Electricity and Magnetism Inductance Transformers Maxwell's Laws

Home , Magnetism

Electricity and Magnetism Inductance Transformers Maxwell’s Laws

Lana Sheridan

De Anza College

Dec 1, 2015 Last time

• Ampere’s law

• Faraday’s law

• Lenz’s law Overview

• induction and energy transfer

• induced electric ﬁelds

• inductance

• self-induction

• RL Circuits Inductors A capacitor is a device that stores an electric ﬁeld as a component of a circuit.

inductor a device that stores a magnetic ﬁeld in a circuit

It is typically a coil of wire. 782 Chapter 26 Capacitance and Dielectrics

▸ 26.2 continued Categorize Because of the spherical symmetry of the sys- ϪQ tem, we can use results from previous studies of spherical Figure 26.5 (Example 26.2) systems to find the capacitance. A spherical capacitor consists of an inner sphere of radius a sur- ϩQ a Analyze As shown in Chapter 24, the direction of the rounded by a concentric spherical electric field outside a spherically symmetric charge shell of radius b. The electric field b distribution is radial and its magnitude is given by the between the spheres is directed expression E 5 k Q /r 2. In this case, this result applies to radially outward when the inner e sphere is positively charged. 820 the field Cbetweenhapter 27 the C spheresurrent and ( aR esistance, r , b). b S Write an expression for the potential difference between 2 52 ? S 782 Chapter 26 Capacitance and DielectricsTable 27.3 CriticalVb TemperaturesVa 3 E d s a the two conductors from Equation 25.3: for Various Superconductors

▸ 26.2 continued Material Tc (K) b b dr 1 b ApplyCategorize the result Because of Example of the spherical 24.3 forsymmetry the electric ofHgBa the sys 2fieldCa- 2Cu 3O8 2 52134 52 Ϫ5Q Vb Va 3 Er dr ke Q 3 2 ke Q Tl—Ba—Ca—Cu—O 125a a r r a outsidetem, wea sphericallycan use results symmetric from previous charge studies distribution of spherical Figure 26.5 (Example 26.2) systems to findS the capacitance. S Bi—Sr—Ca—Cu—O 105 c d and note that E is parallel to d s along a radial line: A spherical capacitor consists of 1 1 a 2 b YBa2Cu3O7 an inner sphere of radius a sur- 92 ϩQ a (1) Vb 2 Va 5 ke Q 2 5 ke Q Analyze As shown in Chapter 24, the directionNb of Ge the rounded by a concentric spherical23.2 782 Chapter 26 Capacitance and Dielectrics 3 b a ab electric field outside a spherically symmetric Nb chargeSn shell of radius b. The electric field18.05 3 a b b distribution is radial and its magnitude is givenNb by the between the spheres is directed 9.46

