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1 Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM Homework: See websie. Table of Conens: 3. Self-inducance in a circui, 3. -Circuis, 4 3.a Charging he coil, 5 3.b Differen mehod o solve he differenial equaion, 6 3.c Discharging he coil, Energy in a Magneic Field, 7 3.3a Energy in a coaxial cable, Muual Inducance, 3.5 Oscillaions in an C circui, 3.5a Solving d.e. by using complex funcions, 3.6 The C circui (wihou exerior power source), 3

2 Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM 3. Self-inducance in a circui. Faraday s law has imporan consequences for elecric circuis. Every circui is a loop, and a ime changing curren will herefore produce a back-emf. Consider a simple circui consising of a resisor, an emf, and a swich. When we urn he swich on he curren increases slowly from o a maximum value. The curren is obviously varying wih ime. The magneic field creaed by his ime varying curren is also increasing wih ime. According Faraday s law his magneic field creaes a ime varying emf and a ime varying induced curren in he same loop, which goes agains he original curren, due o enz s law. curl j () ds I() (3.) ( ) ( ) A I () (3.) l I () l which will induce an emf in he same circui according o Faraday s law: dφ ( ) d I( ) A (3.3) ( ) A l l rae of magneic flux hrough he loop of he circui inducance clockwise () creaed by original curren Iind ind ( ) induced emf I( ) original curren For simplificaion and demonsraion purposes I have muliplied on he lef side in (3.) wih he lengh of he closed pah. Therefore, we see easily ha he magneic field (3.) in his process of self-inducion is proporional o he original imevarying curren. I is his same magneic field which eners Faraday s law on he righ side of (3.3) We see ha he emf of self-inducance is proporional o he rae of change of he original curren. We inroduce he new quaniy, he inducance of he circui, as he proporionaliy facor beween he emf and he rae of change of he original curren. For a coil wih N ighly wound loops he flux hrough N loops is equal o N imes he flux hrough a single loop. An emf is creaed in each single loop, and he oal emf is he sum of all he emfs. There is only one single curren, herefore: dφ (3.4) N The self induced emf is always proporional o he rae of change of he original curren. (Noe ha from now on we avoid he leers l or for lenghs. We reserve he leer for self-inducance.)

3 Dr. Friz Wilhelm page 3 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM From (3.4) we ge: (3.5) NdΦ The change in flux is direcly proporional o he change in curren.when here is no curren here is no flux, and vice versa. Inegraing, we ge NΦ (3.6) NdΦ I (3.7) NΦ I To summarize: he ime varying curren I() in a circui creaes a ime varying magneic field () in he circui (Ampere s law). The ime varying magneic field induces a imevarying back-emf (Faraday s law) and an induced curren in he circui. oh, he self induced emf and he induced curren I ind are direced agains he original emf and he original curren I ha creaed hem. Example 3.: Find he inducance of a uniformly wound very long solenoid having N urns and lengh h. (We use he leer h for lengh insead of which is now reserved for he inducance.) N N wires inercep he surface A () ni I() h A I N NA h I From Ampere s law we know: curl j h NI ecall ha in he calculaion of he magneic field he Ampereian surface for which we calculaed he flux of he curren densiy was a recangular surface wih is long side parallel o he magneic field inside he solenoid. In he figure above we call his surface A. N (3.8) () ni () I () h

4 Dr. Friz Wilhelm page 4 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM In order o calculae he emf induced by he ime changing magneic field we calculae he flux of his magneic field hrough N surfaces creaed by he N urns of wire. Each of hese surfaces is perpendicular o he Ampereian surface. We should call hem Faraday surfaces. The circulaion around each such loop creaes a back-emf, which is equal o he ime changing dφ magneic flux hrough his loop. i. For N loops we ge a oal back-emf: N dφ (3.9) i N i The magneic flux for a single loop of wire is equal o he magneic field inside he solenoid imes he surface of one single loop of wire or A. Therefore, NΦ NA NA N N A (3.) I n Ah nv (3.) I I I h h solenoid nv Volume of he inerior solenoid. y inroducing a ferromagneic maerial ino he solenoid he self-inducance can be grealy increased, because i srenghens he magneic field. This effec ges absorbed in he value for which hen is no he permeabiliy of empy space, and is called μ wihou he.. e us calculae he inducance for he solenoid direcly from Faraday s law: d d N curle da E ds da NA NA h A A A The line inegral of he induced elecric field mus be aken N imes for each loop. For a ighly wound solenoid i is equal o dφ (3.) i N E ds πrne N A The flux-inegral yields N imes he magneic flux hrough a simple surface, which is he circular loop creaed by a single circular loop of wire: d d N (3.3) da NA NA h A Wih he definiion of (3.4) we ge he resul in (3.): N A N N (3.5) NA N ha n V h h h h 3. -Circuis. Solving he equaions of an dc circui during sar-up and shu-down: e us furher sudy wha happens in a dc-circui wih a resisor. (All circuis have a resisance, wheher we explicily include a resisor or no.)

