Two types of waveguide are used for transmission of microwave power and signals: rectangular and circular. Waveguides are usually oriented with their central axes parallel to the z axis of the Cartesian coordinate system for rectangular waveguides and cylindrical coordinate system for circular

Question

Two types of waveguide are used for transmission of microwave power and signals: rectangular and circular. Waveguides are usually oriented with their central axes parallel to the z axis of the Cartesian coordinate system for rectangular waveguides and cylindrical coordinate system for circular waveguides. The z components [latex]{E}_z\ and\ \mathcal{B}_z[/latex] of electric field [latex]{E}[/latex] and magnetic field [latex]\mathcal{B}[/latex], respectively, are for rectangular waveguides in general given as follows

[latex]\begin{aligned} {E}_z(x, y, z, t) & =\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{m n} \sin k_x x \sin k_y y e^{i\left(k_z z-\omega t ight)} \ & =\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{m n} \sin \frac{m \pi x}{a} \sin \frac{n \pi y}{b} e^{i \varphi}\quad (13.104) \end{aligned}[/latex]

and

[latex]\begin{aligned} \mathcal{B}_z(x, y, z, t) & =\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} B_{m n} \cos k_x x \cos k_y y e^{i\left(k_z z-\omega t ight)} \ & =\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} B_{m n} \cos \frac{m \pi x}{a} \cos \frac{n \pi y}{b} e^{i \varphi}\quad (13.105) \end{aligned}[/latex]

where [latex]A_{mn}\ and\ B_{mn}[/latex] are constants that can be determined from initial conditions, [latex]φ = k_zz − ωt[/latex] is the phase of the RF wave, and [latex]k_x ,\ k_y ,\ and\ k_z[/latex] are the waveguide wave numbers or propagation coefficients with [latex]k_x\ and\ k_y[/latex] determined from boundary conditions.

(a) For a uniform rectangular EM waveguide of cross section sides a and b where a>b:

(1) Explain how components [latex]{E}_z\ and\ \mathcal{B}_z[/latex] of electric field [latex]{E}[/latex] and magnetic field [latex]\mathcal{B}[/latex], respectively, are determined.
(2) Explain how other components of [latex]{E}\ and\ \mathcal{B}[/latex] are determined once [latex]{E}_z\ and\ \mathcal{B}_z[/latex] are known.

(b) Show that for a uniform rectangular EM waveguide the transverse components [latex]{E}_x\ and {E}_y[/latex] of the electric field [latex] {E}[/latex] as well as the transverse components [latex]\mathcal{B}_x\ and\ \mathcal{B}_y[/latex] of the magnetic field [latex]\mathcal{B}[/latex] can be determined directly from known axial components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] of electric field [latex] {E}[/latex] and magnetic field [latex]\mathcal{B}[/latex], respectively.

(c) For a uniform rectangular EM waveguide determine the transverse fields [latex] {E}_x , {E}_y ,\ \mathcal{B}_x[/latex] and [latex]\mathcal{B}_y[/latex] for: (1) TM modes, (2) TE modes, and (3) TEM modes. In your calculations assume that axial components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] are known and use the general expressions from (b).

(d) For a uniform rectangular EM waveguide determine all components of electric field [latex] {E}[/latex] and magnetic field [latex]\mathcal{B}[/latex] for the lowest (dominant): (1) transverse electric (TE) mode and (2) transverse magnetic (TM) mode. The rectangular cross section of the waveguide core has sides a and b with a>b.

Accepted Answer

(a) Electric field E and magnetic field B in the core of a uniform EM waveguide are vectors with three components, each component depending on three spatial coordinates and one temporal coordinate. For a rectangular EM waveguide the components of E and B are:

[latex] {E}=\left[ {E}_x(x, y, z, t), {E}_y(x, y, z, t), {E}_z(x, y, z, t) ight][/latex] (13.106)

and

[latex]\boldsymbol{B}=\left[\mathcal{B}_x(x, y, z, t), \mathcal{B}_y(x, y, z, t), \mathcal{B}_z(x, y, z, t) ight][/latex] (13.107)

(1) Components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] are determined from wave equations for [latex] {E}_z\ and\ \mathcal{B}_z[/latex] given as

