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

About the further exercises of Section 5.6

Twenty further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within two groups. Exercises 5.6.1–5.6.10 belong to the lecture: Exercises 5.6.1–5.6.2 are about phasors and the phase velocity, 5.6.3–5.6.5 about the complex permittivity and the wave number, 5.6.6–5.6.7 about the Helmholtz equation and 5.6.8–5.6.10 about power. Exercises 5.6.11–5.6.20 belong to the exercise lecture: reading time-harmonic fields, loops, mixers and receivers, power, and the medium found from measurements. They use only

u = Re {U e^{j ω t}}, sin θ = cos (θ - \frac{π}{2}), \frac{\partial}{\partial t} \to j ω, v_{p} = \frac{ω}{∣\nabla ϕ ∣}, λ = \frac{2 π}{β}, ε_{c} = ε - j \frac{σ}{ω} = ε^{'} - j ε^{''}, tan δ = \frac{ε ^{''}}{ε ^{'}}, k = ω μ ε_{c} = β - j α, \nabla^{2} E + k^{2} E = 0, \nabla \cdot E = 0 (source-free), \nabla \times E = - j ω μ H, \nabla \times H = j ω ε_{c} E, η = \frac{ω μ}{k} = \frac{μ}{ε _{c}}, emf = - \frac{d Φ}{d t},

and, from Section 6.1, the instantaneous and the complex Poynting vectors $P (t) = E (t) \times H (t)$ and $P_{c} = \frac{1}{2} E \times H^{*}$ , with $P_{a v} = Re {P_{c}}$ , and the time-harmonic Poynting relation

- \oint_{S} P_{c} \cdot d S - \int_{V} \frac{1}{2} E \cdot J_{i}^{*} d V = \int_{V} [\frac{1}{2} σ ∣ E ∣^{2} + j 2 ω (w_{m} - w_{e})] d V,

with the time-average energy densities $w_{e} = \frac{1}{4} ε ∣ E ∣^{2}$ and $w_{m} = \frac{1}{4} μ ∣ H ∣^{2}$ (real $ε$ , $μ$ ). For a wave $E_{0} e^{- j k \cdot r}$ every $\nabla$ becomes $- j k$ , so Faraday's law gives $H = k \times E / ω μ$ . A small loop of $N$ turns and area $A$ in a field $H$ of angular frequency $ω$ picks up an emf of amplitude $N ω μ ∣ H \cdot n ∣ A$ , with $n$ its normal.

Exercise 5.6.9

Where the reactive power goes

Take the medium of Exercise 5.6.5 ( $f = 50 MHz$ , $ε_{r}^{'} = 9$ , $σ = 1/30 S/m$ , $μ = μ_{0}$ , $k = 3.628 - j 1.814 m^{- 1}$ ) and the wave $E (z) = E_{0} e^{- j k z} a_{x}$ with $E_{0} = 10 V/m$ . (a) Show that $- d P_{c, z} / d z = \frac{1}{2} σ ∣ E ∣^{2} + j 2 ω (w_{m} - w_{e})$ holds point by point, with $w_{e} = \frac{1}{4} ε^{'} ∣ E ∣^{2}$ and $w_{m} = \frac{1}{4} μ_{0} ∣ H ∣^{2}$ . (b) Evaluate at $z = 0$ : $H$ , $P_{c}$ , $- d P_{c, z} / d z$ , the Joule heat density, $w_{e}$ , $w_{m}$ . (c) Show that $w_{m} / w_{e} = sec δ$ . Which energy dominates in a lossy medium?

Try it yourself first.

Strategy

As in Problem 5.6.5, $H = (k / ω μ_{0}) E_{x} a_{y}$ and $P_{c, z} = \frac{1}{2} ∣ E_{0} ∣^{2} (k / ω μ_{0})^{*} e^{- 2 α z}$ . Differentiate, and use the two real equations hidden in $k^{2} = ω^{2} μ ε^{'} (1 - j tan δ)$ : $β^{2} - α^{2} = ω^{2} μ ε^{'}$ and $2 α β = ω μ σ$ .

