---
title: "Electromagnetic Waves in Plasmas, Conductors and Dielectrics"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 14
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics
---

# Chapter 14 — Electromagnetic Waves in Plasmas, Conductors and Dielectrics

A shortwave broadcast crosses an ocean by bouncing off the upper atmosphere, which reflects it like a mirror — yet the same layer lets television and satellite signals through. A metal spoon is shiny, a metal mesh in the door of a microwave oven holds in the waves while you watch your food through it, and a sheet of glass is transparent to light but opaque to ultraviolet. In each case a wave enters a material, the material’s charges respond, and their response changes the wave: its speed, its [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex), even whether it propagates at all. This chapter treats the three standard models of a medium — a gas of free electrons, an ohmic conductor, and bound electrons — with the one method that handles them all: add the induced current to [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) and read off the [dispersion relation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation).

## 14.1 Waves in a plasma

**Definition 14.1 (Plasma; the free-electron model).**

A *plasma* is a globally neutral gas of free electrons (density $n$, mass $m$, charge $-e$) and positive ions, which, being thousands of times heavier, are taken as fixed. The upper atmosphere (ionosphere, $n \sim 10^{11}$–$10^{12}$ m$^{-3}$), a discharge tube, the solar corona, and — at optical frequencies — the conduction electrons of a metal are plasmas. Collisions are neglected (the wave’s period is short compared with the time between collisions). Its characteristic frequency is the *plasma frequency*

$$
\omega_p = \sqrt{\frac{ne^2}{\varepsilon_0m}} , \qquad
f_p = \frac{\omega_p}{2\pi} \approx 9\,\text{Hz}\times\sqrt{n\,[\text{m}^{-3}]} :
$$

$3\,\mathrm{MHz}$ to $9\,\mathrm{MHz}$ for the ionosphere, $2 \times 10^{15}\,\mathrm{Hz}$ (ultraviolet) for a metal.

**Theorem 14.2 (Dispersion relation of a plasma).**

For a transverse PPH wave $\underline{\vect E} = \underline{\vect E}_0\eu^{\iu(\omega t - kx)}$ in a [plasma](#def-b2-waves-in-media-plasma), the electrons oscillate with $\underline{\vect v} = -e\underline{\vect E}/\iu m\omega$ and carry the [current density](https://one-course.com/books/physics/4/en/chapter/10-charges-currents-and-conduction#def-b2-charges-currents-conduction-densities) $\underline{\vect j} = \underline\gamma\underline{\vect E}$ with $\underline\gamma = ne^2/\iu m\omega$; [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) then give

$$
k^2 = \frac{\omega^2 - \omega_p^2}{c^2} .
$$

Above $\omega_p$ the wave propagates, with $v_\varphi = c/\sqrt{1 - \omega_p^2/\omega^2}
> c$ and $v_g = c\sqrt{1 - \omega_p^2/\omega^2} < c$, $v_\varphi v_g = c^2$; below $\omega_p$, $k = -\iu\kappa$ with $\kappa = \sqrt{\omega_p^2 - \omega^2}/c$: the wave is *evanescent*, penetrates over $1/\kappa$ and is totally reflected ([Chapter 15](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#ch-b2-wave-interfaces)); the [plasma](#def-b2-waves-in-media-plasma) is then a mirror. At $\omega
\gg \omega_p$ the electrons cannot follow and the [plasma](#def-b2-waves-in-media-plasma) is transparent.

**Proof.** Equation of motion $m\,\dd\vect v/\dd t = -e\vect E$ (the magnetic force $ev B \sim evE/c$ is negligible for $v \ll c$, and the ions’ motion is $m/M$ times smaller): $\iu m\omega\underline{\vect v} = -e\underline{\vect E}$. Then $\underline{\vect j} = -ne\underline{\vect v}$. The [plasma](#def-b2-waves-in-media-plasma) stays neutral for a transverse wave ($\vect k\cdot\vect E = 0$ gives $\rho = 0$ by Gauss), so Maxwell–Ampère reads $-\iu\vect k\wedge\underline{\vect B} = \mu_0\underline\gamma\underline{\vect E}
+ \iu\omega\mu_0\varepsilon_0\underline{\vect E}$ and Maxwell–Faraday $\underline{\vect B} = \vect k
\wedge\underline{\vect E}/\omega$; combining, $k^2\underline{\vect E} = \mu_0(\iu\omega\underline\gamma -
\omega^2\varepsilon_0)(-1)\underline{\vect E}$, i.e. $k^2 = \omega^2\mu_0\varepsilon_0 - \iu\omega\mu_0
\underline\gamma = (\omega^2 - ne^2/\varepsilon_0m)/c^2$. The velocities follow as in [Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets). ∎

**Remark 14.3 (Where the energy goes).**

The conductivity is imaginary: $\langle\vect j\cdot\vect E\rangle = 0$, the electrons take energy from the wave during half a cycle and give it back during the next — a lossless medium, in which the wave’s energy is shared between the fields and the electrons’ kinetic energy, and travels at $v_g$. Collisions add a small real part to $\underline\gamma$ and a small absorption: the ionosphere’s D layer absorbs medium waves by day, which is why distant AM stations come in at night.

**Example 14.4 (The ionosphere and radio).**

With $n = 1 \times 10^{12}\,\mathrm{m}^{-3}$ in the F layer, $f_p = 9\,\mathrm{MHz}$: shortwave broadcasts below that are reflected (and can hop around the Earth between the ionosphere and the ground), FM ($100\,\mathrm{MHz}$), television and satellite links pass through. A wave at $3\,\mathrm{MHz}$ penetrates only $1/\kappa = c/\sqrt{\omega_p^2 - \omega^2} = 6\,\mathrm{m}$ into the layer before turning back. A spacecraft re-entering the atmosphere is sheathed in a [plasma](#def-b2-waves-in-media-plasma) of $n \sim 10^{18}$ m$^{-3}$, $f_p \sim 10\,\mathrm{GHz}$: radio contact is lost for minutes.

