# Calculating the internal inductance of a long wire without the concept of flux linkage

I did read the following questions: "Flux linkage inside of a conductor", "Derivation of self-inductance of a long wire" and "Trouble understanding fractional flux linkage"

The answers to them are based on the concept of "flux linkage":

$$\lambda = N \cdot \phi$$

where $$\lambda$$ is the flux linkage; $$N$$ is the number of turns; $$\phi$$ is the magnetic flux

I always thought that flux linkage was an just an "artifice" used in order to avoid explaining integration surfaces like the picture below when calculating the magnetic flux.

If I am correct, there is a way of calculating the internal inductance of a long wire without the concept of flux linkage. The obvious attempt gives the wrong answer (same obtained by the authors of the aforementioned questions) and I have no idea how to make it right.

If you know the magnetic Vector potential $$A$$, you can calculate the magnetic flux in any complex surface like the solenoid you have there you just have to use the fact that $$B=\nabla\times A$$ , then from the definition of magnetic flux and the stokes theorem you have: $$\Phi_M=\iint_{S}\mathbf B\cdot dS = \iint_{S}\mathbf \nabla\times A\cdot dS = \oint_{closed-contour}\mathbf A\cdot dL$$
$$L=\frac{\oint_{closed-contour}\mathbf A\cdot dL}{I}$$