Courtesy of IBM Research Laboratory Q Q ▸ 26.2 continued 2 radially outward when the inner ab Substituteexpression the E 5absolute keQ /r . Invalue this case,of D Vthis into result Equation appliesPb to 26.1: 5 5 7.18 5 sphere is positivelyC charged. Ϫ (26.6) A small permanentCategorize magnet levi- Because of the spherical, symmetry, of the sys- DV V 2 QV k b 2 a the fieldtem, we between can use the results spheres from previous (a studiesr b). of spherical Hg b4.15 a e tated above a disk of the super- Figure 26.5 (Example 26.2) systems to find the capacitance. Sn A spherical capacitor consists of b 3.72 conductor YBa Cu O , which is in S ϩQ a Write2 3 an7 expression for the potential difference betweenAl an inner sphere ofV radius2 Va sur52- E ? d Ss 0 1.19 0 1 2 liquid nitrogenFinalize at 77Analyze TheK. Ascapacitance shown in Chapter depends 24, the direction on a of and the b roundedas expected. by a concentricb sphericalThea potential3 difference between the spheres in Equation the twoelectric conductors field outside from a spherically Equation symmetric 25.3: charge Zn shell of radius b. The electric field a 0.88 b (1) is negativedistribution because is radial and Q its is magnitude positive is given and by b the. a. betweenTherefore, the spheres isin directed Equation 26.6, when we take the absolute value, we change expression E 5 k Q /r 2. In this case, this result applies to radially outward when the inner b b b e sphere is positively charged. dr 1 a 2Apply b tothe bthe field2 result a.between The of the Example resultspheres ( ais ,24.3 ar ,positive bfor). the electric number field. V 2 V 52 E dr 52k Q 5 k Q Today, thousands of superconductorsb a are known,3 r and e as3 Table2 27.3e rillustrates, b a a r a outside a spherically symmetric charge distribution S WHAT WriteIF ? an expressionIf Sthe radius for the potential b of the difference outer between sphere approaches2 52 ?infinity,S what does the capacitance become? and note that E is parallelthe to criticald Ss along temperatures a radial line: V bof Vrecentlya 3 E discoveredd s superconductors are substantiallyc d the two conductors from Equation 25.3: a 1 1 a 2 b higher than initially thought possible.(1) V b Two2 Va 5 kindske Q of2 superconductors5 ke Q are recog- Answer In Equation 26.6, we let b `: b b b a ab nized. The Smore recently identified ones are dressentially1 b ceramics with high criti- Apply the result of Example 24.3 for the electric field V 2 V 52 E dr 52k Q 5 k Q b aab 3 r abe 3 r 2 a ae r b outside a spherically symmetriccal temperatures,charge distribution whereas superconductinga Q a materialsQ sucha ab as those observed by S S D C 5 lim 5 5 5 c4pd P0a 980 SubstituteChapterand note the32 that absolute EInductance is parallel value to d s of along V ainto radial Equation line: S26.1: C 5 5 5 (26.6) Kamerlingh-Onnes areb `metals.ke b 2 If a room-temperature1ke 1b ka e2 b superconductor is ever iden- (1) Vb 2 Va 5 ke Q DV2 5Vbke2Q Va ke b 2 a tified, its effect on technology could beb tremendous.a ab Notice that this expression is the same as Equation 26.2, thea capacitanceb of an isolated spherical conductor. The value of T is sensitive1Q to chemicalQ2 1 2ab composition,0 0 1 pressure,2 and molecular Substitute the absolute value of DV into Equation 26.1:c C 5 5 5 (26.6) ▸ 32.5 continuedFinalize The capacitance depends on a and b as expected.D The 2potential difference2 between the spheres in Equation structure. Copper, silver, andV gold,Vb whichVa k eareb aexcellent conductors, do not exhibit (1) is negative because Q is positive and b . a. Therefore, in Equation 26.6, when we take the absolute value, we change superconductivity. 0 0 1 2 SOLUTI O N a 2 bFinalize to b 2 The a. Thecapacitance result depends is a positive on a and number b as expected.. The potential difference between the spheres in Equation (1) is negative because Q is positiveOne and truly b . a . remarkableTherefore, in Equation feature 26.6, whenof superconductors we take the absolute value, is wethat change once a current is set up WHAT IF ? If the radius b of the outer sphere approaches infinity, what does the capacitance become? Conceptualize Bea sure 2 b to byou 2 a. Thecan result identify isin a positivethem, the number it two 26.3persists. coils without inCombinations the any situation applied andpotential understand ofdifference Capacitors that(because a changing R 5 0). currentSteady cur in -one coil induces a currentAnswerWHA T In inIF Equation? theIf the second radius 26.6, rentsb of coil.we the let outerhave b S sphere been`: approaches observed infinity, to whatpersist does the in capacitance superconducting become? loops for several years with Answer In Equation 26.6,no we letapparent b S `: decay! ab ab a TwoC 5 orlimab more ab capacitorsa5 5often5 4arepP0a combined in electric circuits. We can calculate Categorize We will determine the result using conceptsb S ` discussed in this section, so we categorize this example as a An importantC 5 lim andk 5eusefulb 2 a5 application5k4epPb0a k eof superconductivity is in the development b S ` k b 2 a k b k substitution problem. of superconductingthe equivalente magnets,e capacitancee in which the of magnitudes certain combinations of the magnetic using field methods are described in NoticeNotice that that this this expression expression is theis the same same as Equation as Equation 26.2, the1 capacitance26.2,2 the ofcapacitance an1 2isolated spherical of an conductor.isolated spherical conductor. Capacitor approximatelythis section. ten1 times2 Throughout1 2 greater than this those section, produced we assume by the the best capacitors normal elec to- be combined are Circuit component symbols NB Use Equation 30.17symbol to express the magnetictromagnets. initiallyfield Such in the superconductinguncharged. B 5m magnets i are being considered as a means of 0 , interior of the base solenoid: storing energy. In Superconductingstudying electric magnets circuits, are we currently use a simplified used in medical pictorial magnetic representation called a Combinations of Capacitors resonance26.3circuit imaging, diagram. or MRI, Suchunits, awhich diagramF produce uses high-quality circuit symbols images toof internalrepresent various circuit Battery ϩ 26.3 Combinations NofH CapacitorsBH N H BA NBNH Find the mutual inductance, notingorgans thatTwo the withoutor more magnetic capacitors the need often for are excessivecombinedM 5 in exposure electric 5circuits. of patientsWe can5 calculate mto0 x-rays orA other harm- batterysymbol V Ϫ the equivalentelements. capacitance The of certain circuit combinations symbolsi using are methods connectedi described in by , straight lines that represent the flux F through the handle’s coilful caused radiation. by the mag- BH Capacitor this Twosection.wires or Throughout more between capacitors this section, the often we circuit assume are thecombined elements. capacitors toin be Theelectric combined circuit circuits. are symbols We can forcalculate capacitors, batteries, netic fieldThe ofdirection the base of thesymbol coil is BA: initiallytheand uncharged. equivalent switches capacitance as well as of the certain color combinations codes used using for methods them in described this text in are given in Fig- effective flowCapacitor of positive Inthis studying section. electric Throughout circuits, we use this a simplified section, pictorial we assume representation the capacitors called a to be combined are circuit diagram. Such a diagram uses circuit symbols to represent various circuit Wirelesscapacitorcharge charging is clockwise.Switch Csymbol is usedBattery inOpen ϩa number ofinitially otherure 26.6.