Chapter 32, Plane Electromagnetic Waves V Video Solutions, Electromagnetic Fields and Waves: Including Electric Circuits

Problem 1

(32.1) Reflection from a good conductor
Draw two figures similar to those of Fig. 30-4, showing $E$ and $H$ for an electromagnetic wave incident on a good conductor. You will, of course, have to exaggerate the values of $E_{m}$ and of $\lambda$ in the conductor. Be sure to show the phases correctly. Show $x=, y-, z$-axes on both figures to relate one with the other.

Ameer Said

Numerade Educator

02:57

Problem 2

$(32,1)$ Reflection from a good conductor
Show that, for a good nonmagnetic conductor in air,
(a) $\left|\frac{E_{\text {Rm }}}{E_{l m}}\right|_{\perp}=1-\frac{\delta}{\lambda_{0}} \cos \theta_{f}$
(b) $\left|\frac{E_{\text {Rm }}}{E_{I m}}\right|_{\|}=1-\frac{\delta}{\lambda_{0} \cos \theta_{l}}$
This latter relation is not valid at grazing incidence, where $\cos \theta_{1}$ tends to zero.

A good conductor is a better reflector when $\boldsymbol{E}$ is normal to the plane of incidence. High-quality metallic reflectors have coefficients of reflection of about $90 \%$ near normal incidence in the visible, with unpolarized light.

Mirza Aslam Beig

Numerade Educator

11:45

Problem 3

(32.1) $\left|E_{R m} / E_{I m}\right|$ as a function of the angle of incidence for reflection on a conductor
For a good conductor, $\sigma / \omega \epsilon \geq 50$. Then
$$
\frac{\lambda_{0}}{\delta}=\frac{c}{\omega}\left(\frac{\omega \sigma \mu}{2}\right)^{1 / 2}=c\left(\frac{\sigma \mu \epsilon}{2 \omega \epsilon}\right)^{1 / 2} \geq c(25 \epsilon \mu)^{1 / 2}=5\left(\epsilon, \mu_{r}\right)^{1 / 2}
$$
So $\lambda_{0} / \delta \geq 10$ if $\epsilon_{r}=4$ and $\mu_{r}=1$.
Plot $\left|E_{R m} / E_{l m}\right|_{\perp}$ and $\left|E_{R m} / E_{l m}\right|_{||}$as functions of $\theta_{I}$ for a nonmagnetic good conductor in air and for $\lambda_{0} / \delta=10$. You will find that, when $E$ is in the plane of incidence, there exists a pseudo-Brewster angle for which the amplitude of the reflected wave is minimum.

Prachita Kush

Numerade Educator

03:16

Problem 4

(32.1) Liquid-crystal displays (LCDs)
In liquid-crystal displays the liquid is sandwiched between a transparent multiple electrode in front and a single black electrode in the back. Upon application of a voltage to a portion of the front window, the rodlike molecules of the nematic fluid in that region stand perpendicular to the window, and one can see the black electrode in the back. Elsewhere, the molecules reflect light because their orientations are haphazard.
The transparent multiple electrode is a thin coating either of a semiconducting metal oxide, such as tin oxide, or of gold. The surface resistance (Prob. 4-9) is of the order of 10 to 100 ohms per square. At 600 nanometers, and for gold, $\beta \lambda_{0}=1.29$ and $\alpha \lambda_{0}=2.59 .$ The conductivity of gold in the form of a thin film is $4.26 \times 10^{7}$ siemens/meter.
(a) Calculate the skin depth $\delta$.
(b) By what factor does the amplitude decrease in the gold film if its thickness $s$ is $0.05 \delta ?$
(c) What is the surface resistance?
(d) Calculate the thickness of the film in wavelengths $\lambda_{0}$.
A proper calculation of the transmission would take into account multiple reflections. The effect of multiple reflections is, however, much less than in Prob. $30-10$ because of the attenuation in the film.

