Physcs - lectrcty and Magnetsm ecture - Inductance, Crcuts Y&F Chapter 30, Sect - 4 Inductors and Inductance Self-Inductance Crcuts Current Growth Crcuts Current Decay nergy Stored n a Magnetc Feld nergy Densty of a Magnetc Feld Mutual Inductance Summary Copyrght. Janow Fall 03
Inducton: bascs Magnetc Flux: dφ B B da B nˆ da B nˆ Faraday s aw: A changng magnetc flux through a col of wre nduces an MF n the wre, proportonal to the number of turns, N. nd dφ N B enz s aw: The current drven by an nduced MF creates an nduced magnetc feld that opposes the flux change. B nd & nd oppose changes n Φ B Inducton and energy transfer: The forces on the loop oppose the moton of the loop, and the power requred to sustan moton provdes electrcal power to the loop. P F v Fv P ε ε Blv Transformer prncple: changng current n prmary nduces MF and current n secondary col. A changng magnetc flux creates a non-conservatve electrc feld. ε ds dφ N B Copyrght. Janow Fall 03
Changng magnetc flux nduces electrc felds: trval transformer A thn solenod, cross secton A, n turns/unt length zero feld outsde solenod nsde solenod: B µ 0n Flux through a conductng loop: Φ BA µ 0nA Current vares wth tme, so flux vares and an MF s nduced n loop A : dφ nd µ 0 Current nduced n the loop s: na d nd nd conductng loop, resstance, d/ nd, d/ If d/ s postve, B s growng, then B nd opposes change and nd s Counter-clockwse What makes the current nd flow? B 0 there so t s not the orentz force An nduced electrc feld nd along the loop causes current to flow It s caused drectly by dφ/ lectrc feld lnes are loops that don t termnate on charge. -feld s there even wthout the conductor (no current flowng) -feld s non-conservatve (not electrostatc) as the lne ntegral around a closed path s not zero nd loop nd ds dφ B Generalzed Faradays aw Path must be constant B Copyrght. Janow Fall 03
xample: Fnd the nduced electrc feld nd In the rght fgure, db/ constant, fnd the expresson for the magntude of the nduced electrc feld at ponts wthn and outsde the magnetc feld. Due to symmetry: For r < : So For r > : So loop Φ B nd ds ds ds ds ( πr) BA B( πr (πr) πr dφ db ΦB BA B( π db (πr) π The magntude of nduced electrc feld grows lnearly wth r, then falls off as /r for r> ) ) B r db r B ds µ db 0 enc Copyrght. Janow Fall 03
Self-Inductance: Analogous to nerta ANY magnetc flux change s ressted. Changng current n a sngle col nduces a back MF nd n the same col opposng the current change, an nduced current nd, and a consstent nduced feld B nd. N DISTINGUISH: Mutual-nducton: d / n transformer prmary also nduces MF and current n lnked secondary col (transformer prncple). Self-nducton n a sngle Col: d/ produces back MF due to enz & Faraday aws: Φ nd opposes dφ/ due to current change. nd opposes d/. Changng current n a sngle col causes magnetc feld and flux created by ths current to change n the same sense Flux Change nduces flux opposng the change, along wth opposng MF and current. Ths back emf lmts the rate of current (flux) change n the crcut For ncreasng current, back MF lmts the rate of ncrease For decreasng current, back MF sustans the current Inductance measures opposton to the rate of change of current Copyrght. Janow Fall 03
Defnton of Self-nductance ecall capactance: depends only on geometry It measures energy stored n the feld C Q V Self-nductance depends only on col geometry It measures energy stored n the B feld lnked flux unt current Joseph Henry 797 878 self-nductance SI unt of nductance: number of turns Henry flux through one turn depends N Φ B on current & all N turns cancels current dependence n flux above H. T.m / Ampere Weber / Ampere Volt.sec / Ampere ( Ω.sec) Why choose ths defnton? Cross-multply N Φ d B Take tme dervatve d dφ N B - contans all the geometry s the back MF Another form of Faraday s aw! Copyrght. Janow Fall 03
xample: Fnd the Self-Inductance of a solenod + - N turns Area A ength l Volume V Al Feld: Flux n just one turn: B μ0n ΦB BA μ0 N n NA Apply defnton of self-nductance: where # turns unt ength A N turns contrbute to selfflux through ON turn Depends on geometry only, lke capactance. NΦB N A μ0 μ0n V Proportonal to N! Check: Same f you start wth Faraday s aw for Φ B : dφ B nd N for solenod use Φ B above nd N. μ 0 NA d μ0n A d d Note: Inductance per unt length has same dmensons as µ 0 μ N 0 A T.m [ µ 0 ] A. H m Copyrght. Janow Fall 03
