# 329sp09hw6sol - ECE-329 Spring 2009 Homework 6 Solution...

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Unformatted text preview: ECE-329 Spring 2009 Homework 6 Solution March 2, 2009 1. Verifying vector identity H E- E H = ( E H ) for E = 2 e- z x and H = 4 e- z y. The left-hand side of the identity gives H E- E H = (4 e- z y ) x y z z 2 e- z- (2 e- z x ) x y z z 0 4 e- z = (4 e- z y ) (- 2 e- z y )- (2 e- z x ) (4 e- z x ) =- 8 e- 2 z- 8 e- 2 z =- 16 e- 2 z and the right-hand side gives ( E H ) = ( 2 e- z x 4 e- z y ) = ( 8 e- 2 z z ) = z ( 8 e- 2 z ) =- 16 e- 2 z , thus, the identity is verified. 2. A surface current J s =- J so cos( t ) x flowing on the x- y plane induces electromagnetic waves on both sides of the sheet. The induced fields are given by E = E o cos( t z ) x for z and H = H o cos( t z ) y for z , where E o = o H o . a) The average Poynting vector for the wave propagating in + z-direction is hPi z&amp;gt; = 1 2 E o H o z, while, for the wave propagating in- z-direction, we have hPi z&amp;lt; =- 1 2 E o H o z. Since the average power density provided by the current sheet is 1 W / m 2 , we can write that hPi z&amp;gt; + hPi z&amp;lt; = E o H o = E 2 o o = 1 W m 2 ....
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329sp09hw6sol - ECE-329 Spring 2009 Homework 6 Solution...

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