Notes 9 Spring 2005

Notes 9 Spring 2005 - Notes 9 Spring 2005.doc We see that...

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Notes 9 Spring 2005.doc We see that in very lossy media, i.e. when 100 tan > δ we approximate the angle on the intrinsic impedance as ! ° 45 Let’s see how well that approximation works 1 : ° = φ = δ = ε ° = φ = δ = ε = φ = δ = ε ° = φ = δ = ε ° = φ = δ = ε ° = φ = δ = ε 994 . 44 5000 tan 5000 j 1 9 . 44 500 tan 500 j 1 7 . 44 100 tan 100 j 1 4 . 44 50 tan 50 j 1 6 . 43 20 tan 20 j 1 1 . 42 10 tan 10 j 1 r r r r r r So, approximating the angle at works pretty well for ° 45 100 tan > δ . Recall that we approximate the skin depth in highly conductive media as ω μ σ 2 δ = (see page 7 of Notes 8). If we rearrange this as σ δ = ωµ 2 1 we can obtain an expression for in terms of the skin depth. Note that the approximation for η η in good conductors on page 7 of Notes 8 is ° σδ = ° σ = ° σ ωµ η σ δ 45 2 45 45 2 1 eff So for highly conducting media ° σδ 45 2 η Also, recalling that for good conductors α = β and α = δ 1 we have πδ β π = λ 2 2 ωδ β ω = p v and 1 Note that from page 6 of Notes 8 the exact equation for the intrinsic impedance is () 2 1 2 1 tan j 1 j 1 eff δ ε′ µ = ε′ ω σ ε′ µ = η so that the angle is simply δ tan tan 2 1 . 1
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Notes 9 Spring 2005.doc Furthermore, in terms of the skin depth δ , in a good conductor the electric field can be written as δ δ β α = = z j z 0 x z j z 0 x e e E x ˆ e e E x ˆ E G and the magnetic field as ° δ δ δ δ σδ = = 45 j z j z 0 x z j z 0 x e e e 2 E y ˆ e e η E y ˆ H G Example: A 100 Hz electromagnetic wave is propagating vertically down into seawater, which is characterized by , , and o µ = µ o 81 ε = ε m S 4 = σ . The electric field magnitude measured at the surface of the seawater ( ) is 1 V/m. What is the magnitude of the electric field at a depth of 100m? 0 z = We note that and we calculate the loss tangent: σ = σ eff () ( ) ( ) 100 10 88 . 8 10 85 . 8 81 100 2 4 tan 6 12 eff >> × = × π = ε′ ω σ = ε′ ω σ = δ So the seawater is highly conductive and we can use the appropriate approximation for : α 1 2 7 m 10 97 . 3 4 10 4 100 × = × π π = µσ π = α f Therefore, the electric field magnitude at 100m is ( ) m V 2 100 10 97 . 3 z 0 x 10 88 . 1 e 1 e E E 2 × α × = = = What is the power density at the surface of the seawater? At 100m? Solution: The electric field strength is z j z 0 x e e E x ˆ E β α = G and the magnetic field strength is φ β α η = j z j z 0 x e e e E y ˆ H G . Using these forms yields {} × = H E e 2 1 G G G S 2
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Notes 9 Spring 2005.doc = φ β α β α 0 e e e
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This note was uploaded on 01/20/2012 for the course EE 4460 taught by Professor Czarnecki during the Fall '10 term at LSU.

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Notes 9 Spring 2005 - Notes 9 Spring 2005.doc We see that...

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