HW2_Solution

HW2_Solution - z 1 Ez 0 x Hy 1 3. y Complex Poynting's...

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3. x y z E z H y 1 1 0 Complex Poynting's theorem states that () 2 s fd me PPP jWW ω =++ . s P is the complex power supplied by the sources. * 1 2 f PE H d S ∫∫ G GG w is the complex power leaving the surface enclosing the volume. 2 1 2 d d V σ = ∫∫∫ G x is the time average power loss in the volume. 2 1 4 m WH d V μ = ∫∫∫ G x is the time average magnetic energy stored in the volume. 2 1 4 e WE d V ε = ∫∫∫ G x is the time average electric energy stored in the volume. In this problem 0 s P = since there are no sources in the volume. Therefore Re df PP =− and ( ) 2I m f WW P −= . (a) On the surface at 0 x = 100sin z Ey a π = G G , 6 sin j y He y a = G G , and x dS dxdya = G G . Note that on an closed surface normal always points out. On all the other surfaces tangential E field is zero. Therefore there is no contribution to the surface integral on any other surface other than the surface at 0 x = . Hence ** 0 11 22 x EH d S S = ו = w .
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW2_Solution - z 1 Ez 0 x Hy 1 3. y Complex Poynting's...

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