Physics 1402: Lecture 21 Today s Agenda

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1 ecure 4 Physics 142: ecure 21 Tody s Agend Announcemens: nducion, R circuis Homework 6: due nex Mondy nducion / A curren Frdy's w ds N S B v B B S N B v 1

2 ecure 4 nducion Self-nducnce, R ircuis X X X X X X X X X ε /R V Recp from he ls hper: Frdy's w of nducion N S B v S N B v Time dependen flux is genered y chnge in mgneic field srengh due moion of he mgne Noe: chnging mgneic field cn lso e produced y ime vrying curren in nery loop n ime vrying curren in conducor induce EMF in in h sme conducor? d/d B 2

3 ecure 4 Self-nducnce The inducnce of n inducor ( se of coils in some geomery..eg solenoid, oroid) hen, like cpcior, cn e clculed from is geomery lone if he device is consruced from conducors nd ir. f exr meril (eg iron core) is dded, hen we need o dd some knowledge of merils s we did for cpciors (dielecrics) nd resisors (resisiviy) S UNTS for : Henry Archeypl inducor is long solenoid, jus s pir of prllel ples is he rcheypl cpcior. r << l ong Solenoid: lculion N urns ol, rdius r, engh l l r N urns For single urn, The ol flux hrough solenoid is given y: nducnce of solenoid cn hen e clculed s: This (s for R nd ) depends only on geomery (meril) 3

4 ecure 4 R ircuis A =, he swich is closed nd he curren srs o flow. oop rule: Noe h his eqn is idenicl in form o h for he R circui wih he following susiuions: R: R R: ecure 21, AT 1 A = he swich is hrown from posiion o posiion in he circui shown: 1A Wh is he vlue of he curren long ime fer he swich is hrown? () = () = ε / 2R (c) = 2ε / R 1B Wh is he vlue of he curren immediely fer he swich is hrown? () = () = ε / 2R (c) = 2ε / R 4

5 ecure 4 R ircuis To find he curren s fc of ime, we need o choose n exponenil soluion which sisfies he oundry condiion: We herefore wrie: The volge drop cross he inducor is given y: R ircui (ε on) urren ε/r /R 2/R Mx = ε/r 63% Mx =/R Volge on ε Mx = ε/r 37% Mx =/R V 5

6 ecure 4 R ircuis Afer he swich hs een in posiion for long ime, redefined o e =, i is moved o posiion. oop rule: The pproprie iniil condiion is: The soluion hen mus hve he form: R ircui (ε off) urren ε/r /R 2/R Mx = ε/r 37% Mx =/R Volge on Mx = -ε 37% Mx =/R V -ε 6

7 ecure 4 ε on ε off ε/r /R 2/R ε/r /R 2/R ε V V -ε Review: R ircuis (Time-vrying currens) Dischrge cpcior: iniilly chrged wih Q=ε onnec swich o =. lcule curren nd chrge s funcion of ime. ε R oop heorem onver o differenil equion for q: 7

8 ecure 4 Review: R ircuis (Time-vrying currens) Dischrge cpcior: Tril soluion: ε R heck h i is soluion: Noe h his guess incorpores he oundry condiions:! Dischrge cpcior: Review: R ircuis (Time-vrying currens) urren is found from differeniion: ε R onclusion: pcior dischrges exponenilly wih ime consn τ = R urren decys from iniil mx vlue (= -ε/r) wih sme ime consn 8

9 ecure 4 Dischrging pcior hrge on ε R 2R Mx = ε 37% Mx =R urren q Mx = -ε/r 37% Mx =R -ε/r ε hrging R 2R Dischrging R ε 2R q q ε/r - ε/r 9

10 ecure 4 1

11 ecure 4 Energy of n nducor How much energy is sored in n inducor when curren is flowing hrough i? Sr wih loop rule: Muliply his equion y : From his equion, we cn idenify P, he re which energy is eing sored in he inducor: We cn inegre his equion o find n expression for U, he energy sored in he inducor when he curren = : 11

12 ecure 4 Where is he Energy Sored? lim: (wihou proof) energy is sored in he Mgneic field iself (jus s in he pcior / Elecric field cse). To clcule his energy densiy, consider he uniform field genered y long solenoid: l The inducnce is: r N urns Energy U: We cn urn his ino n energy densiy y dividing y he volume conining he field: Muul nducnce Suppose you hve wo coils wih muliple urns close o ech oher, s shown in his cross-secion We cn define muul inducnce M 12 of coil 2 wih respec o coil 1 s: oil 1 oil 2 B N 1 N 2 cn e shown h : 12

13 ecure 4 nducors in Series Wh is he comined (equivlen) inducnce of wo inducors in series, s shown? Noe: he induced EMF of wo inducors now dds: 1 2 eq Since: And: nducors in prllel Wh is he comined (equivlen) inducnce of wo inducors in prllel, s shown? Noe: he induced EMF eween poins nd e is he sme! 1 2 eq Also, i mus e: We cn define: And finlly: 13

14 ecure 4 onsider he nd R series circuis shown: ircuis R Suppose h he circuis re formed = wih he cpcior chrged o vlue Q. lim is h here is quliive difference in he ime developmen of he currens produced in hese wo cses. Why?? onsider from poin of view of energy! n he R circui, ny curren developed will cuse energy o e dissiped in he resisor. n he circui, here is NO mechnism for energy dissipion; energy cn e sored oh in he cpcior nd he inducor! Q i R R/ ircuis Q i R: curren decys exponenilly i : curren oscilles -i 1 14

15 ecure 4 Oscillions (quliive) Oscillions (quniive) Wh do we need o do o urn our quliive knowledge ino quniive knowledge? Wh is he frequency ω of he oscillions?

16 ecure 4 Begin wih he loop rule: Oscillions (quniive) Q i Guess soluion: (jus hrmonic oscillor!) rememer: where: ω deermined from equion φ, Q deermined from iniil condiions Procedure: differenie ove form for Q nd susiue ino loop equion o find ω. Review: Oscillions Guess soluion: (jus hrmonic oscillor!) Q i where: ω deermined from equion φ, Q deermined from iniil condiions which we could hve deermined from he mss on spring resul: 16

17 ecure 4 ecure 21, AT 2 A = he cpcior hs chrge Q ; he resuling oscillions hve frequency ω. The mximum curren in he circui during hese oscillions hs vlue. Wh is he relion eween ω nd ω 2, he 1A frequency of oscillions when he iniil chrge = 2Q? () ω 2 = 1/2 ω () ω 2 = ω (c) ω 2 = 2 ω ecure 21, AT 2 A = he cpcior hs chrge Q ; he resuling oscillions hve frequency ω. The mximum curren in he circui during hese oscillions hs vlue. 1B Wh is he relion eween nd 2, he mximum curren in he circui when he iniil chrge = 2Q? () 2 = () 2 = 2 (c) 2 = 4 17

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