Homework8_Solutions (1).pdf - Solutions Homework 8 ECE 333 Professor Amir Arbabi Problem 1 H 1 = 2\u02c6 x \u2212 y\u02c6 3\u02c6 z E 2 = 3\u02c6 x 4\u02c6 y \u2212 2\u02c6 z(a Find

Homework8_Solutions (1).pdf - Solutions Homework 8 ECE 333...

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Solutions Homework 8 ECE 333 Professor Amir Arbabi Problem 1 H 1 = 2 ˆ x - ˆ y + 3 ˆ z E 2 = 3 ˆ x + 4 ˆ y - 2 ˆ z (a) Find E 1 , D 1 , H 2 , and B 2 for fig (i) First, for E 1 and D 1 use the relations ˆ n × ( E 1 - E 2 ) = 0 ˆ n · ( D 1 - D 2 ) = ρ s = 0 Using these can break down fields into normal and tangential components. Starting with D 2 n : D 2 n = ( E 2 · ˆ n ) ˆ n = 0 r 2 4 ˆ y = 0 8 ˆ y 1
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Solutions Homework 8 ECE 333 Professor Amir Arbabi we know from relations above, that D 2 n = D 1 n Thus, E 1 n = 1 0 r 1 D 1 n = 0 8 ˆ y 0 3 = 8 3 ˆ y our tangential component can be found as follows: E 2 t = E 2 - E 2 n = 3 ˆ x - 2 ˆ z Since tangential E field is continuous across interface, we can say E 1 t = E 2 t and thus E 1 = E 1 t + E 1 n = 3 ˆ x + 8 3 ˆ y - 2 ˆ z D 1 = 0 [9 ˆ x + 8 ˆ y - 6 ˆ z ] Similar procedure can be followed for B and H : ˆ n × ( H 1 - H 2 ) = J s = 0 ˆ n · ( B 1 - B 2 ) = 0 So with B 1 n = μ H 1 n = ( μ H 1 · ˆ n ) ˆ n = - μ ˆ y = - μ 0 2 ˆ y and B 1 n = B 2 n can find normal component as 2
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Solutions Homework 8 ECE 333 Professor Amir Arbabi H 2 n = 1 μ 0 μ r 2 B 1 n = 1 μ 0 μ r 2 μ 0 μ r 1 H 1 n = - 2 ˆ y Tangential component of H 2 can be found as follows H 1 t = H 1 - H 1 n = 2 ˆ x + 3 ˆ z And with J s we have H 1 t = H 2 t and so H 2 = H 2 t + H 2 n = 2 ˆ x - 2 ˆ y + 3 ˆ z B 2 = μ 0 [2 ˆ x - 2 ˆ y + 3 ˆ z ] (b) Find E 1 , D 1 , H 2 , and B 2 for fig (ii) The procedure here remains the same, but we must adjust our normal vector ˆ n : ˆ n 1 = 1 17 [ - ˆ x + 4 ˆ y ] ˆ n 2 = 1 17 [ ˆ x - 4 ˆ y ] And then, proceeding as in part (a): Starting with D 2 n : D 2 n = ( E 2 · ˆ n 2 ) ˆ n 2 = 0 r 2 - 13 17 ˆ x - 4 ˆ y 17 = 0 - 13 ˆ x + 52 ˆ y 17 3
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Solutions Homework 8 ECE 333 Professor Amir Arbabi
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