“cordless”uncharged. The devices.symbol Onefor the significant capacitor example reflects is the inductivegeometry charging of the most common symbol Ϫ 27.6elements. Electrical The circuit symbols Power are connected by straight lines that represent the used by some manufacturerssymbol of electricwires cars betweenInmodel that studying theavoids for circuit electric a direct elements.capacitor, circuits, metal-to-metal The circuita we pair use symbols ofa simplified parallel forcontact capacitors, pictorialbetweenplates. batteries, Therepresentation the positivecar and thecalledterminal charg a - of the battery I and switches as well as the color codes used for them in this text are given in Fig- ing apparatus. Battery Closedϩ In typicalcircuit iselectric at diagram. the circuits, higher Such energypotential a diagram TET and isuses transferred iscircuit represented symbols by electrical toin represent the transmission circuit various symbol circuitfrom by the longer line. Switch Open ure elements.26.6. The symbol The for circuit the capacitor symbols reflects are the connected geometry of the by most straight common lines that represent the switch Ssymbolsymbol Ϫ a sourcemodel forsuch a capacitor, as a battery a pair of parallelto some plates. device The positive such terminal as a oflightbulb the battery or a radio receiver. Figure 26.6 CircuitClosed symbolsLet’s for is determine atwires the higher between potential an expression the and circuitis represented that elements. inwill the allowcircuit The symbol us circuit to by calculate the symbols longer line. the for capacitors,rate of this batteries, energy capacitors, batteries, and switches. b Figure 26.6 Circuitc symbols fortransfer. and First,Parallel switches consider as Combination well the as simple the color circuit codes in used Figure for 27.11,them inwhere this textenergy are isgiven delivered in Fig - ϩ Notice thatcapacitors,Switch capacitors batteries,Open andare switches. in Parallelure 26.6.Combination The symbol for the capacitor reflects the geometry of the most common Notice that capacitors are in to a resistor.Two (Resistors capacitors are designatedconnected by as the shown circuit in symbol Figure 26.7a are.) Because known theas a parallel combi- ⌬V R Two capacitors connected as shown in Figure 26.7a are known as a parallel combi- resistorblue, batteriesblue, Rsymbol batteries are arein ingreen, green, and and model for a capacitor, a pair of parallel plates. The positive terminal of the battery Ϫ connectingnation of capacitors.wires also Figure have 26.7b resistance,shows a circuit somediagram energyfor this combination is delivered of to the wires and switches areswitches in are red. in red. TheClosed The closed32.5 isOscillations atnation the higher of capacitors.potential in and Figurean is represented LC 26.7b Circuit in shows the circuit a circuit symbol diagram by the longer for line.this combination of a switch can carry current,d whereassome capacitors. to the Theresistor. left plates Unless of the capacitors noted areotherwise, connected to we the shallpositive assume terminal of the resistance of the switch canthe carryopen one current, cannot. whereas the batterycapacitors. by a conducting The wire andleft are plates therefore of both the at the capacitors same electric potential are connected to the positive terminal of ϩ Figure 26.6 Circuit symbolsWhen for a capacitor is connected to an inductor as illustrated in Figure 32.10, the C the open one cannot. wires is smallthe comparedbattery by with a conducting the resistance wire of andthe circuitare therefore element bothso that at the the energy same electric potential Ϫ capacitors, batteries,L and switches. Parallel Combination inductorQ L combinationdelivered to isthe an wires LC iscircuit. negligible. If the capacitor is initially charged and the switch is max Notice that capacitors are in Two capacitors connected as shown in Figure 26.7a are known as a parallel combi- blue, batteries are in green,then and Imagine closed, followingboth the acurrent positive in quantity the circuit of charge and Qthe moving charge clockwise on the aroundcapacitor the oscil - Figure 27.11 A circuit consist- nation of capacitors. Figure 26.7b shows a circuit diagram for this combination of switches are in red. The closedlatecircuit between in Figure maximum 27.11 from positive point a and through negative the battery values. and If resistor the resistance back to point of the a. cir- ing of a resistorswitch of canresistance carry current, R whereas capacitors. The left plates of the capacitors are connected to the positive terminal of S We identify the entire circuit as our system. As the charge moves from a to b through and a batterythe having open onea potential cannot. cuit is zero,the batteryno energy by a conducting is transformed wire and to are internal therefore energy. both at the In same the electricfollowing potential analysis, difference DV across its terminals. the battery, the electric potential energy of the system increases by an amount Q DV Figure 32.10 A simple LC cir- the resistance in the circuit is neglected. We also assume an idealized situation in cuit. The capacitor has an initial which energy is not radiated away from the circuit. This radiation mechanism is charge Q max, and the switch is discussed in Chapter 34. open for t , 0 and then closed at Assume the capacitor has an initial charge Q (the maximum charge) and t 5 0. max the switch is open for t , 0 and then closed at t 5 0. Let’s investigate what happens from an energy viewpoint. When the capacitor is fully charged, the energy U in the circuit is stored in 2 the capacitor’s electric field and is equal to Q max/2C (Eq. 26.11). At this time, the current in the circuit is zero; therefore, no energy is stored in the inductor. After the switch is closed, the rate at which charges leave or enter the capacitor plates (which is also the rate at which the charge on the capacitor changes) is equal to the current in the circuit. After the switch is closed and the capacitor begins to discharge, the energy stored in its electric field decreases. The capacitor’s discharge represents a current in the circuit, and some energy is now stored in the magnetic field of the inductor. Therefore, energy is transferred from the electric field of the capacitor to the magnetic field of the inductor. When the capacitor is fully discharged, it stores no energy. At this time, the current reaches its maximum value and all the energy in the circuit is stored in the inductor. The current continues in the same direction, decreasing in magnitude, with the capacitor eventually becoming fully charged again but with the polarity of its plates now opposite the initial polarity. This process is followed by another discharge until

