Inductance of a single-phase two-wire line

The inductance of a single-phase two-wire line is found by adding the internal and external flux linkages of each conductor due to its own current and the return current.

     I -->                    <-- I
   (  a  )<------- D ------->(  b  )
   radius r1                 radius r2

Step 1: Internal flux linkage

Inside a conductor of radius $r$ at radius $x$, the enclosed current is $I_x = I x^2/r^2$. From Ampere's law, $H_x = \dfrac{Ix}{2\pi r^2}$. The flux in a tube of thickness $dx$ links only the fraction $x^2/r^2$ of the current:

$$\lambda_{int} = \int_0^r \frac{x^2}{r^2}\cdot\frac{\mu_0 I x}{2\pi r^2}\,dx = \frac{\mu_0 I}{8\pi} = \frac{I}{2}\times10^{-7}\ \text{Wb-T/m}$$

Step 2: External flux linkage

Outside the conductor, $H_x = I/2\pi x$, and the flux between radii $D_1$ and $D_2$ links the full current:

$$\lambda_{12} = \frac{\mu_0 I}{2\pi}\ln\frac{D_2}{D_1} = 2\times10^{-7} I\ln\frac{D_2}{D_1}$$

Step 3: Flux linkage of conductor a

Flux due to $I_a = I$ out to a distant point P (distance $D_{1P}$), plus flux due to $I_b = -I$:

$$\begin{aligned} \lambda_a &= 2\times10^{-7}\left[I\left(\frac{1}{4} + \ln\frac{D_{1P}}{r_1}\right) - I\ln\frac{D_{2P}}{D}\right] \end{aligned}$$

As P moves to infinity, $D_{1P}/D_{2P} \to 1$, and using $\tfrac14 = \ln e^{1/4}$:

$$\lambda_a = 2\times10^{-7} I\ln\frac{D}{r_1 e^{-1/4}} = 2\times10^{-7} I\ln\frac{D}{r_1'}$$

where $r_1' = 0.7788 r_1$ is the GMR.

Step 4: Inductances

$$L_a = \frac{\lambda_a}{I} = 2\times10^{-7}\ln\frac{D}{r_1'}\ \text{H/m}, \qquad L_b = 2\times10^{-7}\ln\frac{D}{r_2'}\ \text{H/m}$$

Loop inductance:

$$L = L_a + L_b = 4\times10^{-7}\ln\frac{D}{\sqrt{r_1' r_2'}}\ \text{H/m}$$

For identical conductors ($r_1 = r_2 = r$):

$$L = 4\times10^{-7}\ln\frac{D}{r'}\ \text{H/m}, \quad r' = 0.7788r$$

$L_a$ is the inductance per conductor; the loop inductance is twice that value.