Keshav Singh

Numerade Educator

01:05

Problem 5

(32.1) The surface impedance of a conductor
By definition, the surface impedance of a conductor is equal to the ratio of the tangential components of $\boldsymbol{E}$ and $\boldsymbol{H}$ at the surface, or to $E_{1} / H_{t}$.
(a) Show that the surface impedance of a good conductor is given by
$$
Z_{x}=\left(\frac{\omega \mu}{2 \sigma}\right)^{1 / 2}(1+j)=\frac{1+j}{\sigma \delta}
$$
The quantity $1 / \sigma \delta$ corresponds to the surface resistance of Prob. 4-9. For copper, $1 / \sigma \delta$ is equal to $0.261$ miliohm per square at 1 megahertz, from Table 29-1.
(b) Show that the power dissipated per square meter in the conductor is $H_{t, \mathrm{~ms} s}^{2} / \sigma \delta .$

Now we saw in Prob. $19-4$ that $H_{t}$ is equal to the current per unit width in the conductor. It follows that the power dissipated in the conductor is the same as if the current were uniformly distributed throughout the thickness $\delta$

Dominador Tan

Numerade Educator

01:04

Problem 6

(32.1) Cutting steel plate with a laser beam
Figure $32-6$ shows a laser beam cutting a steel plate,
(a) Why does the beam cut at a faster rate when the $E$ vector lies in the plane of the paper than when it is perpendicular?
(b) Roughly what percentage of the beam power serves to heat the steel in the former case?
(c) Can you explain why the kerf is narrower and more even when $E$ is in the plane of the paper?

If the required kerf is not straight, then the laser should rotate to keep the $\boldsymbol{E}$ vector of the beam parallel to the path. A simpler solution is to use a circularly polarized beam.

Kratika Bhadauria

Numerade Educator

06:54

Problem 7

(32.1) The standing wave at normal incidence on a good conductor Show that the electromagnetic energy density in a plane standing wave at normal incidence on a good conductor is uniform.

Mohit Khurana

Texas A&M University

01:22

Problem 8

(32.1) Multiple reflections in a dielectric plate backed by a conductor
An electromagnetic wave falls at an angle of $\theta_{l}$ on a slab of dielectric tha is backed by a good conductor.
Under what condition is there a single reflected wave?

Varsha Aggarwal

Numerade Educator

03:12

Problem 9

(32.2) The radiation force on a sphere
Calculate the radiation force on a reflecting sphere of radius $R$ in terms of the Poynting vector of the incident radiation.

Averell Hause

Carnegie Mellon University

03:04

Problem 10

(32.2) The radiation force on a cylinder
Calculate the radiation force per unit length on a cylinder of radius $R$ whose axis is perpendicular to the Poynting vector of the incoming radiation.
These monitors have a short response time, of the order of 1 nanosecond. They are made in various sizes, with crystals of the order of 1 centimeter in diameter and a few centimeters long. The crystal absorbs about one-quarter of the pulse energy. The peak power density can be as high as 20 megawatts/centimeter $^{2}$.
Find the ratio $V / \mathscr{F}_{m}$. Set $\mathscr{F}=\mathscr{Y}_{m} \exp (-a x)$ inside the crystal.

Averell Hause

Carnegie Mellon University

03:04

Problem 10

(32.2) The radiation force on a cylinder
Calculate the radiation force per unit length on a cylinder of radius $R$ whose axis is perpendicular to the Poynting vector of the incoming radiation.

Averell Hause

Carnegie Mellon University

07:38

Problem 11

(32.2) Radiation pressure and comet tails
(a) Compare the gravitational and radiation forces exerted by the sun on a spherical particle of radius $a$ whose density is 5000 kilograms/meter $^{3}$. The sun radiates $3.8 \times 10^{26}$ watts. See the Table of Physical Constants at the end of the book. Assume that the particle is black.
(b) Calculate the value of $a$ for which the two forces are equal. You should find that particles smaller than about $0.1$ micrometer in radius are repelled at any distance from the sun. This explains why comet tails that consist of fine particles point away from the sun. Such comets are said to be Type 2. We have disregarded diffraction, which is important when $a \gg \lambda$.

The tails of Type 1 comets are gaseous. They also point away from the sun, but for a different reason. The force then arises from an interaction between this gas and the solar wind (Prob, 28-12), which consists of ionized hydrogen that evaporates from the sun.

Carlos Henrique De Lima

Numerade Educator

02:34

Problem 11

(32.2) Photon-drag radiation monitor
Figure $32-7$ shows a schematic diagram of a photon-drag radiation monitor. These devices are used to monitor the intensity of powerful laser pulses. The beam enters on the left, through an antireflection coating (Prob. 30-14) and a transparent electrode (Prob. 32-4). The body of the monitor is a single crystal of semiconductor that is quite transparent at the wavelength used. The beam exits on the right where there is, again, an antireflection coating and a transparent electrode, in that order.

