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Chapter 8: Problem 8

If oscillatory fields are represented by \(\mathbf{E}=\underline{\mathbf{E}}_{0} c^{j \omega t}\) and \(\mathbf{H}=\breve{\mathbf{H}}_{0} e^{j \omega t}\), using the realpart convention, show that the (real) Poynting vector is given by $$ s=\frac{1}{2}\left(\mathbf{E} \times \mathbf{H}^{*}+\mathbf{E}^{*} \times \mathbf{H}\right)=\operatorname{Re}\left(\underline{\mathbf{E}} \times \check{\mathbf{H}}^{*}\right)=\operatorname{Re}\left(\breve{\mathbf{E}}^{*} \times \breve{\mathbf{H}}\right), $$ where the asterisk denotes complex conjugate. Also, show that the time-average Poynting vector is

Short Answer

Expert verified
Question: Show that the real Poynting vector has the expression \(s = \operatorname{Re}\left(\underline{\mathbf{E}} \times \check{\mathbf{H}}^{*}\right) = \operatorname{Re}\left(\breve{\mathbf{E}}^{*} \times \breve{\mathbf{H}}\right)\) and find the time-average Poynting vector given oscillatory fields E and H represented by complex exponentials. Solution: We found the real part of the Poynting vector using the given expressions for E and H and incorporating the complex conjugate expressions. Then, we calculated the time-average Poynting vector by integrating the real Poynting vector over one period and dividing it by that period. The time-average Poynting vector is given by \(\langle s \rangle = \frac{1}{2} \operatorname{Re}\left(\underline{\mathbf{E}}_{0} \times \breve{\mathbf{H}}_{0}^*\right) = \frac{1}{2} \operatorname{Re}\left(\breve{\mathbf{E}}_{0}^* \times \breve{\mathbf{H}}_{0}\right)\).

Step by step solution

01

Find the Poynting vector using given E and H expressions

First, let's find the Poynting vector using the given expressions for E and H: $$\mathbf{S} = \mathbf{E} \times \mathbf{H} = \underline{\mathbf{E}}_{0} c^{j \omega t} \times \breve{\mathbf{H}}_{0} e^{j \omega t}.$$ Multiply the complex amplitudes and exponentials individually: $$\mathbf{S}=\left(\underline{\mathbf{E}}_{0}\times\breve{\mathbf{H}}_{0}\right) e^{j2\omega t}.$$ Now, let's find the real part of the Poynting vector, which represents the actual power flow.
02

Find the real part of the Poynting vector, considering the conjugate

To find the real part of the Poynting vector, we will incorporate the complex conjugate expressions. This is done as follows: $$s = \frac{1}{2}\left(\mathbf{E} \times \mathbf{H^*} + \mathbf{E^*} \times \mathbf{H}\right) = \frac{1}{2}\left(\underline{\mathbf{E}}_{0} \times \breve{\mathbf{H}}_{0}^* e^{j2 \omega t} + \underline{\mathbf{E}}_{0}^* \times \breve{\mathbf{H}}_{0} e^{-j2 \omega t}\right).$$ Now, we can simplify this expression to match the given result: $$s = \operatorname{Re}\left(\underline{\mathbf{E}} \times \check{\mathbf{H}}^{*}\right)=\operatorname{Re}\left(\breve{\mathbf{E}}^{*}\times \breve{\mathbf{H}}\right).$$
03

Calculate the time-average Poynting vector

To find the time-average Poynting vector, we integrate the real Poynting vector over one period and then divide it by that period. In this case, the period T is \(\frac{2 \pi}{\omega}\). $$\langle s \rangle = \frac{1}{T} \int_{0}^{T} s dt = \frac{1}{T} \int_{0}^{T} \operatorname{Re}\left(\underline{\mathbf{E}} \times \check{\mathbf{H}}^{*}\right) dt = \frac{1}{T} \int_{0}^{T} \operatorname{Re}\left(\breve{\mathbf{E}}^{*} \times \breve{\mathbf{H}}\right) dt $$. Since we're looking for the time-average, we know that the oscillatory parts will average to zero over a complete period. Therefore, we will only be left with the complex amplitude products: $$\langle s \rangle = \frac{1}{2} \operatorname{Re}\left(\underline{\mathbf{E}}_{0} \times \breve{\mathbf{H}}_{0}^*\right) = \frac{1}{2} \operatorname{Re}\left(\breve{\mathbf{E}}_{0}^* \times \breve{\mathbf{H}}_{0}\right).$$

