329fall08hw6sol

# 329fall08hw6sol - ECE-329 Fall 2008 Homework 6 Solution 1...

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ECE-329 Fall 2008 Homework 6 — Solution October 12, 2008 1. Some identities. a) Given E = E x ˆ x + E y ˆ y + E z ˆ z and H = H x ˆ x + H y ˆ y + H z ˆ z , we show the following, H · ∇ × E - E · ∇ × H = H x ± ∂E z ∂y - ∂E y ∂z ² + H y ± ∂E x ∂z - ∂E z ∂x ² + H z ± ∂E y ∂x - ∂E x ∂y ² - E x ± ∂H z ∂y - ∂H y ∂z ² - E y ± ∂H x ∂z - ∂H z ∂x ² - E z ± ∂H y ∂x - ∂H x ∂y ² = ± E y ∂H z ∂x + H z ∂E y ∂x - E z ∂H y ∂x - H y ∂E z ∂x ² + ± E z ∂H x ∂y + H x ∂E z ∂y - E x ∂H z ∂y - H z ∂E x ∂y ² + ± E x ∂H y ∂z + H y ∂E x ∂z - E y ∂H x ∂z - H x ∂E y ∂z ² = ∂x ( E y H z - E z H y ) + ∂y ( E z H x - E x H z ) + ∂z ( E x H y - E y H x ) = ∇ · ( E × H ) . b) We will verify that F = x 2 ˆ x + z ˆ y satisﬁes the following identity ∇ × ∇ × F = ( ∇ · F ) - ∇ 2 F . The left-hand side gives ∇ × ∇ × F = ∇ × ³ ³ ³ ³ ³ ³ ˆ x ˆ y ˆ z ∂x ∂y ∂z x 2 z 0 ³ ³ ³ ³ ³ ³ = ∇ × ( - ˆ x ) = 0 , and the right hand side gives ( ∇ · F ) - ∇ 2 F = ( ∇ · ( x 2 ˆ x + z ˆ y )) - ∇ 2 ( x 2 ˆ x + z ˆ y ) = (2 x ) - x = 2ˆ x - x = 0 , therefore, the identity is veriﬁed. 2. A surface current

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329fall08hw6sol - ECE-329 Fall 2008 Homework 6 Solution 1...

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