Chapter 29, Plane Electromagnetic Waves Ii Video Solutions, Electromagnetic Fields and Waves: Including Electric Circuits

Problem 1

Good conductors
Show that for a good conductor
(a) $\delta / x_{0} \ll 1$,
(b) $\delta^{2} / \mathscr{D}=2 \epsilon /\left(\mu \sigma^{2}\right)$,
(c) $\delta^{2} \mathscr{D}=2 \boldsymbol{K}_{0}^{2} /(\epsilon, \mu,$, .

Narayan Hari

Numerade Educator

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Problem 2

Draw curves for $E$, similar to those of Fig. 29-1, for $\omega t=0$ to $2 \pi$ at intervals of $\pi / 4$.

Victor Salazar

Numerade Educator

04:33

Problem 3

You are asked to design copper bus bars that can carry 5000 amperes at 60 hertz over a distance of 5 meters. The total length of bus is 10 meters. The power dissipation in the line should not exceed 1 kilowatt. Suggest a plausible cross section.

Aja S

Numerade Educator

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Problem 4

(29.1) Heat propagation
It is interesting to draw a parallel between the flow of heat in a thermally conducting medium and the propagation of an electric or magnetic field in an electrically conducting medium.

Let $\Phi$ be the heat flux density in watts per square meter and $\Phi=-\lambda \boldsymbol{V} T$, where $\lambda$ is the thermal conductivity in watts per meter-kelvin and $T$ is the temperature in kelvins. Then, for conservation of energy,
$$
\boldsymbol{\nabla} \cdot \Phi=-\rho c \frac{\partial T}{\partial t}+Q
$$
where $\rho$ is the mass density in kilograms/meter $^{3}, c$ is the specific heat in joules/kilogram-kelvin, and $Q$ is the heat produced within the medium in watts/meter $^{3} .$
For $Q=0$,
$$
\nabla^{2} T-\frac{\rho c}{\lambda} \frac{\partial T}{\partial t}=0
$$
This equation is identical in form to that for an electromagnetic wave in a good conductor, with $\rho c / \lambda$ corresponding to $\mu \sigma$. Its solution for heat flow in one dimension is similar:
$$
\begin{array}{ccc}
\hline \text { PROPERTY } & \text { COPPER } & \text { IRON } \\
\hline \sigma & 5.8 \times 10^{7} & 1.0 \times 10^{7} \\
c & 385 & 460 \\
\rho & 8.9 \times 10^{3} & 7.9 \times 10^{3} \\
\lambda & 41.8 & 6.27 \\
\mu, & 1 & 100 \\
\hline
\end{array}
$$
$$
T=T_{m} \exp \left[j\left(\omega t-\frac{z}{\delta_{\mathrm{th}}}\right)-\frac{z}{\delta_{\mathrm{th}}}\right], \quad \delta_{\mathrm{th}}=\left(\frac{2 \lambda}{\omega \rho c}\right)^{1 / 2}
$$
Compare the velocities of propagation of $T$ and of $\boldsymbol{B}$ in copper and in iron. See Table 29-2.

Lainey Roebuck

Numerade Educator

06:36

Problem 5

(29.1) The surface impedance of a conductor
By definition, the surface impedance of a conductor is the ratio $E_{t} / H_{t}$ at the surface, the subscript $t$ indicating a tangential component. It is shown in Prob. $19-4$ that $H_{t}$ is numerically equal to the current per unit width in the conductor.
(a) Show that the surface impedance of a good conductor is
$$
(1+j)\left(\frac{\omega \mu}{2 \sigma}\right)^{1 / 2} \text { or } \quad \frac{1+j}{\sigma \delta}
$$
where $\sigma$ is the conductivity and $\delta$ is the skin depth. The quantity $1 / \sigma \delta$ is the surface resistance. The surface impedance and the surface resistance are expressed in ohms/square. See Prob. 4-9. For example, the surface resistance of copper at 3 gigahertz is $14.4$ miliohms/square.
(b) Show that, if the tangential magnetic field is $H_{t}$, then the power dissipated per square meter in the conductor is given by $H_{t, \text { rmss }}^{2} /(\sigma \delta) .$ This means that the power dissipated is the same as if the surface current (of density numerically equal to $H_{t}$ ) were distributed uniformly over a thickness $\delta$ of the conductor.

Ameer Said

Numerade Educator

02:36

Problem 6

Induction heating consists in exciting eddy currents in a conductor by exposing it to an alternating magnetic field. The method serves for melting, in an induction furnace, for heating before a forging operation, or for hardening. An induction furnace comprises a crucible surrounded by a coil. The largest furnaces have capacities of tens of tons and powers up to a few megawatts. Once the load has melted, magnetic forces within the liquid provide stirring.

The coil is usually a single layer of water-cooled copper pipe that surrounds the object to be heated. If the object is ferromagnetic, a small part of the heating comes from hysteresis losses.

Induction heating has the advantage of convenience and of not contaminating the metal with combustion gases. Also, by choosing the frequency correctly, it is possible to apply a heat treatment down to a known depth. For example, plowshares require a hard, heat-treated skin that resists abrasion and a soft core that resists breakage.