xample: calculate self-nductance for an deal solenod N 000 turns, radus r 0.5 m, length 0. m l 7 μ 0 N A 4π 0 3 49.4 0 Henrys 6 0 0. 49.4 mll π - (5 0 ) Henrys Ideal nductor (abstracton): Internal resstance r 0 (recall B 0 outsde B μ 0 n nsde (deal solenod) deal battery) Non-deal nductors have nternal resstance: r V nd V IND r measured voltage Drecton of r depends on current Drecton of depends on d/ If current s constant, then nduced 0 Inductor behaves lke a wre wth resstance r Copyrght. Janow Fall 03
Induced MF n an Inductor : Whch statement descrbes the current through the nductor below, f the nduced MF s as shown? A. ghtward and constant. B. eftward and constant. C. ghtward and ncreasng. D. eftward and decreasng.. eftward and ncreasng. ε d Copyrght. Janow Fall 03
enz s aw appled to Back MF + - If s ncreasng: dφ B > 0 opposes ncrease n Power s beng stored n B feld of nductor - + If s decreasng: dφ B < 0 opposes decrease n Power s beng tapped from B feld of nductor What f CUNT s constant? Copyrght. Janow Fall 03
xample: Current I ncreases unformly from 0 to A. n 0. seconds. Fnd the nduced voltage (back MF) across a 50 mh (mll-henry) nductance. + - d defnes postve > 0 means that and toward drecton current the rght s ncreasng Apply: d Substtute: Δ Δt + Amp 0. sec 0 Amp sec 50 mh 0 Amp sec 0.5 Volts Negatve result means that nduced MF s opposed to both d/ and. Copyrght. Janow Fall 03
Inductors n Crcuts The Crcut Inductors, sometmes called cols, are common crcut components. Insulated wre s wrapped around a core. They are manly used n AC flters and tuned (resonant) crcuts. Analyss of seres crcuts: A battery wth MF drves a current around the loop, producng a back MF n the nductor. Derve crcut equatons: apply Krchoff s loop rule, convert to dfferental equatons (as for C crcuts) and solve. New rule: when traversng an nductor n the same drecton as the assumed current, nsert: d Copyrght. Janow Fall 03
Seres crcuts + - a b Inductance & resstance + MF Fnd tme dependent behavor Use oop ule & Juncton ule Treat as an MF along current Gven,, : Fnd,, U for nductor as functons of tme Growth phase, swtch to a. oop equaton: d Decay phase, swtch to b, exclude, oop equaton: d 0 0 d AWAYS through s clockwse and growng: opposes At t 0, rapdly growng current but 0, acts lke a broken wre As t nfnty, large stable current, d/ 0 Back MF 0, /, acts lke an ordnary wre nergy s stored n & dsspated n nergy stored n now dsspated n Current through s stll clockwse, but collapsng now acts lke a battery mantanng current Current at t 0 equals / Current 0 as t nfnty energy depleted Copyrght. Janow Fall 03
crcut: decay phase soluton After growth phase equlbrum, swtch from a to b, battery out Current 0 / ntally stll flowng CW through Inductance tres to mantan current usng stored energy Polarty of reverses versus growth. ventually 0 oop quaton s : Substtute : - ( t) + d Current decays exponentally: 0 b Crcut quaton: d Frst order dfferental equaton wth smple exponental soluton At t 0: large current mples large d /, so s large As t nfnty: current stablzes, d / and current both 0 (t) τ 0 e / t / τ 0 nductve tme constant 0 - + d/ <0 durng decay, opposte to current e 0. 37 Back MF and V decay exponentally: d τ e t / τ V + e t / τ Q(t) C τ τ 3τ t Compare to C crcut, decay Ce t / C capactve tme cons tan t Copyrght. Janow Fall 03
crcut: growth phase soluton oop quaton s : Substtute : (t) - + d 0 Crcut quaton: d + Frst order dfferental equaton agan - saturatng exponental solutons As t nfnty, d / approaches zero, current stablzes at nf / At t 0: current s small, d / s large, back MF opposes battery. Current starts from zero, grows as a saturatng exponental. (t) τ nf / ( t / τ ) e nductve tme nf constant 0 at t 0 n above equaton d/ / fastest rate of change, largest back MF Back MF decays exponentally d t / τ e e τ t / Voltage drop across resstor V - τ nf τ τ 3τ Q(t) C C t capactve e t / C ( e ) tme 0. 63 Compare to C crcut, chargng cons tan t Copyrght. Janow Fall 03
