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Problem 6.18: Solution

This document calculates the ratio of conduction current density to displacement current density for an electromagnetic wave propagating through seawater at different frequencies. It finds that at 1 kHz and 1 MHz, the displacement current is negligible, at 1 GHz neither current is negligible, and at 100 GHz the conduction current is negligible.

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glynfinck
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306 views1 page

Problem 6.18: Solution

This document calculates the ratio of conduction current density to displacement current density for an electromagnetic wave propagating through seawater at different frequencies. It finds that at 1 kHz and 1 MHz, the displacement current is negligible, at 1 GHz neither current is negligible, and at 100 GHz the conduction current is negligible.

Uploaded by

glynfinck
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Problem 6.

18 An electromagnetic wave propagating in seawater has an electric


field with a time variation given by E = ẑE0 cos ω t. If the permittivity of water is
81ε0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction
current density to displacement current density at each of the following frequencies:
(a) 1 kHz
(b) 1 MHz
(c) 1 GHz
(d) 100 GHz
Solution: From Eq. (6.44), the displacement current density is given by
∂ ∂
Jd = D=ε E
∂t ∂t
and, from Eq. (4.67), the conduction current is J = σ E. Converting to phasors and
taking the ratio of the magnitudes,
¯ ¯ ¯ ¯
¯e ¯ ¯ e ¯¯
¯ J ¯ ¯ σE σ
¯ ¯=¯ ¯= .
e ¯ ωεr ε0
Jd ¯ ¯ jωεr ε0 E
¯e

(a) At f = 1 kHz, ω = 2π × 103 rad/s, and


¯ ¯
¯e ¯
¯J¯ 4
¯ ¯= = 888 × 103 .
¯ Jd ¯ 2π × 10 × 81 × 8.854 × 10−12
e 3

The displacement current is negligible.


(b) At f = 1 MHz, ω = 2π × 106 rad/s, and
¯ ¯
¯e ¯
¯J¯ 4
¯ ¯= = 888.
¯e
Jd ¯ 2 π × 10 6 × 81 × 8.854 × 10−12

The displacement current is practically negligible.


(c) At f = 1 GHz, ω = 2π × 109 rad/s, and
¯ ¯
¯e ¯
¯J¯ 4
¯ ¯= = 0.888.
¯e
Jd ¯ 2π × 10 9 × 81 × 8.854 × 10−12

Neither the displacement current nor the conduction current are negligible.
(d) At f = 100 GHz, ω = 2π × 1011 rad/s, and
¯ ¯
¯e ¯
¯J¯ 4
¯ ¯= = 8.88 × 10−3 .
¯e
Jd ¯ 2π × 10 11 × 81 × 8.854 × 10 −12

The conduction current is practically negligible.

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