Electromagnetic Waves in a Conducting Medium: The Complete Theory
This document details the propagation of electromagnetic waves in a material that has conductivity \(\sigma \neq 0\). Unlike in a vacuum, the wave loses energy to the medium (attenuation) and the electric and magnetic fields oscillate out of phase.
1. The Medium Parameters and Maxwell's Equations
We consider a medium characterized by:
- Permittivity (\(\varepsilon\)): Governs the electric displacement field \(\vec{D} = \varepsilon \vec{E}\).
- Permeability (\(\mu\)): Governs the magnetic flux density \(\vec{B} = \mu \vec{H}\).
- Conductivity (\(\sigma\)): Governs the conduction current density \(\vec{J} = \sigma \vec{E}\) (Ohm's Law).
1.1 Maxwell's Equations for Conducting Media
In a linear, isotropic, homogeneous conductor, we start with the general Maxwell equations:
- \(\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon}\) (Gauss's Law)
- \(\nabla \cdot \vec{H} = 0\) (Gauss's Law for Magnetism)
- \(\nabla \times \vec{E} = -\mu \frac{\partial \vec{H}}{\partial t}\) (Faraday's Law)
- \(\nabla \times \vec{H} = \sigma \vec{E} + \varepsilon \frac{\partial \vec{E}}{\partial t}\) (Ampere-Maxwell Law)
Note on Eq. 4: The term \(\sigma \vec{E}\) represents the conduction current \(J_c\), and \(\varepsilon \frac{\partial \vec{E}}{\partial t}\) represents the displacement current \(J_d\).
1.2 Decay of Free Charge (Why \(\rho = 0\))
Before deriving the wave equation, we must establish the behavior of charge density \(\rho\). Using the equation of continuity \(\nabla \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0\) and Ohm's law \(\vec{J} = \sigma \vec{E}\):
\[ \sigma (\nabla \cdot \vec{E}) + \frac{\partial \rho}{\partial t} = 0 \]From Gauss's Law, \(\nabla \cdot \vec{E} = \rho/\varepsilon\), so:
\[ \frac{\sigma}{\varepsilon}\rho + \frac{\partial \rho}{\partial t} = 0 \implies \frac{\partial \rho}{\partial t} = -\frac{\sigma}{\varepsilon}\rho \]Integrating this gives: \(\rho(t) = \rho_0 e^{-(\sigma/\varepsilon)t}\). In a good conductor, \(\sigma/\varepsilon\) is huge (relaxation time is tiny), meaning any initial free charge dissipates instantly to the surface. Therefore, for wave propagation inside the conductor, we set \(\rho = 0\).
2. Derivation of the Wave Equation
Now we derive the differential equation governing the electric field.
Step 1: Take the curl of Faraday's Law (Eq 3):
\[ \nabla \times (\nabla \times \vec{E}) = -\mu \frac{\partial}{\partial t} (\nabla \times \vec{H}) \]Step 2: Use the vector identity \(\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}\). Since \(\rho=0\), \(\nabla \cdot \vec{E} = 0\), so the left side becomes \(-\nabla^2 \vec{E}\).
Step 3: Substitute Ampere's Law (Eq 4) into the right side:
\[ -\nabla^2 \vec{E} = -\mu \frac{\partial}{\partial t} \left( \sigma \vec{E} + \varepsilon \frac{\partial \vec{E}}{\partial t} \right) \]Step 4: Distribute the derivative and rearrange:
This is the Telegrapher's Equation. The term \(\mu\sigma \frac{\partial \vec{E}}{\partial t}\) is the "damping term" responsible for energy loss.
3. The Complex Wave Vector
Assume a plane wave propagating in the \(z\)-direction with frequency \(\omega\):
\[ \vec{E}(z, t) = E_0 e^{i(kz - \omega t)} \hat{x} \]Substituting derivatives into the wave equation:
- \(\nabla^2 \rightarrow \frac{\partial^2}{\partial z^2} \rightarrow (ik)^2 = -k^2\)
- \(\frac{\partial}{\partial t} \rightarrow -i\omega\)
- \(\frac{\partial^2}{\partial t^2} \rightarrow (-i\omega)^2 = -\omega^2\)
The wave equation becomes:
\[ -k^2 E - \mu\varepsilon(-\omega^2)E - \mu\sigma(-i\omega)E = 0 \] \[ k^2 = \mu\varepsilon\omega^2 + i\mu\sigma\omega \]Since \(k\) is a complex number, we express it as \(k = \alpha + i\beta\). Squaring this gives:
\[ k^2 = (\alpha + i\beta)^2 = (\alpha^2 - \beta^2) + i(2\alpha\beta) \]Comparing the real and imaginary parts of \(k^2\):
- Real Part: \(\alpha^2 - \beta^2 = \mu\varepsilon\omega^2\)
- Imaginary Part: \(2\alpha\beta = \mu\sigma\omega\)
3.1 Solving for \(\alpha\) and \(\beta\) (Detailed Algebra)
We solve this system of simultaneous equations. First, square both equations and add them to find \(\alpha^2 + \beta^2\):
\[ (\alpha^2 - \beta^2)^2 + (2\alpha\beta)^2 = (\mu\varepsilon\omega^2)^2 + (\mu\sigma\omega)^2 \] \[ \alpha^4 - 2\alpha^2\beta^2 + \beta^4 + 4\alpha^2\beta^2 = \omega^4\mu^2\varepsilon^2 \left[ 1 + \left(\frac{\sigma}{\omega\varepsilon}\right)^2 \right] \] \[ (\alpha^2 + \beta^2)^2 = \omega^4\mu^2\varepsilon^2 \left[ 1 + \left(\frac{\sigma}{\omega\varepsilon}\right)^2 \right] \]Taking the square root:
\[ \alpha^2 + \beta^2 = \omega^2\mu\varepsilon \sqrt{1 + \left(\frac{\sigma}{\omega\varepsilon}\right)^2} \]Now we have a system for \(\alpha^2\) and \(\beta^2\):
- Sum: \(\alpha^2 + \beta^2 = \omega^2\mu\varepsilon \sqrt{1 + (\frac{\sigma}{\omega\varepsilon})^2}\)
- Diff: \(\alpha^2 - \beta^2 = \omega^2\mu\varepsilon\)
Adding them gives \(2\alpha^2\); subtracting gives \(2\beta^2\). The general solutions are:
4. Good Conductors vs. Dielectrics
Case A: Good Conductor (\(\sigma \gg \omega\varepsilon\))
In a good conductor (like Copper at RF frequencies), the conduction current dominates. The term \(\frac{\sigma}{\omega\varepsilon} \gg 1\). We approximate \(\sqrt{1 + (\dots)^2} \approx \frac{\sigma}{\omega\varepsilon}\) and ignore the \(\pm 1\).