the circuit returns to its original state of maximum charge Q max and the plate polarity shown in Figure 32.10. The energy continues to oscillate between inductor and capacitor. The oscillations of the LC circuit are an electromagnetic analog to the mechanical oscillations of the particle in simple harmonic motion studied in Chapter 15. Much of what was discussed there is applicable to LC oscillations. For example, we investigated the effect of driving a mechanical oscillator with an external force, Inductance

Just like capacitors have a capacitance that depends on the geometry of the capacitor, inductors have an inductance that depends on their structure.

For a solenoid inductor:

2 L = µ0n A`

where n is the number of turns per unit length, A is the cross sectional area, and ` is the length of the inductor.

Units: henries, H.

1 henry = 1 H = 1 T m2 /A Value of µ0: New units

The magnetic permeability of free space µ0 is a constant.

−7 µ0 = 4π × 10 T m / A

It can also be written in terms of henries:

−7 µ0 = 4π × 10 H / m

(Remember, 1 H = 1 T m2 / A) Inductance

However, capacitance is defined as being the constant of proportionality relating the charge on the plates to the potential difference across the plates q = C (∆V ). Inductance also is formally defined this way.

inductance the constant of proportionality relating the magnetic ﬂux linkage (NΦB ) to the current: NΦ NΦ = L I ; L = B B I

ΦB is the magnetic ﬂux through the coil, and I is the current in the coil. Inductance of Solenoid Inductors

Suppose now that the only source of magnetic ﬂux in the solenoid is the ﬂux produced by a current in the wire.

Then the ﬁeld produced within the solenoid is:

B = µ0In

where n is the number of turns per unit length.

That means the ﬂux will be:

◦ ΦB = BA cos(0 ) = BA = µ0InA

where A is the cross sectional area of the solenoid. Inductance of Solenoid Inductors

NΦ L = B I

Replacing N = n`, ΦB = µ0InA: n`(µ In)A L = 0 I

So we conﬁrm our expression for a solenoid inductor:

2 L = µ0n A` Induction from an external ﬂux vs Self-Induction

So far we have thought about the effect of a changing magnetic flux on the E-field and emf produced in some region.

We also deﬁned induction by acknowledging that a current in a solenoid will produce a magnetic ﬂux, and relating that to the current in the coil: NΦB = L I

In general, the magnetic ﬂux ΦB = B · A, could be due not only to the B-ﬁeld produced by the current in the wire, but also have an additional external source.

If it does not, we say L is the self-inductance of the inductor. (This is usually how the symbol L is used.) Self-Induction When the current in the solenoid circuit is changing there is a (self-) induced emf in the coil.

From Faraday’s Law, we have

∆(NΦ ) E = − B ∆t

Since L is a constant for a particular inductor, ∆i E = −L L ∆t (Derivative form:) di E = −L L dt

The emf opposes the change in current. Inductors vs. Resistors

Inductors are a bit similar to resistors.

Resistors resist the ﬂow of current.

Inductors resist any change in current.

If the current is high and lowered, the emf acts to keep the current ﬂowing. If the current is low and increased, the emf acts to resist the increase. 32.1 Self-Induction and Inductance 971

