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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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Key Concepts

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

TE Modes
In electromagnetic wave theory, TE modes or Transverse Electric modes have a characteristic feature where the electric field vector has no component in the direction of wave propagation. Since the electric field is transverse, it is represented as the gradient of a scalar function \(\phi\). This is symbolically expressed as \(-abla \phi\), where \(abla\) denotes the gradient operator.
To further understand TE modes, it is crucial to consider boundary conditions, specifically the tangential-E boundary condition. This boundary condition asserts that the tangential component of the electric field must be continuous across any boundary. When applied to the TE mode, this condition implies that the gradient of the scalar function \(\phi\) in the tangential direction must be the same on both sides of the boundary.
  • The continuity of the tangential electric component \(E_t\) ensures \(-abla \phi_t1 = -abla \phi_t2\).
  • This results in \(abla \phi \cdot \mathbf{n} = 0\), meaning the normal derivative must vanish at the boundary.
This means any change in \(\phi\) normal to the boundary does not contribute to the electric field, emphasizing that the field maintains a purely transverse nature. Understanding this principle is key to mastering TE mode applications.
TM Modes
TM modes, short for Transverse Magnetic modes, are electromagnetic configurations where the magnetic field vector is perpendicular to the direction of wave propagation. Here, like in TE modes, the field components are derived from a scalar function \(\phi\), this time influencing the magnetic field \(\mathbf{H}\) as \(-abla \phi\).
In TM modes, the boundary condition of specific interest is the normal-B boundary condition. This condition emphasizes the need for the normal component of the magnetic flux density (denoted as \(B_n\)) to be consistent on either side of a boundary. It's expressed as continuity in magnetic fields across the medium boundaries.
  • Mathematically, this condition can be represented as \(B_{n1} = B_{n2}\).
  • For practical scenarios with equal permeability, the derivative of \(\phi\) in the normal direction becomes continuous, leading to \(\phi_1 = \phi_2\) across the boundary.
Therefore, in TM modes, the scalar function \(\phi\) itself must vanish at the boundary, ensuring the magnetic field retains a comprehensive transverse arrangement. Understanding TM modes is instrumental for applying these concepts in designing and analyzing waveguides and antennas.
Boundary Conditions
Boundary conditions in electromagnetics dictate how electromagnetic fields behave at interfaces between different media. Such conditions are pivotal as they influence how waves are transmitted or reflected at boundaries.
For TE modes, the tangential-E boundary conditions ensure the continuity of electrical tangents across boundaries. This means \(\phi\)'s normal derivative must vanish, establishing that electric fields adjust smoothly at material interfaces.
In contrast, with TM modes, the normal-B boundary condition focuses on magnetic normality ensuring continuity in the normal component of magnetic fields. This results in the scalar \(\phi\) vanishing at boundaries, supporting consistent magnetic field behavior.
Boundary conditions are crucial because:
  • They ensure field solutions are physically realizable and practical.
  • They are foundational in assessing field distributions in complex structures like waveguides or cavities.
  • They aid in predicting wave behavior in different scenarios, enhancing designs in telecommunications, antenna designs, and signal processing.
Mastering boundary conditions not only aids in grasping TE and TM modes but is also essential for the broader study of electromagnetic wave propagation and its applications.

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Most popular questions from this chapter

Consider two unbounded plane waves whose vector wave numbers \(k_{1}\) and \(\kappa_{2}\left(\left.\right|_{1,2} \mid=\right.\) \(\omega / c)\) define a plane and whose electric fields are polarized normal to the plane. (a) Show that the superposition of these two plane waves is a wave traveling in the direction bisecting the angle \(\alpha\) between \({ }_{1}{ }_{1}\) and \({ }_{k}\) and that the \(\mathbf{E}\) field vanishes on a set of nodal planes spaced \(a=\) \(\lambda_{0} / 2 \sin \frac{1}{2} \alpha\) apart. \((b)\) Show that plane conducting walls can be placed at two adjacent nodal planes without violating the electromagnetic boundary conditions and likewise that a second pair of conducting walls of arbitrary separation \(b\) can be introduced to construct a rectangular waveguide of cross section \(a\) by \(b\), propagating the \(T E_{10}\) mode. Thus establish that the TE \(_{10}\) mode (more generally, the TE \(_{10}\) modes) may be interpreted as the superposition of two plane waves making the angle \(\frac{1}{2} \alpha\) with the waveguide axis and undergoing multiple reflections from the sidewalls. Note: The situation is directly analogous to that discussed in Sec. 2.4. Higherorder TE modes \((m>0)\) and TM modes may be described similarly as a superposition of four plane waves.

An artificial line can be made from a number of lumped inductors and capacitors by arranging them alternately as shown in the diagram. The circuit can be analyzed in terms of an elementary \(T\) section, also shown. Show that in the low-frequency limit the characteristic impedance of the artificial line of arbitrary length is $$ Z_{0}=\left(\frac{L}{C}\right)^{1 / 2} $$ Artifici. ll liac Prob, \(8,1.12\) and that the speed of propagation, in sections per second, is $$ c^{\prime}=\frac{1}{(L C)^{\mathrm{t} / 3}} $$ Further show that the line does not transmit waves at frequencies greater than the culoff frequency $$ \omega_{c}=\frac{2}{(L C)^{1 / 2}} $$ How do \(Z_{4}\) and \(c^{\prime}\) vary with frequency near \(\omega_{c} ?\) What if each inductor has a small amount of stray capacitance in parallel with it? What if adjacent inductors are coupled by mutual inductance? What practical applications might such circuits have?

Use Gauss' and Stokes' theorems (Appendix A) to convert Maxwell's differential equations for vacuum, \((82.1)\) to \((8.2 .4)\), to their integral form $$ \begin{aligned} &\oint_{S} \mathbf{E} \cdot d \mathbf{S}=\frac{q}{\epsilon_{0}} \\ &\oint_{L} \mathbf{E} \cdot d \mathbf{l}=-\frac{d \Phi_{m}}{d t} \\ &\oint_{S} \mathbf{B} \cdot d \mathbf{S}=0 \\ &\oint_{L} \mathbf{B} \cdot d \mathbf{l}=\mu_{0} I+\mu_{0} \frac{d \Phi_{*}}{d t} \end{aligned} $$ † See Sec. \(5.4\) and Prob. 8.2.4. where the closed surface \(S\) contains the net charge \(q\) and the closed line (loop) \(L\) is linked by the net current \(I\), the magnetic flux \(\Phi_{m}=\int \mathbf{B} \cdot d \mathbf{S}\), and the electric flux \(\Phi_{e}=\epsilon_{0} \int \mathbf{E} \cdot d \mathbf{S}\). Note: The corresponding equations for a general electromagnetic medium are developed in Prob. \(8.6 .1 .\)

Show that the time-average Poynting vector for the far fields (8.9.13) of an oscillating electric dipole \((8.9 .12)\) is $$ \underline{E}=\hat{\mathrm{f}} \frac{c k^{4} p_{0}^{2}}{32 \pi^{2} \epsilon_{0}} \frac{\sin ^{2} \theta}{r^{2}} $$ and that the average total power radiated by the oscillating dipole is $$ \left(\frac{d W}{d l}\right)=\frac{c \kappa^{4} p_{0}^{2}}{12 \pi \epsilon_{0}} $$ Why is the sky blue and the sunset red?

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).

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