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

Let us now take the electric field of a plane wave to be of the form

E (r) = E_{0} e^{- j k \cdot r}, (6.1.21)

with $E_{0}$ a constant vector, and find the magnetic field that goes with it from the first Maxwell equation, $\nabla \times E = - j ω μ H$ . The curl of a scalar function times a vector is $\nabla \times (f A) = f \nabla \times A + \nabla f \times A$ (Section 2.2), and here $f = e^{- j k \cdot r}$ , whose gradient is $\nabla e^{- j (k_{x} x + k_{y} y + k_{z} z)} = - j k e^{- j k \cdot r}$ . Hence

H (r) = \frac{j}{ω μ} \nabla \times E (r) = \frac{j}{ω μ} e^{- j k \cdot r} = 0 \nabla \times E_{0} + \frac{j}{ω μ} - j k e^{- j k \cdot r} \nabla e^{- j k \cdot r} \times E_{0} = \frac{k \times E ( r )}{ω μ} . (6.1.22)

With $k = k a_{n}$ and $k = ω μ ε$ the factor is $k / ω μ = ε / μ$ , and

H (r) η = \frac{1}{η} a_{n} \times E (r), = \frac{μ}{ε} [Ω] . (6.1.23)

The quantity $η$ is the intrinsic impedance (or wave impedance) of the medium, written $Z$ in some texts. In the same way, the magnetic field $H (r) = H_{0} e^{- j k \cdot r}$ substituted into the second Maxwell equation, $\nabla \times H = j ω ε E$ , gives $E = - k \times H / ω ε$ , that is

E (r) = - η a_{n} \times H (r) = η H (r) \times a_{n} . (6.1.24)

The divergence equations add the last piece of information. Since $\nabla \cdot (f E_{0}) = \nabla f \cdot E_{0}$ ,

\nabla \cdot E = - j k \cdot E = 0 ⟹ a_{n} \cdot E = 0, and likewise a_{n} \cdot H = 0. (6.1.25)

Neither field has a component along the direction of propagation. Equation (6.1.23) then says that $H$ is $E$ turned by $9 0^{\circ}$ about $a_{n}$ and divided by $η$ .

The uniform plane wave

A plane wave $E = E_{0} e^{- j k \cdot r}$ , $k = k a_{n}$ , in a medium $(μ, ε)$ has

H = \frac{1}{η} a_{n} \times E, E = - η a_{n} \times H, a_{n} \cdot E = a_{n} \cdot H = 0, η = \frac{μ}{ε} . (6.1.26)

$E ⊥ H ⊥ a_{n}$ , the three form a right-handed triad, $a_{E} \times a_{H} = a_{n}$ , and $∣ E ∣/∣ H ∣ = η$ .

In plain words

In a plane wave the electric and the magnetic field lie in the wavefront, at right angles to each other, and their ratio is a property of the medium alone. Know one of them and the direction of travel, and you know the other.

Such waves are called uniform plane waves, or TEM waves (transverse electromagnetic): no field component along $a_{n}$ , and the same amplitude everywhere on a wavefront. Figure 6.1.5 shows the triad on one wavefront.

The triad of a uniform plane wave. \vec E and \vec H lie in the wavefront, perpendicular to each other, and \vec E\times\vec H points along the direction of propagation \vec a_n, which is also the direction of the power flow. — **Figure 6.1.5.** The triad of a uniform plane wave. $E$ and $H$ lie in the wavefront, perpendicular to each other, and $E \times H$ points along the direction of propagation $a_{n}$ , which is also the direction of the power flow.

The uniform plane wave in free space

In vacuum, and to within $0.03 %$ in air, $μ = μ_{0}$ and $ε = ε_{0}$ :

k_{0} η_{0} = ω μ_{0} ε_{0} = \frac{ω}{c} = \frac{2 π}{λ _{0}}, = \frac{μ _{0}}{ε _{0}} = \frac{4 π \times 1 0 ^{- 7}}{1 0 ^{- 9} /36 π} = 120 π \approx 377 Ω. (6.1.27)

An $x$ -polarised wave travelling towards $+ z$ has

E H = E_{0} cos (ω t - k_{0} z) a_{x}, = \frac{E _{0}}{η _{0}} cos (ω t - k_{0} z) a_{y}, (6.1.28)

exactly the wave of Section 5.1. In a lossless medium $η$ is real, so $E$ and $H$ are in phase: they reach their maxima at the same points and the same instants (Figure 6.1.6). In a medium with $μ_{r}$ and $ε_{r}$ , $η = η_{0} μ_{r} / ε_{r}$ .

A uniform plane wave in free space at t=0, \vec E along x (orange) and \vec H along y (blue), travelling towards +z. The two fields are in phase, perpendicular to each other and to \vec a_k=\vec a_z, and on each wavefront (the shaded plane) they are the same everywhere. — **Figure 6.1.6.** A uniform plane wave in free space at $t = 0$ , $E$ along $x$ (orange) and $H$ along $y$ (blue), travelling towards $+ z$ . The two fields are in phase, perpendicular to each other and to $a_{k} = a_{z}$ , and on each wavefront (the shaded plane) they are the same everywhere.

Turn the wave in 3-D and change $f$ , $ε_{r}$ and $μ_{r}$ : $E$ and $H$ stay in phase, $E / H$ follows $η = η_{0} μ_{r} / ε_{r}$ , and $E \times H$ points along the direction of travel.

InteractiveLoading simulation…

Common mistake · The direction of travel of a backward wave

For a wave towards $- z$ , with $e^{+ j β z}$ or $cos (ω t + β z)$ , the unit vector in $H = \frac{1}{η} a_{k} \times E$ is $a_{k} = - a_{z}$ . Writing $+ a_{z}$ out of habit gives an $H$ of the wrong sign, and the wave then carries its power the wrong way. A quick check every time: $E \times H$ must point along the direction read from the exponent (Home exercise 6.1.8).

Quick check

In free space $E = 10 cos (ω t - k_{0} z) a_{y} V/m$ . Find $H$ .

$a_{k} = a_{z}$ and $a_{z} \times a_{y} = - a_{x}$ , so $H = - (10/377) cos (ω t - k_{0} z) a_{x} = - 26.5 cos (ω t - k_{0} z) a_{x} mA/m$ . Check: $a_{y} \times (- a_{x}) = a_{z}$ , the direction of travel.

6.1.3 Uniform plane waves and the intrinsic impedance

The uniform plane wave in free space