Consider a circuit consisting of a switch, a resistor, and a source of emf as shown in Figure 32.1. The circuit diagram is represented in perspective to show the orien- tations of some of the magnetic field lines due to the current in the circuit. When the switch is thrown to its closed position, the current does not immediately jump from zero to its maximum value e/R. Faraday’s law of electromagnetic induction (Eq. 31.1) can be used to describe this effect as follows. As the current increases with time, the magnetic field lines surrounding the wires pass through the loop represented by the circuit itself. This magnetic field passing through the loop causes a magnetic flux through the loop. This increasing flux creates an induced emf in the circuit. The direction of the induced emf is such that it would cause an induced current in the loop (if the loop did not already carry a current), which would establish a magnetic field opposing the change in the original magnetic Brady-Handy Collection, Library of Congress Prints and Photographs Division [LC-BH83-997] field. Therefore, the direction of the induced emf is opposite the direction of the Joseph Henry emf of the battery, which results in a gradual rather than instantaneous increase in American Physicist (1797–1878) the current to its final equilibrium value. Because of the direction of the induced Henry became the first director of emf, it is also called a back emf, similar to that in a motor as discussed in Chapter 31. the Smithsonian Institution and first president of the Academy of Natural This effect is called self-induction because the changing flux through the circuit Science. He improved the design of the and the resultant induced emf arise from the circuit itself. The emf eL set up in this electromagnet and constructed one of case is called a self-induced emf. the first motors. He also discovered the To obtain a quantitative description of self-induction, recall from Faraday’s law phenomenon of self-induction, but he that the induced emf is equal to the negative of the time rate of change of the mag- failed to publish his findings. The unit of inductance, the henry, is named in netic flux. The magnetic flux is proportional to the magnetic field, which in turn his honor. is proportional to the current in the circuit. Therefore, a self-induced emf is always proportional to the time rate of change of the current. For any loop of wire, we can write this proportionality as di e 52L (32.1) L dt where L is a proportionality constant—called the inductance of the loop—that depends on the geometry of the loop and other physical characteristics. If we consider a closely spaced coil of N turns (a toroid or an ideal solenoid) carrying a current i and containing N turns, Faraday’s law tells us that eL 5 2N dFB /dt. Com- bining this expression with Equation 32.1 gives

NFB L 5 (32.2) Inductance of an N-turn coil i where it is assumed the same magnetic flux passes through each turn and L is the After the switch is closed, the inductance of the entire coil. current produces a magnetic flux From Equation 32.1, we can also write the inductance as the ratio through the area enclosed by the loop. As the current increases e toward its equilibrium value, this L 52 L (32.3) di/dt Self-Inductionmagnetic flux changes in time and induces an emf in the loop. Recall that resistance is a measure of the opposition to current as given by Equa- S tion 27.7, R 5 DV/I ; in comparison, Equation 32.3, being of the same mathematical B form as Equation 27.7, shows us that inductance is a measure of the opposition to a S change in current. i The SI unit of inductance is the henry (H), which as we can see from Equation 32.3 is 1 volt-second per ampere: 1 H 5 1 V ? s/A. As shown in Example 32.1, the inductance of a coil depends on its geometry. This ϩ dependence is analogous to the capacitance of a capacitor depending on the geome- R Ϫ e try of its plates as we found in Equation 26.3 and the resistance of a resistor depending on the length and area of the conducting material in Equation 27.10. Inductance i calculations can be quite difficult to perform for complicated geometries, but the Figure 32.1 Self-induction in a examples below involve simple situations for which inductances are easily evaluated. simple circuit. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) halliday_c30_791-825hr.qxd 11-12-2009 12:19we cannot Page 807define an electric potential within the inductor itself, where the flux is changing. However, potentials can still be defined at points of the circuit that are not within the inductor—points where the electric fields are due to charge distributions and their associated electric potentials.Self-Induction L Moreover, we can define a self-induced potential difference VL across an L inductor (between its terminals, which we assume to be outside the region of PART 3 The changing changing flux). For an ideal inductor (its wire has negligible resistance), the mag- 807 current changes 30-9 RL CIRCUITS (a) nitude of VL is equal to the magnitude of the self-induced emf ᏱL. the flux, which If, instead, the wire in the inductor has resistance r,we mentally separate the creates an emf This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) i (decreasing) inductor into a resistance r (which we take to be outside the region of changing that opposes we cannot define an electric potential within the inductor itself, where the flux flux) and an ideal inductor of self-induced emf ᏱL.As with a real battery of emf the change. is changing. However, potentialsᏱ andcan internalstill be defined resistance at points r,the ofpotential the circuit difference that across the terminals of a real are not within the inductor—pointsinductor where then the differs electric from fields the areemf. due Unless to charge otherwise indicated, we assume here L distributions and their associatedthat electric inductors potentials. are ideal. L Moreover, we can define a self-induced potential difference VL across an L L inductor (between its terminals, which we assume to be outside the region of The changing changing flux). For an ideal inductorCHECKPOINT(its wire has negligible 5 resistance), the mag- current changes (a) (b) nitude of VL is equal to the magnitudeThe figure of the shows self-induced an emf Ᏹ emfinduced ᏱL. in a coil. Which of the flux, which L L If, instead, the wire in the inductorthe following has resistance can describe r,we the mentally current through separate the the coil: (a) creates an emf i (decreasing) Fig. 30-14 (a) The current i is increasing, inductor into a resistance r (whichconstant we take and to rightward, be outside (b) the constant region and of leftward, changing (c) in- that opposesand the self-induced emf Ᏹ appears along creasing and rightward,Ᏹ (d) decreasing and rightward, L flux) and an ideal inductor of self-induced emf L.As with a real battery of emf the change. the coil in a direction such that it opposes Ᏹ and internal resistance r,the potential(e) increasing difference and leftward, across (f) the decreasing terminals and of leftward? a real the increase.The arrow representing ᏱL can inductor then differs from the emf. Unless otherwise indicated, we assume here L be drawn along a turn of the coil or along- that inductors are ideal. side the coil. Both are shown. (b) The cur-

