Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Put $m = 1$ , $n = 0$ , $p = 1$ in Eq. (7.8.11). Then $h^{2} = (π / a)^{2}$ , the components $E_{x}$ , $H_{y}$ vanish, and only $E_{y}$ , $H_{x}$ , $H_{z}$ remain. Call the peak electric field $E_{0} = - j ω μ a H_{0} / π$ . With $π / (ω μ d) = λ / (2 d η)$ , which follows from $ω μ = k η = 2 π η / λ$ ,

E_{y} H_{x} H_{z} = E_{0} sin \frac{π x}{a} sin \frac{π z}{d}, = - j \frac{E _{0}}{η} \frac{λ}{2 d} sin \frac{π x}{a} cos \frac{π z}{d}, = j \frac{E _{0}}{η} \frac{λ}{2 a} cos \frac{π x}{a} sin \frac{π z}{d}, (7.8.12)

where $λ = v / f_{101}$ is the wavelength of the filling at resonance. At resonance Eq. (7.8.7) reads $(2 π / λ)^{2} = (π / a)^{2} + (π / d)^{2}$ , that is

(\frac{λ}{2 a})^{2} + (\frac{λ}{2 d})^{2} = 1. (7.8.13)

The two factors are the $sin θ$ and $cos θ$ of the zig-zag angle of the TE $_{10}$ mode (Section 7.3): the mode is two plane waves bouncing between the side walls at the angle $θ$ with $sin θ = λ /2 a$ , and the end walls fold them back once more.

In the time domain, with $E_{0}$ real,

E_{y} H_{x} H_{z} = E_{0} sin \frac{π x}{a} sin \frac{π z}{d} cos ω t, = \frac{E _{0}}{η} \frac{λ}{2 d} sin \frac{π x}{a} cos \frac{π z}{d} sin ω t, = - \frac{E _{0}}{η} \frac{λ}{2 a} cos \frac{π x}{a} sin \frac{π z}{d} sin ω t . (7.8.14)

(a) Oblique view: \vec E (orange) and the \vec H loops (dashed) of the mid-plane y=b/2. — (a) Oblique view: $E$ (orange) and the $H$ loops (dashed) of the mid-plane $y = b /2$ .

(b) Top view: E_y (dots, size \propto|E_y|) and the lines of \vec H. — (a) Oblique view: $E$ (orange) and the $H$ loops (dashed) of the mid-plane $y = b /2$ .

The field breathes

Figure 7.8.5 shows the mode. At $t = 0$ all the energy is electric. A quarter period later $E = 0$ and all the energy is in $H$ . The pattern does not travel: it stays in place and only breathes. The loops of $H$ circle round the region where $E_{y}$ changes fastest, the displacement current $ε \partial E_{y} / \partial t$ (Section 5.2), exactly as the magnetic field of an $L C$ circuit circles round the current that charges the capacitor. In a cavity the "capacitor" is the middle of the box, where $E$ is strong, and the "inductor" is the region near the walls, where $H$ is strong. The lines of $H$ in a plane $y = const$ are the contours of $E_{y}$ , since $H = (j / ω μ) \nabla \times E$ with $\nabla \times (E_{y} a_{y}) = - (\partial E_{y} / \partial z) a_{x} + (\partial E_{y} / \partial x) a_{z}$ , a vector tangent to the curves $E_{y} = const$ .

Excitation

The field pattern tells us how to feed the mode. A probe (the inner conductor of a coaxial line) through the centre of the broad wall couples to the largest $E_{y}$ (Figure 7.8.1). A small loop on an end wall couples to the largest $H_{x}$ , which is there at $x = a /2$ . An iris in an end wall connects the cavity to a guide that carries TE $_{10}$ , whose field has the same transverse pattern.

7.8.6 The dominant TE₁₀₁ mode

Excitation