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A \(150 \mathrm{MHz}\) uniform plane wave in free space is described by \(\mathbf{H}_{s}=\) \((4+j 10)\left(2 \mathbf{a}_{x}+j \mathbf{a}_{y}\right) e^{-j \beta z} \mathrm{~A} / \mathrm{m} .(a)\) Find numerical values for \(\omega, \lambda\), and \(\beta .\) (b) Find \(\mathcal{H}(z, t)\) at \(t=1.5 \mathrm{~ns}, z=20 \mathrm{~cm} .(c)\) What is \(|E|_{\max } ?\)

Short Answer

Expert verified
Question: Calculate the values of \(\omega\), \(\lambda\), and \(\beta\) for a uniform plane wave with a frequency of \(150~\mathrm{MHz}\) in free space. Also, determine the magnetic field \(\mathcal{H}(z, t)\) at \(t = 1.5~\mathrm{ns}\) and \(z = 20~\mathrm{cm}\) and the maximum value of the electric field amplitude, \(|E|_{\max}\). Answer: The values of the parameters are calculated to be: \(\omega = 300\pi\times 10^6~\mathrm{rad/s}\), \(\lambda = 2~\mathrm{m}\), and \(\beta = \pi~\mathrm{rad/m}\). The magnetic field \(\mathcal{H}(z, t)\) at \(t=1.5~\mathrm{ns}\) and \(z=20~\mathrm{cm}\) is \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns}) = -(8+20j)\mathbf{a}_{x} +(20+8j)\mathbf{a}_{y} ~\mathrm{A/m}\). The maximum value of the electric field amplitude is \(|E|_{\max} = \frac{10}{376.7}~\mathrm{V/m}\).

Step by step solution

01

Given Information and Constants

Given, the frequency of the wave is \(f = 150~\mathrm{MHz}\). Also, we have the magnetic field \(\mathbf{H}_{s} = (4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\beta z} \mathrm{~A/m}\). We need to find the values of \(\omega, \lambda\), and \(\beta\). The speed of light in free space is \(c \approx 3 \times 10^8~\mathrm{m/s}\). The vacuum permeability and permittivity are, respectively, \(\mu_0 = 4\pi \times 10^{-7}~\mathrm{N/A^2}\) and \(\varepsilon_0 = 8.854 \times 10^{-12}~\mathrm{F/m}\).
02

Calculate \(\omega\)

First, let's calculate the angular frequency, \(\omega\). We know that \(\omega = 2\pi f\), so: \(\omega = 2\pi(150\times 10^6) = 300\pi\times 10^6~\mathrm{rad/s}\)
03

Calculate \(\lambda\)

Now we can calculate the wavelength, \(\lambda\). The relationship between frequency, wavelength, and the speed of light in free space is \(c = f\lambda\). So we have: \(\lambda = \frac{c}{f} = \frac{3\times 10^8}{150\times 10^6} = 2~\mathrm{m}\)
04

Calculate \(\beta\)

Next, let's calculate the propagation constant, \(\beta\). We know that \(\beta = \frac{2\pi}{\lambda}\), so: \(\beta = \frac{2\pi}{2} = \pi~\mathrm{rad/m}\)
05

Find \(\mathcal{H}(z, t)\) at \(t=1.5~\mathrm{ns}\) and \(z=20~\mathrm{cm}\)

The given magnetic field is in phasor form. We have to convert it to time domain form, which is: \(\mathcal{H}(z, t) = \Re\{\mathbf{H}_{s}e^{j\omega t}\}\) This results in: \(\mathcal{H}(z, t) = \Re\{(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\pi z}e^{j 300\pi\times10^6t} ~\mathrm{A/m}\}\) Now, we will find \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns})\): \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns}) = \Re\{(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\pi (0.2)}e^{j 300\pi\times10^6(1.5\times 10^{-9})} ~\mathrm{A/m}\}\) After evaluating real part, we get, \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns}) = -(8+20j)\mathbf{a}_{x} +(20+8j)\mathbf{a}_{y} ~\mathrm{A/m}\)
06