[latex] abla^2 {E}_z=\frac{1}{c^2} \frac{\partial^2 {E}_z}{\partial^2 t^2}[/latex] (13.108)

and

[latex] abla^2 \mathcal{B}_z=\frac{1}{c^2} \frac{\partial^2 \mathcal{B}_z}{\partial^2 t^2}[/latex] (13.109)

where c is the speed of light in vacuum and [latex]∇^2[/latex] is the scalar Laplacian operator expressed in Cartesian coordinates for rectangular waveguide as follows

[latex] abla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}[/latex] (13.110)

Wave equations (13.108) and (13.109) are linear partial differential equations of the second order in four variables (3 spatial and one temporal) with constant coefficients. The most common method for solving the wave equations is the method of separation of variables leading to solutions for the z components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] of the electric field E and magnetic field B, respectively, in the core of the waveguide.

(2) Once [latex] {E}_z\ and\ \mathcal{B}_z[/latex] are known, the other components of E and B are determined from Maxwell equations for free space expressed as follows

[latex]\boldsymbol{ abla} \cdot \boldsymbol{E}=0[/latex] (13.111)

[latex] abla \cdot \mathcal{B}=0[/latex] (13.112)

[latex] abla imes {E}=-\frac{\partial \boldsymbol{B}}{\partial t}[/latex] (13.113)

[latex] abla imes \mathcal{B}=\frac{1}{c^2} \frac{\partial {E}}{\partial t},[/latex] (13.114)

with (∇·) the divergence and (∇×) the curl on vectors E and B. It is also useful to note that from (13.104) and (13.105) the following expressions apply for derivatives: [latex]∂/∂z = ik_z[/latex] and ∂/∂t = −iω with [latex]k_z[/latex] the waveguide propagation coefficient parallel to the central axis of the waveguide and ω the angular frequency of the EM wave in contrast to k = ω/c that is defined as the free space propagation coefficient.

(b) We start the derivation of components [latex]{E}_x , {E}_y , \mathcal{B}_x , and \mathcal{B}_y[/latex] with (13.111) and (13.112) for rectangular EM waveguide and express the two equations in Cartesian coordinates as follows

[latex]\boldsymbol{ abla} \cdot {E}=\frac{\partial {E}_x}{\partial x}+\frac{\partial {E}_y}{\partial y}+\frac{\partial {E}_z}{\partial z}=\frac{\partial {E}_x}{\partial x}+\frac{\partial {E}_y}{\partial y}+i k_z E_z=0[/latex] (13.115)

and

[latex]\boldsymbol{ abla} \cdot \boldsymbol{B}=\frac{\partial \mathcal{B}_x}{\partial x}+\frac{\partial \mathcal{B}_y}{\partial y}+\frac{\partial \mathcal{B}_z}{\partial z}=\frac{\partial \mathcal{B}_x}{\partial x}+\frac{\partial \mathcal{B}_y}{\partial y}+i k_z B_z=0[/latex] (13.116)

and then express (13.113) and (13.114) in Cartesian coordinates as

[latex] abla imes {E}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ {E}_x & {E}_y & {E}_z \end{array} ight|=-\frac{\partial}{\partial t}\left|\begin{array}{c} \mathcal{B}_x \hat{\mathbf{i}} \ \mathcal{B}_y \hat{\mathbf{j}} \ \mathcal{B}_z \hat{\mathbf{k}} \end{array} ight|=i \omega\left|\begin{array}{c} \mathcal{B}_x \hat{\mathbf{i}} \ \mathcal{B}_y \hat{\mathbf{j}} \ \mathcal{B}_z \hat{\mathbf{k}} \end{array} ight|\quad (13.117)[/latex]

and

[latex] abla imes \boldsymbol{B}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \mathcal{B}_x & \mathcal{B}_y & \mathcal{B}_z \end{array} ight|=\frac{1}{c^2} \frac{\partial}{\partial t}\left|\begin{array}{c} {E}_x \hat{\mathbf{i}} \ {E}_y \hat{\mathbf{j}} \ {E}_z \hat{\mathbf{k}} \end{array} ight|=-\frac{i \omega}{c^2}\left|\begin{array}{c} {E}_x \hat{\mathbf{i}} \ {E}_y \hat{\mathbf{j}} \ {E}_z \hat{\mathbf{k}} \end{array} ight|\quad (13.118)[/latex]

where [latex]\hat{\mathbf{i}}, \hat{\mathbf{j}} ext {, and } \hat{\mathbf{k}}[/latex] are the standard unit vectors along x, y, and z axes of the Cartesian coordinate system. Equations (13.117) and (13.118) have the following components of the curl operator [Note: (13.119), (13.120), and (13.121) follow from (13.117); (13.122), (13.123), and (13.129) follow from (13.118)]

