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?