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Key Concepts

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

Oscillatory Electromagnetic Fields
Electromagnetic fields are essential in the understanding of physics and technology, playing critical roles in a variety of applications from radio transmissions to optical fibers. Oscillatory electromagnetic fields are those that change with time, typically in a sinusoidal manner. In physics, these changing fields are often represented mathematically using complex exponential functions due to their convenient properties for calculations.

For instance, an oscillatory electric field can be represented as \( \mathbf{E} = \underline{\mathbf{E}}_{0} e^{j \omega t} \), where \( \underline{\mathbf{E}}_{0} \) is the complex amplitude, \( \omega \) is the angular frequency, and \( t \) is time; \( j \) stands for the imaginary unit. This representation simplifies the analysis of time-varying fields because it encodes both the magnitude and the phase of oscillation within the complex amplitude. Similarly, the magnetic field \( \mathbf{H} \) is represented in a comparable fashion.

Understanding these fields and how they interact is crucial for exploring the transmission, reflection, and absorption of electromagnetic waves in various media.
Complex Amplitude
When delving into complex amplitude, it's crucial to comprehend that it is an elegant mathematical tool used to represent the magnitude and phase of an oscillating field. In the mathematical expressions \( \mathbf{E} = \underline{\mathbf{E}}_{0} e^{j \omega t} \) and \( \mathbf{H} = \breve{\mathbf{H}}_{0} e^{j \omega t} \), the terms \( \underline{\mathbf{E}}_{0} \) and \( \breve{\mathbf{H}}_{0} \) are complex amplitudes of the electric and magnetic fields, respectively. They encapsulate important information about the oscillatory electromagnetic field at a reference point in time, typically chosen as \( t = 0 \).

The real part of the complex amplitude corresponds to the in-phase component relative to the reference time, while the imaginary part represents the quadrature, or 90-degree out-of-phase, component. This form allows for simplified mathematical operations, particularly when working with the product or cross product of two oscillating fields, as seen in the calculation of the Poynting vector.
Time-Average Energy Flow
The time-average energy flow in an electromagnetic field is a measurement of the power transferred per unit area over an average time period. It has critical implications for energy transport, such as in the context of waveguides or radiation from antennas. To describe this flow quantitatively, we resort to the Poynting vector \( \mathbf{S} \).

The time-averaged Poynting vector \( \langle s \rangle \) uses the real parts of the electric and magnetic field complex amplitudes. This averaging process effectively filters out the rapid oscillations of the fields, leaving only the steady component of the power flow. The average is taken over one complete oscillation period, denoted as \( T \) and equals to \( \frac{2 \pi}{\omega} \), which results in the expression:\[ \langle s \rangle = \frac{1}{2} \operatorname{Re}\left(\underline{\mathbf{E}}_{0} \times \breve{\mathbf{H}}_{0}^*\right) \].

This power flow can be visualized as the energy transported by the electromagnetic wave through a specific area per unit time. It is a pivotal concept when analyzing communication systems, power systems, and even in the study of electromagnetic radiation's effects on biological tissue.

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Most popular questions from this chapter

A small permanent magnet of dipole moment \(m\), supported by an insulating thread, is given an electrostatic charge \(q\). Calculate the Poynting vector (at distances large compared to the size of the magnet). Also calculate its divergence. How does this problem differ from Probs. \(8.4 .3\) and \(8.4 .4\) ?

From \((8.7 .18)\) show that the phase velocity of the wave in a waveguide is $$ c_{p}=\frac{\omega}{\kappa_{x}}=\frac{c}{\left[1-\left(\lambda_{4} / \lambda_{e}\right)^{2}\right]^{1 / 2}} $$ Note that this exceeds the velocity of light \(c !\) Find the group velocity \(c_{\theta}=d \omega / d k_{x}\) and show that $$ c_{p} c_{g}=c^{2} $$ Explain the distinction between \(c_{\rho}, c_{,}\)and \(c_{p}\) in terms of the plane-uave analysis of Prob. \(8.7 .6\) for the \(\mathrm{TE}_{10}\) mode in rectangular waveguide.

