Exercise 5.6.18: Identifying an absorber from its two fields

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 5.6.18

Identifying an absorber from its two fields

Inside a sheet of microwave absorber (a rubber loaded with magnetic and carbon powders), a $1 GHz$ plane wave is measured as

E (z; t) H (z; t) = 100 e^{- 57.2 z} \times cos (2 π \times 1 0^{9} t - 138.1 z) a_{x} V/m, = 0.473 e^{- 57.2 z} \times cos (2 π \times 1 0^{9} t - 138.1 z - 22. 5^{\circ}) a_{y} A/m .

Assuming a real permeability $μ$ and $ε_{c} = ε - j σ / ω$ , find (a) $k$ and $η$ , (b) $μ_{r}$ , $ε_{r}$ and $σ$ , (c) the wavelength in the absorber and the depth at which the power density has dropped by $20 dB$ , (d) $P_{a v}$ at $z = 0$ .

Try it yourself first.

cf. Problems 5.6.7, 5.6.8, 5.6.10