3209-2009-Solutions10

3209-2009-Solutions10 - ECE 3209 — Electromagnetic Fields...

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Unformatted text preview: ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 10 Solutions 1. Cheng, P. 6-39. Mutual inductance between a long wire and loop. The ~ B-field due to a long straight wire (from Ampere’s Law) is, ~ B ( r ) = μ I 2 πr ˆ φ Thus, the flux through the loop is Φ B = Z Z loop ~ B ( d- r cos θ ) · d~a = μ I 2 π Z b Z 2 π rdrdθ d- r cos θ where we have used da = rdrdθ . The integral above is of the form, Z dx a + b cos x = 2 √ a 2- b 2 tan- 1 ‡ √ a 2- b 2 tan( x/ 2) a + b · Thus, Φ B = μ I π Z b rdr n 2 √ d 2 + r 2 ‡ tan- 1 ( ∞ )- tan- 1 (0) ·o = μ I Z b rdr √ d 2 + r 2 Φ b = μ I (- p d 2- r 2 ) fl fl fl b = μ I ( d- p d 2- b 2 ) M = Φ B I = μ ( d- p d 2- b 2 ) 2. Cheng, P. 7-6. Eddy Current Loss (a) For the cylindrical core shown in figure 7-12(a), we can apply Faraday’s Law in integral form and exploit symmetry to find the Faraday-induced electric field in the core: I C ~ E · d ~ ‘ =- d dt Z S ~ B · d~a ⇒ 2 πrE φ =- d dt ‡ B ( t ) · πr 2 · By symmetry the electric field must be circumferentially directed and constant on a circular path (of radius r ) coaxial with the cylinder, thus ~ E =- ˆ φ r 2 dB ( t ) dt =- rωB 2 cos( ωt ) The instantaneous (ohmic) power dissipated in a height h is thus, P L ( t ) = Z V ~ E · ~ J dv = σ Z V | E | 2 dv = σ Z a Z 2 π Z h ‡ rωB 2...
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This note was uploaded on 10/16/2010 for the course ECE 309 taught by Professor Weikle during the Fall '08 term at UVA.

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3209-2009-Solutions10 - ECE 3209 — Electromagnetic Fields...

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