Find \(|E|_{\max}\)

We will use Ampère's law: \(\nabla\times\mathbf{H} = \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t}\). For a plane wave, the relationship between \(\mathbf{H}\) and \(\mathbf{E}\) is: \(\mathbf{E} = \frac{1}{\eta}\mathbf{H}\times\mathbf{a}_{z}\), where \(\eta\) is the intrinsic impedance, and for free space is given by \(\eta = \sqrt{\frac{\mu_0}{\varepsilon_0}}\). So, first calculation of \(\eta\): \(\eta = \sqrt{\frac{4\pi \times 10^{-7}}{8.854 \times 10^{-12}}} = 376.7~\Omega\) Now, let's calculate the phasor electric field: \(\mathbf{E}_{s} = \frac{1}{\eta}(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\beta z}\times(-j\mathbf{a}_{z})\) To find the maximum amplitude, we need the magnitude of the electric field phasor: \(|E|_{\max} = \frac{1}{\eta}|\mathbf{H}_{s}| = \frac{1}{376.7}\sqrt{(4)^2+(10)^2} = \frac{10}{376.7}~\mathrm{V/m}\)

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Key Concepts

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

Angular Frequency
Angular frequency is a concept that describes how fast an object is rotating or, in a more general wave context, the rate at which a wave oscillates in time. It's very similar to regular frequency but expressed in radians per second instead of cycles per second. Angular frequency, denoted by \(\omega\), is calculated using:
  • \(\omega = 2\pi f\)
where \(f\) is the frequency of the wave.
In the context of this exercise, we're given a frequency \(f = 150 ~\mathrm{MHz}\). By plugging this into our formula, we get:
  • \(\omega = 2\pi \times 150 \times 10^6 = 300\pi \times 10^6~\mathrm{rad/s}\)
This formula shows how the wave's energy moves through space and time. Angular frequency is crucial for understanding electromagnetic wave behavior because it details how often a wave completes a full cycle in radians, making it an essential characteristic of any wave, including electromagnetic ones.
Wavelength
Wavelength, denoted by \(\lambda\), is the physical distance between consecutive points of a wave, typically measured from crest to crest or trough to trough. It's inversely related to frequency; as one increases, the other decreases. The formula linking wavelength to frequency and the speed of light \(c\) in free space is:
  • \(\lambda = \frac{c}{f}\)
For a wave in free space, \(c \approx 3 \times 10^8~\mathrm{m/s}\), so using the frequency \(f = 150~\mathrm{MHz}\), we get:
  • \(\lambda = \frac{3\times 10^8}{150\times 10^6} = 2~\mathrm{m}\)
This value indicates that every cycle of the electromagnetic wave spans a length of 2 meters in space. The concept of wavelength is vital since it influences how waves interact with objects and interfere with each other, such as when determining the resolution of optical instruments or the bandwidth of antennas.
Propagation Constant
The propagation constant \(\beta\) is a measure of how an electromagnetic wave propagates through space. It represents the phase change per unit distance along the direction of wave travel. Essentially, \(\beta\) tells us how the wave's phase evolves as it moves forward. The formula for calculating \(\beta\) is:
  • \(\beta = \frac{2\pi}{\lambda}\)
By inserting the wavelength \(\lambda = 2~\mathrm{m}\), we find:
  • \(\beta = \frac{2\pi}{2} = \pi ~\mathrm{rad/m}\)
This indicates that for every meter the wave travels, its phase changes by \(\pi\) radians. The propagation constant is a critical parameter in the study of waveguides, transmission lines, and antenna design.
It helps us understand how much of the wave's energy is lost or gained as it propagates through different media.
Intrinsic Impedance
Intrinsic impedance, symbolized as \(\eta\), is an important parameter in electromagnetism, specifically describing the relationship between the electric and magnetic fields in a wave traveling through a medium. For free space, it defines the natural resistance to electromagnetic wave propagation, mathematically given by:
  • \(\eta = \sqrt{\frac{\mu_0}{\varepsilon_0}}\)
Where \(\mu_0\) is the permeability of free space \(\approx 4\pi \times 10^{-7}~\mathrm{N/A^2}\), and \(\varepsilon_0\) the permittivity \(\approx 8.854 \times 10^{-12}~\mathrm{F/m}\). Evaluating this results in:
  • \(\eta \approx 376.7~\Omega\)
This impedance helps us understand characteristics such as how waves reflect or transmit at boundaries between different media. In practical terms, \(\eta\) allows us to calculate the electric field (\(E\)) when the magnetic field (\(H\)) is known, using the relation:
  • \(\mathbf{E} = \frac{1}{\eta}\mathbf{H}\times \mathbf{a}_{z}\)
This relation is key to describing wave behavior and is used in designing antennas, understanding transmission, and reflection phenomena.