L rent i is decreasing, and the self-induced emf appears in a direction such that it opposes CHECKPOINT 5 the decrease. The figure shows an emf Ᏹ induced30-9 in aRL coil.Circuits Which of (b) L L the following can describe the current through the coil: (a) In Section 27-9 we saw that if we suddenly introduceFig. an emf30-14 Ᏹ into(a) Thea single-loop current i is increasing, constant and rightward, (b) constant and leftward, (c) in- circuit containing a resistor R and a capacitor C,the chargeand the self-inducedon the capacitor emf Ᏹ doesappears along creasing and rightward, (d) decreasing and rightward, L not build up immediately to its final equilibrium valuethe C coilᏱ but in a approaches direction such it that in an it opposes (e) increasing and leftward, (f) decreasing and leftward? exponential fashion: the increase.The arrow representing ᏱL can be drawn along a turn of the coil or along- q ϭ CᏱ(1 Ϫ eϪt/␶C). side the coil. Both are shown.(30-36) (b) The current i is decreasing, and the self-induced emf The rate at which the charge builds up is determinedappears by in the a direction capacitive such that time it opposes 30-9 RL Circuits constant tC,defined in Eq.27-36 as the decrease. t ϭ RC In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into aC single-loop. (30-37) circuit containing a resistor R and aIf capacitor we suddenly C,the remove charge onthe the emf capacitor from this does same circuit, the charge does not not build up immediately to its immediatelyfinal equilibrium fall tovalue zero C butᏱ but approaches approaches zero it in in an an exponential fashion: exponential fashion: Ϫt/␶C q ϭ q0e . (30-38) q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The time constant tC describes the fall of the charge as well as its rise. The rate at which the charge buildsAn upanalogous is determined slowing byof the rise capacitive (or fall) time of the current occurs if we introduce constant tC,defined in Eq.27-36an as emf Ᏹ into (or remove it from) a single-loop circuit containing a resistor R and an inductortC ϭ RC L..When the switch S in Fig.30-15(30-37) is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current If we suddenly remove the emf from this same circuit, the charge does not would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S immediately fall to zero but approaches zero in an exponential fashion: induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b + R the qcurrent,ϭ q eϪt which/␶C. means that it opposes (30-38)the battery emf Ᏹ in polarity. Thus, the 0 – current in the resistor responds to the difference between two emfs, a constant Ᏹ L The time constant t describes the fall of the charge as well as its rise. C due to the battery and a variable Ᏹ (ϭϪL di/dt) due to self-induction.As long as An analogous slowing of the rise (or fall) of the current occurs ifL we introduce an emf Ᏹ into (or remove it from)ᏱL ais single-looppresent, the circuit current containing will be less a resistor than Ᏹ /RR.and an inductor L.When the switch S inAs Fig.30-15 time goes is closed on, the on ratea,for at example,the which the current cur- increases becomes less rapid Fig. 30-15 An RL circuit.When switch rent in the resistor starts to rise.and If the the magnitude inductor were of notthe present, self-induced the current emf, which is proportional to di/dt, S is closed on a,the current rises and approaches a limiting value Ᏹ/R. would rise rapidly to a steady valuebecomes Ᏹ/R.Because smaller. Thus,of the theinductor,however,a current in the circuit self- approaches Ᏹ/R asymptotically.a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Fig. 30-15 An RL circuit.When switch and the magnitude of the self-induced emf, which is proportional to di/dt, S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically. proaches a limiting value Ᏹ/R. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) we cannot define an electric potential within the inductor itself, where the flux is changing. However, potentials can still be defined at points of the circuit that are not within the inductor—points where the electric fields are due to charge distributions and their associated electric potentials. L

Moreover, we can define a self-induced potential difference VL across an L inductor (between its terminals, which we assume to be outside the region of The changing changing flux). For an ideal inductor (its wire has negligible resistance), the mag- current changes (a) nitude of VL is equal to the magnitude of the self-induced emf ᏱL. the flux, which If, instead, the wire in the inductor has resistance r,we mentally separate the creates an emf i (decreasing) inductor into a resistance r (which we take to be outside the region of changing that opposes flux) and an ideal inductor of self-induced emf ᏱL.As with a real battery of emf the change. Ᏹ and internal resistance r,the potential difference across the terminals of a real inductor then differs from the emf. Unless otherwise indicated, we assume here that inductors are ideal. Self-inductance question L L

