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EE_387_200809SP_homework_03_solution

# EE_387_200809SP_homework_03_solution - 1 ri a ℜc = EE 387...

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1 Homework 03 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 r i r o 0 c c rc l A μμ ℜ= 2 c N L = 36 (10 10 )(1.203 10 ) 109.66 110 c NL =ℜ × = 1 2 3 2( ) () (35 50) 10 0.2670 m ci o io lr r rr π =⋅ + =+ × = 2 1 2 2 1 4 32 1 4 42 [( ) ] [(50 35) 10 ] 1.767 10 m co i oi Ar r =− × 74 6 0.2670 (1000)(4 10 )(1.767 10 ) 1.203 10 −− = ×× a. b. 1.

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2 Homework 03 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 r i r o (0.2670 m) (1.5 mm) 0.2655 m c l =− = 74 6 0.2655 (1000)(4 10 )(1.767 10 ) 1.196 10 c π −− ℜ= ×× 3 6 1.5 10 (4 10 )(1.767 10 ) 6.755 10 g × 66 6 1.196 10 6.755 10 7.951 10 cg ℜ=ℜ +ℜ 36 (10 10 )(7.951 10 ) 282 NL =ℜ × = a. b. 2.
3 Homework 03 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 l xo l z i l xo l xi l xi B H B H μ = ⇒= c B Ni l = c c Hl Ni Ni H l = c l B i N = 4(400 100) 1.200 m c l = = 7 11 . 2 (1100)(4 10 ) 100 8.68 A i π = × = a. b. 3.

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4 Homework 03 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 l xo l z i l xo l xi l xi N λ ϕ = 0.5 100 5mWb N = = = 1 2 1 2 (0.5 V s)(2 A) 0.5 J Wi = =⋅ = Li = 0.5 V s 2A 0.25 H L i = = = 2 2 0 2 0 c rc c N L N l A AN l μμ = = = 2 0 c r c Ll μ = a. b. c. d. [ ] 1 2 3 1 2 3 4( ) 4 (200 100) 10
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EE_387_200809SP_homework_03_solution - 1 ri a ℜc = EE 387...

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