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Chapter 11: Problem 29

Consider a left circularly polarized wave in free space that propagates in the forward \(z\) direction. The electric field is given by the appropriate form of Eq. (100). Determine ( \(a\) ) the magnetic field phasor, \(\mathbf{H}_{s} ;(b)\) an expression for the average power density in the wave in \(\mathrm{W} / \mathrm{m}^{2}\) by direct application of Eq. (77).

Short Answer

Expert verified
Based on the given information about the electric field of a left circularly polarized wave propagating in the forward z direction, determine the magnetic field phasor and an expression for the average power density in the wave. Solution: 1. The magnetic field phasor is given by: \(\textbf{H} = \frac{1}{\omega\mu}(E_y\textbf{u}_x - jE_x\textbf{u}_z)e^{j(\omega t - kz)}\) 2. The average power density of the wave is: \(S_{av} = \frac{1}{2}(E_xE_y^* - E_yE_x^*)\textbf{u}_y\) W/m²

Step by step solution

01

(Step 1: Recall the Equation (100) for the electric field phasor)

For a left circularly polarized wave, the electric field phasor, \(\textbf{E}\), is given by Equation (100) as: \(\textbf{E}(\textbf{r}, t) = E_x\textbf{u}_x\ e^{j(\omega t - kz)} + jE_y\textbf{u}_y\ e^{j(\omega t - kz)}\) where \(E_x\) and \(E_y\) are the amplitudes of the electric field components, \(k\) is the wave number, \(\omega\) is the angular frequency, and \(\textbf{u}_x\) and \(\textbf{u}_y\) are the unit vectors along the x and y axes.
02

(Step 2: Find the magnetic field phasor using Faraday's Law)

Faraday's Law relates the electric field to the magnetic field as follows: \(\nabla \times \textbf{E} = -j\omega\mu\textbf{H}\) So, we can write the relation between the magnetic field phasor, \(\textbf{H}\), and the electric field phasor, \(\textbf{E}\), as: \(\textbf{H} = -\frac{1}{j\omega\mu}(\nabla \times \textbf{E})\)
03

(Step 3: Determine the curl of the electric field phasor)

Calculate the curl \(\nabla \times \textbf{E}\): \(\nabla \times \textbf{E} = \Big(\frac{\partial}{\partial y}(jE_y e^{j(\omega t - kz)}) - \frac{\partial}{\partial z}(E_x e^{j(\omega t - kz)})\Big)\textbf{u}_x - \Big(\frac{\partial}{\partial x}(jE_y e^{j(\omega t - kz)})\Big)\textbf{u}_y + \Big(\frac{\partial}{\partial x}(E_x e^{j(\omega t - kz)})\Big)\textbf{u}_z\) Using the chain rule, we have: \(\nabla \times \textbf{E} = (jE_yjk e^{j(\omega t - kz)})\textbf{u}_x - 0\textbf{u}_y - (E_xjk e^{j(\omega t - kz)})\textbf{u}_z\)
04

(Step 4: Calculate the magnetic field phasor)

Substitute the curl of the electric field phasor into the expression for the magnetic field phasor: \(\textbf{H} = -\frac{1}{j\omega\mu}(jE_y k\textbf{u}_x\ e^{j(\omega t - kz)} - E_x k\textbf{u}_z\ e^{j(\omega t - kz)})\) Hence, the magnetic field phasor is given by \(\textbf{H} = \frac{1}{\omega\mu}(E_y\textbf{u}_x - jE_x\textbf{u}_z)e^{j(\omega t - kz)}\)
05

(Step 5: Recall the equation (77) for average power density)

The average power density, \(S_{av}\), is given by Equation (77) as: \(S_{av} = \frac{1}{2}Re(\textbf{E} \times \textbf{H}^*)\)
06

(Step 6: Calculate the cross product of electric and conjugate magnetic fields)

