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329sp09hw7sol - ECE-329 Spring 2009 Homework 7 Solution 1...

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ECE-329 Spring 2009 Homework 7 — Solution March 10, 2009 1. Continuity equation. a) Taking the divergence of Ampere’s 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 = ρ, we show ∇ · J + ∂ρ ∂t = 0 . b) Taking the divergence of J = ( 4 z 2 ˆ x + 3 x 3 y ˆ y + 2 z ( y - y o ) 2 ˆ 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 ∂ρ ∂t = -∇ · J ρ ( r , t ) = - 3 x 3 + 2( y - y o ) 2 t + ρ o C m 3 . Evaluating at r = 0 and considering that ρ o = 0 and y o = 1 , we find ρ ( 0 , 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 σ and permittivity o is ρ ( r , t ) = cos(100 x ) e - σ o t C / m 3 . 1
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ECE-329 Spring 2009 a) Using the continuity equation ∇ · J + ∂ρ ∂t = 0 and noting that J has only a component along x -direction (i.e., J = J x ˆ x ), we can write ∂J x ∂x = - ∂ρ ∂t . Solving this equation we get J = σ 100 o sin(100 x ) e - σ o t ˆ x A m 2 . Then, the associated electric field is E = 1 σ J = 1 100 o sin(100 x ) e - σ o t ˆ x V m .
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