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.
 A: 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:
$$\Phi_M=\iint_{S}\mathbf B\cdot dS = \iint_{S}\mathbf \nabla\times A\cdot dS = \oint_{closed-contour}\mathbf A\cdot dL$$
In other words you first calculate the magnetic vector potential via an already known magnetic field or you solve poisson equation for magnetostatics. Then, you calculate the line integral across the helix or any closed contour of the surface you want calculate the flux.
The magnetic potential A will be proportional to the current I, then the flux will be proportional to the current. Then you calculate the inductance simply by dividing this flux by the current.
This way you dont have a need for flux linkage, you would insert the N into the trajectory of the helix for the closed line integral.
$$L=\frac{\oint_{closed-contour}\mathbf A\cdot dL}{I}$$
