# Why does the inductance of a coaxial cable not depend on the number $N$ of coils?

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

• 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! May 16, 2015 at 21:12
• But then how does asking to calculate the self-inductance make sense? At least in my class we defined it in terms of coils. May 16, 2015 at 21:17
• 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. May 16, 2015 at 21:18
• 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 May 16, 2015 at 21:21
• 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... May 16, 2015 at 21:41

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.