The electric field intensity of a plane wave traveling in free space is given by the following expression Ex-t-y24- cos- t-k0xV-mIn this field- consider a square area 10 cm - 10 cm on a plane x - y - 1- The total time-averaged power in mW passing through the square area is

E(x,t)=â_y24π cos(ω t k_0x)(V/m) Power density P=12 | E |^2η=12(24π )^2120πâ_x Time averaged power passing through square area P_avg=P.ds = [ 12(24π )^21 ...

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Poynting Theorem

The electric ...

Question

The electric field intensity of a plane wave traveling in free space is given by the following expression

$\to E (x, t) = {^a}_{y} 24 π c o s (ω t - k_{0} x) (V / m)$

In this field, consider a square area 10 cm $\times$ 10 cm on a plane x + y = 1. The total time-averaged power (in mW) passing through the square area is _____
53.29

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Solution

The correct option is A 53.29
$\to E (x, t) = {^a}_{y} 24 π c o s (ω t - k_{0} x) (V / m)$

Power density

$\to P = \frac{1}{2} \frac{{| E |}^{2}}{η} = \frac{1}{2} \frac{(24 π)^{2}}{120 π} {^a}_{x}$

Time-averaged power passing through square area

$P_{a v g} = \to P . \to d s$

$= [\frac{1}{2} \frac{(24 π)^{2}}{120 π} {^a}_{x}] [\frac{{^a}_{x} + {^a}_{y}}{\sqrt{2}} \times 10^{- 2}]$

$P_{a v g} = \frac{24 π}{\sqrt{2}} = 53.29 m W$

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The correct option is A 53.29→E(x,t)=^ay24πcos(ωt−k0x)(V/m) Power density →P=12|E|2η=12(24π)2120π^ax Time-averaged power passing through square area Pavg=→P.→ds =[12(24π)2120π^ax][^ax+^ay√2×10−2] Pavg=24π√2=53.29mW