Technopedia

The technologist’s encyclopedia

Electromagnetic media

Electromagnetic media is the study of how matter responds to electromagnetic fields. In vacuum the field equations use two fixed constants, ε0\varepsilon_0 and μ0\mu_0. A medium replaces them with material response functions — permittivity ε\varepsilon (how the material’s charges polarize), permeability μ\mu (how its magnetic moments align), and conductivity σ\sigma (how easily charges flow) — that encode what the material does to the field. These three numbers, the constitutive parameters, set the wave speed, reflection, transmission, absorption, and dispersion in every field regime.

When it is relevant. Whenever an electromagnetic field meets matter: a capacitor filled with dielectric, a signal travelling down a coaxial cable or a PCB trace, light through glass or water, a radar wave hitting metal, an MRI coil in tissue, a fiber-optic cable, an antenna near a ground plane, a microwave oven, a solar cell, or a lens. It is the bridge between the field equations of Electromagnetism and the real materials those fields propagate through.

This page modifies all field classes — Electrostatics, Magnetostatics, Electrodynamics, and Electromagnetic Radiation — by replacing the vacuum constants with material response.

Authoritative sources

1. Constitutive relations

Definition. The three constitutive relations close Maxwell’s equations by relating the auxiliary fields D\mathbf{D} (electric displacement), B\mathbf{B} (magnetic flux density), and J\mathbf{J} (conduction current density) to the fundamental fields E\mathbf{E} and H\mathbf{H}:

D=ε E,B=μ H,J=σ E\mathbf{D} = \varepsilon\,\mathbf{E},\qquad \mathbf{B} = \mu\,\mathbf{H},\qquad \mathbf{J} = \sigma\,\mathbf{E}

In vacuum ε=ε0\varepsilon=\varepsilon_0, μ=μ0\mu=\mu_0, σ=0\sigma=0. In a material these become the medium’s constitutive parameters.

Details

The idea, step by step.

  1. Why two pairs of fields? Maxwell’s equations are most cleanly written with E\mathbf{E} and H\mathbf{H} (the fields that exert forces and do work), but the sources of the fields are charge and current. The auxiliary fields D\mathbf{D} and B\mathbf{B} are introduced so that Gauss’s law and the no-monopole law depend only on free charge and free current: ∇⋅D=ρfree\nabla\cdot\mathbf{D}=\rho_{\text{free}} and ∇⋅B=0\nabla\cdot\mathbf{B}=0. The constitutive relations are the material’s rule for converting between the two descriptions.

  2. The relations are the material’s “personality.” A vacuum has ε0\varepsilon_0, μ0\mu_0, σ=0\sigma=0. A dielectric has a large ε\varepsilon (it polarizes strongly). A ferromagnet has a large μ\mu (its domains align). A metal has a large σ\sigma (charges flow). Everything downstream — wave speed, reflection, loss — is a consequence of these three numbers.

  3. They are not always simple scalars. The linear, isotropic, non-dispersive form above is the simplest case. Real media can make ε\varepsilon, μ\mu, σ\sigma into tensors (anisotropic), frequency-dependent functions (dispersive), or field-dependent (nonlinear). Sections 5 and 8 develop these generalizations.

E, H fundamental D, B, J auxiliary constitutive relations D = ε E B = μ H J = σ E ε, μ, σ are the material's constitutive parameters
  1. Authoritative source. The constitutive framework is standard; see Ellingson, Electromagnetic Properties of Materials and Jackson, Classical Electrodynamics.
Example problem

A material has εr=9\varepsilon_r = 9, μr=1\mu_r = 1, and is a perfect insulator (σ=0\sigma = 0). What are its absolute permittivity and permeability?

Solution ε=εrε0=9×8.85×10−12≈7.97×10−11 F/m\varepsilon = \varepsilon_r \varepsilon_0 = 9 \times 8.85\times10^{-12} \approx 7.97\times10^{-11}\ \text{F/m} μ=μrμ0=1×4π×10−7≈1.26×10−6 H/m\mu = \mu_r \mu_0 = 1 \times 4\pi\times10^{-7} \approx 1.26\times10^{-6}\ \text{H/m}

The relative parameters are the ratios to the vacuum values; the absolute parameters are what you plug into the field equations.

2. Permittivity and polarization

Definition. The permittivity ε\varepsilon measures how strongly a material polarizes in response to an electric field. The relative permittivity (or dielectric constant) is εr=ε/ε0≥1\varepsilon_r = \varepsilon/\varepsilon_0 \ge 1. Polarization P\mathbf{P} is the dipole moment per unit volume, and D=ε0E+P\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}.