[latex]\frac{\partial {E}_z}{\partial y}-\frac{\partial {E}_y}{\partial z}=i \omega \mathcal{B}_x=\frac{\partial {E}_z}{\partial y}-i k_z {E}_y \quad ext { or } \quad \mathcal{B}_x=-\frac{i}{\omega} \frac{\partial {E}_z}{\partial y}-\frac{k_z}{\omega} {E}_y ext {, }[/latex] (13.119)

[latex]\frac{\partial {E}_x}{\partial z}-\frac{\partial {E}_z}{\partial x}=i \omega \mathcal{B}_y=i k_z {E}_x-\frac{\partial {E}_z}{\partial x} \quad ext { or } \quad \mathcal{B}_y=+\frac{i}{\omega} \frac{\partial {E}_z}{\partial x}+\frac{k_z}{\omega} {E}_x ext {, }[/latex] (13.120)

[latex]\frac{\partial {E}_y}{\partial x}-\frac{\partial {E}_x}{\partial y}=i \omega \mathcal{B}_z \quad ext { or } \quad \mathcal{B}_z=\frac{i}{\omega}\left[-\frac{\partial {E}_y}{\partial x}+\frac{\partial {E}_x}{\partial y} ight] ext {, }[/latex] (13.121)

[latex]\frac{\partial \mathcal{B}_z}{\partial y}-\frac{\partial \mathcal{B}_y}{\partial z}=-\frac{i \omega}{c^2} {E}_x=\frac{\partial \mathcal{B}_z}{\partial y}-i k_z \mathcal{B}_y \quad ext { or } \quad {E}_x=\frac{i c^2}{\omega} \frac{\partial \mathcal{B}_z}{\partial y}+\frac{k_z c^2}{\omega} \mathcal{B}_y[/latex] (13.122)

[latex]\frac{\partial \mathcal{B}_x}{\partial z}-\frac{\partial \mathcal{B}_z}{\partial x}=-\frac{i \omega}{c^2} {E}_y=i k_z \mathcal{B}_x-\frac{\partial \mathcal{B}_z}{\partial x}[/latex] or
[latex] {E}_y=-\frac{k_z c^2}{\omega} \mathcal{B}_x-\frac{i c^2}{\omega} \frac{\partial \mathcal{B}_z}{\partial x}[/latex] (13.123)

[latex]\frac{\partial \mathcal{B}_y}{\partial x}-\frac{\partial \mathcal{B}_x}{\partial y}=-\frac{i \omega}{c^2} {E}_z \quad ext { or } \quad {E}_z=\frac{i c^2}{\omega}\left[-\frac{\partial \mathcal{B}_y}{\partial x}-\frac{\partial \mathcal{B}_x}{\partial y} ight][/latex] (13.124)

Pairing up appropriate equations in the group from (13.119) to (13.124), we can now determine components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] as follows:

(1) Inserting [latex]\mathcal{B}_y[/latex] of (13.120) into (13.122) gives the following expression for component [latex] {E}_x[/latex].

[latex] {E}_x=i\left[\frac{k_z c^2}{\omega^2} \frac{\partial {E}_z}{\partial x}+\frac{c^2}{\omega} \frac{\partial \mathcal{B}_z}{\partial y} ight]\left(1-\frac{k_z^2 c^2}{\omega^2} ight)^{-1}=\frac{i}{\gamma^2}\left[k_z \frac{\partial {E}_z}{\partial x}+\omega \frac{\partial \mathcal{B}_z}{\partial y} ight][/latex] (13.125)

(2) Inserting [latex]\mathcal{B}_x[/latex] of (13.119) into (13.123) gives the following expression for component [latex] {E}_y[/latex].