5 Dr. Friz Wilhelm page 5 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM 3.a Charging he capacior: When we close he swich in such a circui, allowing a curren o flow, he curren will no immediaely flow wih is maximum value. I is growing from o is maximum value /. This means of course ha he curren changes wih ime, which again means ha during his ime a back-emf is esablished in he circui according o Faraday s law:. The emf acs agains he original curren according o enz s law. When, afer he curren has reached is maximum value, we remove he original power source, he curren will decrease and again cause a back emf, his ime o oppose he curren s decrease. Swich is closed, curren increases ( ) induced (clockwise) Iind ind ( ) I( ) Jus like he resisance was a measure of he opposiion o he curren V I so is he inducance a measure of he resisance o he change in he original curren. oh quaniies ΔV- I and (energies per uni charge) will reduce he volage supplied by he original ouside power supply. We sill have energy conservaion when we complee he loop of a circui. The power supplied by he ouside source equals he power used by he circui. When he swich is closed in he circui above wih a power supply he curren in he circui increases from o is maximum value. We apply energy conservaion o he whole loop and ge a familiar differenial equaion: We need o solve our equaion for I(). e us firs do his hrough simple separaion of he variables: I I (3.6) I I We ge he soluion of an exponenially increasing curren wih he ime-consan I I e e > τ τ () τ Afer a ime of 4.6τ he curren reaches 99% of is maximum value of. e us check his:

6 Dr. Friz Wilhelm page 6 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM (3.7).99 e.99 e. e ln 4.6τ I, herefore also a ime varying magneic During his ime we have an increasing curren ( ) field, herefore a self-induced emf. ind Therefore, he polariy of he induced emf is opposie o he original emf. While he induced curren is flowing, power is being delivered o he magneic field which builds up in he coil because of P I I U I. A he same ime power is being dissipaed in he resisor as hea according o P I. 3.b Differen mehod o solve he differenial equaion (3.6): (3.8) I + We firs recognize ha he d.e. conains a consan. The general rule for solving d.e. s says o firs solve he d.e. wihou he consan and hen o add a special soluion which saisfies he equaion wih he consan. The d.e. wihou he consan is easily solved: (3.9) I I() Ae A special consan soluion is given by I. You can verify ha by puing ha soluion ino he original d.e. + Our general soluion is herefore obained by adding he special soluion o he general soluion o yield: (3.) I Ae + Now, we apply he iniial condiion ha he curren is a and ge: (3.) A+ A This gives as he same soluion we obained before:(3.) I () e

7 Dr. Friz Wilhelm page 7 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM ( ) induced (ccw) Iind ind ( ) We close he swich afer II 3.c Discharging he coil: When, afer a while, we disconnec he power source, he curren in he circui will no go o in an insan bu will also decrease exponenially. I is again impeded by he induced emf which his ime opposes he decrease of he curren. We can see his again by solving he differenial equaion of he circui (which has been energized previously) afer he power supply has been removed. The swich in he circui is now closed. The curren now decreases because i dissipaes energy in he resisor and no more ouside power is being supplied. < he polariy is again reversed. I + I (3.3) I τ τ I Ie e < τ (The ime consan τ ) Summing his up, we need o inroduce a new self induced emf ino any circui conaining a ime varying curren. This is done by adding an inducor coil ino he circui, usually symbolized by a spiral along he circui, wih he value for is inducance. In his way, he whole analysis ges reduced o wriing down Kirchhoff s rules for he circui as an insananeous equaion. Jus like wih he volage drop across a resisor we have an emf drop across such an inducor coil. As he origin of he volage ulimaely does no maer, we also alk abou a volage drop across an inducor coil: (3.4) V I and ind V I( ) 3.3 Energy in a Magneic Field. We sar again wih an circui conaining an emf. The insananeous Kirchhoff rule reads like he elecric energy conservaion. (We used Kirchhoff's laws so far mosly for dc-currens. They are jus a convenien summary of he laws of charge and energy conservaion, which mus hold for any circui, including hose wih ime varying currens and volages. This is why we call Kirchhoff's rules now "insananeous.") (3.5) I