Details

The idea, step by step.

  1. Polarization is charge separation. In a dielectric, the applied E\mathbf{E} field nudges the electrons slightly off the nuclei, creating tiny dipoles. The polarization P\mathbf{P} is the net dipole moment per unit volume. Bound charges pile up on the surfaces, producing an internal field that opposes the applied field.

  2. Permittivity counts the polarization. The total displacement is D=ε0E+P\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}. For a linear dielectric P=ε0χeE\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} (with χe\chi_e the electric susceptibility), so D=ε0(1+χe)E=εE\mathbf{D} = \varepsilon_0(1+\chi_e)\mathbf{E} = \varepsilon \mathbf{E}. The relative permittivity is εr=1+χe\varepsilon_r = 1 + \chi_e.

  3. Why it matters: capacitance and field reduction. A dielectric between capacitor plates raises the capacitance by εr\varepsilon_r (it stores more charge at the same voltage) and reduces the field inside the material by εr\varepsilon_r for a fixed free charge. Water has εr≈80\varepsilon_r \approx 80, glass εr≈4−−10\varepsilon_r \approx 4--10, silicon εr≈11.7\varepsilon_r \approx 11.7, air εr≈1.0006\varepsilon_r \approx 1.0006.

+ − bound − bound + dipoles align with E; bound charge opposes the applied field
  1. Authoritative source. See Ellingson, Electromagnetic Properties of Materials and Griffiths, Introduction to Electrodynamics.
Example problem

A parallel-plate capacitor has plate area A=0.01 m2A = 0.01\,\text{m}^2 and separation d=1 mmd = 1\,\text{mm}. It is filled with a dielectric of εr=4\varepsilon_r = 4. Find its capacitance.

Solution C=εrε0Ad=4×8.85×10−12×0.011×10−3≈3.54×10−10 F≈354 pFC = \frac{\varepsilon_r \varepsilon_0 A}{d} = \frac{4 \times 8.85\times10^{-12} \times 0.01}{1\times10^{-3}} \approx 3.54\times10^{-10}\ \text{F} \approx 354\ \text{pF}

Without the dielectric the capacitance would be four times smaller (≈88.5 pF\approx 88.5\,\text{pF}) — the dielectric multiplies it by εr\varepsilon_r.

3. Permeability and magnetization

Definition. The permeability μ\mu measures how strongly a material magnetizes in response to a magnetic field. The relative permeability is μr=μ/μ0\mu_r = \mu/\mu_0. Magnetization M\mathbf{M} is the magnetic dipole moment per unit volume, and B=μ0(H+M)\mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M}). Materials are classified by their response: diamagnetic (μr<1\mu_r < 1, weakly repelled), paramagnetic (μr>1\mu_r > 1, weakly attracted), and ferromagnetic (μr≫1\mu_r \gg 1, strongly attracted, with hysteresis and saturation).

Details

The idea, step by step.

  1. Magnetization is aligned magnetic dipoles. Atoms carry tiny magnetic moments (from electron spin and orbital motion). An applied H\mathbf{H} field tends to align them. The magnetization M\mathbf{M} is the net moment per unit volume, and B=μ0(H+M)\mathbf{B} = \mu_0(\mathbf{H}+\mathbf{M}).

  2. Three regimes. In diamagnetic materials (copper, water, bismuth) the induced moments oppose the field, so μr\mu_r is just below 1. In paramagnetic materials (aluminum, oxygen) the moments align weakly with the field, so μr\mu_r is just above 1. In ferromagnetic materials (iron, nickel, cobalt) the moments lock into domains that align cooperatively, giving μr\mu_r of hundreds to tens of thousands — but with hysteresis (the response depends on history) and saturation (it stops growing once all domains align).

  3. Why it matters. High-μ\mu materials concentrate magnetic flux, which is why transformer cores, inductor cores, and MRI magnets use them. Ferromagnetic response is nonlinear and history-dependent, so the simple B=μH\mathbf{B}=\mu\mathbf{H} relation is only an approximation there; it is exact for the linear diamagnetic and paramagnetic cases.

Diamagnetic μr < 1 Paramagnetic μr > 1 Ferromagnetic μr ≫ 1 magnetic moments: oppose (dia), weakly align (para), strongly align (ferro)
  1. Authoritative source. See Ellingson, Electromagnetic Properties of Materials and Griffiths, Introduction to Electrodynamics.
Example problem

A toroidal core of relative permeability μr=2000\mu_r = 2000 carries a coil of N=500N = 500 turns with current I=0.1 AI = 0.1\,\text{A}. The mean magnetic path length is ℓ=0.2 m\ell = 0.2\,\text{m}. Find the magnetic field strength HH and the flux density BB.

