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329sum09hw7sol

# 329sum09hw7sol - ECE-329 Summer 2009 Homework 7 Solution 1...

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ECE-329 Summer 2009 Homework 7 — Solution July 10, 2009 1. Verifying vector identity H · ∇ × E - E · ∇ × H = ∇ · ( E × H ) for E = 2 e - αz ˆ x and H = 4 e - αz ˆ y. Since the left-hand side of the identity gives H · ∇ × E - E · ∇ × H = (4 e - αz ˆ y ) · ˆ x ˆ y ˆ z 0 0 ∂z 2 e - αz 0 0 - (2 e - αz ˆ x ) · ˆ x ˆ y ˆ z 0 0 ∂z 0 4 e - αz 0 = (4 e - αz ˆ y ) · ( - 2 αe - αz ˆ y ) - (2 e - αz ˆ x ) · (4 αe - αz ˆ x ) = - 8 αe - 2 αz - 8 αe - 2 αz = - 16 αe - 2 αz and the right-hand side gives ∇ · ( E × H ) = ∇ · ( 2 e - αz ˆ x × 4 e - αz ˆ y ) = ∇ · ( 8 e - 2 αz ˆ z ) = ∂z ( 8 e - 2 αz ) = - 16 αe - 2 αz , then the identity is verified. 2. Continuity equation. a) 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 ρ ( r , t ) at r = 0 and given that ρ o = 0 and y o = 1 , we find ρ ( 0 , t ) = - 2 t C m 3 .

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329sum09hw7sol - ECE-329 Summer 2009 Homework 7 Solution 1...

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