MAGNETIC FIELDS CURRENT & RESISTANCE

Fiels an Waves I Spring 005 MAGNETIC FIELDS CURRENT & RESISTANCE Name Slutin Sectin Typs Crrecte Multiple Chice 1. (8 Pts). (8 Pts) 3. (8 Pts) 4. (8 Pts) 5. (8 Pts) Ntes: 1. In the multiple chice questins, each questin may have mre than ne crrect answer; circle all f them.. Fr multiple chice questins, yu may a sme cmments t justify yur answer. 3. Make sure yur calculatr is set t perfrm trignmetric functins in raians & nt egrees. Regular Questins 6. (0 Pts) 7. (0 Pts) 8. (Opt 1 0 Pts) 8. (Opt 0 Pts) 8. (Opt 3 0 Pts) Ttal (100 Pts) Sme Cmments an Helpful Inf: In this test, we use tw types f ntatin fr unit vectrs. Keep in min that a$ $ = a$ = y$ a$ z$ y z = a$ r$ r = a$ $ φ = φ a$ $ θ = θ Be sure t shw yur wrk fr the multiple chice questins. Draw pictures fr each prblem t be sure that yu unerstan the prblem statement. Please nte that there are 3 ptins fr the last prblem. Yu shul quickly ecie which ne yu want t an wrk it thrugh. 1 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 MULTIPLE CHOICE QUESTIONS Ecept fr prblem 4, there is nly ne answer t any f these questins. 1. Frce (8 pints) A rectangular lp with a time-invariant current I is place in a unifrm magnetstatic fiel. The lp can rtate abut its ais nrmal t the page. The magnetic fiel eerts a trque n the lp fr a. cases (a) an (b) nly. b. case (c) nly. c. cases (a), (b) an (c) nly.. cases () an (e) nly. e. cases (c), () an (e) nly. K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005. Magnetic Cre (8 Pints) A thin trial cre, mae f a ferrmagnetic material f permeability μ, has an air gap, as shwn in the figure. There is a time-invariant current thrugh the wining. The magnitue f the magnetic fiel intensity vectr in the ferrmagnetic cre with respect t the clck-wise reference irectin is H c. The magnitue f the magnetic fiel intensity vectr in the gap with respect t the same reference irectin is H g f. H = H g g. H g = 0 h. H μ H g c = c i. Hg = μ H μ j. Hg = μ H μ c c 3. Mutual Inuctance (8 Pints) Of the fur mutual psitins f the tw lps shwn, the magnitue f the mutual inuctance between the lps is largest fr the psitin in i. Figure (a) ii. Figure (b) iii. Figure (c) iv. Figure () v. Cannt tell 3 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 4. Fiels an Waves Heres (8 Pints) Ientify which name ges with each equatin. Equatin (c) es nt have a name an ne name ges with all f the equatins. a) b) c) ) r r r r E l = B S t r r r r r r H l = J S + D S t Faraay (a) r r B S = 0 Gauss () r r D S = ρ v Ampere (b) v Mawell All 5. Ampere s Law (8 pints) A cylinrical cnuctr f a circular crss sectin (raius = a) carries a time-invariant current I (I > 0) irecte ut f the page. The line integral f the magnetic flu ensity vectr, r B, alng a clse circular cntur C psitine insie the cnuctr (the cntur raius r is smaller than the cnuctr raius a) is a r a) μ I b) μ I c) greater than μ I ) less than μ I e) less than μ I an psitive f) greater than μ I an negative g) zer 4 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 REGULAR QUESTIONS 6. Bunary Cnitins (0 pints) y μ 5. 6. 7. 8. 9. 10. 11. 1. 13. 14. 15. 