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

Problem 5.6.5

Interactive: Phasor Explorer

Problem 5.6.5

Power balance in a lossy medium

A $100 MHz$ wave travels along $+ z$ in a nonmagnetic medium with $ε_{r} = 4$ and $σ = 1/60 S/m$ . At $z = 0$ its electric field is $E (0; t) = 100 cos (2 π \times 1 0^{8} t) a_{x} V/m$ .

Find the loss tangent, the complex permittivity $ε_{c}$ and the wave number $k = β - j α$ .
Write the phasor $E (z)$ and find $H (z)$ from the phasor form of Faraday's law.
Find the complex Poynting vector and the time-average power density $P_{a v} (z)$ .
Find the time-average power entering and leaving the slab $0 \leq z \leq 0.5 m$ through a cross-section of $1 m^{2}$ , and show that the difference is the Joule heat $\int_{V} \frac{1}{2} σ ∣ E ∣^{2} d V$ .

Try it yourself first.

Strategy

A lossy medium is a lossless one with $ε \to ε_{c}$ : compute $ε_{c}$ , then $k = ω μ_{0} ε_{c}$ . Faraday's law $\nabla \times E = - j ω μ_{0} H$ gives $H$ , the product $\frac{1}{2} E \times H^{*}$ of Eq. (5.6.27) gives the power, and the slab contains no sources, so the power that goes in and does not come out must be the heat.

Bookkeeping

$ω = 2 π \times 1 0^{8} rad/s$ and $μ = μ_{0}$ . $ε = 4 ε_{0} = 4 \times 1 0^{- 9} /36 π F/m$ , so $ω ε = 1/45 S/m$ . In free space $k_{0} = ω / c = 2 π /3 rad/m$ . At $z = 0$ the phasor is $100 a_{x} V/m$ (zero phase).

tan δ ε_{c} = \frac{σ}{ω ε} = \frac{1/60}{1/45} = 0.75, = 4 ε_{0} (1 - j 0.75) = (4 - j 3) ε_{0} .

The square root is exact: $4 - j 3 = 5 e^{- j 36.8 7^{\circ}}$ , and $(3 - j)^{2} /2 = (9 - 1 - j 6) /2 = 4 - j 3$ , so $4 - j 3 = (3 - j) / 2$ . Then

k = ω μ_{0} ε_{c} = k_{0} 4 - j 3 = \frac{2 π}{3} \cdot \frac{3 - j}{2} = π 2 (1 - \frac{j}{3}),

so $β = π 2 = 4.443 rad/m$ and $α = π 2 /3 = 1.481 Np/m$ .

$E (z) = 100 e^{- j k z} a_{x} = 100 e^{- 1.481 z} e^{- j 4.443 z} a_{x} V/m$ . Faraday's law, $\nabla \times E = - j ω μ_{0} H$ , with only $E_{x} (z)$ present:

⟹ \nabla \times E = \frac{\partial E _{x}}{\partial z} a_{y} = - j k E_{x} a_{y} H = \frac{k}{ω μ _{0}} E_{x} a_{y},

\frac{k}{ω μ _{0}} = \frac{k _{0} 4 - j 3}{ω μ _{0}} = \frac{4 - j 3}{η _{0}} = \frac{3 - j}{120 π 2} = 5.93 \times 1 0^{- 3} e^{- j 18.4 3^{\circ}} S,

using $k_{0} / ω μ_{0} = 1/ c μ_{0} = 1/ η_{0}$ .

Answer

H (z) H (z; t) = 0.593 e^{- j 18. 4^{\circ}} e^{- 1.481 z} e^{- j 4.443 z} a_{y} A/m = 0.593 e^{- 1.481 z} \times cos (2 π \times 1 0^{8} t - 4.443 z - 18. 4^{\circ}) a_{y} A/m

With $a_{x} \times a_{y} = a_{z}$ and $∣ e^{- j k z} ∣^{2} = e^{- 2 α z}$ ,

P_{c} = \frac{1}{2} E \times H^{*} = \frac{1}{2} 10 0^{2} (\frac{k}{ω μ _{0}})^{*} e^{- 2 α z} a_{z} = 29.66 e^{j 18.4 3^{\circ}} e^{- 2.962 z} a_{z} = (28.13 + j 9.38) e^{- 2.962 z} a_{z} W/ m^{2},

Answer

P_{a v} (z) = Re {P_{c}} = 28.13 e^{- 2.962 z} a_{z} W/ m^{2}

Through $1 m^{2}$ at $z = 0$ and at $z = 0.5 m$ :

W_{in} W_{out} = 28.13 W, = 28.13 e^{- 2.962 \times 0.5} = 28.13 e^{- 1.481} = 6.40 W .

The Joule heat density is $\frac{1}{2} σ ∣ E ∣^{2} = \frac{1}{2} \cdot \frac{1}{60} \cdot 1 0^{4} e^{- 2 α z} = 83.33 e^{- 2 α z} W/ m^{3}$ , and over the slab of $1 m^{2}$ cross-section

\int_{V} \frac{1}{2} σ ∣ E ∣^{2} d V = \int_{0}^{0.5} 83.33 e^{- 2.962 z} d z = \frac{83.33}{2.962} (1 - e^{- 1.481}) = 21.74 W .

Answer

W_{in} - W_{out} = 28.13 - 6.40 = 21.74 W = \int_{V} \frac{1}{2} σ ∣ E ∣^{2} d V

Turn on "Phasor along z" below and see the phasor shrink by $e^{- α z}$ as it turns: the field falls to $1/ e$ in $1/ α = 0.675 m$ .

InteractiveLoading simulation…

Check

The closed forms of Eq. (5.6.20) give $β = k_{0} ε_{r} /2 [1 + 0.7 5^{2} + 1]^{1/2} = \frac{2 π}{3} 2 \times 1.5 = π 2$ and $α = \frac{2 π}{3} 2 \times 0.5 = π 2 /3$ . The balance also holds point by point: $- d P_{a v} / d z = 2 α P_{a v} = \frac{1}{2} σ ∣ E ∣^{2}$ , because $2 α β = ω μ_{0} σ$ , and indeed $2 \times 1.481 \times 4.443 = 13.16 = 2 π \times 1 0^{8} \times 4 π \times 1 0^{- 7} /60$ .

Remark

This is the time-harmonic Poynting relation of Section 6.1 applied to a slab with no sources inside: the power that does not come out is heat. $77 %$ of the power entering at $z = 0$ stays in the first half metre (Figure 5.6.16). The field falls to $1/ e$ in $1/ α = 0.675 m$ , the power in $1/2 α = 0.338 m$ , and over one wavelength, $λ = 2 m$ , the power drops to $e^{- 4 π /3} = 1.5 %$ . The imaginary part $Im {P_{c}} = 9.38 e^{- 2 α z} W/ m^{2}$ is reactive: $H$ lags $E$ by $18. 4^{\circ} = δ /2$ , so part of the energy sloshes back and forth instead of travelling.

(b) The heat density. — (a) The power flow.

Problem 5.6.5: Power balance in a lossy medium