CHECKPOINT 5 The figure shows an emf EL induced in a coil. The figure shows an emf Ᏹ induced in a coil. Which of (b) L L the following can describe the current through the coil: (a) Fig. 30-14 (a) The current i is increasing, constant and rightward, (b) constant and leftward, (c) in- and the self-induced emf Ᏹ appears along creasing and rightward, (d) decreasing and rightward, L the coil in a direction such that it opposes (e) increasing and leftward, (f) decreasing and leftward?Which of the following can describe the current through the coil: the increase.The arrow representing ᏱL can (A) constant and rightward be drawn along a turn of the coil or along- (B) increasing and rightward side the coil. Both are shown. (b) The cur- (C) decreasing and rightward rent i is decreasing, and the self-induced emf (D) decreasing and leftward appears in a direction such that it opposes 30-9 RL Circuits the decrease. In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into a single-loop circuit containing a resistor R and a capacitor C,the charge on the capacitor does not build up immediately to its final equilibrium value CᏱ but approaches it in an exponential fashion: q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The rate at which the charge builds up is determined by the capacitive time constant tC,defined in Eq.27-36 as tC ϭ RC. (30-37) If we suddenly remove the emf from this same circuit, the charge does not immediately fall to zero but approaches zero in an exponential fashion:

Ϫt/␶C q ϭ q0e . (30-38)

The time constant tC describes the fall of the charge as well as its rise. An analogous slowing of the rise (or fall) of the current occurs if we introduce an emf Ᏹ into (or remove it from) a single-loop circuit containing a resistor R and an inductor L.When the switch S in Fig.30-15 is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Fig. 30-15 An RL circuit.When switch and the magnitude of the self-induced emf, which is proportional to di/dt, S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically. proaches a limiting value Ᏹ/R. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

CHECKPOINT 5 The figure shows an emf EL induced in a coil. The figure shows an emf Ᏹ induced in a coil. Which of (b) L L the following can describe the current through the coil: (a) Fig. 30-14 (a) The current i is increasing, constant and rightward, (b) constant and leftward, (c) in- and the self-induced emf Ᏹ appears along creasing and rightward, (d) decreasing and rightward, L the coil in a direction such that it opposes (e) increasing and leftward, (f) decreasing and leftward?Which of the following can describe the current through the coil: the increase.The arrow representing ᏱL can (A) constant and rightward be drawn along a turn of the coil or along- (B) increasing and rightward side the coil. Both are shown. (b) The cur- (C) decreasing and rightward ← rent i is decreasing, and the self-induced emf (D) decreasing and leftward appears in a direction such that it opposes 30-9 RL Circuits the decrease. In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into a single-loop circuit containing a resistor R and a capacitor C,the charge on the capacitor does not build up immediately to its final equilibrium value CᏱ but approaches it in an exponential fashion: q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The rate at which the charge builds up is determined by the capacitive time constant tC,defined in Eq.27-36 as tC ϭ RC. (30-37) If we suddenly remove the emf from this same circuit, the charge does not immediately fall to zero but approaches zero in an exponential fashion:

Ϫt/␶C q ϭ q0e . (30-38)

1 U = Li 2 B 2

q 1 2 Compare with UE = 2C (or UE = 2 CV ) for the energy stored in a capacitor. Energy Density of a Magnetic Field

Energy is stored in a magnetic ﬁeld!

B2 uB = 2µ0

1 2 Compare with uE = 2 0E for electric ﬁelds. For self-inductance on a coil labeled 1:

N1ΦB,1 = L1i1

For mutual inductance:

N1ΦB,12 = M i2

The ﬂux is in coil 1, but the current that causes the ﬂux is in coil 2.

Mutual Inductance

An inductor can have an induced emf from its own changing magnetic ﬁeld.

It also can have an emf from an external changing ﬁeld.

That external changing ﬁeld could be another inductor. Mutual Inductance

An inductor can have an induced emf from its own changing magnetic ﬁeld.

It also can have an emf from an external changing ﬁeld.

That external changing ﬁeld could be another inductor.

For self-inductance on a coil labeled 1:

N1ΦB,1 = L1i1

For mutual inductance:

N1ΦB,12 = M i2

The ﬂux is in coil 1, but the current that causes the ﬂux is in coil 2. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 814 Mutual Inductance For mutual inductance:

N1ΦB,12 = M i2 814 CHAPTER 30 INDUCTION AND INDUCTANCEThe ﬂux is in coil 1, but the current that causes the ﬂux is in coil 2.

B N 1 1 N 1 Φ 12

N 2 Φ 21 B2 N 2

B1 B2

i i 2 Fig. 30-19 Mutual induction. (a) The 1 : magnetic ﬁeld B1 produced by current i1 in coil 1 extends through coil 2. If i1 is varied 0 0 (by varying resistance R), an emf is induced in coil 2 and current registers on the meter R + – +– R connected to coil 2. (b) The roles of the Coil 1 Coil 2 Coil 1 Coil 2 coils interchanged. (a) (b) which has the same form as Eq. 30-28,

L ϭ N⌽/i, (30-58) the deﬁnition of inductance.We can recast Eq. 30-57 as

M21i1 ϭ N2⌽21. (30-59)

