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

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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 \(\langle\mathbf{S}\rangle\).

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Most popular questions from this chapter

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 \(<\mathbf{S}>\).

Suppose \(\epsilon_{r x}=2.15, \epsilon_{r y}=2.10\), and the wave electric field at input is polarized at \(45^{\circ}\) to the positive \(x\) and \(y\) axes. (a) Determine, in terms of the free space wavelength, \(\lambda\), the shortest length of the material, such that the wave, as it emerges from the output, is circularly polarized. ( \(b\) ) Will the output wave be right or left circularly polarized? Problem \(11.30\) is good background. Suppose that the length of the medium of Problem \(11.31\) is made to be twice that determined in the problem. Describe the polarization of the output wave in this case.

A \(10 \mathrm{GHz}\) uniform plane wave propagates in a lossless medium for which \(\epsilon_{r}=8\) and \(\mu_{r}=2 .\) Find \((a) v_{p} ;(b) \beta ;(c) \lambda ;(d) \mathbf{E}_{s} ;(e) \mathbf{H}_{s} ;(f)\langle\mathbf{S}\rangle .\)

The phasor magnetic field intensity for a \(400 \mathrm{MHz}\) uniform plane wave propagating in a certain lossless material is \(\left(2 \mathbf{a}_{y}-j 5 \mathbf{a}_{z}\right) e^{-j 25 x} \mathrm{~A} / \mathrm{m}\). Knowing that the maximum amplitude of \(\mathbf{E}\) is \(1500 \mathrm{~V} / \mathrm{m}\), find \(\beta, \eta, \lambda, v_{p}\), \(\epsilon_{r}, \mu_{r}\), and \(\mathcal{H}(x, y, z, t) .\)

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

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