Lecture 40 Induction. Review Inductors Self-induction RL circuits Energy stored in a Magnetic Field

Transcription

1 ecure 4 nducon evew nducors Self-nducon crcus nergy sored n a Magnec Feld 1

2 evew nducon end nergy Transfers mf Bv Mechancal energy ransform n elecrc and hen n hermal energy P Fv B v

3 evew eformulaon of Faraday s aw r r ds dφ B ( Faraday' s law) 3

4 nducors and nducance nducor: sore energy n he magnec feld example: a solenod nducance: B r NΦ (nducance defned) Uns: 1 henry 1 H 1 T m / A 4

5 nducance of a Solenod e s calculae from he defnon B r N B l Φ μ n N Φ μ ( nl)( B A) n A ( nl)( B A) ( nl)( μ (nducance n) A of a μ n n l solenod) N l A 5

6 Self-nducance Consder he loop a he rgh, loop. swch closed curren sars o flow n he loop. (s nfne??) V/!!! No. nducance lms d / Therefore, magnec feld produced n he area enclosed by he loop. (B proporonal o ) X X X X X X X X X X X X X X a b Therefore, flux hrough loop ncreases as he curren ncreases Therefore, emf nduced n loop opposng nal drecon of curren flow because opposes ncreasng flux (Faraday s aw) hs emf causes V(a) > V(b) [~lke a baery o oppose he real baery reduce curren flow] Fac: he curren urns on a such a rae o gve V(a)-V(b) baery volage: d d Φ B V ( a) V ( b) Self-nducon: changng curren hrough a loop nducng an opposng emf n ha same loop.

7 Self-nducance The magnec feld produced by he curren n he loop shown s proporonal o ha curren: B The flux, herefore, s also proporonal o he curren. r r ΦB BdS We defne hs consan of proporonaly beween flux and curren o be he nducance,. Φ B Combnng wh Faraday s aw gves he emf nduced by a changng curren: dφ B d d ε ( ) d ε

8 N urns Self-nducance The nducance of an nducor (a se of cols n some geomery; e.g., solenod) hen, can be calculaed from s geomery alone f he devce s consruced from conducors and ar (smlar o he capacance of a capacor). f exra maeral (e.g., ron core) s added, hen we need o add some knowledge of maerals as we dd for capacors (delecrcs) and ressors (ressvy) ΦB ε ( d / ) C Q V C κc ρ A Archeypal nducor s a long solenod, jus as a par of parallel plaes s he archeypal capacor. l r r << l d A d<< A -----

9 Self-nducon Drecon of he self-nduced emf (Fg 3-17): enz aw 9

10 Checkpon 3-5 The fgure shows an emf nduced n a col. Whch of he followng can descrbe he curren hrough he col: (a) consan and rghward, (b) consan and lefward, (c) ncreasng and rghward, (d) decreasng and rghward, (e) ncreasng and lefward, (f) decreasng and lefward. ncreasng (e) decreasng (d) 1

11 evew C crcus where dq q + q( ) C q τ C q / C C (1 e ) C (1 e C q e / τ C / τ C ) (dschargng (chargng C) C) 11

12 Crcus A, he swch s closed and he curren sars o flow. a b oop rule: d ε ε Noe ha hs equaon s dencal n form o ha for he C crcu wh he followng subsuons: Q dq C: ε C C : Therefore, τ C C τ 1 Q C

13 Crcus To fnd he curren as a funcon of me, we need o choose an exponenal soluon whch sasfes he boundary conon: d ( ) ( ) ε We herefore wre: ( / 1 e ) The volage drop across he nducor s gven by: d / V εe ε a ε b τ

14 Crcu (ε on) Curren ε Max ε/ ( / 1 e ) ε/ / / 63% Max a / Volage on d V εe Max ε/ / ε V 37% Max a /

15 Crcus Afer he swch has been n poson a for a long me, redefned o be, s moved o poson b. a b oop rule: + d ε The approprae nal conon s: The soluon hen mus have he form: V ε e / ( d εe ) / ε

