Notes+10+Fall+2004

# Notes+10+Fall+2004 - Notes 10 Fall 2004 Example The...

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Notes 10 Fall 2004.doc Example: The electrical constitutive parameters of moist earth at a frequency of 1 MHz are m S 1 10 = σ , 1 r and , 4 r = µ = ε′ . Assuming that the electric field magnitude of a uniform plane wave at the interface (on the side of the earth) is m V 2 10 3 × , find: a. The distance through which the wave must travel before the magnitude of the electric field reduces to m V 2 10 104 . 1 × . Answer: α = α = × × α × = × α = 368 . 0 ln z z e 2 10 3 2 10 104 . 1 z e 2 10 3 2 10 104 . 1 z e 0 x E E G We see that we must find the attenuation constant, . α Find the loss tangent: 1 6 . 449 10 85 . 8 4 10 2 10 12 6 1 >> = × π = ε′ ε ′ = ε′ ω σ , therefore this is considered a highly conductive medium and we can approximate σµ π = β = α f . 1 m 628 . 0 7 10 4 1 . 0 6 10 = × π π = α m 59 . 1 628 . 0 368 . 0 ln z = = b. The wavelength inside the earth. Answer: m 10 628 . 0 2 2 = π = β π = λ (vs. 300m in free space) c. The phase velocity inside the earth. Answer: s m 6 10 10 628 . 0 6 10 2 2 p v × = π = β π = β ω = f 1

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Notes 10 Fall 2004.doc d. The intrinsic impedance inside the earth. Answer: ε ′ ε′ µ = ε µ = η j 0 Can we neglect one of the terms in the denominator? Recall the loss tangent: 6 . 449 = ε′ ε ′ . So we can neglect ε and ° = × π
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Notes+10+Fall+2004 - Notes 10 Fall 2004 Example The...

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