8 Dr. Friz Wilhelm page 8 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM The energy of he emf will go o he resisor, where i is dissipaed as hea, and he coil, where i is sored in he magneic field. To ge he rae of change of energy in he circui, i.e. is power we simply muliply (3.5) wih I. ( Power I V I ) (3.6) I I + I power delivered o he resisor power delivered o du he inducance (3.7) power delivered o he inducor coil P Inegraing his over ime: du I (3.8) du I I (3.9) ecall ha he energy in he capacior was equal o Q (3.3) UC C( V) C which allowed us o calculae he energy densiy in he empy space beween he plaes of a parallel plae capacior as: (3.3) ue E For he magneic energy inside he space of a solenoidal coil we need from (3.). (3.3) nv For he curren in he coils of he solenoid, expressed by he magneic field we use: (3.33) ni I n Insering and I ino (3.9) gives us : U ( nv) n (3.34) U V U I follows ha he magneic energy densiy u inside a coil is given by : V (3.35) u 3.3a Energy in a coaxial cable: U I

9 Dr. Friz Wilhelm page 9 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM In a coaxial cable he inner hin cylindrical cable carries he curren from he source and he ouer hin cylindrical cable carries he curren back o he source, hus closing a loop. The inside of he inner cylindrical cable has no magneic field (Ampere s law.) On he ouside of he ouer cable here is also no magneic field as he oal curren hrough a cross-secion of he cable is. The only magneic field exiss beween he wo cable surfaces. We have seen earlier ha he magneic field ouside of a conducion wire curls around he wire (inner cylinder) and is equal o I (3.36) ( r) from Ampere's law: π r I π r In order o calculae he energy (3.37) U I conained in he magneic field of he coaxial cable we need o find he inducance of he cable: which is given by : NΦ (3.38) I This is in connecion wih Faraday s law, which uses surfaces perpendiclar o he surface used in deermining he magneic field in Ampere s law. Call hem he Faraday surfaces. We need he surface hrough which he magneic field flows and around which a curren and emf is being creaed: (3.39) E ds da A A I So, we need o calculae he flux of he magneic field r () hrough a recangular crosssecion of he cable (along he cable, no perpendicular o i), in which he magneic field is π r angenial o he concenric circle around he inner cable wih radius r>a (where a is he inner radius of he inside cable.) (Don confuse he coaxial cable wih a solenoid!) The oal Faraday surface is he recangle of wih (b-a) and lengh h. The magneic field is perpendicular o ha surface and varies wih he disance r from he cenral axis of he concenric cylinders. The small area da(r) of he cylindrical surface lies a a disance r from he cener line and has he hickness dr. Is lengh is he lengh of he cable h, hus da() r hdr r h he area of he flux d Φ hdr corresponds o he heigh of he cable (lengh) imes he radial incremen dr. I r () π r

10 Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM ( r) circles around he cenral conducor inerceps area dr h dφ h dr cenral conducor wih curren I inerceps area dr h dφ h dr I h dr π r We ge for he flux: b I Ih b (3.4) Φ hdr ln πr π a For we ge : NΦ N h b (3.4) ln ; N I π a Fo he oal magneic energy sored in he coaxial cable we ge : hi b (3.4) U I ln 4π a a 3.5 Muual Inducance: If we have a ime varying curren I in circui, and a ime varying curren I in an adjacen circui, I will induce an emf in circui and I will induce an emf in circui. We alk abou muual inducance M. If we have a ime-varying magneic field () (from a ime varying curren) hen obviously he field creaes a flux hrough any closed surface, no jus hrough he one responsible for is field (which leads o he self inducance we jus discussed.) If here is anoher closed loop wih a surface close by hen, he mageic field will creae a flux Φ and herefore an induced emf in ha loop also, and hen vice versa.

11 Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM N urns in coil ; wih I and surface A A N urns in coil ; wih I A ( ) The magneic field creaed by he lef loop inerceps he righ loop, creaing a magneic flux Φ of he magneic field hrough he surface A. If we have a ime dependen curren and magneic field he change in flux will creae an emf in he second loop. d N ( A ) M ( ) Φ A Φ (3.43) which provides he definiion of he mual inducance M Obviously, he magneic field creaed by he curren I in he second coil, will creae a flux hrough he firs loop Φ A and herefore also an induced emf. We define he muual inducance accordingly: d N ( A ) M N (3.44) which provides he definiion of he urns; wih I mual inducance M N urns in coil ; wih I I urns ou ha he wo muual inducances are equal and we don need o disinguish beween hem. (3.45) M M M The muual inducance depends on he shape and make of he coil. Thus, each coil creaes a new emf in he oher coil: (3.46) M and M The muual inducance has he same form as he self-inducance. I plays a role in ransformers, which we shall discuss laer. Φ

