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Unformatted text preview: ECE 329 Spring 2010 Homework 8  Solution Due: Mar. 16, 2010 1. a) From the problem, we know that in the free space region z > , the electric ux density is D 1 = 3ˆ a x + 2ˆ a z ( C/m 2 ) . Therefore, the electric eld is E 1 = D 1 1 = D 1 = 3 ˆ a x + 2 ˆ a z ( V/m ) . b) In the free space region z > , the magnetic eld is H 1 = 4ˆ a y 4ˆ a z ( A/m ) . Thus, the magnetic ux density is B 1 = μ 1 H 1 = μ H 1 = 4 μ ˆ a y 4 μ ˆ a z ( Wb/m 2 ) . c) By using the discontinuity condition for the electric ux density on the z = 0 boundary ˆ a z · ( D 1 D 2 ) = ρ S ( C/m 2 ) , we obtain the zcomponent of D 2 D 2 z = D 1 z ρ S = 2 ( 4) = 6 ( C/m 2 ) . On the other hand the xycomponents of E are continuous across the boundary, since ˆ a z × ( E 1 E 2 ) = . Thus, we have E 2 x = E 1 x = 3 ( V/m ) . Therefore, D 2 = 15ˆ a x + 6ˆ a z ( C/m 2 ) , and E 2 = D 2 2 = 3 ˆ a x + 6 5 ˆ a z ( V/m ) . d) By using the discontinuity condition for the magnetic eld on the z = 0 boundary ˆ a z × ( H 1 H 2 ) = J S ( A/m ) , we obtain the xycomponents of H 2 H 2 y = H 1 y ( J S x ) = 4 ( 3) = 1 ( A/m ) ....
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This note was uploaded on 04/20/2010 for the course ECE ECE 329 taught by Professor Goddart during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
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