Internal inductance independent of geometry and current

The internal inductance of a round conductor works out to $\mu_0/8\pi$ H/m, which contains neither the radius nor the current. So it does not depend on the conductor's size or the current.

Derivation

Consider a long round conductor of radius $r$ carrying current $I$, uniformly distributed (DC or low frequency).

        _______
      /   ...   \      x  = radius of tube
     |  ( x )dx  |     dx = thickness
      \_________/      r  = conductor radius

1. Field inside. At radius $x < r$, the enclosed current is $I_x = I\dfrac{x^2}{r^2}$. By Ampere's law, $H_x \cdot 2\pi x = I_x$:

$$H_x = \frac{Ix}{2\pi r^2}, \qquad B_x = \mu_0 H_x = \frac{\mu_0 I x}{2\pi r^2}$$

2. Flux in a tube. Flux per metre length in a tube of thickness $dx$:

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

3. Partial linkage. This flux links only the current inside radius $x$, i.e. a fraction $x^2/r^2$ of the total:

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

4. Integrate from $0$ to $r$:

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

5. Internal inductance.

$$L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = \frac12\times10^{-7}\ \text{H/m}$$

Conclusion

• The radius $r$ cancels in step 4, so $L_{int}$ does not depend on the conductor size (geometry).
• The current $I$ cancels in step 5, so it does not depend on the current.
• It depends only on the permeability of the conductor material ($\mu_r = 1$ for copper and aluminium), giving $0.05$ mH/km.

This is why the internal flux can be included simply by replacing $r$ with $r' = re^{-1/4} = 0.7788r$.