Kowshik Dey

Numerade Educator

07:38

Problem 12

The tails of Type 1 comets are gaseous. They also point away from the sun, but for a different reason. The force then arises from an interaction between this gas and the solar wind (Prob. $28-12)$, which consists of ionized hydrogen that evaporates from the sun.

Carlos Henrique De Lima

Numerade Educator

06:35

Problem 12

Radiation pressure in the semiconductor propels the charge carriers to the right. If the carriers are electrons, the electrodes become charged, as in the figure, and the voltage $V$ is a measure of the beam power.
These monitors have a short response time, of the order of 1 nanosecond. They are made in various sizes, with crystals of the order. of 1 centimeter in diameter and a few centimeters long. The crystal absorbs about one-quarter of the pulse energy. The peak power density. can be as high as 20 megawatts/centimeter $^{2}$.
Find the ratio $V / \mathscr{S}_{m}$. Set $\mathscr{Y}=\mathscr{S}_{m} \exp (-a x)$ inside the crystal.

Keshav Singh

Numerade Educator

01:38

Problem 13

$(32.2 .2)$ Radiation pressure with $\boldsymbol{E}$ in the plane of incidence
Show that the radiation pressure on a nonmagnetic good conductor, when $\boldsymbol{E}$ lies in the plane of incidence, is the same as in Sec. 32.2.1. In this instance there is both a magnetic force within the conductor and an electric force on the surface charges. Use Gauss's law to find $\sigma_{f}$.

Suhas Katkar

Numerade Educator

05:10

Problem 14

(32.2.3) The levitation of transparent particles in a laser beam
Figure $32-8$ shows a simplified diagram of a device for levitating transparent particles in a laser beam. The particles can range from 1 to 100 micrometers in diameter. The light intensity is maximum on the axis of the beam and tapers off on either side. The particle stays on the axis of the beam at a fixed height.
(a) Show qualitatively that, if the particle strays away from the axis, it suffers a restoring force. The axis is therefore a position of equilibrium. You can show this by sketching the paths of two rays that enter the
particle from below, one to the left of the particles' center and one to the right. Refraction deflects the rays and hence changes their momenta.
(b) Show that for a given beam intensity the vertical position of the beam is also stable. A reflecting particle is ejected laterally.

Kathleen Tatem

Numerade Educator

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Problem 15

(32.3) The angular momentum of an electrically charged permanent magnet
The field of an electrically charged permanent magnet possesses an angular momentum because $\boldsymbol{E}$ is radial, while $\boldsymbol{H}$ points approximately in the $\hat{\theta}$ direction, so that $\boldsymbol{E} \times \boldsymbol{H}$ is azimuthal.
We first calculate the value of the momentum from the known values of $\boldsymbol{E}$ and $\boldsymbol{H}$, and we then show that its existence follows from the law of conservation of momentum.
Imagine a conducting sphere of radius $R$ whose magnetization $M$ is uniform. You may take for granted that outside the sphere the magnetic field is the same as that of a small magnetic dipole of moment ${ }^{4} \pi R^{3} M$ situated at the center. The sphere carries a charge $Q$.
(a) Find the angular momentum of the field.
(b) Calculate the value of the angular momentum $L$ for $R=$ 20 millimeters and $M=10^{6}$ amperes/meter when the sphere is charged to a potential of 1000 volts. Could the sphere be useful as a gyroscope?
(c) Now let us start with an uncharged sphere and gradually deposit charge on it by means of an axial ion beam. Charge flows in at the north pole and distributes itself uniformly over the surface of the sphere. The magnetic field exerts a torque $T_{\operatorname{mog}}$ on the charging current. To prevent the sphere from turning, the support exerts an opposing mechanical torque $T_{\text {mect }}$ such that
$$
T_{\text {mech }}=-T_{\operatorname{mag}}=\frac{d L}{d t}
$$
Choose polar coordinates with the north pole at $\theta=0$. Show that the downward surface current density at $\theta$ is
$$
\alpha=\frac{1+\cos \theta}{4 \pi R \sin \theta} \frac{d Q}{d t}
$$
(d) Now show that $T_{\text {mas }}=-d L / d t$, as above.
We have calculated the torque exerted by the magnet on the current. There is no torque exerted by the current on the magnet for the following reason. The comnonent Jacob Jason Quintero in "NUMERADE ...

Lainey Roebuck

Numerade Educator