![Left: the dispersion relation of a plasma — no propagation below _p, a hyperbola approaching the light line above. Right: shortwaves below the plasma frequency are reflected by the ionosphere and hop around the Earth; higher frequencies escape to space.](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/fig-60d49cf67023.svg)

![Left: the dispersion relation of a plasma — no propagation below _p, a hyperbola approaching the light line above. Right: shortwaves below the plasma frequency are reflected by the ionosphere and hop around the Earth; higher frequencies escape to space.](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/fig-de346e6acb0c.svg)

*Left: the [dispersion relation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) of a [plasma](#def-b2-waves-in-media-plasma) — no propagation below $\omega_p$, a hyperbola approaching the light line above. Right: shortwaves below the [plasma frequency](#def-b2-waves-in-media-plasma) are reflected by the ionosphere and hop around the Earth; higher frequencies escape to space.*

## 14.2 Waves in an ohmic conductor: the skin effect

**Theorem 14.5 (Skin effect).**

In an ohmic conductor ($\vect j = \gamma\vect E$, $\gamma$ real, at frequencies with $\omega\tau \ll 1$ and $\varepsilon_0\omega \ll \gamma$, i.e. below $\sim1 \times 10^{16}\,\mathrm{Hz}$ for copper) the [displacement current](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) is negligible and the fields obey the diffusion equation $\Delta\vect E = \mu_0\gamma\,\partial_t\vect E$. A PPH wave entering the conductor at $x = 0$ has

$$
\underline k^2 = -\iu\mu_0\gamma\omega , \qquad \underline k = \frac{1 - \iu}\delta , \qquad
\delta = \sqrt{\frac2{\mu_0\gamma\omega}} ,
$$

so that $\underline E = E_0\,\eu^{-x/\delta}\eu^{\iu(\omega t - x/\delta)}$: it is damped over the *[skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex)* $\delta$, with a phase that also turns by one radian per $\delta$ — a wavelength $2\pi\delta$, a [phase velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) $\omega\delta \ll
c$. The current $\vect j = \gamma\vect E$ is confined to a layer of thickness $\delta$ under the surface, and $\vect B$ lags $\vect E$ by $45{}^{\circ}$ with $|B| = |E|\sqrt2/\omega\delta$. For copper, $\delta = 9.2\,\mathrm{mm}$ at $50\,\mathrm{Hz}$, $0.21\,\mathrm{mm}$ at $100\,\mathrm{kHz}$, $2.1\,\text{µ}\mathrm{m}$ at $1\,\mathrm{GHz}$.

**Proof.** $\operatorname{\vect{curl}}\vect B = \mu_0\gamma\vect E$ (displacement dropped) and $\operatorname{\vect{curl}}\vect E = -\partial_t\vect B$: taking the curl of the second, $-\Delta\vect E = -\mu_0\gamma\partial_t\vect E$ (the conductor is neutral, $\operatorname{div}
\vect E = 0$). For $\eu^{\iu(\omega t - kx)}$: $k^2 = -\iu\mu_0\gamma\omega$, and $\sqrt{-\iu} = (1 - \iu)/\sqrt2$ gives $\underline k = (1 - \iu)/\delta$ (the root that decays into the conductor). Faraday: $\underline B = \underline k\underline E/\omega = (1 - \iu)
\underline E/\omega\delta$, of modulus $\sqrt2E/\omega\delta$ and phase $-\pi/4$. ∎

![Left: the field inside a conductor — a damped oscillation, dead within a few skin depths. Right: the skin depth of copper and of sea water against frequency: millimetres at mains frequency, micrometres at gigahertz; the sea lets only the longest radio waves in.](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/fig-6828fb2798a7.svg)

![Left: the field inside a conductor — a damped oscillation, dead within a few skin depths. Right: the skin depth of copper and of sea water against frequency: millimetres at mains frequency, micrometres at gigahertz; the sea lets only the longest radio waves in.](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/fig-564add435c96.svg)

*Left: the field inside a conductor — a damped oscillation, dead within a few [skin depths](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex). Right: the [skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) of copper and of sea water against frequency: millimetres at mains frequency, micrometres at gigahertz; the sea lets only the longest radio waves in.*

**Example 14.6 (Consequences of the skin effect).**

(i) A wire of radius $a \gg \delta$ carries its current in a ring of thickness $\delta$: its resistance rises from $1/\gamma\pi a^2$ to about $1/\gamma
2\pi a\delta$ per metre — for a $1\,\mathrm{mm}$ copper wire, four times the DC value at $1\,\mathrm{MHz}$, a hundred times at $1\,\mathrm{GHz}$. High-frequency coils use "Litz" wire (many insulated strands) or silver-plated tubes; the power grid’s conductors are stranded with an aluminium sheath. (ii) A metal box attenuates a wave by $\eu^{-d/\delta}$ per thickness $d$: a $1\,\mathrm{mm}$ aluminium enclosure blocks $1\,\mathrm{MHz}$ by $100\,\mathrm{dB}$ but $50\,\mathrm{Hz}$ hardly at all — low-frequency magnetic fields need iron. (iii) Submarines receive only very-long-wave radio ($\delta = 2.3\,\mathrm{m}$ at $10\,\mathrm{kHz}$ in sea water); mines are detected by the eddy currents a coil induces in them ([Chapter 11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations), the diffusion time $\mu_0\gamma d^2$).

## 14.3 Waves in a dielectric: the bound electron

**Proposition 14.7 (Elastically bound electron; complex index).**

In an insulator each electron is bound to its atom: a displacement $\vect r$ from equilibrium calls the restoring force $-m\omega_0^2\vect r$ and a damping $-m\Gamma\,\dd\vect r/\dd t$ (the energy radiated or handed to the lattice). Driven by the wave, $m\ddot{\vect r} = -m\omega_0^2\vect r - m\Gamma\dot{\vect r}
- e\vect E$; the displaced electrons give the medium the *polarization* (dipole moment per unit volume) $\vect P = -ne\vect r = \varepsilon_0\underline\chi(\omega)
\underline{\vect E}$ with the *susceptibility*

$$
\underline\chi(\omega) = \frac{\omega_p^2}{\omega_0^2 - \omega^2 + \iu\Gamma\omega} ,
\qquad \omega_p^2 = \frac{ne^2}{\varepsilon_0m} ,
$$

and the bound current $\vect j = \partial_t\vect P$. [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) then give

$$
\underline k^2 = \frac{\omega^2}{c^2}\,(1 + \underline\chi) = \frac{\omega^2}{c^2}\,\underline n^2 ,
\qquad \underline n = n' - \iu n'' ,
$$

with $n'$ the refractive index ([phase velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) $c/n'$) and $n''$ the extinction (the intensity decays as $\eu^{-2n''\omega x/c}$). Far below a resonance ($\omega \ll \omega_0$, $\Gamma$ negligible), $n'^2 \approx 1 + \omega_p^2/\omega_0^2 +
\omega_p^2\omega^2/\omega_0^4$: the index *rises* with frequency — normal dispersion, Cauchy’s law $n \approx A + B/\lambda^2$ — and absorption is negligible; near $\omega_0$ the medium absorbs, and just above it $n'$ falls with $\omega$ (anomalous dispersion).

