EE325_quiz_6_sol_Yilmaz

EE325_quiz_6_sol_Yilmaz - NAME Solution I EE325-Fall 2009...

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NAME: _____________ Solution I EE325-Fall 2009 Quiz 6 1) The magnetic field strength is given as φ ρρ ρ ⎢⎥ =− 1 ˆ sin( ) cos( ) A/m Ha . a) Find the current density. (Hint: ρφ ρ ρρ φ ∂∂∂ ∇× = ˆˆ ˆ 1 z z z HHH aa a H ) (3 points) ρ ρ ρ ρ ρ =∇× + = 1 ˆ ( (sin( ) cos( ))) 1 ˆ(c o s () c o s () s in ()) ) ˆ sin( ) z z z J H a a a b) Find the total current passing through the region ρ < 1 . (Hint: sin( ) sin( ) cos( ) xx d x x ) (5 points) Alternative 1: π φ ρ πρ ρ ρ π ρ ρ ρ π = = == = ∫∫ ∫∫ i i 12 00 1 1 0 0 sin( ) 2 sin( ) 2 (sin( ) cos( )) 2 (sin(1) cos(1)) S zz I dd d Jds Alternative 2: Ampere’s integral law: ππ φ ρ ρ ρ ρ φ π = ⎤⎡ ⋅= = = ⎥⎢ ⎦⎣ v 22 C0 0 1 ˆ sin( ) cos( ) 2 sin(1) cos(1) Hd l 2) If ρ v S v V dv Dd s and v C 0 Ed l , then a) ∇⋅ = ? D Why? (1 point) ρ = v D (point form of Gauss’s law, can be shown by using the definition of divergence and taking limit of the above expression as the volume shrinks to zero) b) = ? E Why? (1 point) = 0 E ( point
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EE325_quiz_6_sol_Yilmaz - NAME Solution I EE325-Fall 2009...

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