Solution

Ampère’s law around the toroid gives H=NI/ℓH = NI/\ell:

H=500×0.10.2=250 A/mH = \frac{500 \times 0.1}{0.2} = 250\ \text{A/m}

Then B=μH=μrμ0HB = \mu H = \mu_r \mu_0 H:

B=2000×4π×10−7×250≈0.63 TB = 2000 \times 4\pi\times10^{-7} \times 250 \approx 0.63\ \text{T}

The high-μ\mu core concentrates the flux: the same coil in air (μr=1\mu_r=1) would give only B≈3.1×10−4 TB \approx 3.1\times10^{-4}\,\text{T}.

4. Conductivity and Ohm’s law

Definition. The conductivity σ\sigma (units S/m) measures how easily free charges move in response to an electric field. Ohm’s law in local form is J=σE\mathbf{J} = \sigma \mathbf{E}. Its reciprocal is the resistivity ρ=1/σ\rho = 1/\sigma. A conductor has large σ\sigma (copper ≈5.8×107 S/m\approx 5.8\times10^7\,\text{S/m}); an insulator has σ≈0\sigma \approx 0.

Details

The idea, step by step.

  1. Conduction is mobile charge. In a metal, a sea of free electrons drifts under E\mathbf{E}. The current density is proportional to the field: J=σE\mathbf{J} = \sigma \mathbf{E}. This is the microscopic Ohm’s law; the circuit form V=IRV = IR is its integral consequence.

  2. Conductivity is the source of loss. Wherever σ>0\sigma > 0, the field does work on the charges at a rate J⋅E=σE2\mathbf{J}\cdot\mathbf{E} = \sigma E^2 per unit volume, which is dissipated as heat (Joule heating). This is why a signal in a lossy medium attenuates — the energy is converted to thermal energy. See Poynting’s theorem for the energy bookkeeping.

  3. The spectrum of materials. Conductors (copper, aluminum) have σ∼107−−108 S/m\sigma \sim 10^7--10^8\,\text{S/m}. Semiconductors (silicon) range over many orders of magnitude depending on doping. Electrolytes and biological tissue are weak conductors (σ∼0.1−−1 S/m\sigma \sim 0.1--1\,\text{S/m}). Good insulators (glass, dry air) have σ≪10−12 S/m\sigma \ll 10^{-12}\,\text{S/m}.

insulator (σ ≈ 0) conductor (σ ~ 10⁸ S/m) glass tissue silicon copper σ sets Joule loss: power density = σ E²
  1. Authoritative source. See Ellingson, Electromagnetic Properties of Materials and Griffiths, Introduction to Electrodynamics.
Example problem

A copper wire of cross-sectional area A=1 mm2A = 1\,\text{mm}^2 carries a current density J=106 A/m2J = 10^6\,\text{A/m}^2. With σ=5.8×107 S/m\sigma = 5.8\times10^7\,\text{S/m}, find the electric field in the wire and the power dissipated per unit length.

Solution

From J=σE\mathbf{J} = \sigma \mathbf{E}:

E=Jσ=1065.8×107≈0.017 V/mE = \frac{J}{\sigma} = \frac{10^6}{5.8\times10^7} \approx 0.017\ \text{V/m}

The power per unit length is the Joule heating P/ℓ=J2/σ⋅AP/\ell = J^2/\sigma \cdot A:

Pℓ=J2σ A=(106)25.8×107×10−6≈17.2 W/m\frac{P}{\ell} = \frac{J^2}{\sigma}\,A = \frac{(10^6)^2}{5.8\times10^7}\times 10^{-6} \approx 17.2\ \text{W/m}

This is the microscopic form of I2RI^2 R: the field does work on the electrons, and the metal converts it to heat.

5. Lossy media and skin depth

Definition. A real medium is described by a complex permittivity εc=ε−j σ/ω\varepsilon_c = \varepsilon - j\,\sigma/\omega (or equivalently a complex conductivity), which folds conduction loss into the wave equation. In a good conductor the wave decays exponentially with depth; the skin depth is the distance over which the field falls by 1/e1/e:

δ=1πfμσ\delta = \frac{1}{\sqrt{\pi f \mu \sigma}}
Details

The idea, step by step.