**Proof.** Complex amplitudes: $(\omega_0^2 - \omega^2 + \iu\Gamma\omega)\underline{\vect r} = -e\underline{\vect E}
/m$; $\underline{\vect P} = -ne\underline{\vect r}$. Then, as for the [plasma](#def-b2-waves-in-media-plasma), $k^2 = \omega^2\mu_0
\varepsilon_0 - \iu\omega\mu_0\underline\gamma$ with $\underline{\vect j} = \iu\omega\underline{\vect P} = \iu\omega\varepsilon_0
\underline\chi\underline{\vect E}$, i.e. $\underline\gamma = \iu\omega\varepsilon_0\underline\chi$: $k^2 = (\omega^2/c^2)
(1 + \underline\chi)$. (The [plasma](#def-b2-waves-in-media-plasma) is the case $\omega_0 = 0$, $\Gamma = 0$.) Expansion: $1/(\omega_0^2 - \omega^2) \approx (1 + \omega^2/\omega_0^2)/\omega_0^2$. ∎

![The bound-electron (Lorentz) model: the real part of the index rises with frequency below the resonance (normal dispersion: violet is slower than red in glass), the imaginary part peaks at the resonance (absorption), and the index falls through the resonance (anomalous dispersion).](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/fig-65ff1852dab3.svg)

*The bound-electron (Lorentz) model: the real part of the index rises with frequency below the resonance (normal dispersion: violet is slower than red in glass), the imaginary part peaks at the resonance (absorption), and the index falls through the resonance (anomalous dispersion).*

**Example 14.8 (Glass, water, air).**

Glass has its electronic resonances in the ultraviolet ($\omega_0 \sim 10^{16}$ rad/s): in the visible it is transparent and dispersive, $n = 1.51$ at $700\,\mathrm{nm}$, $1.53$ at $400\,\mathrm{nm}$, enough for a prism to spread a rainbow and for a fibre’s pulses to spread ([Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets)); in the ultraviolet it absorbs — you do not tan behind a window. Water has, besides, a rotational resonance of its polar molecules near $20\,\mathrm{GHz}$ (a Debye relaxation rather than a sharp line): it absorbs microwaves strongly, which heats the food and blinds radar in rain, and has $n'' \approx 0$ in a narrow visible window — the window through which eyes evolved. Air, with $n - 1 = 2.9 \times 10^{-4}$, still bends the setting Sun by half a degree.

**Remark 14.9 (One equation, three media).**

Every linear medium enters [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) through its induced current $\underline{\vect j} = \underline\gamma(\omega)\underline{\vect E}$, and the [dispersion relation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) is always $\underline k^2 = \omega^2/c^2 - \iu\omega\mu_0\underline\gamma(\omega)$:

| medium | $\underline\gamma(\omega)$ | $\underline k^2$ |
| --- | --- | --- |
| [plasma](#def-b2-waves-in-media-plasma) | $ne^2/\iu m\omega$ (imaginary) | $(\omega^2 - \omega_p^2)/c^2$: cut-off |
| ohmic conductor | $\gamma$ (real) | $-\iu\mu_0\gamma\omega$: [skin effect](#thm-b2-waves-in-media-skin) |
| dielectric | $\iu\omega\varepsilon_0\underline\chi(\omega)$ | $(\omega^2/c^2)(1 + \underline\chi)$: index |

A metal is all three in turn: ohmic below $1 \times 10^{13}\,\mathrm{Hz}$, a [plasma](#def-b2-waves-in-media-plasma) (reflecting) in the visible, transparent in the far ultraviolet.

**Method 14.10 (Waves in a medium).**

(1) Write the equation of motion of the charges and get $\underline{\vect j} =
\underline\gamma\underline{\vect E}$. (2) Insert into Maxwell–Ampère with the PPH ansatz: $\underline k^2 = \omega^2/c^2 - \iu\omega\mu_0\underline\gamma$. (3) Take the root with $\operatorname{Im}k \le 0$ (decay in the direction of travel), split into real (propagation, $v_\varphi = \omega/k'$) and imaginary ([attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex)) parts. (4) Identify the regimes in frequency — propagating, evanescent, absorbing — and the characteristic frequency ($\omega_p$, $1/\mu_0\gamma\delta^2$, $\omega_0$). (5) Find $\vect B$ from Faraday and, if needed, the Poynting flux and the dissipated power.

## 14.4 Exercises

**Exercise 14.1 ★.**

[Plasma](#def-b2-waves-in-media-plasma) frequencies for $n = 1 \times 10^{11}\,\mathrm{m}^{-3}$ (ionosphere by night), $1 \times 10^{12}\,\mathrm{m}^{-3}$ (by day), $1 \times 10^{18}\,\mathrm{m}^{-3}$ (re-entry sheath), $8.5 \times 10^{28}\,\mathrm{m}^{-3}$ (copper), $1 \times 10^{5}\,\mathrm{m}^{-3}$ (interstellar space). For each, which of the following passes: a $1\,\mathrm{MHz}$ AM station, a $100\,\mathrm{MHz}$ FM station, a $1.5\,\mathrm{GHz}$ GPS signal, visible light?

**Solution of Exercise 14.1.**

$f_p \approx 9\sqrt n$ Hz: $2.8\,\mathrm{MHz}$, $9\,\mathrm{MHz}$, $9\,\mathrm{GHz}$, $2.6 \times 10^{15}\,\mathrm{Hz}$ ($\lambda_p = 115\,\mathrm{nm}$), $2.8\,\mathrm{kHz}$. AM at $1\,\mathrm{MHz}$: only through interstellar space. FM: through the ionosphere, not the sheath. GPS: everything but the sheath. Light: everything but copper.