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Most popular questions from this chapter

A linearly polarized uniform plane wave, propagating in the forward \(z\) direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along \(y\left(\epsilon_{r y}\right)\) differs from that seen by waves polarized along \(x\left(\epsilon_{r x}\right) .\)

Voltage breakdown in air at standard temperature and pressure occurs at an electric field strength of approximately \(3 \times 10^{6} \mathrm{~V} / \mathrm{m}\). This becomes an issue in some high-power optical experiments, in which tight focusing of light may be necessary. Estimate the lightwave power in watts that can be focused into a cylindrical beam of \(10 \mu \mathrm{m}\) radius before breakdown occurs. Assume uniform plane wave behavior (although this assumption will produce an answer that is higher than the actual number by as much as a factor of 2 , depending on the actual beam shape).

In an anisotropic medium, permittivity varies with electric field direction, and is a property seen in most crystals. Consider a uniform plane wave propagating in the \(z\) direction in such a medium, and which enters the material with equal field components along the \(x\) and \(y\) axes. The field phasor will take the form: $$\mathbf{E}_{s}(z)=E_{0}\left(\mathbf{a}_{x}+\mathbf{a}_{y} e^{j \Delta \beta z}\right) e^{-j \beta z}$$ where \(\Delta \beta=\beta_{x}-\beta_{y}\) is the difference in phase constants for waves that are linearly polarized in the \(x\) and \(y\) directions. Find distances into the material (in terms of \(\Delta \beta\) ) at which the field is ( \(a\) ) linearly polarized and \((b)\) circularly polarized. (c) Assume intrinsic impedance \(\eta\) that is approximately constant with field orientation and find \(\mathbf{H}_{s}\) and \(<\mathbf{S}>\).

The planar surface \(z=0\) is a brass-Teflon interface. Use data available in Appendix \(\mathrm{C}\) to evaluate the following ratios for a uniform plane wave having \(\omega=4 \times 10^{10} \mathrm{rad} / \mathrm{s}:\left(\right.\) a) \(\alpha_{\text {Tef }} / \alpha_{\text {brass }} ;(b) \lambda_{\text {Tef }} / \lambda_{\text {brass }} ;\) (c) \(v_{\text {Tef }} / v_{\text {brass }}\).

Given a \(100-\mathrm{MHz}\) uniform plane wave in a medium known to be a good dielectric, the phasor electric field is \(\mathcal{E}_{s}=4 e^{-0.5 z} e^{-j 20 z} \mathbf{a}_{x} \mathrm{~V} / \mathrm{m}\). Determine (a) \(\epsilon^{\prime} ;(b) \epsilon^{\prime \prime} ;(c) \eta ;(d) \mathbf{H}_{s} ;(e)\langle\mathbf{S}\rangle ;(f)\) the power in watts that is incident on a rectangular surface measuring \(20 \mathrm{~m} \times 30 \mathrm{~m}\) at \(z=10 \mathrm{~m}\).

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