Find the cross product of \(\textbf{E}\) and \(\textbf{H}^*\): \(\textbf{E} \times \textbf{H}^* = (E_y^* \textbf{u}_y + jE_x^* \textbf{u}_z)e^{-j(\omega t - kz)} \times \frac{1}{\omega\mu}(E_y^*\textbf{u}_x + jE_x^*\textbf{u}_z)e^{-j(\omega t - kz)}\) Using the cross product rule, we have: \(\textbf{E} \times \textbf{H}^* = \frac{1}{\omega\mu}(jE_xE_y^* - jE_yE_x^*)e^{-j2(\omega t - kz)}\textbf{u}_y\)
07

(Step 7: Find the average power density)

The average power density is given by, \(S_{av} = \frac{1}{2}Re(jE_xE_y^* - jE_yE_x^*)\textbf{u}_y\) Therefore, the average power density of the given left circularly polarized wave is: \(S_{av} = \frac{1}{2}(E_xE_y^* - E_yE_x^*)\textbf{u}_y\) W/m²

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Most popular questions from this chapter

A uniform plane wave has electric field \(\mathbf{E}_{s}=\left(E_{y 0} \mathbf{a}_{y}-E_{z 0} \mathbf{a}_{z}\right) e^{-\alpha x} e^{-j \beta x} \mathrm{~V} / \mathrm{m} .\) The intrinsic impedance of the medium is given as \(\eta=|\eta| e^{j \phi}\), where \(\phi\) is a constant phase. \((a)\) Describe the wave polarization and state the direction of propagation. \((b)\) Find \(\mathbf{H}_{s} \cdot(c)\) Find \(\mathcal{E}(x, t)\) and \(\mathcal{H}(x, t) .(d)\) Find \(<\mathbf{S}>\) in \(\mathrm{W} / \mathrm{m}^{2} \cdot(e)\) Find the time-average power in watts that is intercepted by an antenna of rectangular cross-section, having width \(w\) and height \(h\), suspended parallel to the \(y z\) plane, and at a distance \(d\) from the wave source.

Consider the power dissipation term, \(\int \mathbf{E} \cdot \mathbf{J} d v\), in Poynting's theorem (Eq. (70)). This gives the power lost to heat within a volume into which electromagnetic waves enter. The term \(p_{d}=\mathbf{E} \cdot \mathbf{J}\) is thus the power dissipation per unit volume in \(\mathrm{W} / \mathrm{m}^{3}\). Following the same reasoning that resulted in Eq. (77), the time-average power dissipation per volume will be \(=(1 / 2) \mathcal{R} e\left\\{\mathbf{E}_{s} \cdot \mathbf{J}_{s}^{*}\right\\} .\) (a) Show that in a conducting medium, through which a uniform plane wave of amplitude \(E_{0}\) propagates in the forward \(z\) direction, \(\left.

A hollow tubular conductor is constructed from a type of brass having a conductivity of \(1.2 \times 10^{7} \mathrm{~S} / \mathrm{m} .\) The inner and outer radii are 9 and \(10 \mathrm{~mm}\), respectively. Calculate the resistance per meter length at a frequency of (a) \(\mathrm{dc} ;\) (b) \(20 \mathrm{MHz}\); (c) \(2 \mathrm{GHz}\).

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 \(8<\rho<20 \mathrm{~mm}, 0<\phi<2 \pi\).

Small antennas have low efficiencies (as will be seen in Chapter 14 ), and the efficiency increases with size up to the point at which a critical dimension of the antenna is an appreciable fraction of a wavelength, say \(\lambda / 8 .(a) \mathrm{An}\) antenna that is \(12 \mathrm{~cm}\) long is operated in air at \(1 \mathrm{MHz}\). What fraction of a wavelength long is it? \((b)\) The same antenna is embedded in a ferrite material for which \(\epsilon_{r}=20\) and \(\mu_{r}=2,000\). What fraction of a wavelength is it now?
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