$- d P_{c, z} / d z = 2 α P_{c, z} = α ∣ E_{0} ∣^{2} e^{- 2 α z} (β + j α) / ω μ_{0}$ . Its real part is $α β ∣ E ∣^{2} / ω μ_{0} = \frac{1}{2} σ ∣ E ∣^{2}$ , because $2 α β = ω μ_{0} σ$ . Its imaginary part is $α^{2} ∣ E ∣^{2} / ω μ_{0}$ , and on the other side

2 ω (w_{m} - w_{e}) = \frac{ω}{2} (μ_{0} \frac{∣ k ∣ ^{2}}{ω ^{2} μ _{0}^{2}} - ε^{'}) ∣ E ∣^{2} = \frac{β ^{2} + α ^{2} - ω ^{2} μ _{0} ε ^{'}}{2 ω μ _{0}} ∣ E ∣^{2} = \frac{α ^{2}}{ω μ _{0}} ∣ E ∣^{2} .

$ω μ_{0} = 2 π \times 5 \times 1 0^{7} \times 4 π \times 1 0^{- 7} = 40 π^{2} = 394.8 Ω/ m$ and $k / ω μ_{0} = (3.628 - j 1.814) /394.8 = 1.027 \times 1 0^{- 2} e^{- j 26.5 7^{\circ}} S$ , so

H (0) P_{c} (0) = 0.1027 e^{- j 26.5 7^{\circ}} a_{y} A/m, = \frac{1}{2} \times 100 \times 1.027 \times 1 0^{- 2} e^{+ j 26.5 7^{\circ}} a_{z} = (0.4594 + j 0.2297) a_{z} W/ m^{2},

- \frac{d P _{c, z}}{d z}_{0} \frac{1}{2} σ E_{0}^{2} = 2 α P_{c, z} (0) = 1.667 + j 0.833 W/ m^{3}, = \frac{1}{2} \cdot \frac{1}{30} \cdot 100 = 1.667 W/ m^{3},

w_{e} w_{m} 2 ω (w_{m} - w_{e}) = \frac{1}{4} \times 9 ε_{0} \times 100 = 1.989 nJ/ m^{3}, = \frac{1}{4} μ_{0} \times 0.102 7^{2} = 3.316 nJ/ m^{3}, = 0.833 W/ m^{3} .

\frac{w _{m}}{w _{e}} = \frac{μ _{0} ∣ k ∣ ^{2} / ω ^{2} μ _{0}^{2}}{ε ^{'}} = \frac{∣ k ∣ ^{2}}{ω ^{2} μ _{0} ε ^{'}} = ∣1 - j tan δ ∣ = sec δ,

here $5/3$ , and indeed $3.316/1.989 = 1.667$ .

Answer

P_{c} (0) = (0.459 + j 0.230) a_{z} W/ m^{2}, \frac{w _{m}}{w _{e}} = sec δ = \frac{5}{3} - \frac{d P _{c, z}}{d z} = heat 1.667 + j 2 ω (w_{m} - w_{e}) 0.833 W/ m^{3}

Check

The ratio $Im / Re$ of $P_{c}$ is $0.2297/0.4594 = 0.5 = α / β$ , and the imaginary part of the divergence is $2 α \times 0.2297 = 0.833 W/ m^{3}$ , the same number as $2 ω (w_{m} - w_{e})$ .

Remark

In a lossless medium $w_{m} = w_{e}$ and the plane wave carries no reactive power. In a lossy one the conduction current adds to the magnetic field ( $∣ H ∣$ grows with $∣ k ∣$ ) but not to the electric energy, so the magnetic energy exceeds the electric one by the factor $sec δ$ : $5/3$ here, and $≫ 1$ in a good conductor, where almost all the stored energy is magnetic. The reactive power that flows into each slice of the medium feeds exactly this surplus, $2 ω (w_{m} - w_{e})$ , back and forth every half period.

cf. Problem 5.6.5

Exercise 5.6.9: Where the reactive power goes