Question

A current Iflows down a wire of radius a.

(a) If it is uniformly distributed over the surface, what is the surface current density K?

(b) If it is distributed in such a way that the volume current density is inversely

proportional to the distance from the axis, what is J(s)?

Short answer

(a) The surface current density corresponding to a current Iflowing down a wire of radius ais$\frac{I}{2\pi a}$.

(b) The volume current density corresponding to a current Iflowing down a wire of radius ais $\frac{I}{2\pi as}$.

Step-by-step solution

Given data

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

Surface and volume current density

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

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

Here, the integration is over the cross sectional area.

Surface current density in wire of radius   

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

$K=\frac{I}{2\pi a}$

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

Volume current density in wire of radius  

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),

$$\begin{gathered} I=k{\int }_0^a\frac{1}{s}2\pi sds \\ =k2\pi a \\ k=\frac{I}{2\pi a} \end{gathered}$$

Substitute this in equation (3) to get

$J=\frac{I}{2\pi as}$

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