xample: For growth phase fnd back MF as a functon of tme t / τ Use growth phase soluton (t) ( e ) At t 0: current 0 (t 0) ( 0 e ) 0 Back MF s ~ to rate of change of current d (-)(-) t / τ Dervatve : e where τ τ d t / τ e ; At t 0 : 5V + - τ S 0.H τ Ω 0.H Ω 0. sec Back MF equals the battery potental causng current to be 0 at t 0 drop across 0 acts lke a broken wre at t 0 - - 0.37 V 0 After a very long (nfnte) tme: Current stablzes, back MF0 5A acts lke an ordnary wre at t nfnty Copyrght. Janow Fall 03
Current through the battery - : The three loops below have dentcal nductors, resstors, and batteres. ank them n terms of current through the battery just after the swtch s closed, greatest frst. A.I, II, III. B.II, I, III. C.III, I, II. D.III, II, I..II, III, I. I. II. III. Hnt: what knd of wre does act lke? ( t / τ e ) τ / nductve tme constant (t) nf nf eq eq Copyrght. Janow Fall 03
Current through the battery - 3: The three loops below have dentcal nductors, resstors, and batteres. ank them n terms of current through the battery a long tme after the swtch s closed, greatest frst. A. I, II, III. B. II, I, III. C. III, I, II. D. III, II, I.. II, III, I. I. II. III. Hnt: what knd of wre does act lke? ( t / τ e ) τ / nductve tme constant (t) nf nf eq eq Copyrght. Janow Fall 03
Summarzng crcuts growth phase ε When t s large: Inductor acts lke a wre. When t s small (zero), 0. Inductor acts lke an open crcut. The current starts from zero and ncreases up to a maxmum of ε / wth a tme constant gven by τ Inductve tme constant Compare: The voltage across the resstor s V ε( e The voltage across the nductor s ε τ C C ε + ε Capactve tme constant t / t / t / V V ( e ) e ) ε ε ( e t / ) Copyrght. Janow Fall 03
Summarzng crcuts decay phase The swtch s thrown from a to b Krchoff s oop ule for growth was: d ε 0 Now t s: d + 0 The current decays exponentally: Voltage across nductor: d ε e t / Voltage across resstor also decays: ε t / V e ε d t / t / V + + e e ε V (V) Copyrght. Janow Fall 03
nergy stored n nductors ecall: Capactors store energy n ther electrc felds U U electrostatc potental Q CV C energy u u electrostatc U Volume energy ε0 densty Inductors also store energy, but n ther magnetc felds Magnetc P Dervaton consder power nto or from nductor du d U du d Power B B B UB magnetc potental energy u B magnetc energy densty U B U B B u B Volume µ 0 U B grows as current ncreases, absorbng energy When current s stable, U B and u B are constant U B dmnshes when current decreases. It powers the persstent MF durng the decay phase for the nductor Copyrght. Janow Fall 03 derved usng p-p capactor derved usng solenod
Sample problem: energy storage n magnetc feld of an nductor durng growth phase a) At equlbrum (nfnte tme) how much energy s stored n the col? (Col acts lke a wre) 0.35 U x 53 x 0 3 x (34.3) 34.3 A V 53 mh 0.35 Ω U 3 J b) How long (t / ) does t take to store half of ths energy? At t / : U B U / / / t e / τ ( e t / / / / τ take natural log of both sdes ) t / - τ ln( / ). 3 τ τ / 53 x 0 0.5 sec. -3 / 0.35 Copyrght. Janow Fall 03
Mutual Inductance xample: a par of co-axal cols d/ n the frst col nduces current n the second col, n addton to self-nduced effects. M depends on geometry only, as dd and C Changng current n prmary ( ) creates varyng flux through col nduced MF n col Defnton: number of turns n col cross-multply M mutual nductance N Φ tme dervatve d d M Φ N N Φ M flux through one turn of col due to all N turns of col current n col d M M contans all the geometry s MF nduced n by Another form of Faraday s aw! The smaller col radus determnes how much flux s lnked, so.. d M M M M proof not obvous Copyrght. Janow Fall 03
Calculatng the mutual nductance M B et col (outer) be a short loop of N turns, not a long Solenod Feld µ 0N nsde loop (assume near unform) center large col N turns radus (prmary) small col N turns radus (secondary) Flux through oop depends on area A & B Φ B A μ0n π (for each loop n col ) If current n oop s changng: dφ μ0nn π d d nduced voltage n loop N M M μ0nn M π smaller radus ( ) determnes the lnkage Summarzng results for mutual nductance: - M M d M N Φ NΦ M NΦ Copyrght. Janow Fall 03
Summary: ecture Chapter 30 Inducton II Crcuts Copyrght. Janow Fall 03