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Chapter 8: Problem 1

Develop Poynting's theorem for the general material medium of relative permittivity \(\kappa_{\&}\) and permeability \(\kappa_{m}\) introduced in Prob. \(8.2 .2\); i.e., substitute the Maxwell curl equations \((8.2 .18)\) and \((8.2 .20)\) in the expansion of \(\nabla \cdot(\mathbf{E} \times \mathbf{H})\) to obtain $$ \oint_{S}(\mathbf{E} \times \mathbf{H}) \cdot d \mathbf{S}+\int_{V}\left(\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}+\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}\right) d v+\int_{V} \mathbf{E} \cdot \mathbf{J} d v=0 $$ from which it follows that the Poynting vector is $$ s=\mathbf{E} \times \mathbf{H} $$ and the energy density is $$ \begin{aligned} \left(W_{1}\right)_{\text {leld }} &=\frac{1}{2} \mathbf{E} \cdot \mathbf{D}+\frac{1}{2} \mathbf{H} \cdot \mathbf{B} \\ &=\frac{1}{2} \kappa_{\ell} \epsilon_{0} E^{2}+\frac{1}{2 \kappa_{\mathrm{m}} \mu_{0}} B^{2} \end{aligned} $$ What restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\) are necessary to obtain \((8.4 .24)\) ?

Short Answer

Expert verified
In summary, the Poynting vector for a general material medium is given by \(\mathbf{E} \times \mathbf{H}\) and the energy density is given by \(\frac{1}{2} \kappa_{\&} \epsilon_{0} E^{2}+\frac{1}{2 \kappa_{m} \mu_{0}} B^{2}\). To obtain equation (8.4.24), \(\frac{1}{2} \epsilon_{0} E^{2}+\frac{1}{2 \mu_{0}} B^{2}\), we need the restrictions on relative permittivity \(\kappa_{\&}\) and permeability \(\kappa_{m}\) to be both equal to 1.

Step by step solution

01

Develop Poynting's theorem for the general material medium

We are given the Maxwell curl equations \((8.2 .18)\) and \((8.2 .20)\) and have to substitute these in the expansion of \(\nabla \cdot(\mathbf{E} \times \mathbf{H})\). The Maxwell curl equations are: $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$ $$ \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} $$ We know that \(\mathbf{D} = \kappa_{\&} \epsilon_0 \mathbf{E}\) and \(\mathbf{B} = \kappa_{m} \mu_0 \mathbf{H}\). Substituting these relations in the above equations, we get: $$ \nabla \times \mathbf{E} = -\frac{\partial (\kappa_{m}\mu_0 \mathbf{H})}{\partial t} $$ $$ \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial (\kappa_{\&} \epsilon_0 \mathbf{E})}{\partial t} $$ Now, we take the divergence of the Poynting vector: $$ \nabla \cdot (\mathbf{E} \times \mathbf{H}) = \nabla \cdot (\mathbf{E} \times (\frac{1}{\kappa_{m}\mu_0} \mathbf{B})) $$ Substituting the Maxwell curl equations, we obtain Poynting's theorem as: $$ \oint_{S}(\mathbf{E} \times \mathbf{H}) \cdot d \mathbf{S} + \int_{V} \left(\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t} + \mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}\right) d v + \int_{V} \mathbf{E} \cdot \mathbf{J} d v = 0 $$
02

Find the Poynting vector and energy density

From the Poynting's theorem, we can find the Poynting vector as: $$ s = \mathbf{E} \times \mathbf{H} $$ The energy density can be found by considering the two components in the above integral expression: $$ \begin{aligned} \left(W_{1}\right)_{\text {leld }} &=\frac{1}{2} \mathbf{E} \cdot \mathbf{D}+\frac{1}{2} \mathbf{H} \cdot \mathbf{B} \\ &=\frac{1}{2} \kappa_{\ell} \epsilon_{0} E^{2}+\frac{1}{2 \kappa_{\mathrm{m}} \mu_{0}} B^{2} \end{aligned} $$
03

Restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\)

Now, we need to find the restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\) necessary to obtain equation \((8.4 .24)\). Equation \((8.4 .24)\) is given by: $$ \frac{1}{2} \epsilon_{0} E^{2}+\frac{1}{2 \mu_{0}} B^{2} $$ Comparing this equation with the energy density equation derived above, we can see that the restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\) are given by: $$ \kappa_{\&} = 1 $$ $$ \kappa_{m} = 1 $$ We need these restrictions for the energy density equation to match with equation \((8.4 .24)\).