  1. Loss appears as an imaginary part. In a sinusoidal field of angular frequency ω\omega, the conduction current σE\sigma \mathbf{E} is in phase with E\mathbf{E} while the displacement current jωεEj\omega\varepsilon\mathbf{E} is 90∘90^\circ out of phase. Combining them gives an effective complex permittivity εc=ε−jσ/ω\varepsilon_c = \varepsilon - j\sigma/\omega. The loss tangent tan⁡δ=σ/(ωε)\tan\delta = \sigma/(\omega\varepsilon) compares the two: large when conduction dominates (good conductor), small when displacement dominates (dielectric).

  2. The wave decays inside a conductor. Substituting εc\varepsilon_c into the wave equation gives a complex propagation constant γ=α+jβ\gamma = \alpha + j\beta, where α\alpha is the attenuation (Np/m). In a good conductor (σ≫ωε\sigma \gg \omega\varepsilon), α≈β≈πfμσ\alpha \approx \beta \approx \sqrt{\pi f \mu \sigma}, and the field is E(z)=E0e−z/δE(z) = E_0 e^{-z/\delta} with δ=1/α\delta = 1/\alpha.

  3. Skin depth sets the penetration. The field (and current) are confined to a thin surface layer of thickness δ\delta. For copper at f=1 GHzf = 1\,\text{GHz}, δ≈2.1 μm\delta \approx 2.1\,\mu\text{m}; at 60 Hz60\,\text{Hz} it is ≈8.5 mm\approx 8.5\,\text{mm}. This is why high-frequency currents flow on the surface (the skin effect), why RF shielding works, and why thick conductors are wasteful at high frequency.

conductor surface δ (field = E₀/e) |E| depth z E(z) = E₀ e^(−z/δ), δ = 1/√(π f μ σ)
  1. Authoritative source. See Jackson, Classical Electrodynamics, and Ellingson, Electromagnetic Properties of Materials.
Example problem

Compute the skin depth of copper (σ=5.8×107 S/m\sigma = 5.8\times10^7\,\text{S/m}, μ≈μ0\mu \approx \mu_0) at f=1 GHzf = 1\,\text{GHz}.

Solution δ=1πfμ0σ=1π×109×4π×10−7×5.8×107\delta = \frac{1}{\sqrt{\pi f \mu_0 \sigma}} = \frac{1}{\sqrt{\pi \times 10^9 \times 4\pi\times10^{-7} \times 5.8\times10^7}} πfμ0σ≈2.27×1015 s−2⇒δ≈14.77×107≈2.1×10−6 m=2.1 μm\pi f \mu_0 \sigma \approx 2.27\times10^{15}\ \text{s}^{-2} \qquad\Rightarrow\qquad \delta \approx \frac{1}{4.77\times10^{7}} \approx 2.1\times10^{-6}\ \text{m} = 2.1\ \mu\text{m}

At 1 GHz1\,\text{GHz} the field penetrates only about two micrometres into copper — a thin skin, hence the name.

6. Wave speed and impedance in a medium

Definition. In a lossless medium the wave speed and the intrinsic impedance are

v=1με=cμrεr,η=με=η0μr/εrv = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}},\qquad \eta = \sqrt{\frac{\mu}{\varepsilon}} = \frac{\eta_0}{\sqrt{\mu_r/\varepsilon_r}}

where η0=μ0/ε0≈377 Ω\eta_0 = \sqrt{\mu_0/\varepsilon_0} \approx 377\,\Omega is the impedance of free space.

Details

The idea, step by step.

  1. The medium slows the wave. The same wave-equation derivation that gives c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0} in vacuum gives v=1/μεv = 1/\sqrt{\mu\varepsilon} in a medium. Since μ≥μ0\mu \ge \mu_0 and ε≥ε0\varepsilon \ge \varepsilon_0, the wave is always slower than cc. For a dielectric with μr≈1\mu_r \approx 1, v=c/εrv = c/\sqrt{\varepsilon_r}.

  2. The impedance sets the E/B ratio. In a plane wave the ratio of the electric to magnetic field is the intrinsic impedance η=μ/ε\eta = \sqrt{\mu/\varepsilon}. In vacuum η0≈377 Ω\eta_0 \approx 377\,\Omega. In a dielectric with εr=4\varepsilon_r = 4, η=377/2≈189 Ω\eta = 377/2 \approx 189\,\Omega. This impedance is what determines reflection at an interface (Section 7).

