Electric current > Current. Current density. Current in vacuum > Problem 8.1.20

Statement

$8.1.20.$

a. Current I is applied to point A of the medium, and current I is withdrawn
from point B. Assuming that each point of the medium independently creates
a stationary spherically symmetric current field, determine the surface cur-
rent density in the plane of symmetry of points A and B. What is the total
current through this plane? How will the solution change if a current I is
applied to point B?
b. Determine the current density distribution over the ground surface if there
is a point source with current I at a depth h from its surface.

Solution

Part (a): Source A (+I) and sink B (–I)

Positions:

$ A at (-d/2, 0, 0)$
$B at (+d/2, 0, 0)$

Separation d.

Individual fields:

Source A: $\displaystyle \mathbf{J}_A = \frac{I}{4\pi r_A^3}\mathbf{r}_A$

Sink B (outgoing current): $\displaystyle \mathbf{J}_B = -\frac{I}{4\pi r_B^3}\mathbf{r}_B$

Plane of symmetry: x = 0, perpendicular to line AB at its midpoint.

By symmetry, on the plane x = 0:

$r_A = r_B = r = \sqrt{(d/2)^2 + y^2 + z^2}$
· The transverse (y, z) components of
$\mathbf{J}_A $and $\mathbf{J}_B$ cancel.

The x-components add constructively.

$J_x(0, y, z) = \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) + \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) = \frac{I d}{4\pi r^3}$
$J_y = J_z = 0$.

Thus, the volume current density in the symmetry plane is purely perpendicular to the plane:

$\boxed{\mathbf{J} = \frac{I d}{4\pi \big[(d/2)^2 + \rho^2\big]^{3/2}}\; \hat{x}}, \qquad \rho = \sqrt{y^2 + z^2}$.

Total current through the plane:
Integrate over the entire plane x = 0:

$I_{\text{plane}} = \int_0^\infty J_x(\rho)\, 2\pi\rho$, $d\rho
= \int_0^\infty \frac{I d}{4\pi (\frac{d^2}{4}+\rho^2)^{3/2}}, 2\pi\rho, d\rho.$

With a = $d/2, the integral equals 1/a = 2/d$. Then:

$I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I$.

$\boxed{I_{\text{total}} = I}$

All the current leaving A crosses the symmetry plane toward B.

If B also carries a current I (both are sources):
Now
$\mathbf{J}_B = +\frac{I}{4\pi r_B^3}\mathbf{r}_B. On the plane x = 0$ the x-components oppose:

$J_{Ax} = \frac{I d}{8\pi r^3},\quad J_{Bx} = -\frac{I d}{8\pi r^3}$.

Result: $\mathbf{J} = 0$
on the plane, total current = 0. No net flow by symmetry.

Part (b): Point source at depth h below the Earth's surface

Model: Conducting half-space (z > 0), insulating surface at z = 0 (no current flow into the air). Real source at $(0,0,h)$ with current I.

Method of images:
To satisfy$ J_z = 0 at z = 0$, an identical image source of current I is added at $(0,0,-h)$, considering the full space.

Current density on the surface (z = 0):

Vector from real source:

$\mathbf{r}_1 = (x, y, -h), r_1 = \sqrt{\rho^2 + h^2}$.

Vector from image source: $\mathbf{r}_2 = (x, y, h), r_2 = r_1 = r$

$\mathbf{J}_{\text{surf}} = \frac{I}{4\pi r^3}\mathbf{r}_1 + \frac{I}{4\pi r^3}\mathbf{r}_2$
$= \frac{I}{4\pi r^3}(2x, 2y, 0)$
$= \frac{I}{2\pi (\rho^2 + h^2)^{3/2}}\, (x\hat{x} + y\hat{y})$

It is radially outward from the point directly above the source (\rho = 0). Magnitude:

$\boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.$

The surface current density is purely horizontal and decreases with radial distance $\rho.$

Answer

$\boxed{I_{\text{total}} = I}$
$\mathbf{J} = 0$
on the plane, total current = 0. No net flow by symmetry.

$ \boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.$