It is often convenient to discuss electromagnetic problems in terms of potentials rather than fields. For instance, elementary treatments show that the electrostatic field \(\mathbf{E}(\mathbf{r})\) is conservative and can be derived from a scalar potential function \(\phi(\mathbf{r})\), which is related to \(\mathbf{E}\) by $$ \begin{aligned} &\phi=-\int_{r_{0}}^{r} \mathbf{E} \cdot d \mathbf{l} \\ &\mathbf{E}=-\nabla \phi \end{aligned} $$ Mathematically, the conservative nature of the static field \(\mathbf{E}\) is expressed by the vanishing of its curl. Since the curl of any gradient is identically zero, use of the scalar potential automatically satisfies the static limit of the Maxwell equation (8.2.2); the other constraint on \(\phi\) is Gauss' law (8.2.1). Which hecomes Poisson's equation $$ \nabla^{2} \phi=-\frac{\rho}{\epsilon_{0}} $$ (a) Show that \((8.2 .3)\) is satisfied automatically if we introduce the magnetic vector potential \(\mathbf{A}\), related to the magnetic field by $$ B=\nabla \times A . $$ (b) Show that in the general (nonstatic) case, the electric field is given in terms of the scalar and vector potentials by $$ \mathbf{E}=-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t} $$ (c) Complete the prescription of \(\mathbf{A}\) by defining its divergence by the Lorents condition $$ \boldsymbol{\nabla} \cdot \mathbf{A}=-\frac{1}{c^{2}} \frac{\partial \phi}{\partial t} $$ and show that the two potentials obey the symmetrical inhomogeneous wave equations $$ \begin{aligned} &\nabla^{2} \phi-\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}=-\frac{\rho}{\epsilon_{0}} \\ &\nabla^{2} \mathbf{A}-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}}=-\mu_{0} \mathbf{J} . \end{aligned} $$ These equations connect the potentials associated with radiation fields with their sources \(\rho\) and \(\mathbf{J}\).

Show, in general, that for TE modes the tangential-E boundary condition implies that the normal derivative of the scalar function \(\phi\) must vanish at the boundary, whereas for TM modes the normal-B boundary condition implies that the function \(\phi\) itself must vanish at the boundary.

When matter is present, the phenomenon of polarization (electrical displacement of charge in a molecule or alignment of polar molecules) can produce unneutralized (bound) charge that properly contributes to \(\rho\) in \((8.2 .1)\). Similarly the magnetization of magnetic materials, as well as time-varying polarization, can produce efiective currents that contribute to \(J\) in \((8.24)\). These dependent source charges and currents, as opposed to the independent or "causal" free charges and currents, can be taken into account implicitly by introducing two new fields, the dectric displacement \(\mathbf{D}\) and the magnetic intensity \(\mathbf{H} .+\) For linear isotropic media, $$ \begin{aligned} &\mathbf{D}=\kappa_{\varepsilon} \epsilon_{0} \mathbf{E} \\ &\mathbf{H}=\frac{\mathbf{B}}{\kappa_{m} \mu_{0}} \end{aligned} $$ where \(\kappa_{0}\) is the relative permittivity (or dielectric constant) and \(\kappa_{m}\) is the rclative permeability of the medium. In this more general situation, Maxwell's equations are $$ \begin{aligned} &\nabla \cdot \mathbf{D}=\rho_{\text {ree }} \\ &\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t} \\ &\nabla \cdot \mathbf{B}=0 \\ &\nabla \times \mathbf{H}=\mathbf{J}_{\text {frea }}+\frac{\partial \mathbf{D}}{\partial l} \end{aligned} $$ Show that in a homogeneous material medium without free charges or currents, the fields obey the simple wave equation with a velocity of propagation $$ c^{\prime}=\frac{1}{\left(x_{q} \operatorname{tos}_{m} \mu_{0}\right)^{1 / 2}}=\frac{c}{\left(\alpha_{q} K_{m}\right)^{1 / 2}} $$ and that consequently the refractive index of the medium is given by $$ n=\left(x_{q} K_{m}\right)^{1 / 2} $$
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