# Inductance of Single Conductor of Transmission Line

## Internal Inductance

Consider a long conductor carrying current “I” and having radius of conductor “r” as shown in figure below.

According to Ampere’s law which states that the line integral of the magnetic field intensity (H) around a closed path (a closed loop or a closed curve) is equal to the total current (Ienc) passing through the enclosed surface.

Let “Hx” is the magnetic field intensity at point “x”. then\\\;\\\oint H_{x}ds=I_{x}\\\;\\where “Ix ” is the current enclosed by the flux around which “Hx ” is measured.

Thus H_{x}(2\pi x) = I_{x}\\\;\\H_{x}=\frac{I_{x}}{2\pi x}\;\;\;\;\;\;\;\;\;\;(1)

Current density at any point on the cross section is \\\;\\\sigma = \frac{I}{\pi r^{2}}=\frac{I_{x}}{\pi x^{2}}\\\;\\I_{x}=\frac{\pi x^2}{\pi r^2}I=\frac{x^2}{r^2}I\;\;\;\;\;\;\;\;\;\;(2)

From equation (1) and (2)\\\;\\H_{x}=\frac{x}{2 \pi r^2}I\\\;\\We know, Magnetic flux density \\\;\\B_{x}= \mu H_x

Putting the value of Magnetic field intensity (Hx ) in the above equation\\\;\\B_{x}=\frac{\mu xI}{2 \pi r^{2}}

The flux through a cylindrical shell at distannce “x” from the center with radial thickness “dx” and length one meter is given by

d \phi_{x} = B_{x}(dx\times 1) \\\;\\ d \phi_{x} = B_{x}dx \\\;\\putting the value of Bx in the above equation.\\\;\\d \phi_{x}= \frac{\mu x I}{2 \pi r^{2}}dx

Flux linkage due to d\phi_{x}\\\;\\d \lambda_{x} = \frac{x^{2}}{r^{2}}d\phi_{x} = \frac{\mu Ix^{3}}{2 \pi r^{4}}dx

Total flux linkage (\lambda ) due to all flux inside the conductor is \\\;\\\lambda_{in} = \int_{0}^{r}d \lambda_{x}=\frac{\mu I}{2\pi r^{4}} \int_{0}^{r}x^{3}dx=\frac{\mu I}{8\pi}

Inductance due to internal flux linkage \\\;\\L_{in}=\frac{\lambda _{in}}{I}=\frac{\mu}{8\pi}=\frac{\mu_0\mu_r}{8\pi}=0.5\times10^-7\mu_r \;\;\;\;H/m

External Inductance

Consider the figure shown below

Magnetic field intensity at point “x” is \\\;\\H_{x}=\frac{I}{2 \pi x}\\\;\\Where “I” is the current of conductor

Magnetic flux density at point “x” is \\\;\\B_{x}=\mu H_{x}= \frac{\mu I}{2 \pi x}\\\;\\Flux through a cylinder shell at distance “x” of radial thickness “dx” and length one meter is \\\;\\d \phi_{x} = B_{x}dx=\frac{\mu I}{2 \pi x}dx

Flux linkage due to flux d \phi_{x} is \\\;\\d \lambda_{x}= d \phi_{x} = \frac{\mu I}{2 \pi x}dx

## Free Electrical Design Course

Free Electrical Design Course

WhatsApp GroupÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Telegram Group

This will close in 0 seconds