Internal inductance independent of geometry

Statement: the internal inductance of a round conductor is $L_{int} = \frac{\mu_0}{8\pi} = 0.5\times10^{-7}$ H/m. It does not contain the conductor radius or the spacing, so it is independent of the conductor geometry (size and arrangement).

Derivation

Consider a long solid round conductor of radius r carrying current I with uniform current density ($\mu_r = 1$).

1. Current enclosed within radius x (x < r):

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

2. Field intensity by Ampere's law:

$$2\pi x H_x = I_x \;\Rightarrow\; H_x = \frac{xI}{2\pi r^2}, \qquad B_x = \frac{\mu_0 xI}{2\pi r^2}$$

3. Flux in a tubular element of thickness dx, per metre length:

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

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

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

5. Integrate from 0 to r:

$$\lambda_{int} = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}$$

6. Inductance:

$$L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = 0.5\times10^{-7}\ \text{H/m}$$

Why it is independent of geometry

• The $r^4$ in the denominator (from the field and partial-linkage terms) cancels exactly with the $r^4/4$ from the integral. So the radius drops out: a thin conductor and a thick conductor have the same internal inductance per metre.
• Internal flux exists only inside the conductor, so it does not depend on the spacing to other conductors or on the arrangement (horizontal, vertical, triangular).
• Only the permeability of the conductor material matters. For copper and aluminium $\mu = \mu_0$.

The spacing and radius affect only the external inductance, $2\times10^{-7}\ln\frac{D}{r}$. Combining both parts gives the familiar form with GMR:

$$L = 2\times10^{-7}\left(\frac14 + \ln\frac{D}{r}\right) = 2\times10^{-7}\ln\frac{D}{r'}, \quad r' = re^{-1/4} = 0.7788r$$