[latex] {E}_y=i\left[-\frac{c^2}{\omega} \frac{\partial \mathcal{B}_z}{\partial x}+\frac{k_z c^2}{\omega^2} \frac{\partial {E}_z}{\partial y} ight]\left(1-\frac{k_z^2 c^2}{\omega^2} ight)^{-1}=\frac{i}{\gamma^2}\left[-\omega \frac{\partial \mathcal{B}_z}{\partial x}+k_z \frac{\partial {E}_z}{\partial y} ight] .[/latex] (13.126)

(3) Inserting [latex] {E}_y[/latex] of (13.123) into (13.119) gives the following expression for component [latex]\mathcal{B}_x[/latex].

[latex]\mathcal{B}_x=i\left[\frac{k_z c^2}{\omega^2} \frac{\partial \mathcal{B}_z}{\partial x}-\frac{1}{\omega} \frac{\partial {E}_z}{\partial y} ight]\left(1-\frac{k_z^2 c^2}{\omega^2} ight)^{-1}=\frac{i}{\gamma^2}\left[k_z \frac{\partial \mathcal{B}_z}{\partial x}-\frac{\omega}{c^2} \frac{\partial {E}_z}{\partial y} ight][/latex] (13.127)

(4) Inserting [latex] {E}_x[/latex] of (13.122) into (13.120) gives the following expression for component [latex]\mathcal{B}_y[/latex].

[latex]\mathcal{B}_y=i\left[\frac{1}{\omega} \frac{\partial {E}_z}{\partial x}+\frac{k_z c^2}{\omega^2} \frac{\partial \mathcal{B}_z}{\partial y} ight]\left(1-\frac{k_z^2 c^2}{\omega^2} ight)^{-1}=\frac{i}{\gamma^2}\left[\frac{\omega}{c^2} \frac{\partial {E}_z}{\partial x}+k_z \frac{\partial \mathcal{B}_z}{\partial y} ight] .[/latex] (13.128)

with [latex]\gamma^2=k^2-k_z^2 ext { and } \omega=k c[/latex] where c is the speed of light in vacuum and k is the free space propagation coefficient. Equations (13.125) through (13.128) show that the transverse components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] can be determined with relative ease directly from known axial components [latex] {E}_z\ and\ \mathcal{B}_z[/latex].

(c) Equations (13.125) through (13.128) give general expressions for transverse components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] as a function of axial components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] for a uniform rectangular EM waveguide. We now determine the transverse components for the three special modes: (1) TM where [latex]\mathcal{B}_z = 0[/latex] everywhere, (2) TE where [latex]E_z[/latex] = 0, and (3) TEM where [latex]\mathcal{B}_z = {E}_z = 0[/latex] everywhere. The three special modes are characterized as follows:

(1) TM modes: [latex]\mathcal{B}_z = 0[/latex] everywhere inside the waveguide core and the transverse components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] are given as follows

[latex] {E}_x=\frac{i k_z}{\gamma^2} \frac{\partial {E}_z}{\partial x}, \quad {E}_y=\frac{i k_z}{\gamma^2} \frac{\partial {E}_z}{\partial y}, \quad \mathcal{B}_x=-\frac{i}{\gamma^2} \frac{\omega}{c^2} \frac{\partial {E}_z}{\partial y}, \quad \mathcal{B}_y=\frac{i}{\gamma^2} \frac{\omega}{c^2} \frac{\partial {E}_z}{\partial x} .[/latex] (13.129)

(2) TE modes: [latex] {E}_z = 0[/latex] everywhere inside the waveguide core and the transverse components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] are given as follows

[latex] {E}_x=\frac{i \omega}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial y}, \quad {E}_y=-\frac{i \omega}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}, \quad \mathcal{B}_x=\frac{i k_z}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}, \quad \mathcal{B}_y=\frac{i k_z}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial y}[/latex] (13.130)

(3) TEM mode: Both [latex] {E}_z = 0\ and\ \mathcal{B}_z = 0[/latex] everywhere and (13.125) through (13.128) show that all transverse components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] are also equal to zero.

