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Chapter 8: Problem 6

A plane electromagnetic wave, with momentum density \(\mathrm{p}_{1}\), is incident on a plane absorbing surface at an angle \(\theta\) with respect to the normal. \((a)\) Show that the normal force per unit area, i.e., the pressure, is $$ p=\bar{W}_{1} \cos ^{2} \theta \text {. } $$ (b) Show that if waves are incident on the surface at all angles, the pressure is $$ p=\frac{1}{3} \bar{W}_{1} . $$ (c) If the surface has a power reflection coefficient \(R(\theta)\), how are these results affected?

Short Answer

Expert verified
Answer: The pressure on the plane absorbing surface when the wave is incident at an angle θ is given by \(p=\bar{W_1} \cos^2θ\). When waves are incident on the surface at all angles, the pressure is given by \(p=\frac{1}{3} \bar{W}_1\). When the surface has a power reflection coefficient R(θ), the pressure on the surface is affected by the fraction of the incident wave's power that is absorbed and can be expressed as \(p_{total, R}=\frac{1}{A} \int_0^{\pi/2} (1-R(\theta)) \bar{W_1} \cos^2θ \cdot A \sin\theta d\theta\). The expression for \(p_{total, R}\) depends on the reflection coefficient and it would be different for different surfaces.

Step by step solution

01

State the relationship between the momentum density and the intensity of the wave

The relationship between the momentum density \(p_1\) and the intensity \(I\) of the wave is given by \(p_1=\frac{I}{c}\). The intensity of the wave is a measure of power per unit area.
02

Calculate the intensity and power incident on the surface

Let's consider an area element \(A\) on the absorbing surface. The intensity incident on the area element is \(I = p_1 c \cos\theta\). The power incident on the area element is given by \(W_1 = I A = p_1 c A \cos\theta\).
03

Determine the pressure on the surface

The pressure on the surface due to the electromagnetic wave is given by \(p = \frac{W_1 \cos\theta}{A}\) which simplifies to \(p=\overline{W_1} \cos^2\theta\). Thus, we have shown that the pressure on the plane absorbing surface when the wave is incident at an angle \(\theta\) is given by \(p=\overline{W_1} \cos^2\theta\). #b)# Determine the pressure when waves are incident on the surface at all angles
04

Set up the integral for the pressure on the surface

We will integrate the pressure over all possible angles, from \(0\) to \(\pi/2\). $$ p_{total} = \frac{1}{A} \int_0^{\pi/2} \overline{W_1} \cos^2\theta \cdot A \sin\theta d\theta $$
05

Evaluate the integral for the pressure

To simplify and evaluate the integral, we can replace \(\cos^2\theta\) with \(\frac{1+\cos 2\theta}{2}\) and perform integration: $$ p_{total} = \overline{W_1} \int_0^{\pi/2} \left(\frac{1+\cos 2\theta}{2}\right) \sin\theta d\theta = \overline{W_1} \left(\frac{1}{2}\int_0^{\pi/2} \sin\theta d\theta + \frac{1}{2}\int_0^{\pi/2} \sin\theta\cos 2\theta d\theta\right) $$ After evaluating the integrals, we get: $$ p_{total} = \frac{1}{3} \overline{W_1} $$ Hence, we have shown that when waves are incident on the surface at all angles, the pressure is given by \(p=\frac{1}{3} \bar{W}_{1}\). #c)# Investigate the effect of power reflection coefficient \(R(\theta)\) on these results
06

Describe the effect of the reflection coefficient

When the surface has a power reflection coefficient \(R(\theta)\), a fraction \(R(\theta)\) of the incident wave's power is reflected, while the rest \((1-R(\theta))\) is absorbed by the surface.
07