Metals can also be heat treated with a laser beam, but usually to a depth of only a fraction of a millimeter

Here is a simple example of induction heating. A steel rod of circular cross section lies inside a solenoid that applies an axial and tangential magnetic field $H_{t}$. We showed in Prob. $19-4$ that the net surface current density is equal to $H_{t}$. A wave penetrates normally to the conductor, and the power dissipation in the conductor is the same as if the current were distributed uniformly over a thickness equal to the skin depth. See the preceding problem.

The solenoid has 100 turns per meter and carries a current of 600 amperes rms at 100 kilohertz.

Calculate the skin depth and the power $P^{\prime}$ dissipated in the iron per square meter. Set $\sigma=10^{7}$ and $\mu_{r}=100$. Neglect end effects, and neglect the fact that the relative permeability decreases to unity when the steel becomes red-hot.

Morgan Cheatham

Numerade Educator

02:50

Problem 7

Show that, in a plasma, $v B \ll E$, or $E / B \gg v$, where $v$ is the velocity of an electron.

Suzanne W.

Numerade Educator

02:03

Problem 8

(a) Find the value of $\mathscr{D}$, as defined in Sec. 29.1, for a low-density plasma.
(b) Show that the value of $k^{2}$ that we found for a conductor in $\mathrm{Sec}$, 28.2.3 agrees with that of Sec. 29.2.5.

The values of $\alpha$ and of $\beta$ that we found in Sec. $28.5 .1$ are not valid, however, for imaginary values of $\mathscr{D}$.

Chai Santi

Numerade Educator

01:29

Problem 9

Two uniform plane electromagnetic waves of equal amplitudes propagate in the ionosphere where the free electron density is $N$ per cubic meter. One wave has a circular frequency $\omega_{1}$ and a corresponding wavelength $\lambda_{1} ;$ the other has a slightly different circular frequency $\omega_{2}$ and a wavelength $\lambda_{2}$.
(a) At a given time $t$ there exist values of $z$ for which the two waves are in phase and other values of $z$ for which they are opposite in phase. What is the distance between the maxima?
(b) What is their velocity? This is the group velocity $v_{g}$.
(c) Show that, in the limit, $v_{g}=1 /(d k / d \omega)$.
(d) Calculate the phase velocities and the group velocity for $f_{1}=5.3$ megahertz, $f_{2}=5.4$ megahertz, and $N=5 \times 10^{10}$ electrons/meter $^{3}$.
(e) Calculate the distance and the number of waves between two minima.

Dominador Tan

Numerade Educator

01:01

Problem 10

The $\omega-\beta$ diagram is a curve of $\omega$ as a function of $\beta$. The ratio $\omega / \beta$ is equal to the phase velocity, while the slope $d \omega / d \beta$ is equal to the group velocity.

Ajay Singhal

Numerade Educator

10:26

Problem 11

Pulsars are stars that have suffered gravitational collapse, or neutron stars, and that rotate rapidly while emitting a narrow beam of radiation. The pulse lengths, at the earth, are of the order of 1 millisecond, and the periods of the order of 1 second. Neutron stars consist mostly of neutrons with some electrons and some ions. Their masses are of the order of that of the sun, but their radii are only of the order of 10 kilometers.

Within a few months after the discovery of pulsars, distance estimates were obtained in the following manner. It was observed that the arrival time of a pulse depends on the frequency of observation, the arrival time being later at lower frequencies. This delay is attributed to dispersion in the interstellar medium, which is ionized hydrogen with an electron density $N$ of about $10^{5} /$ meter $^{3}$.
(a) Show that, if $\omega^{2} \gg \omega_{p}^{2}$, a plot of the time delay $\Delta t$ as a function of $1 /\left(f^{2}\right)-1 /\left[(f+\Delta f)^{2}\right]$ is a straight line whose slope is a measure of the distance to the pulsar.
(b) In the case of pulsar CP 0328, arrival times measured at 151,408 , and 610 megahertz gave the following results: between 610 and 408 megahertz, the delay was $0.367$ second; between 408 and 151 megahertz, the delay was $4.18$ seconds.

Find the distance to CP 0328 in parsecs where 1 parsec is $3.086 \times 10^{16}$ meters. It is the distance from which the radius of the earth's orbit, $1.495 \times 10^{11}$ meters, subtends an angle of $1^{\prime}$.

The fact that such plots give straight lines passing through the origin indicates that the assumption that $\omega^{2} \gg \omega_{p}^{2}$ is correct. The delay therefore occurs over large distances in interstellar space, and not inside the pulsar itself.

Similar methods are used to reduce satellite ranging errors due to the ionosphere.

Averell Hause

Carnegie Mellon University

02:55

Problem 12

The energy densities, the Poynting vector, and the group velocity in a plasma

An electromagnetic wave travels in a low-density plasma. The electric field strength is $E_{m} \cos \omega t$.
(a) Calculate the sum of the electric, magnetic, and kinetic energy densities.
(b) Calculate the Poynting vector.
(c) Calculate the group velocity.
(d) Find the relations between these quantities.

Keshav Singh

Numerade Educator