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Key Concepts

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

Maxwell's equations
Maxwell's equations are the backbone of electromagnetic theory. They are a set of four equations that describe how electric and magnetic fields interact. In this problem, we focus on the curl equations:
  • The first equation, \( abla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \), shows how a changing magnetic field \( \mathbf{B} \) generates an electric field \( \mathbf{E} \).
  • The second, \( abla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} \), indicates that changing electric displacement \( \mathbf{D} \) and current density \( \mathbf{J} \) create a magnetic field \( \mathbf{H} \).
These equations help in bridging the electric field with changes in magnetic fields and vice versa, enabling us to explore electromagnetic wave propagation.
Poynting vector
The Poynting vector, denoted as \( \mathbf{S} = \mathbf{E} \times \mathbf{H} \), represents the directional energy flux of an electromagnetic field. Imagine it as a flow of energy per unit area per unit time. Here, the cross product of electric \( \mathbf{E} \) and magnetic field intensity \( \mathbf{H} \) gives a vector that points in the direction energy travels.
  • This vector is crucial for determining energy transfer in systems, helping us map how energy moves through space.
  • It's key in diverse applications like wireless charging and power flow in circuits.
Understanding the Poynting vector aids in visualizing how energy is distributed in an electromagnetic field.
Energy density
Energy density gives an insight into where energy is stored in an electromagnetic field. It is quantified as:\[\left(W_{1}\right)_{\text{leld }} = \frac{1}{2} \mathbf{E} \cdot \mathbf{D} + \frac{1}{2} \mathbf{H} \cdot \mathbf{B}\]This breaks down into two parts:
  • \( \frac{1}{2} \mathbf{E} \cdot \mathbf{D} \) represents the energy stored in the electric field.
  • \( \frac{1}{2} \mathbf{H} \cdot \mathbf{B} \) signifies the energy in the magnetic field.
  • For a general medium, this becomes \( \frac{1}{2} \kappa_{\ell} \epsilon_{0} E^{2} + \frac{1}{2 \kappa_{\mathrm{m}} \mu_{0}} B^{2} \).
This concept helps us evaluate how much energy fields like light or radio waves can store and transmit.
Permittivity and permeability
Permittivity \( (\epsilon) \) and permeability \( (\mu) \) are inherent properties of materials that describe how they respond to electric and magnetic fields, respectively.
  • Permittivity \( \epsilon \) relates to a material's ability to permit electric field lines to pass through it. It is modified by the dimensionless factor \( \kappa_{\&} \), indicating how the material amplifies or decreases the electrical component.
  • Permeability \( \mu \) defines how a material supports or opposes the formation of magnetic fields within it, influenced by \( \kappa_{m} \).
  • In free space (a vacuum), the relative permittivity and permeability are both 1, as used in the restrictions \( \kappa_{\&} = 1 \) and \( \kappa_{m} = 1 \).
Understanding these properties is vital for designing materials that control electromagnetic fields effectively.
Divergence theorem
The divergence theorem is a mathematical principle that connects a flux flowing out of a closed surface to the divergence within the volume enclosed by the surface. It can transform the volume integrals of vector fields into surface integrals.
  • In Poynting's theorem, it allows the transition from a local form (point-wise balance of energy) to an integral form (global balance over a region).
  • This theorem helps in converting computations involving complex field expressions into more manageable boundary analyses.
  • It's used extensively in engineering and physics for simplifying and solving equations involving fluxes.
By applying the divergence theorem, we tie the local behavior of fields to observable quantities over surfaces.

