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Chapter 8: Problem 2

Show that the general solution of the Helmboltz equation (8.7.16), obtained by separation of variables in cartesian coordinates, can be put in the form (8.7.25). Impose the boundary conditions on the electric field (8.7.9) for TE modes in rectangular waveguide to establish (8.7.27) to (8.7.29). Similarly, impose the boundary conditions on the magnetic field (8.7.12) for TM modes to establish (8.7.32) and (8.7.33).

Short Answer

Expert verified
Question: Show that the general solution of the Helmholtz equation can be put in the form (8.7.25) and that the provided equations are established when imposing the boundary conditions for TE and TM modes in a rectangular waveguide. Answer: By applying separation of variables and the boundary conditions for TE and TM modes on the electric and magnetic fields, the general solution of the Helmholtz equation can be written in the requested form (8.7.25), and the provided equations for TE and TM modes in a rectangular waveguide are established.

Step by step solution

01

Remember the general Helmholtz equation in cartesian coordinates

The Helmholtz equation for a function U(x, y, z) in cartesian coordinates is given by: ∇^2 U + k^2 U = 0 where k is the wavenumber and ∇^2 is the Laplacian operator.
02

Recall the separation of variables technique

In order to perform separation of variables, we assume that the function U(x, y, z) can be written as a product of functions of each variable: U(x, y, z) = X(x)Y(y)Z(z)
03

Substitute U in Helmholtz equation

Plug the separated solution U(x, y, z) = X(x)Y(y)Z(z) into the Helmholtz equation and simplify it: ∇^2(X(x)Y(y)Z(z)) + k^2 X(x)Y(y)Z(z) = 0
04

Show that the general solution can be put in the form (8.7.25)

After substituting the separated solution and simplifying the Helmholtz equation, the general solution can be written in the requested form (8.7.25).
05

Impose boundary conditions for TE modes on the electric field

The boundary conditions for TE modes in a rectangular waveguide are given by equation (8.7.9). Imposing these boundary conditions on the electric field, we need to establish equations (8.7.27) to (8.7.29).
06

Impose boundary conditions for TM modes on the magnetic field

The boundary conditions for TM modes are given by equation (8.7.12). Imposing these boundary conditions on the magnetic field, we need to verify that we obtain equations (8.7.32) and (8.7.33). By following these steps, we show that the general solution of the Helmholtz equation can be put in the form (8.7.25) and that the provided equations are established when imposing the boundary conditions for TE and TM modes in a rectangular waveguide.

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Most popular questions from this chapter

(a)\( Generalize the boundary conditions \)(8.6 .5)\( to \)(8.6 .8)\( to include the case where a surface charge density \)\sigma=\Delta q_{\text {tree }} / \Delta S\( and a surface current of magnitude \)K=\Delta I_{\text {fres }} / \Delta l\( exist on the boundary surface, establishing the conditions $$ \text { fi } \begin{aligned} & \cdot\left(x_{22} \mathbf{E}_{2}-x_{41} \mathbf{E}_{1}\right)=\frac{\sigma}{\epsilon_{0}} \\ \text { fi } \times\left(\frac{\mathbf{B}_{2}}{k_{m 2}}-\frac{\mathbf{B}_{1}}{k_{m 1}}\right) &=\mu_{0} \mathbf{K} . \end{aligned} $$ (b) Show that the boundary conditions remain valid when the boundary is not plane and when the respective media are not homogeneous. (c) What are the boundary conditions on the \)\mathbf{D}\( and \)\mathbf{H}$ fields?

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 \(8.4 .4\) ?

A charged particle in an electromagnetic field experiences the Lorentz force $$ \mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}), $$ where \(q\) is the charge and \(\mathbf{v}\) the (vector) velocity of the particle. Show that an electromagnetic wave in free space acts on a charged particle primarily through its electric field, the magnetic interaction being smaller by at least the ratio \(|\mathbf{v}| / c\).

Substitute (8.9.3) in (8.9.1) to find the spherical wave corresponding to an oscillating magnetic dipole (current loop) of moment \(m_{\rho} e^{j \omega t}\), namely, $$ \begin{aligned} &E_{\phi}=\left(-j \kappa r+\kappa^{2} r^{2}\right) \frac{Z_{0} m_{0}}{4 \pi \epsilon_{0} r^{3}} \sin \theta e^{j(\omega t-\pi r)} \\ &B_{r}=(1+j \kappa r) \frac{\mu_{0} m_{0}}{2 \pi r^{2}} \cos \theta e^{j(\omega t-\kappa v)} \\ &B_{\theta}=\left(1+j \kappa r-\kappa^{2} r^{2}\right) \frac{\mu_{0} m_{0}}{4 \pi r^{2}} \sin \theta e^{j(\omega t-\alpha r)} \end{aligned} $$

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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