16. μ 1 The magnetic fiel in regin 1 is H = H ( a + 1 ay) r H = H( a$ + 1000a$ y) r $ $. The magnetic fiel in regin is. Assuming that ne f these regins is free space, what is the permeability μ f the ther regin? (10) The tw bunary cnitins are Ht1 = H1 = H = Ht = H = Han B = n1 μ H = 1 y1 μ H = 1 B = n μh 1000H y = μ. Thus, μ1h = μ1000han μ = μ, μ1 = 1000μ Ientify which regin is free space (air), regin 1 r regin. (10) Regin is air an regin 1 is the magnetic material with μ1 = 1000μ 5 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 7. Ampere s Law (0 pints) y σ = σ a A lng, straight, sli cylinrical cnuctr with a raius f a is shwn abve. The surruning meium is free space. There is a ttal current I carrie by this cnuctr irecte int the page. What is the current ensity vectr? (6) The current ensity is J r I = $ z. The current is int the page an the z-ais pints ut π a f the page. Thus, the current ensity vectr is negative. What is the magnetic fiel intensity vectr H r insie the cnuctr (r<a)? (7) r r πr H l = Hφ πr = Iencl = I πa. Slving fr H r, we have H r Ir = $ φ πa What is the magnetic fiel intensity vectr H r utsie the cnuctr (r>a)? (7) r r Outsie f the wire, H l = Hφ πr = Iencl = I. Slving fr H r, we have H r I = $ φ πr 6 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 8. (Optin 1) Magnetic Circuits (0 pints) Tw winings are wrappe arun the magnetic cre shwn belw. The cil at the left carries a current I 1 an the cil at the right carries a current I. 3.. g.. w. w The permeability f the cre is μ which is very much larger than μ. There are N 1 turns arun the left leg an N turns arun the right leg. The with an epth f all legs are eactly the same an equal t w. The height f the cre is an the ttal with f the cre is 3 as shwn. Fr the first few questins, assume that the gap g es nt eist. It will be ae in at the en. Als, assume that the current in the right han cil (I ) is zer. a. Fin the reluctance f the cre seen by the cil at the left. (3) The circuit iagram fr the reluctance lks like 5R NI R 3R where R 0 = since each f the legs f length have the same reluctance. Thus, the μ w 3R ttal reluctance seen by the surce is Rttal = 5R+ R 3R = 5R+ = 575. R 4R 7 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 b. Fin the ttal flu linke by all the turns f this cil. (3) NNI N 1 1 1 Iμw Λ= N1ψ 1 = = R 575. ttal c. Fin the inuctance f this cil. (3) Λ NNI N 1 1 1 μw L = = = I IR 575. ttal. Fin the flu pruce by this cil that links all the turns f the secn cil. (3) The flu f part b ivies by the usual current ivier relatinship. The ttal flu linking R NN1I NN1Iμw all f the turns f the secn cil is given by N ψ 1 = = 4R 4R 4(. 575) ttal e. Fin the mutual inuctance between the tw cils. (3) The mutual inuctance is M N ψ1 R NN1 NN1μw = = = I 4R 4R 4(. 575) ttal f. Nw a in the gap an repeat questin a abve. Make any reasnable apprimatins. (3) Since the gap reluctance will be much larger than R, it will appear as an pen circuit in parallel with 3R. Thus the ttal reluctance seen by the surce will be 8R. g. Will the self inuctance f the left cil increase, ecrease r stay the same when the gap is ae? Be sure t justify yur answer. () Since the reluctance will be larger (8>5.75), the flu will be smaller fr the same current an, thus, the inuctance will be smaller. 8 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 8. (Optin ) Faraay s Law (0 Pints) In this prblem we will aress ey current heating, which is use etensively in manufacturing t eliver heat t a cnucting material. It is als the reasn why the can in the can crusher an the cin in the cin flipper are heate by their interactin with the crushing r launching cil. Basically, currents inuce in cnuctrs heat the cnuctrs because f their finite resistance. First etermine the inuctance f an N-turn sleni, each turn carrying a current I, which we will