Chapter 8

A plane electromagnetic wave, with momentum density \(\mathrm{p}_{1}\), is incident on a plane absorbing surface at an angle \(\theta\) with respect to the normal. \((a)\) Show that the normal force per unit area, i.e., the pressure, is $$ p=\bar{W}_{1} \cos ^{2} \theta \text {. } $$ (b) Show that if waves are incident on the surface at all angles, the pressure is $$ p=\frac{1}{3} \bar{W}_{1} . $$ (c) If the surface has a power reflection coefficient \(R(\theta)\), how are these results affected?

Postulate wave fields of the form $$ \begin{aligned} &\mathbf{E}=\mathbf{i} f(z-c t)+\mathbf{j} g(z-c t)+\mathbf{k} h(z-c t) \\ &\mathbf{B}=\mathbf{i} q(z-c t)+\mathbf{j} r(z-c t)+\mathbf{k} s(z-c t) \end{aligned} $$ where \(f, g, h, q, r, s\) are arbitrary (nonsinusoidal) functions, independent of \(x\) and \(y .\) Show that such waves are a solution of the wave equations \((8.2 .8)\) and \((82.9)\) and that Maxwell's equations (8.2.1) to \((8.2 .4)\) require $$ \begin{aligned} &h=s=0 \\ &f=c r \\ &g=-c q \end{aligned} $$ that is, that only two of the six functions are really arbitrary.

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.

(a) Use Ampère's law (8.6.4) to prove that the current density \(J(z)\) in the conductor of Fig. \(8.6 .2\) is related to the net magnetic field just outside the conductor by $$ B_{\text {outeide }}=\mu_{0} \int_{0}^{\infty} J(z) d z=\mu_{0} K, $$ where the integral symbolized by \(K\) has the dimensions of a surface current density, namely, amperes per meter. \((b)\) Consider an artificial model whereby the surface current \(K\) is distributed uniformly in the skin layer \(0

Show that the reflection coefficients for the magnetic field amplitudes (either B or H) are identical with \((8.6 .28)\) and \((8.6 .36)\), while the transmission coefficients differ from (8.6.29) and \((8.6 .37)\) by the ratio of the wave impedances of the two media, \((8.5 .18)\) or \((8.5 .19)\). Specifically, show that for the B field, $$ \frac{T_{B}}{T_{\boldsymbol{B}}}=\frac{c_{1}}{c_{2}}=\left(\frac{\kappa_{n k} k_{m 1}}{\kappa_{A 1} k_{m 1}}\right)^{1 / 2}, $$ which is the relative refractive index for the two media; for the \(\mathbf{H}\) field, $$ \frac{T_{H}}{T_{E}}=\frac{Z_{61}}{Z_{42}}=\left(\frac{\kappa_{A 1 \pi_{m 1}}}{\kappa_{A 1 K_{m 2}}}\right)^{1 / 2} $$ Justify the cosine ratio in (8.6.39).

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.

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

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.

Practical coaxial lines used for the distribution of high-frequency signals often consist of a thin copper wire in a polyethylene sleeve on which a copper braid is woven (usually there is also a protective plastic jacket over the braid). Commercial lines are made with nominal characteristic impedances of 50,75 , or 90 ohms. A common 50 -ohm variety has a center conductor of diameter 0035 in. The dielectric constant of polyethylene is \(2.3\) What is the nominal (inside) diameter of the copper braid? What are the capacitance and inductance per foot? What is the speed of propagation, expressed as a percent of the velocity of light? Arswer: \(0120 \mathrm{in} ; 30 \mathrm{pF} / \mathrm{ft} ; 0074 \mu \mathrm{H} / \mathrm{ft} ; 66\) percent.