| ### Statement | | ### Statement |
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| $8.1.20.$ | | $8.1.20.$ |
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| a. Current I is applied to point A of the medium, and current I is withdrawn | | a. Current I is applied to point A of the medium, and current I is withdrawn |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 | | from point B. Assuming that each point of the medium independently creates |
| a stationary spherically symmetric current field, determine the surface cur- | | 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 | | 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 | | current through this plane? How will the solution change if a current I is |
| applied to point B? | | applied to point B? |
| b. Determine the current density distribution over the ground surface if there | | 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. | | is a point source with current I at a depth h from its surface. |
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| ### Solution | | ### Solution |
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| Part (a): Source A (+I) and sink B (–I) | | Part (a): Source A (+I) and sink B (–I) |
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| Positions: | | Positions: |
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| $ A at (-d/2, 0, 0)$ | | $ A at (-d/2, 0, 0)$ |
| $B at (+d/2, 0, 0)$ | | $B at (+d/2, 0, 0)$ |
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| Separation d. | | Separation d. |
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| Individual fields: | | Individual fields: |
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| Source A: $\displaystyle \mathbf{J}_A = \frac{I}{4\pi r_A^3}\mathbf{r}_A$ | | Source A: $\displaystyle \mathbf{J}_A = \frac{I}{4\pi r_A^3}\mathbf{r}_A$ |
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| Sink B (outgoing current): $\displaystyle \mathbf{J}_B = -\frac{I}{4\pi r_B^3}\mathbf{r}_B$ | | Sink B (outgoing current): $\displaystyle \mathbf{J}_B = -\frac{I}{4\pi r_B^3}\mathbf{r}_B$ |
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| Plane of symmetry: x = 0, perpendicular to line AB at its midpoint. | | Plane of symmetry: x = 0, perpendicular to line AB at its midpoint. |
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| By symmetry, on the plane x = 0: | | By symmetry, on the plane x = 0: |
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| $r_A = r_B = r = \sqrt{(d/2)^2 + y^2 + z^2}$ | | $r_A = r_B = r = \sqrt{(d/2)^2 + y^2 + z^2}$ |
| · The transverse (y, z) components of | | · The transverse (y, z) components of |
| $\mathbf{J}_A $and $\mathbf{J}_B$ cancel. | | $\mathbf{J}_A $and $\mathbf{J}_B$ cancel. |
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| The x-components add constructively. | | The x-components add constructively. |
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| $J_x(0, y, z) = \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) + \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) = \frac{I d}{4\pi r^3}$ | | $J_x(0, y, z) = \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) + \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) = \frac{I d}{4\pi r^3}$ |
| $J_y = J_z = 0$. | | $J_y = J_z = 0$. |
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| Thus, the volume current density in the symmetry plane is purely perpendicular to the plane: | | Thus, the volume current density in the symmetry plane is purely perpendicular to the plane: |
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| $\boxed{\mathbf{J} = \frac{I d}{4\pi \big[(d/2)^2 + \rho^2\big]^{3/2}}\;\hat{x}}, \qquad \rho = \sqrt{y^2 + z^2}$. | | $\boxed{\mathbf{J} = \frac{I d}{4\pi \big[(d/2)^2 + \rho^2\big]^{3/2}}\;\hat{x}}, \qquad \rho = \sqrt{y^2 + z^2}$. |
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| Total current through the plane: | | Total current through the plane: |
| Integrate over the entire plane x = 0: | | Integrate over the entire plane x = 0: |
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| $I_{\text{plane}} = \int_0^\infty J_x(\rho)\, 2\pi\rho\$, $d\rho | | $I_{\text{plane}} = \int_0^\infty J_x(\rho)\, 2\pi\rho\$, $d\rho |
| = \int_0^\infty \frac{I d}{4\pi (\frac{d^2}{4}+\rho^2)^{3/2}}\, 2\pi\rho\, d\rho.$ | | = \int_0^\infty \frac{I d}{4\pi (\frac{d^2}{4}+\rho^2)^{3/2}}\, 2\pi\rho\, d\rho.$ |
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| With a = $d/2, the integral equals 1/a = 2/d$. Then: | | With a = $d/2, the integral equals 1/a = 2/d$. Then: |
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| $I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I$. | | $I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I$. |
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| $\boxed{I_{\text{total}} = I}$ | | $\boxed{I_{\text{total}} = I}$ |
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| All the current leaving A crosses the symmetry plane toward B. | | All the current leaving A crosses the symmetry plane toward B. |
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| If B also carries a current I (both are sources): | | If B also carries a current I (both are sources): |
| Now | | Now |
| $\mathbf{J}_B = +\frac{I}{4\pi r_B^3}\mathbf{r}_B. On the plane x = 0$ the x-components oppose: | | $\mathbf{J}_B = +\frac{I}{4\pi r_B^3}\mathbf{r}_B. On the plane x = 0$ the x-components oppose: |
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| $J_{Ax} = \frac{I d}{8\pi r^3},\quad J_{Bx} = -\frac{I d}{8\pi r^3}$. | | $J_{Ax} = \frac{I d}{8\pi r^3},\quad J_{Bx} = -\frac{I d}{8\pi r^3}$. |
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| Result: $\mathbf{J} = 0$ | | Result: $\mathbf{J} = 0$ |
| on the plane, total current = 0. No net flow by symmetry. | | on the plane, total current = 0. No net flow by symmetry. |
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| --- | | --- |
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| Part (b): Point source at depth h below the Earth's surface | | Part (b): Point source at depth h below the Earth's surface |
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| 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. |
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| Method of images: | | 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. |
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| Current density on the surface (z = 0): | | Current density on the surface (z = 0): |
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| Vector from real source: | | Vector from real source: |
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| $\mathbf{r}_1 = (x, y, -h), r_1 = \sqrt{\rho^2 + h^2}$. | | $\mathbf{r}_1 = (x, y, -h), r_1 = \sqrt{\rho^2 + h^2}$. |
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| Vector from image source: $\mathbf{r}_2 = (x, y, h), r_2 = r_1 = r$ | | Vector from image source: $\mathbf{r}_2 = (x, y, h), r_2 = r_1 = r$ |
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| $\mathbf{J}_{\text{surf}} = \frac{I}{4\pi r^3}\mathbf{r}_1 + \frac{I}{4\pi r^3}\mathbf{r}_2$ | | $\mathbf{J}_{\text{surf}} = \frac{I}{4\pi r^3}\mathbf{r}_1 + \frac{I}{4\pi r^3}\mathbf{r}_2$ |
| $= \frac{I}{4\pi r^3}(2x, 2y, 0)$ | | $= \frac{I}{4\pi r^3}(2x, 2y, 0)$ |
| $= \frac{I}{2\pi (\rho^2 + h^2)^{3/2}}\,(x\hat{x} + y\hat{y})$ | | $= \frac{I}{2\pi (\rho^2 + h^2)^{3/2}}\,(x\hat{x} + y\hat{y})$ |
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| It is radially outward from the point directly above the source (\rho = 0). Magnitude: | | It is radially outward from the point directly above the source (\rho = 0). Magnitude: |
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| $\boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.$ | | $\boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.$ |
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| 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 $\rho.$ |
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| #### Answer | | #### Answer |
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| $\boxed{I_{\text{total}} = I}$ | | $\boxed{I_{\text{total}} = I}$ |
