Exercise 5.5.3: Wave or diffusion? Three lossy materials

About the further exercises of Section 5.5

Ten further exercises in the spirit of Problems 5.5.1–5.5.6, the home exercises and the worked examples, graded from easy to harder. Exercises 5.5.1–5.5.2 are about travelling pulses, Exercises 5.5.3–5.5.5 about the wave equation itself, Exercises 5.5.6–5.5.8 about potentials and gauges, and Exercises 5.5.9–5.5.10 about retarded potentials. They use only the equations listed at the head of this subsection. All media are nonmagnetic.

Exercise 5.5.3

Wave or diffusion? Three lossy materials

The source-free wave equation of a conducting medium contains the friction term $μ σ \partial E / \partial t$ . Consider (i) fresh lake water, $ε_{r} = 80$ , $σ = 5 mS/m$ , probed by a radar at $50 MHz$ , (ii) muscle tissue, $ε_{r} = 52.7$ , $σ = 1.74 S/m$ , at the $2.45 GHz$ of a microwave oven, and (iii) a doped silicon wafer, $ε_{r} = 11.7$ , $σ = 10 S/m$ , under a $2.4 GHz$ antenna. (a) Find the time $τ = 2 ε / σ$ of Section 5.5.7 and $ω τ$ . (b) Find $σ / (ω ε)$ and the frequency $f_{t}$ at which the two time-derivative terms of the equation have equal amplitudes. Which behaviour dominates? (c) Compute $α = \frac{σ}{2} μ / ε$ and $1/ α$ , and compare with the exact $α$ of a sinusoidal wave (Home exercise 5.5.1). Where does the simple formula fail?

Try it yourself first.

Strategy

For a signal of angular frequency $ω$ the term $μ ε \partial^{2} / \partial t^{2}$ has the amplitude $μ ε ω^{2}$ and the term $μ σ \partial / \partial t$ the amplitude $μ σ ω$ . Their ratio is $σ / (ω ε) = 2/ (ω τ)$ , and the two are equal at $f_{t} = σ / (2 π ε)$ , the crossover frequency, Eq. (5.2.11) in Section 5.2. The simple $α = 1/ (v τ)$ assumes $ω τ ≫ 1$ .

a), b)

With $ε = ε_{r} ε_{0}$ :

	(i) lake water, 50 MHz	(ii) muscle, 2.45 GHz	(iii) silicon, 2.4 GHz
$τ = 2 ε / σ$	283 ns	0.536 ns	20.7 ps
$ω τ$	88.9	8.24	0.312
$σ / (ω ε)$	0.0225	0.243	6.41
$f_{t} = σ / (2 π ε)$	1.13 MHz	594 MHz	15.4 GHz

Lake water at $50 MHz$ is a good dielectric: a wave with slow damping. Muscle at $2.45 GHz$ is in between, still wave-like, but the loss is far from negligible. The silicon wafer at $2.4 GHz$ is on the conductor side ( $f < f_{t}$ ): the conduction current dominates and the field diffuses rather than travels.

The simple formula and the exact one, $α = ω \frac{μ ε}{2} [1 + (σ / ω ε)^{2} - 1]$ :

	(i) lake water	(ii) muscle	(iii) silicon
$α = \frac{σ}{2} μ / ε$	0.105 Np/m	45.2 Np/m	551 Np/m
$1/ α$	9.49 m	2.21 cm	1.81 mm
exact $α$	0.105 Np/m	44.9 Np/m	285 Np/m
exact $1/ α$	9.49 m	2.23 cm	3.51 mm

Answer

(i) wave, 1/ α = 9.49 m, (ii) lossy wave, 1/ α \approx 2.2 cm, (iii) diffusion, 1/ α = 3.51 mm (simple formula off by a factor 1.9)

Check

For (iii) the diffusion (good-conductor) length $2/ (ω μ σ) = 3.25 mm$ is much closer to the exact $3.51 mm$ than the dielectric formula. For (i) the two values agree to three digits because $(σ / ω ε)^{2} = 5 \times 1 0^{- 4}$ .

Remark

The $2 cm$ of muscle is why a microwave oven heats a thick piece of meat from the outside in, and why $2.45 GHz$ reaches deeper than higher frequency bands would. A lake can be sounded with a $50 MHz$ radar through several metres of fresh water, but a sea, with almost a thousand times larger $σ$ , cannot.

cf. Problem 5.5.4