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Chapter 8: Problem 7

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.

Short Answer

Expert verified
Answer: For TE modes, the normal derivative of the scalar function φ must vanish at the boundary. For TM modes, the scalar function φ itself must vanish at the boundary.

Step by step solution

01

1.Definition of TE and TM modes in terms of scalar function φ

For TE modes, the electric field components, E, are transverse and can be represented as the gradient of a scalar function φ. Mathematically, this can be written as: E = -∇φ For TM modes, the magnetic field components, H, are transverse and can be represented as the gradient of a scalar function φ. Mathematically, this can be written as: H = -∇φ Where the minus sign indicates that the field opposes the increase in potential function.
02

2. The tangential-E boundary condition for TE modes

The tangential-E boundary condition states that the tangential component of electric field (Et) must be continuous across any boundary. Mathematically, this can be written as: Et1 = Et2 Which means that the tangential component of the electric field on one side of the boundary is equal to the tangential component of the electric field on the other side of the boundary. Since E = -∇φ, we can also write it as: -∇φ1_t = -∇φ2_t
03

3. Derivation of φ boundary condition for TE modes

Now let's take the normal derivative of the function φ, which can be written as ∇φ • n, where n is a unit vector normal to the boundary. So, let's evaluate ∇φ • n. ∇φ • n = ∇(−E) • n = (∇×E) • n Since the curl of a gradient is always zero, we get: ∇φ • n = 0 So, for TE modes, the normal derivative of the scalar function φ must vanish at the boundary.
04

4. The normal-B boundary condition for TM modes

The normal-B boundary condition states that the normal component of the magnetic flux density (Bn) must be continuous across any boundary. Mathematically, this can be written as: Bn1 = Bn2 Where B = μH and μ is the permeability of the medium. Since H = -∇φ, we can also write it as: -μ1∇φ1_n = -μ2∇φ2_n
05

5. Derivation of φ boundary condition for TM modes

Now let's consider a boundary of two media having same permeability (μ1 = μ2 = μ), the equation becomes: -∇φ1_n = -∇φ2_n Or, ∇φ1_n = ∇φ2_n This means that the normal derivative of the scalar function φ1 is equal to the normal derivative of the scalar function φ2, which implies φ1 = φ2 + constant. Since the magnetic field strength H is related to the gradient of φ, the constant does not affect H. Hence, at the boundary, φ1 = φ2 = φ. So, for TM modes, the scalar function φ itself must vanish at the boundary. In conclusion, for TE modes, the normal derivative of the scalar function φ must vanish at the boundary, and for TM modes, the scalar function φ itself must vanish at the boundary.

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Most popular questions from this chapter

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.

A waveguide becomes a resonant cavity upon placing conducting walls at the two ends. Show that a resonance occurs when the length \(L\) is an integral number \(n\) of guide halfwavelengths \(\lambda_{e} / 2 ;\) specifically, \(\left(\frac{\omega}{c}\right)^{2}=\left(\frac{l \pi}{a}\right)^{2}+\left(\frac{m \pi}{b}\right)^{2}+\left(\frac{n \pi}{L}\right)^{2} \quad\) rectangular parallelepiped \(\left(\frac{\omega}{c}\right)^{2}=\left(\frac{u_{l m}}{a}\right)^{2}+\left(\frac{n \pi}{L}\right)^{2} \quad\) right circular cylinder. Cavity modes, requiring three integral indices, are named \(\mathrm{TE}_{l m n}\) or \(\mathrm{TM}_{l m n} . \mathrm{Make}\) a mode chart for cylindrical cavities by plotting loci of resonances on a graph of \((d / L)^{2}\) against \((f d)^{2}\), where \(d \equiv 2 a, f \equiv \omega / 2 \pi . \dagger\)

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 \((8.4 .24)\) ?

Use the general form of Maxwell's equations (8.2.17) to (8.2.20), together with Gauss' and Stokes' theorems, to obtain the corresponding integral equations for a material medium, $$ \begin{aligned} &\oint_{s} \mathbf{D} \cdot d \mathbf{S}=q_{t r e t} \\ &\oint_{L} \mathbf{E} \cdot d \mathrm{l}=-\frac{d \Phi_{m}}{d t} \\ &\oint_{S} \mathbf{B} \cdot d \mathbf{S}=0 \\ &\oint_{L} \mathbf{H} \cdot d \mathbf{l}=I_{\text {froe }}+\frac{d \Phi_{\bullet}}{d t} \end{aligned} $$ where \(\Phi_{m}=\int \mathrm{B} \cdot d \mathrm{~S}\) and \(\Phi_{*}=\int \mathrm{D} \cdot d \mathrm{~S}\) are the magnetic and electric fluxes linking the closed line \(L\). These equations are generalizations of \((8.2 .11)\) to \((8.2 .14)\).

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).
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