# Inductance in Single Conductor Power Transmission Line

## Reason of Transmission Line Inductance

Generally, electric power is transmitted through the transmission line with AC high voltage and current. High valued alternating current while flowing through the conductor sets up magnetic flux of high strength with alternating nature. This high valued alternating magnetic flux makes a linkage with other adjacent conductors parallel to the main conductor. Flux linkage in a conductor happens internally and externally. Internally flux linkage is due to self-current and externally flux linkage due to external flux. Now the term inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N number of turn is linked by flux Φ due to current I, then, But for transmission line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the transmission line inductance.### Calculation of Inductance of Single Conductor

#### Calculation of Internal Inductance due to Internal Magnetic Flux of a Conductor

Suppose a conductor is carrying current I through its length l, x is the internal variable radius of the conductor and r is the original radius of the conductor. Now the cross-sectional area with respect to radius x is πx^{2}square – unit and current I

_{x}is flowing through this cross-sectional area. So the value of I

_{x}can be expressed in term of original conductor current I and cross-sectional area πr

^{2}square – unit

Now consider small thickness dx with the 1m length of the conductor, where H_{x} is the magnetizing force due to current I_{x} around the area πx^{2}.
And magnetic flux density B_{x} = μH_{x}, where μ is the permeability of this conductor. Again, µ = µ_{0}µ_{r}. If it is considered that the relative permeability of this conductor µ_{r} = 1, then µ = µ_{0}. Hence, here B_{x} = μ_{0} H_{x}.
dφ for small strip dx is expressed by

Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux. Therefore the flux linkage is

Now, the total flux linkage for the conductor of 1m length with radius r is given by Hence, the internal inductance is

## External Inductance due to External Magnetic Flux of a Conductor

Let us assume, due to skin effect conductor current I is concentrated near the surface of the conductor. Consider, the distance y is taken from the center of the conductor making the external radius of the conductor. H_{y}is the magnetizing force and B

_{y}is the magnetic field density at y distance per unit length of the conductor. Let us assume magnetic flux dφ is present within the thickness dy from D

_{1}to D

_{2}for 1 m length of the conductor as per the figure. As the total current I is assumed to flow in the surface of the conductor, so the flux linkage dλ is equal to dφ. But we have to consider the flux linkage from conductor surface to any external distance, i.e. r to D