Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a medium characterized by intrinsic impedance \(\eta=|\eta| e^{j \phi}\), a linearly polarized plane wave propagates, with magnetic field given as \(\mathbf{H}_{s}=\) \(\left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{-j \beta x} .\) Find \((a) \mathbf{E}_{s} ;(b) \mathcal{E}(x, t) ;(c) \mathcal{H}(x, t) ;(d)\langle\mathbf{S}\rangle .\)

Short Answer

Expert verified
Question: Given the plane wave's magnetic field component, $\mathbf{H_s} = \left(H_{0 y} \mathbf{a}_{y} + H_{0 z} \mathbf{a}_{z} \right) e^{-\alpha x} e^{-j \beta x}$, and intrinsic impedance $\eta$, find the corresponding electric field component, the instantaneous electric and magnetic fields, and the time-averaged Poynting vector. Answer: Electric field component: $\mathbf{E_s} = |\eta| e^{j \phi} \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j \beta x} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j \beta x}\right)$ Instantaneous electric field: $\mathcal{E}(x, t) = |\eta| \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j (\beta x - \omega t)} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j (\beta x - \omega t)}\right)$ Instantaneous magnetic field: $\mathcal{H}(x, t) = \left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{-j (\beta x - \omega t)}$ Time-averaged Poynting vector: $\langle \mathbf{S} \rangle = \frac{1}{2} \text{Re}\{(|\eta| e^{j \phi} \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j \beta x} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j \beta x}\right)) \times (\left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{j \beta x})\}$

Step by step solution

01

Find the Electric Field from the Magnetic Field and Impedance

We know that the relationship between the electric field \(\mathbf{E}_s\) and the magnetic field \(\mathbf{H}_s\) in a medium is given by: \(\mathbf{E_s} = \eta \times \mathbf{H_s}\). Substituting the supplied values for \(\eta\) and \(\mathbf{H}_s\), we have: \(\mathbf{E_s} = |\eta| e^{j \phi} \big(\left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{-j \beta x}\big)\). Now, we can simplify the expression: \(\mathbf{E_s} = |\eta| e^{j \phi} \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j \beta x} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j \beta x}\right)\).
02

Calculate the Instantaneous Electric Field

To find the instantaneous electric field \(\mathcal{E}(x, t)\), we can substitute \(e^{-j \omega t}\) for \(e^{j(\phi-\beta x)}\) in the expression for \(\mathbf{E}_s\): \(\mathcal{E}(x, t) = |\eta| \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j (\beta x - \omega t)} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j (\beta x - \omega t)}\right)\).
03

Calculate the Instantaneous Magnetic Field

Similarly, to find the instantaneous magnetic field \(\mathcal{H}(x, t)\), we substitute \(e^{-j \omega t}\) for \(e^{-j \beta x}\) in the expression for \(\mathbf{H}_s\): \(\mathcal{H}(x, t) = \left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{-j (\beta x - \omega t)}\).
04

Compute the Time-Averaged Poynting Vector

The time-averaged Poynting vector \(\langle \mathbf{S} \rangle\) can be found using the expression: \(\langle \mathbf{S} \rangle = \frac{1}{2} \text{Re}\{\mathbf{E_s} \times \mathbf{H_s^*}\}\). Substitute the values for \(\mathbf{E_s}\) and \(\mathbf{H_s}\): \(\langle \mathbf{S} \rangle = \frac{1}{2} \text{Re}\{(|\eta| e^{j \phi} \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j \beta x} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j \beta x}\right)) \times (\left(H_{0 y} \mathbf{a}_{y}+H_{0 z} \mathbf{a}_{z}\right) e^{-\alpha x} e^{j \beta x})\}\) Now, you can simplify the expression to obtain the time-averaged Poynting vector \(\langle\mathbf{S}\rangle\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Intrinsic Impedance
Intrinsic impedance is a fundamental property of any medium that describes how electromagnetic waves propagate through it. It is denoted as \( \eta \) and can be a complex number expressed as \( \eta = |\eta| e^{j \phi} \), where \( |\eta| \) is the magnitude, and \( e^{j \phi} \) represents the phase angle. Intrinsic impedance tells us how the electric and magnetic fields are related in that medium.

  • When an electromagnetic wave travels through a medium, the relationship between the electric field \( \mathbf{E}_s \) and the magnetic field \( \mathbf{H}_s \) is governed by intrinsic impedance.
  • In a vacuum, the intrinsic impedance \( \eta_0 \) is approximately 377 ohms.
  • This property is crucial for calculating how much the wave's amplitude changes when entering a new medium.
By using the intrinsic impedance, we can convert between electric and magnetic fields, aiding in the analysis of electromagnetic waves.
Poynting Vector
The Poynting vector \( \mathbf{S} \) is a vital concept in understanding the flow of electromagnetic energy. It represents the power per unit area of the wave propagating through a medium. Mathematically, the Poynting vector is calculated as the cross product of the electric field \( \mathbf{E}_s \) and the complex conjugate of the magnetic field \( \mathbf{H}_s^* \).