use t inuce the currents in a cnucting shell. Wires a Nn-Magnetic Cre fr Wining the Cil b Cnucting Shell a. Given that the sleni has N turns an is wun t cmpletely cver a nnmagnetic cre (like a tilet paper tube) with ne thin layer f wire, fin the magnetic fiel r B. Assume that yu can neglect fringing an that the raius f the sleni is a an its length is. Recall that this type f sleni is calle an air cre sleni. (4) The magnetic fiel fr the sleni is slve fr in many places. It is given by r NI B = z$ μ 9 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 b. Determine the inuctance f this sleni using either the flu meth r energy meth. Please inicate which meth yu are using. (4) Λ N r r μ N a Frm the flu meth, the inuctance is L = = B S = I I π NI B Hv a r r μ π μ N πa Frm the energy meth, L = = = I I c. A cnucting shell is place insie the sleni, as shwn. This shell has a raius f b, a length an a thickness Δ. Fr current flwing arun the circumference f the cyliner (in theφ -irectin) what is the resistance f the shell? Assume that it has a cnuctivityσ. (4) r R = π σδ. If the current riving the sleni varies sinusially, etermine the ttal current inuce in the shell. Assume It () = I csω t. Hint: etermine the inuce emf first. (4) The inuce emf is given by the time erivative f the flu linke. Nte that the number f turns in the secnary, in this case, is nly 1 since the cnucting shell has nly ne turn. Λ= r r μ Nπb B S = I csωt s that Λ μ Nπb N b emf = = I ωt = ω μ π cs I sinωt t t emf N b Iinuce = = σ Δ ωμ π I sin ωt R πb e. Determine the pwer elivere t the cnucting shell. (4) This can be answere either with pwer r average pwer. Either are OK. N b P = VI = I t ω μ π σ Δ sinω πb Fr average pwer, leave ut the sin term an ivie by. 10 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005 8. (Optin 3) Unifrm Plane Waves in Lssless an Lssy Materials (0 Pints) A unifrm plane wave is prpagating in a material with prperties similar t thse f istille water. That is,ε = 81ε. The frequency f the wave is 100MHz. The average pwer ensity f the wave is 10 Watts per square meter. a. Determine the angular frequencyω, the prpagatin cnstant β, the wavelength λ, an the intrinsic impeanceη fr this wave. (4) ω = πf = π10 8 ω π β = ω με = 9 λ = μ 10π η = = c β ε 9 b. Write bth the electric an magnetic fiels in phasr ntatin. (4) 1 E m η = 10 E = 0η E = E e m m jβz H E m = e η jβz c. Nw sme pllutants are ae t the water which nt change the real part f ε but result in a small imaginary cmpnent s thatεc = ε' jε". In particular,ε" = 00. ε'. Fin the ecay cnstant α an the cmple intrinsic impeanceη c. (4) ε" ε" γ = jω μεc = jω με' 1 j = jω με' 1 j ε' ε' β = jω μ ε' ε ω μ εε α = ω μ ε " '" j ' j = ε' ε' μ ηc = = ε c μ 1 = ε' ε" 1 j ε' μ 1 μ = 1+ ε' ε" ε' 1 j ε' ε" j ε' 11 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005

Fiels an Waves I Spring 005. Write the phasr frm f the electric an magnetic fiels. (4) αz E = E e e m jβz H E α = e e η m z jβz e. Determine the average pwer elivere t a cubic meter f this material. (4) This is the same calculatin as n the HW prblem. 1 Em Save e αz = We can apprimate the cmple intrinsic impeance with the * η c lssless value, since the lss is nt large. η meter is then: ( α z α ) = ( ) c μ. The pwer elivere per cubic ε' 1 e 10Watts 1 e 10Watts since the epth is equal t 1. ( ) 8 ( 310 ) ( ) () α ω μ εε 8 '" π10 9 π 9 = = 00. = 00. = 006. π ε' 3 01. π ( ) 1 e 10Watts = 31. Watts 1 K. A. Cnnr Rensselaer Plytechnic Institute 5 April 005