Modify the pressure expression due to reflection

We need to multiply the pressure on the surface by the fraction of the incident wave's power that is absorbed: $$ p_{total, R}=\frac{1}{A} \int_0^{\pi/2} (1-R(\theta)) \overline{W_1} \cos^2\theta \cdot A \sin\theta d\theta $$ Evaluating this integral will show how the pressure is affected by the reflection coefficient \(R(\theta)\). The expression for \(p_{total, R}\) depends on the reflection coefficient and it would be different for different surfaces.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Density
Momentum density in the context of electromagnetic waves is an intriguing concept that bridges physics and mathematics. It is defined as the amount of momentum an electromagnetic wave carries per unit volume. When dealing with a plane electromagnetic wave, the momentum density, denoted as \(p_1\), is directly proportional to the intensity of the wave, which brings us to the precise relationship \(p_1 = \frac{I}{c}\). Here, \(I\) represents the electromagnetic wave intensity, and \(c\) is the speed of light in a vacuum, a fundamental constant.

Understanding momentum density is crucial in explaining the pressure exerted by the wave upon interaction with surfaces. When a wave strikes a surface, its momentum density translates to a force, and consequently, a pressure is exerted. This concept is key to comprehending phenomena such as radiation pressure and has implications in areas ranging from solar sails to laser cooling techniques.
Power Reflection Coefficient
The power reflection coefficient, often denoted as \(R(\theta)\), is an essential factor in electromagnetics that quantifies the fraction of incident electromagnetic power reflected by a surface. It depends on the material properties, the angle of incidence, and the frequency of the incident wave. The values of \(R(\theta)\) range from 0 to 1, with 0 indicating that all the incident power is absorbed and 1 indicating total reflection.

When a surface has a specific power reflection coefficient, it affects how electromagnetic waves interact with it. For example, in the textbook exercise, considering \(R(\theta)\) leads to a modification in the earlier derived expressions for pressure. Instead of all the power of the wave contributing to the pressure, only the portion \(1-R(\theta)\) that is absorbed by the surface contributes. This illustrates why surfaces with different reflection coefficients will exert different pressures under the same electromagnetic wave conditions.
Electromagnetic Wave Intensity
Electromagnetic wave intensity, denoted \(I\), is defined as the power transferred per unit area in the direction of wave propagation. It is an important concept because it determines how much energy is carried by a wave and therefore how much energy is available for interaction with materials. Intensity is measured in watts per square meter (\(W/m^2\)).

In practice, understanding wave intensity is crucial in applications such as laser technology, where precise control of energy delivery is required, or in solar power, where the amount of energy harnessed depends on the intensity of sunlight. In the context of the exercise, knowing that the intensity of an electromagnetic wave is responsible for the pressure exerted on surfaces directs us towards an appreciation of the profound impact that electromagnetic phenomena have, not just conceptually, but in real-world applications.

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Most popular questions from this chapter

(a) From \((8.7 .10)\) and \((8.7 .13)\), show that the guided ave impedance \(\left(E_{x}{ }^{2}+E_{y}{ }^{2}\right)^{1 / 2} /\) \(\left(H_{x}{ }^{2}+H_{z}{ }^{2}\right)^{1 / 2}\) is \(Z_{\mathrm{TE}}=\frac{Z_{0}}{\left[1-\left(\lambda_{0} / \lambda_{e}\right)^{2}\right]^{1 / 2}} \quad\) TE modes \(Z_{\text {T? }}=Z_{0}\left[1-\left(\lambda_{0} / \lambda_{c}\right)^{2}\right]^{1 / 2} \quad\) TM modes, where \(Z_{0}\) is the unbounded wave impedance \((8.3 .10)\) or, more generally, \((8.3 .12) .(b)\) For the TE o dominant mode in rectangular waveguide, show that the peak potential difference between opposite points in the cross section is $$ V_{0}=\left[\int_{0}^{b} E_{y}\left(x=\frac{1}{2} a\right) d y\right]_{p e s k}=b E_{0} $$ and that the peak axial current flowing in the top wall is $$ I_{0} \equiv\left[\frac{1}{\mu_{0}} \int_{0}^{a} B_{x}(y=b) d x\right]_{\text {pak }}=\frac{2 a E_{0}}{\pi Z_{\mathrm{TE}}} $$ Since the result of Prob. 8.7.10b can be written $$ \bar{P}=\frac{a b}{4} \frac{E_{0}{ }^{2}}{Z_{\mathrm{TB}}} $$ we can define three other (mode-dependent) waveguide impedances as follows: $$ \begin{aligned} &Z_{V, I}=\frac{V_{0}}{I_{0}}=\frac{\pi}{2}\left(\frac{b}{a} Z_{\mathrm{TE}}\right) \\ &Z_{P, V}=\frac{V_{0}{ }^{2}}{2 P}=2\left(\frac{b}{a} Z_{\mathrm{TE}}\right) \\\ &Z_{P, I}=\frac{2 \bar{P}}{I_{0}{ }^{2}}=\frac{\pi^{2}}{8}\left(\frac{b}{a} Z_{\mathrm{TE}}\right) \end{aligned} $$ which differ by small numerical factors. Only systems supporting a TEM mode (e.g., Sec. 8.1), have a unique impedance.

