91Ó°ÊÓ

There is a wire of radius a carrying a current I.

The surface current density if the current is uniformly distributed over a surface is

$\mathbf{K}\mathbf{=}\frac{\mathbf{I}}{\mathbf{L}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, L is the cross sectional length of the surface.

The total current in terms of volume current density is

$\mathbf{I}\mathbf{=}\mathbf{∫}\mathbf{}\mathbf{Jda}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, the integration is over the cross sectional area.

From equation (1), the surface current density on the surface of the wire of radius a is

$K=\frac{I}{2\mathrm{Ï€}a}$

Thus, the uniform surface current density is $\frac{I}{2\mathrm{Ï€}a}$.

Let the volume current density be

$J=\frac{k}{s}.....\left(3\right)$

Here, k is a constant and is the radial component of polar coordinate.

From equation (2),

$$\begin{gathered} I=k{∫}_0^a\frac{1}{s}2\mathrm{π}sds \\ =k2\mathrm{π}a \\ k=\frac{I}{2\mathrm{π}a} \end{gathered}$$

Substitute this in equation (3) to get

$J=\frac{I}{2\mathrm{Ï€}as}$

Thus, the volume current density is $\frac{I}{2\mathrm{Ï€}as}$.