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

With a perfect conductor at $x = 0$ , $Γ = - 1$ , and Eq. (6.3.18) becomes

E_{1} (x) = E_{0}^{i} (e^{- j β_{1} x} - e^{j β_{1} x}) a_{y} = - 2 j E_{0}^{i} sin β_{1} x a_{y}, (6.3.21)

H_{1} (x) = \frac{E _{0}^{i}}{η _{1}} (e^{- j β_{1} x} + e^{j β_{1} x}) a_{z} = \frac{2 E _{0}^{i}}{η _{1}} cos β_{1} x a_{z} . (6.3.22)

For a real $E_{0}^{i}$ the instantaneous fields follow from $Re {\cdot e^{j ω t}}$ , with $Re {- j e^{j ω t}} = sin ω t$ :

E_{1} (x, t) H_{1} (x, t) = 2 E_{0}^{i} sin β_{1} x sin ω t a_{y}, = \frac{2 E _{0}^{i}}{η _{1}} cos β_{1} x cos ω t a_{z} . (6.3.23)

Each field is now a product of a function of $x$ and a function of $t$ : the pattern does not move. The electric field vanishes at all times on the planes $x = - m λ_{1} /2$ ( $m = 0, 1, 2, \dots$ ), the nodes of $E$ , one of which is the conductor surface. The magnetic field vanishes on the planes $x = - (2 m + 1) λ_{1} /4$ , halfway between them (Figure 6.3.5).

Pure standing wave in front of a perfect conductor

$E_{1}$ and $H_{1}$ of Eq. (6.3.23) are $9 0^{\circ}$ apart in space and in time, so

\frac{1}{2} Re {E_{1} \times H_{1}^{*}} = 0, E_{1} \times H_{1} = \frac{( E _{0}^{i} ) ^{2}}{η _{1}} sin 2 β_{1} x sin 2 ω t a_{x}, (6.3.24)

and the conductor carries the surface current ( $a_{n} = - a_{x}$ points out of the conductor, and $H = 0$ inside it)

J_{S} = a_{n} \times H_{1} (0, t) = - a_{x} \times \frac{2 E _{0}^{i}}{η _{1}} cos ω t a_{z} = \frac{2 E _{0}^{i}}{η _{1}} cos ω t a_{y} . (6.3.25)

In plain words

A perfect mirror sends everything back. The two waves build a stationary pattern in which the energy only sloshes back and forth between electric and magnetic form inside each quarter wavelength, with no net flow. The wall current, twice the incident $H$ , is what radiates the reflected wave.

(a) E_y at \omega t=\frac\pi2 (dark), \frac\pi3, \frac\pi6 and -\frac\pi6 (dashed). — (a) $E_{y}$ at $ω t = \frac{π}{2}$ (dark), $\frac{π}{3}$ , $\frac{π}{6}$ and $- \frac{π}{6}$ (dashed).

(b) H_z at \omega t=0 (dark), \frac\pi6 and \frac\pi3. — (a) $E_{y}$ at $ω t = \frac{π}{2}$ (dark), $\frac{π}{3}$ , $\frac{π}{6}$ and $- \frac{π}{6}$ (dashed).

Energy that only sloshes

At $ω t = π /2$ the field is all electric, stored around the antinodes of $E$ . A quarter period later it is all magnetic, stored around the antinodes of $H$ , a quarter wavelength away. The instantaneous Poynting vector Eq. (6.3.24) carries the energy between the two, changing direction every quarter period and every quarter wavelength, and its time average is zero: a standing wave transports no power. Problem 6.3.2 puts numbers on it.

Common mistake · Placing the first node of

E

a quarter wavelength from the wall

The tangential $E$ must vanish on a perfect conductor, so the wall itself is a node of $E$ and the first node in front of it is at $λ_{1} /2$ . It is the magnetic field that has its maximum on the wall (and with it the surface current) and its first node at $λ_{1} /4$ .

6.3.4 Standing wave in front of a perfect conductor