  3. Why it matters. The speed sets the wavelength in the medium (λ=v/f\lambda = v/f), which governs the size of antennas, transmission lines, and optical paths. The impedance mismatch between two media is the root cause of reflection — the reason a lens needs an anti-reflection coating and a transmission line needs impedance matching.

vacuum: v = c medium: v = c/√(μr εr) same frequency, shorter wavelength in the medium (λ = v/f)
  1. Authoritative source. See Maxwell (1865) and Griffiths, Introduction to Electrodynamics.
Example problem

A dielectric has εr=9\varepsilon_r = 9 and μr=1\mu_r = 1. Find the wave speed and the intrinsic impedance.

Solution v=cεrμr=3.0×1089=1.0×108 m/sv = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{3.0\times10^8}{\sqrt{9}} = 1.0\times10^8\ \text{m/s} η=η0εr/μr=3779≈126 Ω\eta = \frac{\eta_0}{\sqrt{\varepsilon_r/\mu_r}} = \frac{377}{\sqrt{9}} \approx 126\ \Omega

The wave travels at one-third the speed of light, and the impedance is one-third that of free space.

7. Boundary conditions and reflection

Definition. At an interface between two media, the tangential components of E\mathbf{E} and H\mathbf{H} are continuous (in the absence of surface charge/current). For a plane wave at normal incidence, the reflection coefficient (electric field) is

Γ=η2−η1η2+η1\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}

and the transmission coefficient is T=1+Γ=2η2/(η2+η1)T = 1 + \Gamma = 2\eta_2/(\eta_2+\eta_1). The Fresnel equations generalize this to oblique incidence and both polarizations.

Details

The idea, step by step.

  1. Fields must match at the boundary. Maxwell’s equations require the tangential E\mathbf{E} and H\mathbf{H} to be continuous across an interface. An incident wave arriving at the boundary therefore splits into a reflected wave (back into medium 1) and a transmitted wave (into medium 2), and the sum of incident and reflected fields must equal the transmitted field at the surface.

  2. The coefficient is an impedance ratio. Enforcing the boundary conditions gives Γ=(η2−η1)/(η2+η1)\Gamma = (\eta_2-\eta_1)/(\eta_2+\eta_1). If the impedances match (η2=η1\eta_2 = \eta_1), there is no reflection (Γ=0\Gamma = 0). If medium 2 is a perfect conductor (η2=0\eta_2 = 0), the wave is fully reflected with a sign flip (Γ=−1\Gamma = -1).

  3. Power reflection is ∣Γ∣2|\Gamma|^2. The fraction of incident power reflected is ∣Γ∣2|\Gamma|^2. For a wave going from air (η1≈377 Ω\eta_1 \approx 377\,\Omega) into a dielectric with εr=4\varepsilon_r = 4 (η2≈189 Ω\eta_2 \approx 189\,\Omega), Γ≈−0.33\Gamma \approx -0.33 and ∣Γ∣2≈0.11|\Gamma|^2 \approx 0.11 — about 11% of the power is reflected. This is why glass reflects a little and why anti-reflection coatings (a thin quarter-wave layer) are used to cancel it.

medium 1 (η₁) medium 2 (η₂) incident reflected (Γ) transmitted (T) Γ = (η₂ − η₁)/(η₂ + η₁)
  1. Authoritative source. See Maxwell (1865) and Griffiths, Introduction to Electrodynamics.
Example problem

A plane wave in air (η1≈377 Ω\eta_1 \approx 377\,\Omega) strikes normally a dielectric slab with εr=4\varepsilon_r = 4, μr=1\mu_r = 1. Find the reflection coefficient and the fraction of power reflected.

Solution

The slab impedance is η2=η0/εr=377/2≈189 Ω\eta_2 = \eta_0/\sqrt{\varepsilon_r} = 377/2 \approx 189\,\Omega. Then

Γ=η2−η1η2+η1=189−377189+377≈−0.33\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1} = \frac{189 - 377}{189 + 377} \approx -0.33

The power reflection is

∣Γ∣2≈0.11|\Gamma|^2 \approx 0.11

so about 11% of the incident power is reflected and 89% is transmitted. The negative sign means the reflected electric field is inverted.

8. Advanced media

Definition. Real and engineered media go beyond the scalar, constant parameters of Sections 1–7:

Details

The idea, step by step.

  1. Anisotropy: direction matters. In a crystal the atomic arrangement is not the same in every direction, so D\mathbf{D} is not generally parallel to E\mathbf{E}. The relation becomes D=ε E\mathbf{D} = \boldsymbol{\varepsilon}\,\mathbf{E} with ε\boldsymbol{\varepsilon} a tensor. This is the origin of birefringence: an ordinary light beam splits into two polarized rays travelling at different speeds.

