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Chapter 9: Problem 27

Reflected wave ** A sinusoidal wave is reflected at the surface of a medium whose properties are such that half the incident energy is absorbed. Consider the field that results from the superposition of the incident and the reflected wave. An observer stationed somewhere in this field finds the local electric field oscillating with a certain amplitude \(E\). What is the ratio of the largest such amplitude noted by any observer to the smallest amplitude noted by any observer? (This is called the voltage standing wave ratio, or, in laboratory jargon, VSWR.)

Short Answer

Expert verified
The voltage standing wave ratio (VSWR) is 3.

Step by step solution

01

Analyze the given

It is mentioned that half of the incident energy is absorbed by the medium. Therefore, the reflected wave has half the amplitude of the incident wave. The superposition of the two waves is what creates the observed field.
02

Identify the largest amplitude

The largest amplitude observed will be when the incident and reflected wave are in phase and superpose constructively. Thus, the maximum amplitude will be the sum of their amplitudes, i.e., \[E_{max} = E + 0.5E = 1.5E\].
03

Identify the smallest amplitude

The smallest amplitude observed will be when the incident and reflected wave are in anti-phase and superpose destructively. Thus, the minimum amplitude will be the absolute difference of their amplitudes, i.e., \[E_{min} = E - 0.5E = 0.5E\].
04

Compute the ratio

The VSWR or the ratio of the largest amplitude observed to the smallest amplitude observed is given as \[VSWR = E_{max}/E_{min} = (1.5E)/(0.5E) = 3\]. Note: VSWR is a unitless quantity.

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Key Concepts

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

Reflected Wave
When a wave encounters a boundary or a change in medium properties, part of its energy is typically reflected back. This part is known as the reflected wave. In this exercise, half of the incident wave's energy is absorbed by the medium.
As a result, the reflected wave has a lower amplitude than the incident wave. The amplitude of a reflected wave, in this case, is half of that of the original wave. Understanding the nature of reflection is essential for analyzing wave behaviors, especially in applications where wave interference, such as in radio frequency technologies, is significant.
Incident Wave
Waves traveling toward a boundary or a medium change are called incident waves. The incident wave鈥檚 characteristics determine the behavior of the reflected wave upon striking the barrier. In this scenario, the incident wave interacts with a medium that absorbs half of its energy. This energy absorption affects the amplitude of the reflected wave.
The incident wave's amplitude serves as a reference point in the study of interference patterns, such as constructive and destructive superposition. By assessing how much energy is reflected versus absorbed, we can predict changes in wave behavior, playing a pivotal role in designing systems like antennas and sensors.
Superposition
Wave superposition is a fundamental principle stating that when two or more waves overlap, the resultant wave is the sum of the individual waves. This principle is crucial when analyzing fields formed by reflected and incident waves.
The superposition leads to interference patterns. These include constructive interference, where waves build upon each other, and destructive interference, where they cancel out. In the problem, superposition determines the electric field's oscillating amplitude observed by an onlooker, varying between maximum and minimum values. Understanding this concept helps in predicting the resulting wave conditions when multiple waves intersect.
Amplitude
Amplitude refers to the magnitude or height of the wave, essentially indicating its intensity. It represents how much energy the wave carries. In wave interference, due to superposition, the amplitude can vary significantly.
As described in the exercise, the maximum amplitude occurs when the incident and reflected waves are in phase (constructive interference), leading to an amplitude of 1.5 times the original incident amplitude. Conversely, the minimum amplitude occurs when these waves are out of phase (destructive interference), being only half of the incident wave amplitude. Understanding amplitude changes is vital in technologies like telecommunications, where signal strength and clarity are paramount.

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Most popular questions from this chapter

Energy flow in a capacitor ** A capacitor is charged by having current flow in a thin straight wire from the middle of one circular plate to the middle of the other (as opposed to wires coming in from infinity, as in the example in Section 9.6.2). The electric field inside the capacitor increases, so the energy density also increases. This implies that there must be a flow of energy from somewhere. As in Problem 9.10, this "somewhere" is the wire. Verify that the flux of the Poynting vector away from the wire equals the rate of change of the energy stored in the field. (Of course, we would need to place a battery somewhere along the wire to produce the current flow, and this battery is where the energy flow originates. See Galili and Goihbarg (2005).)

Angular momentum paradox A setup consists of three very long coaxial cylindrical objects: a nonconducting cylindrical shell with radius \(a\) and total (uniform) charge \(Q\), another nonconducting cylindrical shell with radius \(b>\) \(a\) and total (uniform) charge \(-Q\), and a solenoid with radius \(R>b ;\) see Fig. 9.13. (This setup is a variation of the setup in Boos (1984).) The current in the solenoid produces a uniform magnetic field \(B_{0}\) in its interior. The solenoid is fixed, but the two cylinders are free to rotate (independently) around the axis. They are initially at rest. Imagine that the current in the solenoid is then decreased to zero. (If you want to be picky about keeping the system isolated from external torques, you can imagine the current initially flowing in a superconductor which becomes a normal conductor when heated up.) The changing \(B\) field inside the solenoid will induce an \(E\) field at the locations of the two cylinders. (a) Find the angular momentum gained by each cylinder by the time the magnetic field has decreased to zero. (b) You should find that the total change in angular momentum of the cylinders is not zero. Does this mean that angular momentum isn't conserved? If it is conserved, verify this quantitatively. You may assume that the two cylinders are massive enough so that they don't end up spinning very quickly, which means that we can ignore the \(B\) fields they generate. Hint: See Problem 9.11.

An electromagnetic wave ** Here is a particular electromagnetic field in free space: $$ \begin{array}{ll} E_{x}=0, \quad E_{y}=E_{0} \sin (k x+\omega t), & E_{z}=0 \\ B_{x}=0, \quad B_{y}=0, & B_{z}=-\left(E_{0} / c\right) \sin (k x+\omega t) \end{array} $$ (a) Show that this field can satisfy Maxwell's equations if \(\omega\) and \(k\) are related in a certain way. (b) Suppose \(\omega=10^{10} \mathrm{~s}^{-1}\) and \(E_{0}=1 \mathrm{kV} / \mathrm{m}\). What is the wavelength? What is the energy density in joules per cubic meter, averaged over a large region? From this calculate the power density, the energy flow in joules per square meter per second.

Energy flow from a wire ** A very thin straight wire carries a constant current \(I\) from infinity radially inward to a spherical conducting shell with radius \(R\). The increase in the charge on the shell causes the electric field in the surrounding space to increase, which means that the energy density increases. This implies that there must be a flow of energy from somewhere. This "somewhere" is the wire. Verify that the total flux of the Poynting vector away from a thin tube surrounding the wire equals the rate of change of the energy stored in the electric field. (You can assume that the radius of the wire is much smaller than the radius of the tube, which in turn is much smaller than the radius of the shell.)

Due to the contradiction between Eqs. (9.2) and (9.5), we know that there must be an extra term in the \(\nabla \times \mathbf{B}\) relation, as we found in Eq. (9.10). Call this term \(\mathbf{W}\). In the text, we used the Lorentz transformations to motivate a guess for \(\mathbf{W}\). Find \(\mathbf{W}\) here by taking the divergence of both sides of \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mathbf{W}\). Assume that the only facts you are allowed to work with are (1) \(\nabla \cdot \mathbf{E}=\rho / \epsilon_{0}\), (2) \(\nabla \cdot \mathbf{B}=0\), (3) \(\nabla \cdot \mathbf{J}=-\partial \rho / \partial t\), and (4) \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\) in the case of steady currents.
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