12 Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM 3.5 Oscillaions in an C circui. e us consider a simple circui wih a capacior C, a coil, and a swich. The capacior is originally fully charged and he swich is open. When we close he swich a ime-varying curren runs hrough he circui, which will induce a back emf in he coil. The mahemaical expression of his ime varying charge is he soluion of a differenial equaion. We bear in mind ha he energy of he capacior decreases as he energy of he coil increases, and vice versa. We can use he energy approach, noicing ha he oal energy in he circui remains consan as here is no resisance in he circui: U C Q C U I I Q () (3.47) U + I ( ) consan C The ime derivaive is because here is no dissipaion of energy in a resisor : Q dq Q d Q + I + (3.48) C C (afer dividing by IdQ/) This is a differenial equaion of he second order in Q, jus like he d.e. of a spring: (3.49) k Q + Q x x C + m We arrive a he same equaion by simply wriing down Kirchhoffs rules for a circui wihou Q and powersupply: Q + Q Q + Q C C C 3.5a Solving d.e. by using complex funcions: This is a good ime o review how we solve such equaions wih complex numbers. You can check ou he paper ch3complex Oscillaions on he websie (Do i, you will need i!). We will make exensive us of his in he following wo chapers. I is no covered in your ex book. Use he complex rial soluion (3.5) ˆ iα Q( ) Qe (cos sin ) Q ω+ i ω Use he fac ha he derivaive of an exponenial funcion urns ino a muliplicaion : dqˆ ( ) d Qe iα i α Q ˆ ( ) Q (3.5) dq ˆ ( ) d i Qe α ( i α) Q ˆ( ) α Q ˆ Wih his equaion (3.49) becomes : (3.5) ( α + ω ) Qˆ ( ) α ± ω Using he iniial condiion ha Q()Q we ge he familiar soluion o (3.49):

13 Dr. Friz Wilhelm page 3 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM (3.53) Q Qcos ω; wih ω C We could have also jus wrien down Kirchhoffs rules for his circui : Q d Q Q + (3.54) C C Q + Q Q ( ) Qcos( ω+ ϕ) wih ω C C Compare hese equaions and soluions wih he equaions for he oscillaions of a spring : k (3.55) x+ x x+ ωx x xcos( ω+ ϕ) m The poenial energy of he spring corresponds o Q (3.56) kx C The kineic energy of he spring corresponds o he energy of he coil : (3.57) mv mx Q I 3.6 The C circui (wihou exerior power source). If we inroduce a resisor ( V I ) ino he circui above (no emf) we ge he siuaion analogous o ha of he spring wih a damping facor b ( F bv ). The circui now loses energy according o I, jus like he spring loses energy according o bv. Saring wih a fully charged capacior, corresponds o saring he spring wih is maximum poenial energy. y using he same process as before we ge he d.e. Q d Q dq Q (3.58) I + + C C (3.59) Q + Q Q + C iα Use he rial soluion: Q Q e Q iαq; and Q α Q This leads o he quadraic equaion in α (earn how o do his!): (3.6) α + iα + ω i ± + 4ω (3.6) α i ± ω (3.6) ( ) ± i i Q Qe Qe ω α Qe ( cosω± sinω) ω

14 Dr. Friz Wilhelm page 4 of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM (3.63) Q + I + C Q + Q + Q C This equaion corresponds o he spring equaion: (3.64) b k x + x x m + m Obviously, we ge he same soluion which is a damped oscillaion : b/ m b cos ω ; ω ω ; ω b k m (3.65) x xe m m Comparing he wo differenial equaions we jus need o esablish a correspondens beween he consans: b (3.66) m k ω m C Thus : (3.67) Q ( ) Qe cos ω; ω C Energy is los by his circui o he une of bv in erms of he spring s consans. This ranslaes o he C circui as: (3.68) Q I jus as we expeced. e us double check his by saring wih he insananeous energy of a circui which is given Q by:(3.47) U + I C In he case of he C circui his energy is consan. In he case of he C circui we know ha energy is los in he resisor: (3.69) du d Q Q Q ( + I ) QQ + II I + II I + I C C C C I Q ooking a he original differenial equaion I I we see ha he erm in parenheses C is equal o I. Therefore, he change in oal energy of he circui is equal o he energy loss o hea or inernal energy: (3.7) du Q I + I I C I ω