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.
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 Hx
Hx is constant at all points that are at a distance x from the center of the conductor. Therefore Hx is constant over the concentric circular path with a radius of x and is tangent to it. Denoting the current enclosed by Ix we can then write
If we now assume that the current density is uniform over the entire conductor, we can write
Substituting (1.6) in (1.5) we get
Assuming a relative permeability of 1, the flux density at a distance of x from the center of the conductor is given by
Where µ0 is the permeability of the free space and is given by 4π X 10-7 H/m.