**Exercise 14.2 ★.**

[Skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) of copper ($\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$) at $50\,\mathrm{Hz}$, $1\,\mathrm{kHz}$, $1\,\mathrm{MHz}$, $1\,\mathrm{GHz}$; of aluminium ($3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$) at $1\,\mathrm{MHz}$; of sea water ($5\,\mathrm{S}/\mathrm{m}$) at $10\,\mathrm{kHz}$ and $1\,\mathrm{GHz}$; of wet soil ($1 \times 10^{-2}\,\mathrm{S}/\mathrm{m}$) at $1\,\mathrm{MHz}$. Which radio frequencies reach a submarine at $10\,\mathrm{m}$?

**Solution of Exercise 14.2.**

Copper: $9.2\,\mathrm{mm}$, $2.1\,\mathrm{mm}$, $65\,\text{µ}\mathrm{m}$, $2.1\,\text{µ}\mathrm{m}$; aluminium at $1\,\mathrm{MHz}$: $84\,\text{µ}\mathrm{m}$; sea water: $2.3\,\mathrm{m}$ at $10\,\mathrm{kHz}$, $7\,\mathrm{mm}$ at $1\,\mathrm{GHz}$; soil: $5\,\mathrm{m}$ at $1\,\mathrm{MHz}$. A submarine at $10\,\mathrm{m}$ needs $\delta
\gtrsim 10\,\mathrm{m}$: below about $500\,\mathrm{Hz}$.

**Exercise 14.3 ★.**

A glass obeys $n = 1.50 + 4.5 \times 10^{3}\,\mathrm{nm}^{2}/\lambda^2$. Index at $400\,\mathrm{nm}$, $550\,\mathrm{nm}$, $700\,\mathrm{nm}$; phase velocities; group index $n_g = n -
\lambda\,\dd n/\dd\lambda$ at $550\,\mathrm{nm}$; angular spread of a white beam refracted at $45{}^{\circ}$ incidence into the glass.

**Solution of Exercise 14.3.**

$n = 1.528$, $1.515$, $1.509$; $v = 1.963$, $1.980$, $1.987 \times 10^{8}\,\mathrm{m}/\mathrm{s}$; $n_g = n +
2B/\lambda^2 = 1.545$. Refraction angles $\arcsin(0.707/n)$: $27.57{}^{\circ}$ and $27.93{}^{\circ}$: a spread of $0.36{}^{\circ}$.

**Exercise 14.4 ★.**

A $5\,\mathrm{MHz}$ wave meets a layer with $n = 2 \times 10^{11}\,\mathrm{m}^{-3}$: is it reflected? Penetration depth; same at $2\,\mathrm{MHz}$; minimum frequency that crosses a layer of $n = 1.5 \times 10^{12}\,\mathrm{m}^{-3}$. Why do shortwave bands "open" and "close" with the hour and the solar cycle?

**Solution of Exercise 14.4.**

$f_p = 4.0\,\mathrm{MHz}$: $5\,\mathrm{MHz}$ passes; $2\,\mathrm{MHz}$ is reflected, penetrating $c/\sqrt{\omega_p^2 - \omega^2} = 14\,\mathrm{m}$. For $n = 1.5 \times 10^{12}\,\mathrm{m}^{-3}$: $11\,\mathrm{MHz}$. The density follows the Sun’s ionizing flux — hour, season, and the eleven-year cycle — so the highest usable frequency moves with them.

**Exercise 14.5 ★★.**

*[AC resistance](#ex-b2-waves-in-media-skinuses).* A copper wire of radius $a = 1.0\,\mathrm{mm}$. (a) DC resistance per metre. (b) At $1\,\mathrm{MHz}$: $\delta$; approximate resistance per metre with the current confined to a ring of thickness $\delta$; ratio to DC. (c) At $100\,\mathrm{MHz}$. (d) Why does a coil of $N$ turns of this wire have a quality factor that stops improving with $N$ at high frequency, and what is Litz wire?

**Solution of Exercise 14.5.**

(a) $1/\gamma\pi a^2 = 5.3\,\mathrm{m}\Omega/\mathrm{m}$. (b) $\delta = 65\,\text{µ}\mathrm{m}$; $R \approx 1/\gamma2\pi a\delta
= 41\,\mathrm{m}\Omega/\mathrm{m}$, $7.7$ times DC. (c) $\delta = 6.5\,\text{µ}\mathrm{m}$, $0.41\,\Omega/\mathrm{m}$, $77$ times. (d) $R \propto \sqrt f$ grows as fast as $L\omega$ once the skin dominates, and every turn adds resistance; Litz wire bundles many insulated strands thinner than $\delta$, so the whole copper conducts.

**Exercise 14.6 ★★.**

*Shielding.* A box of aluminium ($\gamma = 3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$) $1.0\,\mathrm{mm}$ thick. [Attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) of the field amplitude crossing the wall, in dB, at $50\,\mathrm{Hz}$, $10\,\mathrm{kHz}$, $1\,\mathrm{MHz}$ (the factor $\eu^{-d/\delta}$, reflection at the surfaces neglected). Why is the mains’ magnetic field shielded with iron (high permeability) rather than copper?

**Solution of Exercise 14.6.**

$\delta_{\text{Al}} = 11.9$, $0.84$, $0.084\,\mathrm{mm}$: $d/\delta = 0.08$, $1.2$, $12$: $0.7$, $10$, $103\,\mathrm{dB}$. At $50\,\mathrm{Hz}$ the conductor is transparent; iron of high $\mu_r$ diverts the flux (magnetic shielding) and also shrinks $\delta$ by $\sqrt{\mu_r}$.

**Exercise 14.7 ★★.**

*Energy in a [plasma](#def-b2-waves-in-media-plasma).* For the transverse wave of [Theorem 14.2](#thm-b2-waves-in-media-plasmadispersion): (a) electron velocity amplitude and the kinetic energy density $\tfrac12nm\langle v^2\rangle$. (b) Mean electric and magnetic energy densities. (c) Show that the total energy density is $\tfrac12\varepsilon_0E_0^2(1 + \dots)$ — compute the bracket — and that $\langle\Pi\rangle$ divided by it equals $v_g$. (d) What fraction of the energy is in the electrons at $\omega = 2\omega_p$?

