Phasor domain expression for the electric field intensity of a uniform plane wave propagating in a source-free, simple and lossy medium in n = ⸠direction is given by Ē(2) = E, c¯a² -jßz -az (Volts/m) The lossy medium of this problem is known to be a good conductor with ơ = 5.8×10’ S/m and µ = Ho . It is also given that f= 5.8 MHz and E(z=0) = 1 (Volts / m). O Find the time-average Poynting's vector P in terms of coordinate variable z. ) Compute, separately, the time average powers P and P, passing through the surfaces S, and S, , respectively. Compare (P Plte P found in nart (e) Comment on

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1-
Phasor domain expression for the electric field intensity of a uniform plane wave propagating in a
source-free, simple and lossy medium in în = ⸠direction is given by
Ē(2) = E, e¯az-ißz
(Volts/m)
The lossy medium of this problem is known to be a good conductor with o = 5.8×10' S/m and
µ = Ho. It is also given that f= 5.8 MHz and E(z = 0) = 1 (Volts / m).
a) Find the time-average Poynting's vector P in terms of coordinate variable z.
b) Compute, separately, the time average powers P and P, passing through the surfaces S,
and S2, respectively.
c) Compare ( P - P,) to P, found in part (c). Comment on the result.
-10 cm-
S,
S;
at z=8
10 cm-
Transcribed Image Text:1- Phasor domain expression for the electric field intensity of a uniform plane wave propagating in a source-free, simple and lossy medium in în = ⸠direction is given by Ē(2) = E, e¯az-ißz (Volts/m) The lossy medium of this problem is known to be a good conductor with o = 5.8×10' S/m and µ = Ho. It is also given that f= 5.8 MHz and E(z = 0) = 1 (Volts / m). a) Find the time-average Poynting's vector P in terms of coordinate variable z. b) Compute, separately, the time average powers P and P, passing through the surfaces S, and S2, respectively. c) Compare ( P - P,) to P, found in part (c). Comment on the result. -10 cm- S, S; at z=8 10 cm-
☺ penetration degth of 8:
36 ५2५
= 2.745X 10m.
Jz
wrrent density
we know is a is londuct vit and ? is electrie field then
Dissinati ve
power la - ½ EJ *
* dv
Pa: !
Pa =
dv * e 202
Transcribed Image Text:☺ penetration degth of 8: 36 ५2५ = 2.745X 10m. Jz wrrent density we know is a is londuct vit and ? is electrie field then Dissinati ve power la - ½ EJ * * dv Pa: ! Pa = dv * e 202
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