Chapter 28, Plane Electromagnetic Waves I Video Solutions, Electromagnetic Fields and Waves: Including Electric Circuits

Problem 1

A general theorem for electromagnetic fields in free space
(a) Show that, for any electromagnetic field in a vacuum $$\boldsymbol{\nabla} \cdot\left(\boldsymbol{E} \times \dot{\boldsymbol{E}}+c^{2} \boldsymbol{B} \times \dot{\boldsymbol{B}}\right)=-\frac{\partial}{\partial t}(\dot{\boldsymbol{E}} \cdot \boldsymbol{B}-\boldsymbol{E} \cdot \dot{\boldsymbol{B}})$$
where the dots above $\boldsymbol{E}$ and $\boldsymbol{B}$ indicate partial differentiation with respect to time.
This equation has the form of a conservation law. The spatial density of the conserved quantity appears between parentheses on the right. This is expressed in volts squared per cubic meter. The flux density of this conserved quantity is the quantity between parentheses on the left.
(b) Show that the flux vanishes in the field of a linearly polarized wave.
(c) Show that, in the field of a circularly polarized wave, the flux does not vanish, that it is proportional to the frequency, and that it contrary to the direction of propagation if the $\boldsymbol{E}$ and $\boldsymbol{B}$ vectors rotate clockwise for an observer who looks at the source, and in the direction of propagation if $\boldsymbol{E}$ and $\boldsymbol{B}$ rotate in the opposite direction.

Dominador Tan

Numerade Educator

01:26

Problem 2

The phase index of refraction, often called the phase index, is $c / v_{p}$, where $v_{p}$ is the phase velocity. The group index is similarly $c / v_{g}$, where $v_{R}=1 /(d \beta / d \omega)$ is the group velocity (Sec. 29.2.6).
Show that $m=n+\omega d n / d \omega$.

Nick Johnson

Numerade Educator

08:11

Problem 3

The skin depth as a function of frequency in low-conductivity materials
(a) Plot on a single graph the log-log curves of the skin depth as a function of the frequency from $f=1$ to $f=10^{5}$, for $\sigma$ equal to $10^{-2}, 10^{-4}$, $10^{-6}$ and for $\epsilon_{r}=1$ and $\epsilon_{r}=10 .$ Set $\mu_{r}=1 .$ The skin depth will vary from about 10 to $10^{6}$ meters.
(b) Show that, in nonmagnetic good conductors for which $\sigma \geq 50 \omega \epsilon$, $\delta \approx 503 /(f \sigma)^{1 / 2}$
(c) Show that, in nonmagnetic poor conductors for which $\sigma \leqslant 0.1 \omega \epsilon$, $\delta \approx 5.3 \times 10^{-3} \epsilon_{r}^{1 / 2} / \sigma$

Prachita Kush

Numerade Educator

01:19

Problem 4

At optical frequencies $\left(f \approx 10^{15}\right.$ hertz) and above, the values of $\sigma, \epsilon$, and $\mu$ bear no relation to the values measured at lower frequencies. For metals, both $\beta$ and $\alpha$ are of the order of $3 / \lambda_{0}$, within approximately a factor of 10 either way, and $\beta \neq \alpha .$ For aluminum at $\lambda_{0}=650$ nanometers, $\beta \lambda_{0}=1.3$ and $\alpha \lambda_{0}=7.11$
(a) Calculate $\lambda$ and $\delta$.
(b) Calculate $\lambda$ and $\delta$ from Sec. $28.5 .1$ and Table $29-1 .$

Pankaj Kumawat

Numerade Educator

02:44

Problem 5

Show that, in a medium that is only slightly conducting $(\mathscr{D} \ll 1)$,
(a) $\lambda=\left(1-\mathscr{D}^{2} / 8\right) \lambda_{\dot{2}=0}$
(b) $\delta=2^{1 / 2} \chi_{P=0} / D$
(c) $Z=\left(1-\mathscr{D}^{2} / 4\right) \exp j[\arctan (\mathscr{D} / 2)] Z_{s=0}$