**Solution of Exercise 14.7.**

(a) $v_0 = eE_0/m\omega$; $\tfrac12nm\langle v^2\rangle = ne^2E_0^2/4m\omega^2 = \tfrac14\varepsilon_0
E_0^2\,\omega_p^2/\omega^2$. (b) $\tfrac14\varepsilon_0E_0^2$ and $\tfrac14\varepsilon_0E_0^2(1 - \omega_p^2/
\omega^2)$. (c) The sum is $\tfrac12\varepsilon_0E_0^2$ (the bracket is $1$); $\langle\Pi
\rangle = E_0^2k/2\mu_0\omega$, and the ratio is $c^2k/\omega = v_g$. (d) $\tfrac14\cdot\tfrac14
/\tfrac12 = 1/8$.

**Exercise 14.8 ★★.**

*Pulsar dispersion.* Radio pulses from a pulsar cross a distance $L$ of interstellar [plasma](#def-b2-waves-in-media-plasma) ($n \ll$ anything, $\omega \gg \omega_p$). (a) Show that the group delay is $t \approx (L/c)(1 + \omega_p^2/2\omega^2)$. (b) The delay between the arrivals at $f_1$ and $f_2$ is $\Delta t = \dfrac{e^2}{8\pi^2
\varepsilon_0mc}\,nL\Bigl(\dfrac1{f_1^2} - \dfrac1{f_2^2}\Bigr)$: compute the constant. (c) A pulsar at $L = 1\,\mathrm{kpc} = 3.1 \times 10^{19}\,\mathrm{m}$ through $n = 3 \times 10^{4}\,\mathrm{m}^{-3}$: delay between $400\,\mathrm{MHz}$ and $1.4\,\mathrm{GHz}$. (d) Astronomers measure $\Delta t$ and know $n$: what do they get? (The dispersion measure, the standard distance gauge for pulsars.)

**Solution of Exercise 14.8.**

(a) $1/v_g = (1/c)(1 - \omega_p^2/\omega^2)^{-1/2} \approx (1 + \omega_p^2/2\omega^2)/c$. (b) $\Delta t
= (L\omega_p^2/2c)(1/\omega_1^2 - 1/\omega_2^2)$ with $\omega_p^2 = ne^2/\varepsilon_0m$ and $\omega =
2\pi f$: constant $e^2/8\pi^2\varepsilon_0mc = 1.34 \times 10^{-7}\,$ (SI). (c) $nL = 9.3 \times 10^{23}\,\mathrm{m}^{-2}$, $(1/f_1^2 - 1/f_2^2) = 5.7 \times 10^{-18}\,\mathrm{s}^{2}$: $\Delta t = 0.72\,\mathrm{s}$. (d) The column $nL$ (the dispersion measure), hence the distance for a model of $n$.

**Exercise 14.9 ★★.**

*A spectral line.* In a dilute gas ($\underline n \approx 1 + \underline\chi/2$) the [Lorentz model](#prop-b2-waves-in-media-lorentz) gives $n'' = \tfrac12\operatorname{Im}\underline\chi$. (a) Show that near $\omega_0$, $n'' \approx \dfrac{\omega_p^2}{4\omega_0}\,\dfrac{\Gamma/2}{(\omega_0 - \omega)^2 + \Gamma^2/4}$, a Lorentzian of full width $\Gamma$. (b) Absorption coefficient $\alpha = 2n''\omega/c$ at the centre. (c) Sodium vapour, $n = 1 \times 10^{17}\,\mathrm{m}^{-3}$, $\lambda_0 = 589\,\mathrm{nm}$, $\Gamma = 6 \times 10^{7}\,\mathrm{s}^{-1}$: $\alpha$ at the centre and the thickness that absorbs $90\%$ (ignore the Doppler broadening, which in fact widens the line a hundredfold). (d) Sketch $n' - 1$ across the line; where is the dispersion anomalous?

**Solution of Exercise 14.9.**

(a) $\operatorname{Im}\underline\chi = \omega_p^2\Gamma\omega/[(\omega_0^2 - \omega^2)^2 + \Gamma^2\omega^2]$ with $\omega_0^2 - \omega^2 \approx 2\omega_0(\omega_0 - \omega)$: $n'' = \tfrac12\operatorname{Im}\underline\chi$ is the stated Lorentzian. (b) $n''(\omega_0) = \omega_p^2/2\omega_0\Gamma$, $\alpha = 2n''\omega_0/c =
\omega_p^2/\Gamma c$. (c) $\omega_p^2 = 3.2 \times 10^{20}\,\mathrm{s}^{-2}$: $\alpha = 1.8 \times 10^{4}\,\mathrm{m}^{-1}$; $\ln10/\alpha = 0.13\,\mathrm{mm}$. (d) $n' - 1 \propto (\omega_0 - \omega)/[(\omega_0 - \omega)^2 +
\Gamma^2/4]$: positive below, negative above, decreasing within $\pm\Gamma/2$ of the line — anomalous there.

**Exercise 14.10 ★★★.**

*Joule heating in the skin.* A wave $E_0\eu^{\iu\omega t}$ (along $y$) enters a conductor filling $x > 0$. (a) Write $\underline E(x, t)$ and deduce $\underline B(x, t)$ (along $z$) from Faraday. (b) Mean [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) at the surface: show $\langle\Pi_x(0)\rangle = E_0^2/2\mu_0\omega\delta$. (c) Mean Joule power per unit area, $\int_0^\infty\tfrac12\gamma|\underline E|^2\dd x$: show it equals (b). (d) Express the result as $\tfrac12R_sH_0^2$ with $H_0 = B_0/\mu_0$ the surface magnetic field and $R_s = 1/\gamma\delta$ the *surface resistance*; value for copper at $1\,\mathrm{GHz}$, and the loss of a microwave cavity whose walls carry $H_0 = 100\,\mathrm{A}/\mathrm{m}$.

