Internal inductance independent of current

Statement: the internal inductance of a straight round conductor is $L_{int} = \frac{\mu_0}{8\pi} = \frac{1}{2}\times10^{-7}$ H/m. It contains no current term, so it is independent of the current.

Derivation

Take a long round conductor of radius r carrying current I, with uniform current density (skin effect neglected) and $\mu_r = 1$.

        .---------.
      /   .-----.   \     r = conductor radius
     |   /   x   \   |    x = radius of the circular path
     |  |  <--dx  |  |    dx = thin tubular element
     |   \       /   |
      \   '-----'   /
        '---------'

1. Current enclosed by a circle of radius x < r:

$$I_x = \frac{\pi x^2}{\pi r^2}I = \frac{x^2}{r^2}I$$

2. Ampere's law around that circle:

$$H_x(2\pi x) = I_x \;\Rightarrow\; H_x = \frac{x}{2\pi r^2}I, \qquad B_x = \mu_0H_x = \frac{\mu_0 x I}{2\pi r^2}$$

3. Flux in the tube of thickness dx and length 1 m:

$$d\phi = B_x\,dx = \frac{\mu_0 x I}{2\pi r^2}dx$$

4. Flux linkage. This flux links only the fraction $x^2/r^2$ of the total current:

$$d\lambda = \frac{x^2}{r^2}d\phi = \frac{\mu_0 I x^3}{2\pi r^4}dx$$

5. Total internal flux linkage:

$$\lambda_{int} = \int_0^r \frac{\mu_0 I x^3}{2\pi r^4}dx = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}\ \text{Wb-turn/m}$$

6. Internal inductance:

$$L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = \frac{1}{2}\times10^{-7}\ \text{H/m}$$

Conclusion

The flux linkage $\lambda_{int}$ is directly proportional to I, so the ratio $\lambda/I$ is a constant. The current cancels and $L_{int} = 0.5\times10^{-7}$ H/m for any value of current. This is true as long as μ is constant, which holds for non-magnetic conductors such as copper and aluminium. (The same result also shows that $L_{int}$ does not depend on the radius r.)

Physical reason: doubling I doubles both the field H at every point and the flux linkage, so their ratio stays the same.