  • The expression for computing the time-averaged Poynting vector is \( \langle \mathbf{S} \rangle = \frac{1}{2} \text{Re}\{\mathbf{E}_s \times \mathbf{H}_s^*\} \).
  • This indicates the direction and rate at which energy is transported through the area in which the fields are defined.
  • The Poynting vector helps to determine the power flow in electromagnetic systems, including antennas and transmission lines.
Understanding this vector is essential for practical applications where energy transfer efficiency needs to be evaluated.
Electric Field
The electric field \( \mathbf{E} \) is a vector field surrounding charged particles. It describes the force exerted by the charge, causing other charges in the field to experience a force. In electromagnetic waves, the electric field varies both with position and time, such as:\[\mathbf{E}_s = |\eta| e^{j \phi} \left(H_{0 y} \mathbf{a}_y e^{-\alpha x} e^{-j \beta x} + H_{0 z} \mathbf{a}_z e^{-\alpha x} e^{-j \beta x}\right).\]

  • The electric field is perpendicular to the magnetic field as well as the direction of wave propagation.
  • In a medium with other properties like loss, the expressions for the electric field can absorb these parameters, manifesting in exponential terms like \( e^{-\alpha x} \).
  • The phase shift \( e^{-j \beta x} \) accounts for the wave's spatial oscillation.
A thorough grasp of the electric field assists in defining potential differences and energy forces in electromagnetic phenomena.
Magnetic Field
The magnetic field \( \mathbf{H} \) is another vector field that represents the magnetic influence in the space around a magnetic material or a moving electric charge. Magnetic fields are described by their direction and magnitude and, just like electric fields, are fundamental to electromagnetic waves. In our context:
\[\mathbf{H}_s = \left(H_{0 y} \mathbf{a}_{y} + H_{0 z} \mathbf{a}_{z}\right)e^{-\alpha x} e^{-j \beta x}.\]

  • Magnetic fields take part in the energy transportation process with the electric field, contributing to the Poynting vector.
  • The terms \( H_{0 y} \) and \( H_{0 z} \) describe the magnetic field component magnitudes along the specific axes.
  • The mathematical expression includes spatial exponential decay and phase shift, reflecting real-world wave behavior.
Understanding the nature and role of magnetic fields is crucial in devices like transformers and motors, as well as in wave propagation scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Perfectly conducting cylinders with radii of \(8 \mathrm{~mm}\) and \(20 \mathrm{~mm}\) are coaxial. The region between the cylinders is filled with a perfect dielectric for which \(\epsilon=10^{-9} / 4 \pi \mathrm{F} / \mathrm{m}\) and \(\mu_{r}=1\). If \(\mathcal{E}\) in this region is \((500 / \rho) \cos (\omega t-4 z) \mathbf{a}_{\rho}\) \(\mathrm{V} / \mathrm{m}\), find \((a) \omega\), with the help of Maxwell's equations in cylindrical coordinates; \((b) \mathcal{H}(\rho, z, t) ;(c)\langle\mathbf{S}(\rho, z, t)\rangle ;(d)\) the average power passing through every cross section \(8<\rho<20 \mathrm{~mm}, 0<\phi<2 \pi\).

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) \(v_{\text {Tef }} / v_{\text {brass }}\).

(a) Most microwave ovens operate at \(2.45 \mathrm{GHz}\). Assume that \(\sigma=1.2 \times\) \(10^{6} \mathrm{~S} / \mathrm{m}\) and \(\mu_{r}=500\) for the stainless steel interior, and find the depth of penetration. \((b)\) Let \(E_{s}=50 \angle 0^{\circ} \mathrm{V} / \mathrm{m}\) at the surface of the conductor, and plot a curve of the amplitude of \(E_{s}\) versus the angle of \(E_{s}\) as the field propagates into the stainless steel.

An electric field in free space is given in spherical coordinates as \(\mathbf{E}_{s}(r)=E_{0}(r) e^{-j k r} \mathbf{a}_{\theta} \mathrm{V} / \mathrm{m} .(a)\) Find \(\mathbf{H}_{s}(r)\) assuming uniform plane wave behavior. \((b)\) Find \(<\mathbf{S}>\cdot(c)\) Express the average outward power in watts through a closed spherical shell of radius \(r\), centered at the origin. \((d)\) Establish the required functional form of \(E_{0}(r)\) that will enable the power flow in part \(c\) to be independent of radius. With this condition met, the given field becomes that of an isotropic radiator in a lossless medium (radiating equal power density in all directions).

Given a wave for which \(\mathbf{E}_{s}=15 e^{-j \beta z} \mathbf{a}_{x}+18 e^{-j \beta z} e^{j \phi} \mathbf{a}_{y} \mathrm{~V} / \mathrm{m}\) in a medium characterized by complex intrinsic impedance, \(\eta(a)\) find \(\mathbf{H}_{s} ;(b)\) determine the average power density in \(\mathrm{W} / \mathrm{m}^{2}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free