Inductance of Two Conductors Transmission Line

Let A be the cross-section of a conductor. The current I is flowing in the inward direction with respect to the reference plane. Suppose there is another conductor at a distance D. Here, current I is flowing outward with respect to the reference plane. Now, at conductor A, the magnetic field intensity at any point inside the conductor can be expressed as,

$$H=\frac{Ir}{2\pi R^2}$$
Where, r is the distance from the center of the conductor. The magnetic flux density at distance r may be written as
$$B=\mu_0 H =\frac{I\mu_0}{2\pi R^2}r$$
So, the flux density varies with r. Therefore, the total flux lines passing through a circular strip of cross-section of thickness dr will be$$d\phi=B\,dr$$

We have already told,$$I’=\frac{I}{R^2}r^2$$So, we can write$$d\phi=B\,dr = \frac{I\mu_0}{2\pi R^2} \cdot \frac{r^2}{R^2} \cdot r\,dr$$$$= \frac{I\mu_0 r^3dr}{2\pi R^4}$$
Therefore, the total internal flux linkage will be$$\phi_i=\int_0^R d\phi$$$$= \int_0^R \frac{I\mu_0 r^3}{2\pi R^4}dr$$$$= \frac{I\mu_0}{2\pi R^4} \int_0^R r^3dr$$$$= \frac{I\mu_0}{2\pi R^4} \left[\frac{r^4}{4}\right]_0^R$$$$= \frac{I\mu_0}{2\pi R^4} \cdot \frac{R^4}{4}$$$$= \frac{\mu_0 I}{8\pi}$$
Here, you can see that the flux linkage inside the conductor is independent of the radius of the conductor.

Flux Linkage Outside the Conductor

Now, we shall derive the expression for flux linkage outside the conductor. Since one conductor carries current opposite to the other conductor, there will be no flux line that encloses both conductors. Because of that, for any distance greater than D, the total enclosed current is zero. So, there will be flux linkage only within a distance less than or equal to D. Therefore, the external flux linkage only exists from R to D for any of the conductors.
$$H_e=\frac{I}{2\pi r}$$$$B=\mu_0 H_e = \frac{\mu_0 I}{2\pi r}$$
Now, assume the flux density between these two conductors is uniform for a distance dr. If we consider one unit length of conductor,$$d\phi=B\,dr\times 1 = B\,dr$$
Since we are calculating for one conductor only, the number of turns enclosed by this flux is one.
$$d\phi=B\,dr$$$$= \frac{\mu_0 I}{2\pi r}dr$$
Therefore, the total external flux will be
$$\phi_e = \int_R^{D-R} \frac{\mu_0 I}{2\pi r}dr$$ $$= \frac{\mu_0 I}{2\pi} \int_R^{D-R}\frac{dr}{r}$$$$= \frac{\mu_0 I}{2\pi} [\ln r]_R^{D-R}$$$$= \frac{\mu_0 I}{2\pi} [\ln(D-R)-\ln R]$$$$= \frac{\mu_0 I}{2\pi} \ln\frac{D-R}{R}$$Now $R$R is much smaller than D.

So,$$D-R \approx D$$Therefore,$$\phi_e = \frac{\mu_0 I}{2\pi} \ln\frac{D}{R}$$

Total Flux Linkage

Therefore, the total flux linkage is$$\phi_i+\phi_e = \frac{\mu_0 I}{8\pi} + \frac{\mu_0 I}{2\pi} \ln\frac{D}{R}$$
Therefore, the total flux linkage due to two conductors is
$$2(\phi_i+\phi_e) = 2 \left[ \frac{\mu_0 I}{8\pi} + \frac{\mu_0 I}{2\pi} \ln\frac{D}{R} \right]$$

Inductance of Two Conductors Transmission Line

Now we know that$$L I=\phi N$$$$L=\frac{\phi}{I} \qquad [N=1]$$

Hence,$$L= \frac{ 2\left[ \frac{\mu_0 I}{8\pi} + \frac{\mu_0 I}{2\pi} \ln\frac{D}{R} \right] }{I}$$$$= \frac{\mu_0}{4\pi} + \frac{\mu_0}{\pi} \ln\frac{D}{R}$$$$= \frac{\mu_0}{4\pi} \left[ 1+4\ln\frac{D}{R} \right]$$Now,$$\mu_0=4\pi\times10^{-7} \text{ H/m}$$
Therefore, inductance per unit length of the transmission conductor of a two-wire system is
$$L= \frac{\mu_0}{4\pi} \left[ 1+4\ln\frac{D}{R} \right]$$$$= \frac{4\pi\times10^{-7}}{4\pi} \left[ 1+4\ln\frac{D}{R} \right]$$$$= \left( 1+4\ln\frac{D}{R} \right) \times10^{-7}$$Now,$$4\ln e^{1/4}=1$$Therefore,$$L= \left[ 4\ln e^{1/4} + 4\ln\frac{D}{R} \right] \times10^{-7}$$$$= 4\times10^{-7} \ln\left(\frac{D}{R}e^{1/4}\right) \text{ H/m}$$$$= 4\times10^{-7} \ln\left(\frac{D}{Re^{-1/4}}\right) \text{ H/m}$$Let$$R’=Re^{-1/4}$$Then$$L= 4\times10^{-7} \ln\frac{D}{R’} \text{ H/m}$$Here, R′ is the radius of a fictitious conductor. Obviously, it has no internal flux linkage, but it has the same inductance as the actual conductor.$$e^{-1/4}=0.7788$$
The multiplying factor 0.7788 used to adjust the radius in order to account for internal flux linkage applies only to solid round conductors.