are they maybe just simplifying the problem by focusing only on one of the turns of the coil?

  • $\begingroup$ A coax cable is no coil. The direction of the current is axial not azimuthal. Do not use this analogy, and it will not hurt you! $\endgroup$ May 16, 2015 at 21:12
  • $\begingroup$ But then how does asking to calculate the self-inductance make sense? At least in my class we defined it in terms of coils. $\endgroup$
    – user45453
    May 16, 2015 at 21:17
  • $\begingroup$ Even a straight wire has a self inductance. I have to admit if $N\Phi = LI$ was the definition you saw this will not make sense. $\endgroup$ May 16, 2015 at 21:18
  • $\begingroup$ Ok, I see. But then the method itself used in the book doesn't make sense right? I've seen online that it can be solved using magnetic energy but I haven't studied that yet $\endgroup$
    – user45453
    May 16, 2015 at 21:21
  • 1
    $\begingroup$ Ok nevermind I think I understand now: In the book, they are considering that the inner and outer conductors are connected at the ends of the cable, making it one large conducting loop, i.e., a giant coil. So in this conditions the formula seems to make sense and $N=1$ is justified, I think... $\endgroup$
    – user45453
    May 16, 2015 at 21:41

1 Answer 1


The inductance of a coaxial cable does depend on the number of coils. But in addition to that, a coaxial cable will have a nonzero inductance even if it is perfectly straight, because there is a magnetic field inside the cable whenever it is carrying current, and this takes energy to set up, so you cannot instantaneously increase the current from zero to $I$ without performing work to set up the magnetic field.

Your book is calculating this straight-wire self inductance.If you coil up the cable then you will have that as well, but it will be very small because the cable has zero net current, and the magnetic field inside the loop will be very small - due to small asymmetries and imperfections.

In practice, you never really coil up coaxial wire, which is used for communications because of it shielding properties. The only reason you intentionally coil up wires and run current through them is when you actively want the inductance to be large, and coaxial cable will be terrible at that.

  • $\begingroup$ I'd suggest emphasizing the points in the second paragraph, the first paragraph is misleading otherwise. $\endgroup$ May 16, 2015 at 23:08

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