Saman Zulfiqar

Numerade Educator

01:49

Problem 6

Alternate expressions for the characteristic impedance of a conducting medium
Show that the characteristic impedance $E / H$ of a conducting medium is also given by these two other expressions: $Z=\frac{\beta+j \alpha}{\left(\sigma^{2}+\omega^{2} \epsilon^{2}\right)^{1 / 2}}=\left(\frac{j \omega \mu}{\sigma+j \omega \epsilon}\right)^{1 / 2}$

Surendra Kumar

Numerade Educator

00:45

Problem 7

The characteristic impedance of ultra-low-loss polyethylene
The ultra-low-loss polyethylene that serves as insulator in submarine coaxial cables has a loss angle of 50 microradians and a relative permittivity of $2.26 .$ Calculate its characteristic impedance at 45 megahertz.

Hast Aggarwal

Numerade Educator

06:20

Problem 8

The Poynting vector in the field of a resistive wire carrying a current
A long, straight wire of radius $a$ and resistance $R^{\prime}$ ohms/meter carries a current $I .$
(a) Calculate the Poynting vector at the surface, and explain.
(b) Calculate the Poynting vector both outside and inside the wire. Explain.

Nicholas Majtenyi

Numerade Educator

01:41

Problem 9

The Poynting vector in a capacitor
A thin, air-insulated parallel-plate capacitor has circular plates of radius $R$, separated by a distance $s$. A constant current $I$ charges the plates through thin wires along the axis of symmetry.
(a) Find the value of $E$ between the plates as a function of the time. Assume a uniform $E$. Show the direction of $\boldsymbol{E}$ on a figure.
(b) The magnetic field is the sum of two terms, $H_{w}$, related to the current in the wire, and $H_{p}$, related to the current in the plates. The latter current deposits charges on the inside surfaces of the plates.
Find $H_{w}, H_{p}$, and $H$. Use cylindrical coordinates with the $z$-axis along the wire and in the direction of the current. To calculate $H_{p}$, apply Ampère's circuital law to each plate. You should find that the magnetic fields tend to infinity as $\rho \rightarrow 0$. This is simply because we have assumed infinitely thin wires and plates. Show the directions of $\boldsymbol{H}_{w}, \boldsymbol{H}_{p}$, and $\boldsymbol{H}$ on your figure.
(c) Do $\boldsymbol{E}$ and $\boldsymbol{H}$ satisfy Maxwell's equations? You should find that one of our assumptions is incorrect
(d) Find $\boldsymbol{E} \times \boldsymbol{H}$.
(e) Find the electric and magnetic energy densities inside a radius $\rho$. You should find that the magnetic energy density is negligible if $\rho^{2} / t^{2} \ll c^{2}$. This condition applies because we have assumed that the capacitor charges up slowly. If it charged very quickly, then there would be a wave of $\boldsymbol{E}$ and $\boldsymbol{H}$ in the capacitor, $\boldsymbol{E}$ would not be uniform, and the above calculation would be invalid.
(f) Now relate the Poynting vector at $\rho$ to the electric energy inside $\rho$.
(g) Draw a sketch showing $\boldsymbol{E}, \boldsymbol{H}$, and $\boldsymbol{E} \times \boldsymbol{H}$ vectors at various points inside and around the capacitor.

Dominador Tan

Numerade Educator

08:12

Problem 10

The Poynting vector in a solenoid
A long solenoid of radius $R$ and $N$ ' turns per meter carries a current $I$.
(a) The current increases. Calculate $\mathscr{S}=\boldsymbol{E} \times \boldsymbol{H}$. (See example in Sec.
19.1.)
Sketch a cross section of the solenoid, showing the direction of the current and of $\boldsymbol{S}$. Explain.
(b) Repeat with a decreasing current.