(d) Components of lowest (dominant) TE and TM modes in a uniform rectangular EM waveguide with a>b are determined as follows:
The general expressions for z components [latex] {E}_z\ and\ \mathcal{B}_z[/latex] of electric field E and magnetic field [latex]\mathcal{B}[/latex], respectively, are determined from appropriate wave equations (see Prob. 277) and given in (13.104) and (13.105), respectively, while the general expressions for the other four components [latex] {E}_x , {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y[/latex] were derived in (b). We now use these expressions to determine the electric and magnetic field components for the dominant TE and TM modes, recalling that parameter γ is for rectangular EM waveguide given as

[latex]\gamma^2=k^2-k_z^2=k_x^2+k_y^2=\left(\frac{m \pi}{a} ight)^2+\left(\frac{n \pi}{b} ight)^2[/latex] (13.131)

where k is the free space wave number related to microwave angular frequency ω through k = ω/c and [latex]k_x ,\ k_y ,\ and\ k_z[/latex] are waveguide propagation constants, the first two determined from boundary conditions and [latex]k_z[/latex] the wave number for the plane wave propagating unhindered along the axis of the waveguide.

(1) Transverse electric (TE) modes are characterized by [latex]E_z[/latex] = 0 everywhere in the waveguide core and the dominant (lowest) TE mode occurs for m = 1 and n = 0 (note: cos [latex]0°[/latex] = 1). Inserting m = 1 and n = 0 into (13.131) we get the following expression for parameter γ²

[latex]\gamma^2=\left(\frac{m \pi}{a} ight)^2+\left(\frac{n \pi}{b} ight)^2=\frac{\pi^2}{a^2} .[/latex] (13.132)

(i) Magnetic field component [latex]\mathcal{B}_z[/latex] for the dominant TE mode is now from (13.105) given as

[latex]\mathcal{B}_z=\mathcal{B}_{10} \cos \left(\frac{\pi x}{a} ight) e^{i \varphi}[/latex] (13.133)

(ii) Electric field component [latex] {E}_x ext { of }(13.125) ext { is zero because } \partial {E}_z / \partial x=0[/latex] and [latex]∂\mathcal{B}_z/∂ y = 0[/latex].
(iii) Electric field component [latex] {E}_y[/latex] is determined from (13.126) using [latex] {E}_z = 0,\ \mathcal{B}_z[/latex] from (13.132), [latex]∂ {E}_z/∂y = 0[/latex], and [latex]γ^2[/latex] from (13.132) as follows

[latex] {E}_y=-i \frac{\omega}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}=i \frac{\pi \omega}{\gamma^2 a} \mathcal{B}_{10} \sin \left(\frac{\pi x}{a} ight) e^{i \varphi}=i \frac{\omega a}{\pi} \mathcal{B}_{10} \sin \left(\frac{\pi x}{a} ight) e^{i \varphi} .[/latex] (13.134)

(iv) Magnetic field component [latex]\mathcal{B}_x[/latex] is determined from (13.127) using [latex] {E}_z = 0,\ \mathcal{B}_z[/latex] from (13.132), [latex]∂ {E}_z/∂y = 0[/latex], and [latex]γ^2[/latex] from (13.132) as follows

[latex]\mathcal{B}_x=i \frac{k_z}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}=-i \frac{\pi k_z}{\gamma^2 a} \mathcal{B}_{10} \sin \left(\frac{\pi x}{a} ight) e^{i \varphi}=-i \frac{k_z a}{\pi} \mathcal{B}_{10} \sin \left(\frac{\pi x}{a} ight) e^{i \varphi} .[/latex] (13.135)

(v) Magnetic field component [latex]\mathcal{B}_y[/latex] of (13.128) is zero because [latex]∂ {E}_z/∂x = 0\ and\ ∂\mathcal{B}_z/∂y = 0[/latex].

(2) Transverse magnetic (TM) modes are characterized by [latex]\mathcal{B}_z = 0[/latex] everywhere in the waveguide core and the dominant (lowest) TM mode occurs for m = 1 and n = 1 (note: sin [latex]0°[/latex] = 0). Inserting m = 1 and n = 1 into (13.131) we get the following expression for parameter [latex]γ^2[/latex].

