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### Statement
−$8.1.20.$ [Insert the problem 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+ a. Current I is applied to point A of the medium, and current I is withdrawn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)
+$ 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
+ 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.
+$r_A = r_B = r = \sqrt{(d/2)^2 + y^2 + z^2}$
+· The transverse (y, z) components of
+$\mathbf{J}_A $and $\mathbf{J}_B$ cancel.
−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}
+The x-components add constructively.
−J_y = J_z = 0.
+$J_x(0, y, z) = \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) + \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) = \frac{I d}{4\pi r^3}$
+$J_y = J_z = 0$.
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}.
+$\boxed{\mathbf{J} = \frac{I d}{4\pi \big[(d/2)^2 + \rho^2\big]^{3/2}}\;\hat{x}}, \qquad \rho = \sqrt{y^2 + z^2}$.
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.
+$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:
+With a = $d/2, the integral equals 1/a = 2/d$. Then:
−I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I.
+$I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I$.
−\boxed{I_{\text{total}} = 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:
+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}.
+$J_{Ax} = \frac{I d}{8\pi r^3},\quad J_{Bx} = -\frac{I d}{8\pi r^3}$.
−Result: \mathbf{J} = 0 on the plane, total current = 0. No net flow by symmetry.
+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.
+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.
+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.
+Vector from real source:
−\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}).
+$\mathbf{r}_1 = (x, y, -h), r_1 = \sqrt{\rho^2 + h^2}$.
+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}} }.
+$\boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.$
−The surface current density is purely horizontal and decreases with radial distance \rho.
+The surface current density is purely horizontal and decreases with radial distance $\rho.$
#### Answer
−[Insert a concise answer or boxed result]
+$\boxed{I_{\text{total}} = I}$
+$\mathbf{J} = 0$
+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}} }.$