Vishal Gupta

Numerade Educator

07:20

Problem 11

Figure $28-5$ shows a highly simplified diagram of a proton accelerator. A gas discharge within the source $S$ ionizes hydrogen gas to produce protons. Some of the protons emerge through a hole and are focused into a beam $B$ of radius $R_{1}$ inside a conducting tube of radius $R_{2}$. The source is at a potential $V$, and the target is grounded.
To avoid needless complications, we assume that the charge density in the beam is uniform. We also assume that the velocity of the protons is much less than $c: v^{2} \ll c^{2}$.
Calculate, in terms of the current $I$ and the velocity $v$ :
(a) the electric energy per meter $\mathscr{E}_{E}^{\prime}$;
(b) the magnetic energy per meter $\mathscr{E}_{m}^{\prime}$;
(c) the energy flux associated with the Poynting vector $P_{P}$;
(d) the kinetic power $P_{k}$, or the flux of kinetic energy, disregarding $P_{P}$.
The existence of this Poynting vector is interesting. Because of the radial $\boldsymbol{E}$, the voltage inside the beam is slightly positive. So the protons are not accelerated to the full voltage $V$, and the kinetic energy in the beam is slightly lower than $V I$. Most of the power flows down the tube as kinetic energy, and the rest flows as electromagnetic energy. The total power at any point along the tube and on the target is $V I$.
(e) Find the numerical values of these quantities for a 1.00milliampere, $1.00-\mathrm{MeV}$ (megaelectronvolt) proton beam, with $R_{1}=1.00$ ) millimeter and $R_{2}=50.00$ millimeters.

Mahnoor Amin

Numerade Educator

01:33

Problem 12

The solar wind is formed of highly ionized, and hence highly conducting, hydrogen that evaporates from the surface of the sun. In the plane of the earth's orbit, the magnetic field of the sun is approximately radial, pointing outward in certain regions and inward in others. Since the sun rotates (period of 27 days), while the plasma has a radial velocity, the lines of $\boldsymbol{B}$ are Archimedes spirals. This is the garden-hose effect. At the earth, the lines of $\boldsymbol{B}$ form an angle of about $45^{\circ}$ with the sun-earth direction.
At the orbit of the earth the solar wind has a density of about $10^{7}$ proton masses per cubic meter and a velocity of about $4 \times 10^{5}$ meters/second. The magnetic field of the sun is about $5 \times 10^{-9}$ tesla.
(a) Show that, in a neutral $(\rho=0)$ plasma of conductivity $\sigma$ and velocity $v$, Maxwell's equations become $$
\begin{aligned}
\boldsymbol{\nabla} \cdot \boldsymbol{E}=0, & \boldsymbol{\nabla} \times \boldsymbol{E}=-\frac{\partial \boldsymbol{B}}{\partial t}, \quad \boldsymbol{\nabla} \cdot \boldsymbol{B}=0 \\
\boldsymbol{\nabla} \times \boldsymbol{B}=\mu_{0}\left[\sigma(\boldsymbol{E}+\boldsymbol{v} \times \boldsymbol{B})+\epsilon_{0} \frac{\partial \boldsymbol{E}}{\partial t}\right]
\end{aligned}
$$
In a medium of infinite conductivity $\sigma, \boldsymbol{E}=-\boldsymbol{v} \times \boldsymbol{B}$. This is a satisfactory approximation for the solar wind.
(b) Show that the component of the plasma velocity $v$ that is normal to $\boldsymbol{B}$ is given by $\boldsymbol{v}_{\perp}=[\boldsymbol{B} \times(\boldsymbol{v} \times \boldsymbol{B})] / B^{2}$.
(c) Show that the Poynting vector is given by $\mathscr{S}=B^{2} \boldsymbol{v}_{\perp} / \mu_{0}$, or about 6 microwatts/meter $^{2}$. This is about $4 \times 10^{-9}$ times the Poynting vector of solar radiation, which is about $1.4$ kilowatts/meter $^{2}$. The Poynting vector of the solar wind is normal to the local $\boldsymbol{B}$.
(d) Show that the kinetic, magnetic, and electric energy densities of the solar wind are related as follows: $\mathscr{E}_{k}^{\prime} \gg \mathscr{E}_{m}^{\prime} \gg \mathscr{E}_{E}^{\prime}$