  2. Dispersion: frequency matters. Because polarization and magnetization cannot respond instantaneously, ε\varepsilon and μ\mu depend on frequency. The result is dispersion: different colours travel at different speeds, which is why a prism spreads white light into a spectrum. The group velocity (the speed of a pulse) differs from the phase velocity, and near absorption resonances the dispersion is strong.

  3. Nonlinearity: amplitude matters. At high field strengths (lasers), the polarization is no longer proportional to E\mathbf{E}. The χ(2)\chi^{(2)} term mixes frequencies: two photons of frequency ω\omega can combine into one of 2ω2\omega (second-harmonic generation), the basis of green laser pointers.

  4. Metamaterials: engineer the parameters. By arranging subwavelength metallic or dielectric elements, one can make an effective medium with negative ε\varepsilon or μ\mu. A medium with both negative (a double-negative or left-handed medium) has a negative refractive index: light bends the “wrong” way, enabling negative refraction, cloaking, and superlenses that beat the diffraction limit.

n > 0 metamaterial, n < 0 incident normal refraction negative refraction with n < 0 the refracted ray bends to the same side as the incident ray
  1. Authoritative source. See Jackson, Classical Electrodynamics, and the review literature on metamaterials (e.g. Pendry, Negative Refraction, 2000).
Example problem

A dispersive glass has εr=2.25\varepsilon_r = 2.25 at the red end of the spectrum and εr=2.56\varepsilon_r = 2.56 at the blue end (with μr=1\mu_r = 1). Find the refractive index and the speed of light for each colour.

Solution

The refractive index is n=εrμr=εrn = \sqrt{\varepsilon_r \mu_r} = \sqrt{\varepsilon_r}.

Red: n=2.25=1.50n = \sqrt{2.25} = 1.50, so v=c/1.50=2.0×108 m/sv = c/1.50 = 2.0\times10^8\,\text{m/s}.

Blue: n=2.56=1.60n = \sqrt{2.56} = 1.60, so v=c/1.60=1.875×108 m/sv = c/1.60 = 1.875\times10^8\,\text{m/s}.

Blue light travels slower than red in the glass — this frequency dependence (dispersion) is exactly what makes a prism separate colours.

Interactive: wave speed, reflection, and skin depth

Drag the sliders to change the medium’s relative permittivity εr\varepsilon_r, relative permeability μr\mu_r, and conductivity σ\sigma, and watch the wave speed, the normal-incidence reflection coefficient, and the skin depth respond.

Part of Electromagnetism.

Quizz

Test yourself

Problem. A uniform plane wave at f=1 GHzf = 1\,\text{GHz} travels in air and strikes normally a large copper plate (σ=5.8×107 S/m\sigma = 5.8\times10^7\,\text{S/m}, μ≈μ0\mu \approx \mu_0).

  1. What is the skin depth of copper at this frequency?
  2. What fraction of the incident power is reflected?
  3. How far into the copper does the field penetrate before dropping to 1/e1/e of its surface value?
Solution
  1. The skin depth is
δ=1πfμ0σ=1π×109×4π×10−7×5.8×107≈2.1×10−6 m=2.1 μm\delta = \frac{1}{\sqrt{\pi f \mu_0 \sigma}} = \frac{1}{\sqrt{\pi \times 10^9 \times 4\pi\times10^{-7} \times 5.8\times10^7}} \approx 2.1\times10^{-6}\ \text{m} = 2.1\ \mu\text{m}
  1. Copper is a very good conductor, so its intrinsic impedance η2=μ/(σ−jωε)\eta_2 = \sqrt{\mu/(\sigma - j\omega\varepsilon)} is tiny (on the order of a fraction of an ohm) compared with η1≈377 Ω\eta_1 \approx 377\,\Omega. The reflection coefficient is therefore Γ≈(η2−η1)/(η2+η1)≈−1\Gamma \approx (\eta_2 - \eta_1)/(\eta_2 + \eta_1) \approx -1, and the reflected power fraction ∣Γ∣2≈1|\Gamma|^2 \approx 1 — essentially all the power is reflected. This is why metal is an excellent RF shield.

  2. By definition, the field falls to 1/e1/e of its surface value at one skin depth, so the penetration distance is δ≈2.1 μm\delta \approx 2.1\,\mu\text{m}. The wave is confined to a skin only a couple of micrometres thick.