16 Crcu (ε off) Curren ε Max ε/ e / ε/ / / Skech curves! 37% Max a / V Volage on Max -ε d εe / V 37% Max a / -ε

17 ε on ε off ε/ / / ε/ / / ε ( / 1 e ) ε e / ε V V d εe / V V d εe / -ε

18 18 Checkpon 3-6 The fgure shows hree crcus wh dencal baeres, nducors, and ressors. ank he crcus accordng o he curren hrough he baery (a) jus afer he swch s closed and (b) a long me laer, greaes frs. e / ) (1 / τ τ ) (1 lm ) ( / 1 ser e a τ ser 1 3 (), (3), (1) e b ser 1 ) (1 lm ) ( / 1 τ par / 3 (), (3), (1)

19 nergy of an nducor How much energy s sored n an nducor when a curren s flowng hrough? Sar wh loop rule: Mulply hs equaon by : ε + ε + From hs equaon, we can denfy P, he rae a whch energy s beng sored n he nducor: du P We can negrae hs equaon o fnd an expresson for U, he energy sored n he nducor when he curren : U U du d d d a ε b 1 U d

20 Where s he nergy Sored? Clam: (whou proof) energy s sored n he magnec feld self (jus as n he capacor / elecrc feld case). To calculae hs energy densy, consder he unform feld generaed by a long solenod: N l B μ l The nducance s: nergy U: 1 1 μ N l πr N l U μ πr 1 πr B l μ We can urn hs no an energy densy by dvdng by he volume conanng he feld: r N urns U u πr l 1 B μ

21 Checkpon 3-7 The able lss he number of urns per un lengh, curren, and crossseconal area for hree solenods. ank he solenods accordng o he magnec energy densy whn hem, greaes frs. u B 1 μ n 1 ( a) ub μ (n1 ) 1 μ n1 ( b) 1 ub μ ( n1 ) (1 ) μ n1 ( c) 1 1 ub μ ( n1 ) 1 μ n a and b e, hen c 1

22 Sample Problem 3-7 A col has an nducance of 53 mh and a ressance of.37 Ω. U B 1 (a) f a 1 V emf s appled across he col, how much energy s sored n he magnec feld afer he curren has bul up o s equlbrum value? /τ lm (1 e ) e / τ (1 ) τ / U B (53 1 H )(34.3 A) 31 J (b) Afer how many me consans wll half hs equlbrum energy be sored n he magnec feld U B U B / τ / 1 τ (1 e ) e 1.93 ln τ

23 Sample Problem ub B μ A long coaxal cable (Fg. 3-) consss of wo hn-walled concenrc conducng cylnders wh rad a and b. The nner cylnder carres a seady curren, he ouer cylnder provdng he reurn pah for he curren. The curren ses up a magnec feld beween he wo cylnders. (a) Calculae he energy sored n he magnec feld for a lengh l of he cable. du B 1 ub du B ubdv U B dv ubdv B dv μ r r B ds μ enc B( μ π r) μ B B π r μ 4π 1 r U B b b μ 1 μ l 1 μ ( l π rdr) dr 4π r 4π r 4π a a μ l U B ln 1 μ 4π b a l [ ln r] b a 3

24 Muual nducance Demo: curren s nduced n one col when he curren s changed n a neghborng col We can descrbe hs effec quanavely n erms of he muual nducance, he rao of flux hrough he loop o curren n oppose loop. b a Φ ab Φba M b a Φ ab flux hrough col a due o curren n col b Noe M has hs symmery (no obvous perhaps, bu rue) Change curren n loop a changes B a change flux hrough loop b nduces curren n b o oppose hs loop b produces a feld B b n he oppose drecon as B a. The same hng happens wh jus one loop. Ths s called selfnducance.

25 Applcaons of Muual nducance Transformers (sll o come) Change one AC volage no anoher Arpor Meal Deecors Pulsed curren pulsed magnec feld nduces emf n meal Ferromagnec meals draw n more B larger muual nducance larger emf mf curren (how much, how long lass, depends on he ressvy of he maeral) Decayng curren produces decayng magnec feld nduces curren n recever cols Magnude & duraon of sgnal depends on he composon and geomery of he meal objec. V N V 1 N1 ε V 1 N 1 (prmary) ron N (secondary) 5 V

26 Applcaons of Muual nducance Pacemakers s no easy o change he baery! nsead, use an exernal AC supply. Alernang curren alernang B alernang Ф B wearer nsde nduces AC curren o power pacemaker ~ 6

27 Applcaons of Muual nducance 7