**Subject :**Power system Analysis

## Internal Inductance

**Internal inductance**

The cross - section of a cylindrical conductor is show in figure is give below.

__Cross - section of a cylindrical conductor__

Ampere's law states that the magnetomotive force (mmf) in ampere-turns around a closed path is equal to the net current in amperes enclosed by the path, The conductor has a radius of *r* and carries a current *I ,* then we get the following expression.

[Eq. (1.4)]

Where *H* is the magnetic field intensity in At/m, *s* is the distance along the path in meter and *I* is the current in ampere.

Here the field intensity at a distance *x* from the center of the conductor by *H _{x}*

*H _{x }*is constant at all points that are at a distance

*x*from the center of the conductor. Therefore

*H*is constant over the concentric circular path with a radius of

_{x}*x*and is tangent to it. Denoting the current enclosed by

*I*

*we can then write*

_{x}[Eq. (1.5)]

If we now assume that the current density is uniform over the entire conductor, we can write

[Eq. (1.6)]

Substituting (1.6) in (1.5) we get

[Eq. (1.7)]

Assuming a relative permeability of 1, the flux density at a distance of *x* from the center of the conductor is given by

[Eq. (1.8)]

Where *µ _{0}* is the permeability of the free space and is given by 4π X 10

^{-7}H/m.