EE 333 Electricity and Magnetism, Fall 2009 Homework #11 solution

Transcription

1 EE 333 Eetriity and Magnetim, Fa 009 Homework #11 oution 4.4. At the interfae between two magneti materia hown in Fig P4.4, a urfae urrent denity J S = 0.1 ŷ i fowing. The magneti fied intenity H in region i given by H = 3ˆx + 9ẑ. Determine the magneti fux B 1 and B in region 1 and, repetivey. We have two boundary equation, ( ˆn B1 B ) ( = 0 ˆn H1 H ) = J From the firt equation we get impy that B 1z = B z. And, B 1z = B z = 3µ 0 H z = 7µ 0. From the eond equation we ee that (ine ˆn = ẑ that ˆx ŷ ẑ H 1x H x H 1y H y H 1z H z = ŷj y Whih redue to H 1y H y = 0 H 1x H x = J y So H 1y = H y = 0, and H 1x = H x + J y = = 3.1. Now ompute the magneti fied omponent, B 1x = 5µ 0 H 1x = 15.5µ 0, and B 1y = 0. The magneti fied in region 1 i then B 1 = 15.5µ 0ˆx + 7µ 0 ẑ In region we have B x = 3µ 0 H x = 9µ 0, and B y = 0, o overa we get B = 9µ 0ˆx + 7µ 0ˆx 4.5. Conider the probem of determining the magneti vetor potentia A inide and outide an infinite iruar yindria oenoid of radiu a. The oenoid ha N turn per unit ength and the urrent in the winding i I. (a) Ue the ur reation between A and the magneti fied B = A and Stoke theorem to how that where i the area enired by. A d = B d (b) Baed on ymmetry onideration, eet uitabe ontour for A inide and outide the oenoid to how that A = { µ0niρ µ 0NIa ˆφ ˆφ ρ < a a < ρ 1

2 (a) Stoke theorem ay that F = F d S If we et F = A we get A = A d = B d QED (b) To do thi probem we firt need to determine the magneti fied. From ymmetry onideration we ee that at eah radia ditane from the enter of the oenoid, ρ, the magneti fied mut point aong the z-axi independent of the azimutha ange. Thi i due to the rotationa ymmetry of the oenoid. B d = µ 0 J d with any retanguar ontour that extend from beyond one ide of the oenoid to beyond the oppoite ide of the oenoid. Ony the part of the ontour aong the axi ontribute, and ine thoe ontribution mut be oppoite (yet ame fied), and the tota urrent through the ontour i zero, the fied outide the oenoid mut be zero. Next onider another retanguar whih extend from inide the oenoid to outide it. It extend ditane L aong the axi of the oenoid. Ony the ontour path aong the axi of the oenoid, inide the oenoid, ha non-zero magneti fied ontribution. We then get or LB z = µ 0 LNI The magneti fied of a oenoid i then B = B z = µ 0 NI { ẑµ 0 NI inide 0 outide Now we are ready to ompute A. Pik a ontour whih i iruar and goe in the righthand diretion around the z-axi at a ditane ρ. Note that ine B = ẑb z, A = ˆφA φ. Inide the oenoid we then have πρa φ = πρ B z = πρ µ 0 NI

3 or wherea outide the oenoid we have A φ = µ 0ρNI or Putting it together we get πρa φ = πa B z = πa µ 0 NI A φ = µ 0a NI 0 ρni {ˆφµ ρ a A = a < ρ ˆφ µ 0a NI 4.9. Ue the reut of equation 4.99a for the magneti fux at a far point from a iruar urrent oop to determine approximatey the mutua indutane between two thin oaxia iruar ring of radii a and b. Aume that the ditane d between the two ring i muh arger than a and b. The magneti fied from a ma urrent oop a arrying urrent I i On the axi it redue to B = µ 0Ia ( ) oθˆr + in θˆθ 4r 3 B z = µ 0Ia r 3 The amount of fux through a ma oaxia oop of radiu b at ditane d i then The indutane i then ψ = πb B z = µ 0πIa b d 3 L = ψ I = µ 0πa b d 3 Note that it i ymmetri. I.e. ame reut whether a or b i generating the fied with urrent I. 7.. Conider the foowing votage and urrent ditribution: v(z, t) = v o o β (z ut) i(z, t) = v o o β (z ut) 3

4 where β i a ontant, u = 1 and Z o =. By diret ubtitution, verify that v(z, t) and i(z, t) atity the tranmiion ine equation 7.10 to The tranmiion ine equation that we are to verify are Chek the firt equation dv dz =di dt di dz =dv dt d v v dz =d dt d i i dz =d dt Chek the eond equation v o β in β (z ut) =βu v o in β (z ut) 1 = u 1 = 1 1 = Chek the 3rd equation v o β in β (z ut) =v 0 βu inβ (z ut) 1 =u Z 0 = = 4

5 v o β oβ (z ut) =v o β u o (z ut) 1 =u 1 = ( ) Chek the 4th equation v o β oβ (z ut) = v o β u oβ (z ut) 1 =u 1 = ( ) 5