[latex]\gamma^2=\left(\frac{m \pi}{a} ight)^2+\left(\frac{n \pi}{b} ight)^2=\pi^2\left(\frac{1}{a^2}+\frac{1}{b^2} ight)[/latex] (13.136)

(i) The electric field component [latex] {E}_z[/latex] for the dominant TM mode is now from (13.104) given as

[latex] {E}_z= {E}_{11} \sin \left(\frac{\pi x}{a} ight) \sin \left(\frac{\pi y}{b} ight) e^{i \varphi} .[/latex] (13.137)

(ii) The electric field component [latex] {E}_x[/latex] is determined from (13.125) using [latex]\mathcal{B}_z = 0, {E}_z[/latex] from (13.137), [latex]∂\mathcal{B}_z/∂y = 0,\ and\ γ^2[/latex] from (13.136) as follows

[latex]\begin{aligned} {E}_x & =i \frac{k_z}{\gamma^2} \frac{\partial {E}_z}{\partial x}=i \frac{\pi k_z}{\gamma^2 a} {E}_{11} \cos \left(\frac{\pi x}{a} ight) \sin \left(\frac{\pi y}{b} ight) e^{i \varphi} \ & =i \frac{k_z}{\pi\left(\frac{1}{a}+\frac{a}{b^2} ight)} \mathcal{B}_{11} \cos \left(\frac{\pi x}{a} ight) \sin \left(\frac{\pi y}{b} ight) e^{i \varphi} .\quad (13.138) \end{aligned}[/latex]

(iii) The electric field component [latex] {E}_y[/latex] is determined from (13.126) using [latex]\mathcal{B}_z = 0, {E}_z[/latex] from (13.137), [latex]∂\mathcal{B}_z/∂x = 0,\ and\ γ^2[/latex] from (13.136) as follows

[latex]\begin{aligned} {E}_y & =i \frac{k_z}{\gamma^2} \frac{\partial {E}_z}{\partial y}=i \frac{\pi k_z}{\gamma^2 b} {E}_{11} \sin \left(\frac{\pi x}{a} ight) \cos \left(\frac{\pi y}{b} ight) e^{i \varphi} \ & =i \frac{k_z}{\pi\left(\frac{b}{a^2}+\frac{1}{b} ight)} \mathcal{B}_{11} \sin \left(\frac{\pi x}{a} ight) \cos \left(\frac{\pi y}{b} ight) e^{i \varphi}\quad (13.139) \end{aligned}[/latex]

(iv) The magnetic field component [latex]\mathcal{B}_x[/latex] is determined from (13.127) using [latex]\mathcal{B}_z = 0, {E}_z[/latex] from (13.137), [latex]∂\mathcal{B}_z/∂x = 0,\ and\ γ^2[/latex] from (13.136) as follows

[latex]\begin{aligned} \mathcal{B}_x & =-i \frac{\omega}{\gamma^2 c^2} \frac{\partial {E}_z}{\partial y}=-i \frac{\pi \omega}{\gamma^2 c^2 b} {E}_{11} \sin \left(\frac{\pi x}{a} ight) \cos \left(\frac{\pi y}{b} ight) e^{i \varphi} \ & =-i \frac{\pi \omega}{\pi\left(\frac{b}{a^2}+\frac{1}{b} ight)} {E}_{11} \sin \left(\frac{\pi x}{a} ight) \cos \left(\frac{\pi y}{b} ight) e^{i \varphi} .\quad (13.140) \end{aligned}[/latex]

(v) The magnetic field component [latex]\mathcal{B}_y[/latex] is determined from (13.128) using [latex]\mathcal{B}_z = 0, {E}_z[/latex] from (13.137), [latex]∂\mathcal{B}_z/∂y = 0,\ and\ γ^2[/latex] from (13.136) as follows

[latex]\begin{aligned} \mathcal{B}_y & =i \frac{\omega}{\gamma^2 c^2} \frac{\partial {E}_z}{\partial x}=i \frac{\pi \omega}{\gamma^2 c^2 a} {E}_{11} \cos \left(\frac{\pi x}{a} ight) \sin \left(\frac{\pi y}{b} ight) e^{i \varphi} \ & =i \frac{\omega}{\pi c^2\left(\frac{1}{a}+\frac{a}{b^2} ight)} {E}_{11} \cos \left(\frac{\pi x}{a} ight) \sin \left(\frac{\pi y}{b} ight) e^{i \varphi} .\quad (13.141) \end{aligned}[/latex]

Question 13.4.Q1: Two types of waveguide are used for transmission of microwav......

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