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

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Unformatted text preview: ECE-329 Fall 2008 Homework 6 — Solution October 5, 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 + H y ∂E x ∂z- ∂E z ∂x + H z ∂E y ∂x- ∂E x ∂y- E x ∂H z ∂y- ∂H y ∂z- E y ∂H x ∂z- ∂H z ∂x- E z ∂H y ∂x- ∂H x ∂y = E y ∂H z ∂x + H z ∂E y ∂x- E z ∂H y ∂x- H y ∂E z ∂x + E z ∂H x ∂y + H x ∂E z ∂y- E x ∂H z ∂y- H z ∂E x ∂y + E x ∂H y ∂z + H y ∂E x ∂z- E y ∂H x ∂z- H x ∂E y ∂z = ∂ ∂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 = ∇ × ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z x 2 z = ∇ × (- ˆ x ) = , 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 )- 2ˆ x = 2ˆ x- 2ˆ x = , therefore, the identity is verified....
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This note was uploaded on 04/14/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

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

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