**Solution of Exercise 14.10.**

(a) $\underline E = E_0\eu^{-x/\delta}\eu^{\iu(\omega t - x/\delta)}$, $\underline B = \underline k\underline E/\omega
= (1 - \iu)\underline E/\omega\delta$. (b) $\langle\Pi_x\rangle = \tfrac12\operatorname{Re}(\underline E
\underline B^*)/\mu_0 = E_0^2/2\mu_0\omega\delta$ at $x = 0$. (c) $\int_0^\infty\tfrac12\gamma E_0^2
\eu^{-2x/\delta}\dd x = \gamma E_0^2\delta/4 = E_0^2/2\mu_0\omega\delta$ using $\gamma = 2/\mu_0\omega\delta^2$. (d) $H_0 = \sqrt2E_0/\mu_0\omega\delta$, so $\tfrac12R_sH_0^2 = E_0^2/\gamma\mu_0^2\omega^2\delta^3 =
E_0^2/2\mu_0\omega\delta$. Copper at $1\,\mathrm{GHz}$: $R_s = 1/\gamma\delta = 8\,\mathrm{m}\Omega$; $\tfrac12 \times 8 \times 10^{-3} \times 10^4 = 40\,\mathrm{W}/\mathrm{m}^{2}$.

**Exercise 14.11 ★★★.**

*Why metals shine.* Treat silver’s conduction electrons ($n =
5.9 \times 10^{28}\,\mathrm{m}^{-3}$) as a collisionless [plasma](#def-b2-waves-in-media-plasma) at optical frequencies. (a) $\omega_p$ and the corresponding wavelength. (b) At $500\,\mathrm{nm}$: $k$ is imaginary — penetration depth; the wave is totally reflected (no absorption in this model): silver is a mirror. (c) At $200\,\mathrm{nm}$? Hence the "ultraviolet transparency" of alkali metals. (d) Real silver reflects $95\%$, gold looks yellow: what does the model miss (think of collisions, and of bound electrons whose resonances lie in the visible for gold)?

**Solution of Exercise 14.11.**

(a) $\omega_p = 1.4 \times 10^{16}\,\mathrm{rad}/\mathrm{s}$, $\lambda_p = 2\pi c/\omega_p = 140\,\mathrm{nm}$. (b) $\omega = 3.8 \times 10^{15}\,\mathrm{rad}/\mathrm{s}
< \omega_p$: $1/\kappa = c/\sqrt{\omega_p^2 - \omega^2} = 23\,\mathrm{nm}$, total reflection. (c) Still below $\omega_p$: reflecting; transparency only beyond $140\,\mathrm{nm}$ (in sodium, $210\,\mathrm{nm}$). (d) Collisions give a real part to $\underline\gamma$ and a few percent of absorption; bound-electron resonances in the visible (gold, copper) absorb the blue and colour the metal.

**Exercise 14.12 ★★★.**

*Water in the microwave.* The orientation of water’s polar molecules gives a susceptibility $\underline\chi = \chi_s/(1 + \iu\omega\tau)$ (Debye relaxation), $\chi_s \approx 80$, $\tau = 8\,\mathrm{ps}$, on top of a constant $\chi_\infty \approx 4$. (a) Real and imaginary parts of $\underline\varepsilon_r = 1 + \chi_\infty
+ \underline\chi$ at $2.45\,\mathrm{GHz}$ and at $20\,\mathrm{GHz}$. (b) [Complex index](#prop-b2-waves-in-media-lorentz) and the absorption length $1/(2n''\omega/c)$ of the intensity at $2.45\,\mathrm{GHz}$: how deep does an oven cook? (c) Power absorbed per unit volume for a field amplitude $E_0$ in the water: $\tfrac12\omega\varepsilon_0\varepsilon''E_0^2$; value for $E_0 = 2\,\mathrm{kV}/\mathrm{m}$. (d) Why is $2.45\,\mathrm{GHz}$ used rather than the $20\,\mathrm{GHz}$ of maximum absorption?

**Solution of Exercise 14.12.**

(a) $\omega\tau = 0.12$: $\underline\varepsilon_r = 84 - 9.7\iu$; at $20\,\mathrm{GHz}$, $\omega\tau = 1$: $45 - 40\iu$. (b) $\underline n \approx 9.2 - 0.53\iu$; $c/2n''\omega = 1.8\,\mathrm{cm}$: the oven cooks the outer two centimetres, conduction does the rest. (c) $\tfrac12 \times
1.54 \times 10^{10} \times 8.85 \times 10^{-12} \times 9.7 \times 4 \times 10^6 = 2.6\,\mathrm{MW}/\mathrm{m}^{3}$. (d) At $20\,\mathrm{GHz}$ the absorption length would be a millimetre: only the skin would cook; $2.45\,\mathrm{GHz}$ penetrates, and is a free band.

![A shortwave transmitting station at dusk: its waves, a few megahertz below the plasma frequency of the ionosphere, are about to be reflected back to the ground a thousand kilometres away.](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/img-460d76a3f215.jpg)

*A shortwave transmitting station at dusk: its waves, a few megahertz below the [plasma frequency](#def-b2-waves-in-media-plasma) of the ionosphere, are about to be reflected back to the ground a thousand kilometres away.*

![The aurora australis photographed from the International Space Station: particles from the Sun, guided by the Earth’s magnetic field, excite the plasma of the upper atmosphere — the ionosphere whose plasma frequency reflects radio waves (NASA).](https://one-course.com/images/onecourse/chapters/physics-4/b2-waves-in-media/img-cef765dafbe1.jpg)

*The aurora australis photographed from the International Space Station: particles from the Sun, guided by the Earth’s magnetic field, excite the [plasma](#def-b2-waves-in-media-plasma) of the upper atmosphere — the ionosphere whose [plasma frequency](#def-b2-waves-in-media-plasma) reflects radio waves (NASA).*

## 14.5 Problem: The ionosphere, the oven door and the prism

**Problem 14.1.**

Weekend problem — three materials seen by a wave: the plasma that bends GPS signals, the metal that keeps microwaves in, and the glass that spreads white light

**Part I — The ionosphere and GPS.** Electron density in the F layer $n = 1 \times 10^{12}\,\mathrm{m}^{-3}$; the total column of electrons along a vertical path is $N_T = \int n\,\dd z = 2 \times 10^{17}\,\mathrm{m}^{-2}$ by day. GPS: $f_1 = 1575\,\mathrm{MHz}$, $f_2 = 1228\,\mathrm{MHz}$.