If oscillatory fields are represented by \(\mathbf{E}=\underline{\mathbf{E}}_{0} c^{j \omega t}\) and \(\mathbf{H}=\breve{\mathbf{H}}_{0} e^{j \omega t}\), using the realpart convention, show that the (real) Poynting vector is given by $$ s=\frac{1}{2}\left(\mathbf{E} \times \mathbf{H}^{*}+\mathbf{E}^{*} \times \mathbf{H}\right)=\operatorname{Re}\left(\underline{\mathbf{E}} \times \check{\mathbf{H}}^{*}\right)=\operatorname{Re}\left(\breve{\mathbf{E}}^{*} \times \breve{\mathbf{H}}\right), $$ where the asterisk denotes complex conjugate. Also, show that the time-average Poynting vector is

Show that the time-average Poynting vector for the far fields (8.9.13) of an oscillating electric dipole \((8.9 .12)\) is $$ \underline{E}=\hat{\mathrm{f}} \frac{c k^{4} p_{0}^{2}}{32 \pi^{2} \epsilon_{0}} \frac{\sin ^{2} \theta}{r^{2}} $$ and that the average total power radiated by the oscillating dipole is $$ \left(\frac{d W}{d l}\right)=\frac{c \kappa^{4} p_{0}^{2}}{12 \pi \epsilon_{0}} $$ Why is the sky blue and the sunset red?

Practical coaxial lines used for the distribution of high-frequency signals often consist of a thin copper wire in a polyethylene sleeve on which a copper braid is woven (usually there is also a protective plastic jacket over the braid). Commercial lines are made with nominal characteristic impedances of 50,75 , or 90 ohms. A common 50 -ohm variety has a center conductor of diameter 0035 in. The dielectric constant of polyethylene is \(2.3\) What is the nominal (inside) diameter of the copper braid? What are the capacitance and inductance per foot? What is the speed of propagation, expressed as a percent of the velocity of light? Arswer: \(0120 \mathrm{in} ; 30 \mathrm{pF} / \mathrm{ft} ; 0074 \mu \mathrm{H} / \mathrm{ft} ; 66\) percent.

Consider \(\mathbf{E}\) and \(\mathbf{B}\) wave fields whose only dependence on \(z\) and \(t\) is included in the factor \(e^{i\left(\omega t-x_{1} \theta\right)}\). Further assume TE waves such that \(E_{z}=0\). Write out Maxwell's curl equations \((82.2)\) and \((8.2 .4)\) in cartesian components and show \((a)\) that all four transverse field components can be obtained from \(B_{t}\) by first-order partial differentiation and \((b)\) that \(B_{*}\) must be a solution of the Helmholtz equation \((8.7 .16)\). Thus the scalar function \(\phi\) of the text may be interpreted as proportional to \(B_{z}\) for TE waves or proportional to \(E_{s}\) for TM waves.
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