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Most popular questions from this chapter

Consider an inhomogeneous dielectric medium, i.e., one for which the dielectric constant is a function of position, \(\kappa_{e}=\kappa_{e}(x, y, z)\). Show that the fields obey the wave equations $$ \begin{aligned} &\nabla^{2} \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ &\nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}}=-\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$ where, in general, the terms on the right-hand sides couple the cartesian components of the fields. Now introduce the special case that the permittivity changes only in the direction of propagation (the \(z\) direction, say) and show that for monochromatic plane waves the equations become $$ \begin{aligned} &\frac{d^{2} \mathbf{E}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{\theta}(z) \mathbf{E}=0 \\ &\frac{d^{2} \mathbf{B}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{e}(z) \mathbf{B}=\frac{1}{\kappa_{e}(z)} \frac{d x_{e}}{d z} \frac{d \mathbf{B}}{d z} \end{aligned} $$ Approximate solution of this type of equation is discussed in Sec. \(9.1 .\)

Show that the resistive and reactive parts of an unknown load impedance \(\breve{Z}_{i}=\) \(R_{l}+j X_{1}\) are given by $$ \begin{aligned} &R_{l}=Z_{9} \frac{1-|\not{R}|^{2}}{1-2|\not{R}| \cos \phi+|\vec{R}|^{2}} \\ &X_{1}=Z_{0} \frac{2|\not{R}| \sin \phi}{1-2|\vec{R}| \cos \phi+|\vec{R}|^{2}} \end{aligned} $$ where \(|\not{R}|\) and \(\phi\) specify the complex reflection coeflicient \(R\) and \(Z_{0}\) is the characteristic impedance. Note: See Prob. 1.4.3.

Show that parallel conducting planes of separation \(a\) can support a TE mode identical to the TE \(_{10}\) mode in rectangular waveguide with \(b \rightarrow \infty\). Show further that the parallel planes can also support a TM mode that has no direct analog in rectangular waveguide \((b\) finite) but is of the form of \((8.7 .32)\) and \((8.7 .33)\) with \(m \rightarrow 0, \sin m \pi y / b \rightarrow 1 .\)

(a) Show that the skin depth \(\delta\) can be put in the form $$ \delta=\left(\frac{\lambda_{0}}{\pi Z_{\mu} \pi_{m} g}\right)^{1 / 2} $$ where \(Z_{0}\) is the free-space wave impedance \((8.3,10)\) and \(\lambda_{0}=2 \pi c / \omega\) is the vacuum wavelength. (b) Evaluate s for copper, for waves having wavelengths in vacuum of \(5,000 \mathrm{~km}(60-\mathrm{Hz}\) power line); \(100 \mathrm{~m}\) ( \(\sim\) AM broadcast band); \(1 \mathrm{~m}\) ( \(\sim\) television and FM broadcast); \(3 \mathrm{~cm}\) ( \(\sim\) radar); \(500 \mathrm{~nm}\) ( \(\sim\) visible light). How does the size of the skin depth affect the technology of these various applications" (c) The electrical conductivity of sea water is about \(4 \mathrm{mhos} / \mathrm{m}\). How would you communicate by radio with a submarine \(100 \mathrm{~m}\) below the surface?

For normal incidence and nonmagnetic materials show that the power coefficients \((8.6 .38)\) and \((8.6 .39)\) reduce to $$ \begin{aligned} &R_{p}=\left(\frac{n-1}{n+1}\right)^{x} \\ &T_{p}=\frac{4 n}{(n+1)^{2}} \end{aligned} $$ where \(n=c_{1} / c_{2}=Z_{01} / Z_{02}\) is the relative refractive index. Account for the difference in sign between the amplitude reflection coefficients \((8.6 .28)\) and \((8.6 .36)\) at normal incidence (see footnote, page 102). Compare with equations (1.9.6), (4.2.15), and (5.6.13).
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