1. [Plasma frequency](#def-b2-waves-in-media-plasma) of the F layer; does GPS pass? Does a $5\,\mathrm{MHz}$ shortwave?
2. Show that for $\omega \gg \omega_p$ the [group velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#prop-b2-dispersion-wave-packets-group) is $v_g \approx c(1 -  \omega_p^2/2\omega^2)$ and the [phase velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) $c(1 + \omega_p^2/2\omega^2)$ .
3. Extra group delay of the GPS signal crossing the layer: show $\Delta t = \dfrac{e^2}{8\pi^2\varepsilon_0mc}\,\dfrac{N_T}{f^2}$ and compute it for $f_1$ ; the range error $c\,\Delta t$ .
4. The receiver uses both frequencies: from the two delays it removes the error. Express $N_T$ in terms of $\Delta t_1 - \Delta t_2$ ; what precision on the delay difference gives $N_T$ to $1\%$ ?
5. How many extra carrier cycles does the delay of question 3 represent at $f_1$ ?
6. The phase of the carrier is *advanced* by the same amount as the group is delayed: why (signs of the two corrections)? Which one matters for a receiver that counts carrier cycles?
7. At night $N_T$ falls tenfold: error for a single-frequency receiver by day and by night.
8. A solar flare raises $n$ in the D layer ( $80\,\mathrm{km}$ ) to $1 \times 10^{10}\,\mathrm{m}^{-3}$ with many collisions: which radio services are hit, and why does GPS only slightly degrade?

**Part II — The oven door.** The cavity walls and the door mesh are steel ($\gamma = 1.0 \times 10^{6}\,\mathrm{S}/\mathrm{m}$); the mesh holes are $1.0\,\mathrm{mm}$ wide in a sheet $0.5\,\mathrm{mm}$ thick; $f = 2.45\,\mathrm{GHz}$, $P = 800\,\mathrm{W}$.

9. [Skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) in the steel at $2.45\,\mathrm{GHz}$ ; is the wall "thick"?
10. Surface resistance $R_s = 1/\gamma\delta$ ; with a surface magnetic field of amplitude $H_0 = 30\,\mathrm{A}/\mathrm{m}$ on the walls ( $1\,\mathrm{m}^{2}$ ), power lost in the walls, $\tfrac12R_sH_0^2$ per unit area; fraction of $P$ .
11. In a hole of width $a \ll \lambda$ the field cannot propagate (the hole is a waveguide below cut-off, [Chapter 16](https://one-course.com/books/physics/4/en/chapter/16-guided-waves-and-cavities#ch-b2-guided-waves) ): it decays as $\eu^{-\pi z/a}$ across the thickness $z$ . [Attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) of the amplitude across the $0.5\,\mathrm{mm}$ sheet, in dB.
12. Why can you see through the mesh (wavelength of light against the hole size) while the microwaves cannot get out?
13. The glass window is a dielectric with $\underline\varepsilon_r = 4 - 0.02\iu$ : index, and the intensity [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) length in it; is the glass heated?
14. A metal fork in the oven: the field at its tips is enhanced; with the [skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) of question 8, estimate the [current density](https://one-course.com/books/physics/4/en/chapter/10-charges-currents-and-conduction#def-b2-charges-currents-conduction-densities) at the surface for $H_0 = 300\,\mathrm{A}/\mathrm{m}$ (use $j \approx H_0/\delta$ at the surface) and the local heating $j^2/\gamma$ — why sparks?

**Part III — The wire at radio frequency.** A copper wire of radius $a = 0.5\,\mathrm{mm}$ ($\gamma = 6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$) in a $13.56\,\mathrm{MHz}$ induction heater coil carries $20\,\mathrm{A}$ rms.

15. [Skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) ; effective conducting section; [AC resistance](#ex-b2-waves-in-media-skinuses) per metre against DC.
16. Power dissipated per metre of coil wire; temperature the wire would reach with convective cooling $h = 20\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K})$ .
17. The coil induces currents in a steel workpiece ( $\gamma =  1 \times 10^{6}\,\mathrm{S}/\mathrm{m}$ , $\mu_r \approx 1$ when hot): [skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) there; why does induction heating heat only a surface layer, and how is that used for hardening gears?
18. To reduce the coil’s loss the wire is replaced by a $3\,\mathrm{mm}$ copper tube with water inside: new resistance per metre; does the inside of the tube carry current?
19. At what frequency would the [skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) in copper equal the wire radius? Below it, what approximation replaces the "thick-wire" formula?

**Part IV — The prism.** A glass has $n(\lambda) = 1.500 + 5.0 \times 10^{3}\,\mathrm{nm}^{2}/\lambda^2$, with its resonance at $\lambda_0 = 100\,\mathrm{nm}$.

20. Check that the Cauchy form follows from the [Lorentz model](#prop-b2-waves-in-media-lorentz) far below resonance, and that $\omega_p^2/\omega_0^2 = n^2(\infty) - 1$ : value of $\omega_p^2/\omega_0^2$ and of the effective $n$ for this glass.
21. Index and [phase velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) at $400\,\mathrm{nm}$ and $700\,\mathrm{nm}$ ; group index at $550\,\mathrm{nm}$ ; delay between the two colours over $1\,\mathrm{km}$ of such glass in a fibre.
22. A $60{}^{\circ}$ prism at minimum deviation for $550\,\mathrm{nm}$ : deviation (use $n\sin(A/2) = \sin((A + D)/2)$ ); angular distance between $400\,\mathrm{nm}$ and $700\,\mathrm{nm}$ (differentiate); on a screen $2\,\mathrm{m}$ away, the width of the spectrum.
23. Why is the glass opaque at $200\,\mathrm{nm}$ , and what is its colour if a trace of iron adds a weak resonance at $700\,\mathrm{nm}$ ?
24. In the Lorentz picture, why is blue deviated more than red by the prism?
25. In one table give, for the F layer, the steel, the copper and the glass at their working frequencies, the nature of $\underline\gamma$ and the fate of the wave.

