329sp09hw7sol

329sp09hw7sol - ECE-329 Spring 2009 Homework 7 Solution...

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Unformatted text preview: ECE-329 Spring 2009 Homework 7 Solution March 10, 2009 1. Continuity equation. a) Taking the divergence of Amperes law we have ( H ) = J + D t . Since ( H ) = 0 , we can write J + D t = J + D t = 0 . Finally, using Gauss law D = , we show J + t = 0 . b) Taking the divergence of J = ( 4 z 2 x + 3 x 3 y y + 2 z ( y- y o ) 2 z ) A / m 2 we get J = x 4 z 2 + y 3 x 3 y + z 2 z ( y- y o ) 2 = 3 x 3 + 2( y- y o ) 2 . Since J is time independent, we have that t =- J ( r ,t ) =- 3 x 3 + 2( y- y o ) 2 t + o C m 3 . Evaluating at r = and considering that o = 0 and y o = 1 , we find ( ,t ) =- 2 t C m 3 . c) Since the units of J x , J y , and J z are A / m 2 , we get [ J x ] = h 4 z 2 i = A m 2 [4] = A m 4 , [ J y ] = h 3 x 3 y i = A m 2 [3] = A m 6 , [ J z ] = h 2 z ( y- y o ) 2 i = A m 2 [2] = A m 5 . 2. In HW6, we found that the charge density in a conductor with conductivity and permittivity o is ( r ,t ) = cos(100 x ) e- o t C / m 3 . 1 ECE-329 Spring 2009 a) Using the continuity equation J + t = 0 and noting that J has only a component along x-direction (i.e., J = J x x ), we can write J x x =- t ....
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329sp09hw7sol - ECE-329 Spring 2009 Homework 7 Solution...

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