\[ \alpha \approx \beta \approx \omega \sqrt{\frac{\mu\varepsilon}{2}} \sqrt{\frac{\sigma}{\omega\varepsilon}} = \sqrt{\frac{\omega^2\mu\sigma}{2\omega}} = \sqrt{\frac{\omega\mu\sigma}{2}} \]Case B: Good Dielectric (\(\sigma \ll \omega\varepsilon\))
Using the Binomial expansion \(\sqrt{1+x} \approx 1 + x/2\):
- \(\alpha \approx \omega\sqrt{\mu\varepsilon}\) (Same as vacuum speed)
- \(\beta \approx \frac{\sigma}{2} \sqrt{\frac{\mu}{\varepsilon}}\) (Small attenuation)
5. Skin Depth (\(\delta\))
Substituting \(k = \alpha + i\beta\) back into the wave equation \(E = E_0 e^{i((\alpha + i\beta)z - \omega t)}\):
\[ E(z, t) = E_0 e^{-\beta z} e^{i(\alpha z - \omega t)} \]The amplitude decays exponentially as \(e^{-\beta z}\). The Skin Depth \(\delta\) is the distance where amplitude drops to \(1/e\) (37%).
\[ \delta = \frac{1}{\beta} = \sqrt{\frac{2}{\omega\mu\sigma}} = \frac{1}{\sqrt{\pi f \mu \sigma}} \]6. Intrinsic Impedance and Phase Lag
Using Faraday's Law \(\nabla \times \vec{E} = -\mu \frac{\partial \vec{H}}{\partial t}\), we can derive the magnetic field. For a plane wave \(E_x\), the magnetic field is \(H_y\). Their ratio is the Intrinsic Impedance \(\eta\):
\[ \eta = \frac{E_0}{H_0} = \frac{\omega\mu}{k} = \frac{\omega\mu}{\alpha + i\beta} \]To find the phase, we write \(\eta\) in polar form \(|\eta| e^{i\phi}\). The angle \(\phi\) is:
\[ \tan \phi = \frac{\text{Imaginary part of denom}}{\text{Real part of denom}} = \frac{\beta}{\alpha} \]However, mathematically from the complex number \(k\):
\[ \eta = \sqrt{\frac{i\omega\mu}{\sigma + i\omega\varepsilon}} \]The phase angle \(\phi\) of impedance is defined by \(\tan(2\phi) = \frac{\sigma}{\omega\varepsilon}\).
For a Good Conductor: \(\sigma/\omega\varepsilon \to \infty\), so \(2\phi = 90^\circ \implies \phi = 45^\circ\).
Result: \(H\) lags \(E\) by \(45^\circ\). The magnitude is \(|\eta| = \sqrt{\frac{\omega\mu}{\sigma}}\).
7. Energy Analysis
We compare the average stored energy density in Electric (\(u_e\)) vs Magnetic (\(u_m\)) fields.
\[ u_e = \frac{1}{4}\varepsilon E_0^2 \quad , \quad u_m = \frac{1}{4}\mu H_0^2 \]Taking the ratio and substituting \(H_0 = E_0/|\eta|\):
\[ \frac{u_m}{u_e} = \frac{\mu}{\varepsilon} \frac{1}{|\eta|^2} = \frac{\mu}{\varepsilon} \left( \frac{\sigma}{\omega\mu} \right) = \frac{\sigma}{\omega\varepsilon} \]Since \(\sigma \gg \omega\varepsilon\) for a conductor, \(u_m \gg u_e\). The energy in a conducting medium is almost entirely magnetic.
8. Summary of Key Formulas
| Quantity | Formula (General) | Good Conductor Approx |
|---|---|---|
| Attenuation (\(\beta\)) | \(\omega \sqrt{\frac{\mu\varepsilon}{2}} [\sqrt{1 + (\frac{\sigma}{\omega\varepsilon})^2} - 1]^{1/2}\) | \(\sqrt{\pi f \mu \sigma}\) |
| Skin Depth (\(\delta\)) | \(1/\beta\) | \(\frac{1}{\sqrt{\pi f \mu \sigma}}\) |
| Intrinsic Impedance (\(\eta\)) | \(\sqrt{\frac{i\omega\mu}{\sigma + i\omega\varepsilon}}\) | \(\sqrt{\frac{\omega\mu}{\sigma}} \angle 45^\circ\) |