Mayukh Banik

Numerade Educator

04:49

Problem 13

In an induction motor, the stator generates a magnetic field that is perpendicular to, and that rotates about, the axis of symmetry (Prob. 18-6). The rotor is a cylinder of laminated iron (Prob. 25-7), with copper bars parallel to the axis and set in grooves in the cylindrical surface. Copper rings at each end of the rotor connect all the copper bars. The rotor is not connected to the source of electric current that feeds the motor.
As we shall see, the rotor tends to follow the rotating magnetic field. Figure $28-6$ shows the principle of operation. To simplify the analysis, we suppose that the rotor is stationary and that a rotating electromagnet, represented here by its poles $\mathrm{N}$ and $\mathrm{S}$, provides the rotating magnetic field.
(a) Draw a larger figure with wide air gaps, showing the direction of the induced currents in the bars and the direction of $\boldsymbol{E}$ in the air gaps.
(b) The current in the rotor generates a magnetic field. Add arrows showing the direction of that $\boldsymbol{H}$, inside the rotor and in the air gaps.
(c) Now show Poynting vectors $\boldsymbol{E} \times \boldsymbol{H}$ in the air gaps. The field feeds power into the rotor.
(d) Now draw another figure showing the currents in the bars and a line of $\boldsymbol{B}$ for the sum of the magnetic fields.
(e) Show the direction of the magnetic forces on the bars.

Sachin Rao

Numerade Educator

04:02

Problem 14

Figure $28-7$ shows, in simplified form, a cross section of a transformer secondary. The field $\boldsymbol{B}$ inside the core $C$, and the leakage field $\boldsymbol{H}_{l}$ outside, both result from the currents in the windings and from the equivalent currents in the core. The secondary winding is $W$. Assume that $\boldsymbol{B}$ and $\boldsymbol{H}_{\text {i }}$ increase.
(a) Draw a larger figure and show, at one point between $C$ and $W$, vectors $\boldsymbol{A}, \partial \boldsymbol{A} / \partial t$, and $\boldsymbol{E}=-\partial \boldsymbol{A} / \partial t$, disregarding the current in $W$. Show a vector $\boldsymbol{E}$ at one point outside $W$.
(b) Assume that the impedance of the secondary is a pure resistance $R$. Show the direction of the current $I$ in $W$.
(c) Show the direction of its field $\boldsymbol{H}$ at points between $C$ and $W$ and outside $W$.
(d) Show vectors $\boldsymbol{E} \times \boldsymbol{H}$.
(e) How would the directions of the $\boldsymbol{E} \times \boldsymbol{H}$ vectors be affected if $\boldsymbol{B}$ and $\boldsymbol{H}_{l}$ decreased?
(f) What is the time-averaged value of a vector $\boldsymbol{E} \times \boldsymbol{H}_{l}$ ?
(g) Now let us calculate the power flow into the secondary. Assume that the secondary is a long solenoid of $N$ turns and of length $L$. Disregard $\boldsymbol{H}_{l}$ and set $\Phi=\Phi_{m} \exp j \omega t$ in the core. Integrate the Poynting vector over a cylindrical surface situated between the core and the winding, and show that the power flowing into the winding is $\left(N \omega \Phi_{\mathrm{ms}}\right)^{2} / R=V_{\mathrm{rms}}^{2} / R$, where $V$ is the voltage induced in the secondary winding.

Sikandar Baig

Numerade Educator