**Solution of Problem 14.1.**

**1.** $f_p = 9\,\mathrm{MHz}$: GPS passes, the shortwave is reflected.

**2.** $k = (\omega/c)\sqrt{1 - \omega_p^2/\omega^2}$: $v_\varphi = \omega/k \approx c(1 + \omega_p^2/
2\omega^2)$, $v_g = \dd\omega/\dd k = c^2k/\omega \approx c(1 - \omega_p^2/2\omega^2)$.

**3.** $\Delta t = \int(1/v_g - 1/c)\dd z = \int\omega_p^2\dd z/2c\omega^2 = e^2N_T/2\varepsilon_0
mc\omega^2 = (e^2/8\pi^2\varepsilon_0mc)N_T/f^2 = 1.34 \times 10^{-7} \times 2 \times 10^{17}/2.48 \times
10^{18} = 11\,\mathrm{ns}$: $3.2\,\mathrm{m}$.

**4.** $N_T = (\Delta t_1 - \Delta t_2)/K(1/f_1^2 - 1/f_2^2)$; the difference is $7\,\mathrm{ns}$: $1\%$ needs $0.07\,\mathrm{ns}$.

**5.** $v_\varphi$ exceeds $c$ by as much as $v_g$ falls short: the carrier phase arrives early while the modulation arrives late. A carrier-phase receiver applies the correction with the opposite sign.

**6.** $3.2\,\mathrm{m}$ by day, $0.3\,\mathrm{m}$ by night.

**7.** $f_p = 0.9\,\mathrm{MHz}$ with collisions: medium and short waves are absorbed (a radio blackout); GPS at $1.5\,\mathrm{GHz}$ feels an absorption $\propto 1/\omega^2$, negligible.

**8.** $f_1\Delta t = 1.575 \times 10^9 \times 1.1 \times 10^{-8} = 17$ cycles.

**9.** $\delta = \sqrt{2/\mu_0\gamma\omega} = 10\,\text{µ}\mathrm{m}$: a millimetre wall is a hundred [skin depths](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex).

**10.** $R_s = 1/\gamma\delta = 0.1\,\Omega$; $\tfrac12R_sH_0^2 = 45\,\mathrm{W}/\mathrm{m}^{2}$: $45\,\mathrm{W}$, $6\%$.

**11.** $\eu^{-\pi \times 0.5} = 0.21$: $-14\,\mathrm{dB}$ in amplitude, $-27\,\mathrm{dB}$ in power (and the tiny hole couples little to begin with).

**12.** Light, $\lambda = 0.5\,\text{µ}\mathrm{m}$, is two thousand times smaller than the holes and passes; the microwave, $\lambda = 12\,\mathrm{cm}$, is a hundred times larger.

**13.** $\underline n = 2 - 0.005\iu$; $c/2n''\omega = 1.9\,\mathrm{m}$: the glass is hardly warmed.

**14.** $j \approx H_0/\delta = 3 \times 10^{7}\,\mathrm{A}/\mathrm{m}^{2}$, $j^2/\gamma = 9 \times 10^{8}\,\mathrm{W}/\mathrm{m}^{3}$ in a ten-micrometre layer: tips heat in milliseconds, emit electrons, ionize the air — sparks.

**15.** $\delta = 18\,\text{µ}\mathrm{m}$; section $2\pi a\delta = 5.6 \times 10^{-8}\,\mathrm{m}^{2}$ against $\pi a^2 = 7.9 \times 10^{-7}\,\mathrm{m}^{2}$: $R_{\text{ac}} = 0.30\,\Omega/\mathrm{m}$, fourteen times DC.

**16.** $RI^2 = 120\,\mathrm{W}/\mathrm{m}$; surface $\pi d = 3.1 \times 10^{-3}\,\mathrm{m}^{2}$ per metre: $\Delta T \approx 1900\,\mathrm{K}$ — it would melt; such coils are water-cooled tubes.

**17.** $\delta = 0.14\,\mathrm{mm}$: the induced current, and the heat, stay in a tenth of a millimetre; a short pulse followed by a quench hardens the surface of a gear while the core stays tough.

**18.** $R = 1/\gamma2\pi a\delta = 0.10\,\Omega/\mathrm{m}$, three times less; the inside carries nothing — hence a tube, with the coolant where the copper would be wasted.

**19.** $\delta = a$ at $f = 1/\pi\mu_0\gamma a^2 = 17\,\mathrm{Hz}$; below, the current is uniform and the DC formula holds.

**20.** $n^2 \approx 1 + \omega_p^2/\omega_0^2 + (\omega_p^2/\omega_0^2)(\lambda_0/\lambda)^2$: Cauchy’s form with $n(\infty)^2 - 1 = \omega_p^2/\omega_0^2 = 1.25$ for $n(\infty) = 1.5$.

**21.** $n = 1.531$ and $1.510$, $v = 1.96$ and $1.99 \times 10^{8}\,\mathrm{m}/\mathrm{s}$; $n_g(550) =
1.550$; $n_g(400) - n_g(700) = 0.063$: $210\,\mathrm{ns}$ over a kilometre.

**22.** $\sin((A + D)/2) = 1.5165 \times 0.5$: $D = 38.6{}^{\circ}$; $\dd D = 2\sin
(A/2)\dd n/\cos((A + D)/2) = 1.53\,\dd n$: $\Delta D = 1.85{}^{\circ}$, $6.5\,\mathrm{cm}$ at $2\,\mathrm{m}$.

**23.** At $200\,\mathrm{nm}$ the frequency approaches the resonance and $n''$ grows: opaque. A weak resonance at $700\,\mathrm{nm}$ absorbs the red: green glass.

**24.** Blue is closer to the ultraviolet resonance: the bound electrons respond more, $n$ is larger, and the deviation too.

**25.** F layer at $1.5\,\mathrm{GHz}$: imaginary $\underline\gamma$, propagation with a small group delay. Steel at $2.45\,\mathrm{GHz}$: real $\underline\gamma$, [skin effect](#thm-b2-waves-in-media-skin), reflection. Copper at $13.56\,\mathrm{MHz}$: the same, a skin of $18\,\text{µ}\mathrm{m}$. Glass in the visible: $\iu\omega\varepsilon_0\chi$, a real index, a transparent [dispersive medium](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation).