329fall08hw4sol

# 329fall08hw4sol - ECE-329 Fall 2008 Homework 4 — Solution...

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Unformatted text preview: ECE-329 Fall 2008 Homework 4 — Solution 1. Curl and divergence exercises. a) V = x ˆ x + y ˆ y • Vector field x-axis y-axis- 2- 1 1 2- 2- 1 1 2 • Curl ∇ × V = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z x y = . • Divergence ∇ · V = ∂ ∂x ( x ) + ∂ ∂y ( y ) = 1 + 1 = 2 . b) V =- y ˆ x + x ˆ y • Vector field x-axis y-axis- 2- 1 1 2- 2- 1 1 2 • Curl ∇ × V = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z- y x = 2 ˆ z. 1 ECE-329 Fall 2008 • Divergence ∇ · V = ∂ ∂x (- y ) + ∂ ∂y ( x ) = 0 . c) Curl and divergence properties. i. ∇ × V 6 = implies the field strength varies across the direction of the field V . ii. ∇ · V 6 = 0 implies the field strength varies along the direction of the field V . 2. Verifying vector calculus identities, ∇ × ( ∇ f ) = and ∇ · ( ∇ × A ) = 0 . a) The gradient of a scalar field f is defined as ∇ f = ∂f ∂x ˆ x + ∂f ∂y ˆ y + ∂f ∂z ˆ z. Taking the curl, we obtain ∇ × ∇ f = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z ∂f ∂x ∂f ∂y ∂f ∂z = ∂ ∂y ∂f ∂z- ∂ ∂z ∂f ∂y ˆ x + ∂ ∂z ∂f ∂x- ∂ ∂x ∂f ∂z ˆ y + ∂ ∂x ∂f ∂y- ∂ ∂y ∂f ∂x ˆ z = . b) The curl of a vector field A = A x ˆ x + A y ˆ y + A z ˆ z is defined as ∇ × A = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z A x A y A z = ∂A z ∂y- ∂A y ∂z ˆ x + ∂A x ∂z- ∂A z ∂x ˆ y + ∂A y ∂x- ∂A x ∂y ˆ z. Taking the divergence, we obtain ∇ · ( ∇ × A ) = ∂ ∂x ∂A z ∂y- ∂A y ∂z + ∂ ∂y ∂A x ∂z- ∂A z ∂x + ∂ ∂z ∂A y ∂x- ∂A x ∂y = ∂ 2 A z ∂x∂y- ∂ 2 A y ∂x∂z + ∂ 2 A x ∂y∂z- ∂ 2 A z ∂y∂x + ∂ 2 A y ∂z∂x- ∂ 2 A x ∂z∂y = 0 . 3. The magnetic field B in regions where the current density J and the displacement current ∂ D ∂t are both zero satisfies ∇ · B = 0 (Gauss’ law) and ∇ × B = (Ampere’s law). Below, we will verify if the following vector fields can be realized as magnetic fields in such regions. a) F 1 = (2 x + 3 y ) ˆ x + (3 x- 2 y ) ˆ y • Divergence ∇ · F 1 = ∂ ∂x (2 x + 3 y ) + ∂ ∂y (3 x- 2 y ) = 2- 2 = 0 ....
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329fall08hw4sol - ECE-329 Fall 2008 Homework 4 — Solution...

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