If we cause i1 to vary with time by varying R, we have di d⌽ M 1 ϭ N 21 . (30-60) 21 dt 2 dt The right side of this equation is, according to Faraday’s law, just the magnitude of the emf Ᏹ2 appearing in coil 2 due to the changing current in coil 1.Thus, with a minus sign to indicate direction, di Ᏹ ϭϪM 1 , (30-61) 2 21 dt which you should compare with Eq. 30-35 for self-induction (Ᏹ ϭϪL di/dt). Let us now interchange the roles of coils 1 and 2, as in Fig. 30-19b;that is,we set up a current i2 in coil 2 by means of a battery, and this produces a magnetic ﬂux ⌽12 that links coil 1.If we change i2 with time by varying R,we then have,by the argument given above, di Ᏹ ϭϪM 2 . (30-62) 1 12 dt Thus, we see that the emf induced in either coil is proportional to the rate of change of current in the other coil.The proportionality constants M21 and M12 seem to be different. We assert, without proof, that they are in fact the same so that no subscripts are needed. (This conclusion is true but is in no way obvious.) Thus, we have

M21 ϭ M12 ϭ M, (30-63) and we can rewrite Eqs. 30-61 and 30-62 as

di Ᏹ ϭϪM 1 (30-64) 2 dt

di and Ᏹ ϭϪM 2 . (30-65) 1 dt halliday_c30_791-825hr.qxd 11-12-2009 Mutual 12:19 Page 814 Inductance mutual inductance

N Φ N Φ M = 1 B,12 = 2 B,21 i i 814 CHAPTER 30 INDUCTION AND INDUCTANCE 2 1

B N 1 1 N 1 Φ 12

N 2 Φ 21 B2 N 2

B1 B2

which has the same form as Eq. 30-28,

L ϭ N⌽/i, (30-58) the deﬁnition of inductance.We can recast Eq. 30-57 as

M21i1 ϭ N2⌽21. (30-59)

If we cause i1 to vary with time by varying R, we have di d⌽ M 1 ϭ N 21 . (30-60) 21 dt 2 dt The right side of this equation is, according to Faraday’s law, just the magnitude

of the emf Ᏹ2 appearing in coil 2 due to the changing current in coil 1.Thus, with a minus sign to indicate direction, di Ᏹ ϭϪM 1 , (30-61) 2 21 dt which you should compare with Eq. 30-35 for self-induction (Ᏹ ϭϪL di/dt). Let us now interchange the roles of coils 1 and 2, as in Fig. 30-19b;that is,we set up a

current i2 in coil 2 by means of a battery, and this produces a magnetic ﬂux ⌽12 that links coil 1.If we change i2 with time by varying R,we then have,by the argument given above, di Ᏹ ϭϪM 2 . (30-62) 1 12 dt Thus, we see that the emf induced in either coil is proportional to the rate of

change of current in the other coil.The proportionality constants M21 and M12 seem to be different. We assert, without proof, that they are in fact the same so that no subscripts are needed. (This conclusion is true but is in no way obvious.) Thus, we have

M21 ϭ M12 ϭ M, (30-63) and we can rewrite Eqs. 30-61 and 30-62 as

di Ᏹ ϭϪM 1 (30-64) 2 dt

di and Ᏹ ϭϪM 2 . (30-65) 1 dt Mutual Inductance

N1ΦB,12 = M i2 Considering the rate of change of both sides with time, and using ∆ΦB Faraday’s Law E = − ∆t ,

∆i E = −M 2 1 ∆t

and

∆i E = −M 1 2 ∆t

A change of current in one coil causes a Mutual Inductance Applications

If there is a changing current in one coil, an emf can be induced in the other coil.

The current can be transferred to a whole diﬀerent circuit that is no directly connected.

This can be used for wireless charging.

It is also used in transformers: devices that change the voltage and current of a power supply.

For either of those applications to work, there must be a constantly changing current. halliday_c31_826-860hr.qxd 11-12-2009 13:11 Page 847

PART 3 31-10 POWER IN ALTERNATING-CURRENT CIRCUITS 847

Sample Problem

Current amplitude, impedance, and phase constant

In Fig. 31-7, let R ϭ 200 ⍀, C ϭ 15.0 mF, L ϭ 230 mH, We then ﬁnd f ϭ 60.0 Hz, and Ᏹ ϭ 36.0 V. (These parameters are those d m Ᏹ 36.0 V used in the earlier sample problems above.) I ϭ m ϭ ϭ 0.164 A. (Answer) Z 219 ⍀ (a) What is the current amplitude I? (b) What is the phase constant f of the current in the halliday_c31_826-860hr.qxd 11-12-2009 13:11 Page 847 KEY IDEA circuit relative to the driving emf?

The current amplitude I depends on the amplitude Ᏹm of the KEY IDEA driving emf and on the impedance Z of the circuit, accord- The phase constant depends on the inductive reactance, the ing to Eq. 31-62 (I ϭ Ᏹ /Z). m capacitive reactance,PART and 3 the resistance of the circuit, Calculations: So, we need to31-10 find ZPOWER,which INdepends ALTERNATING-CURRENT on resis- according CIRCUITS to Eq. 31-65.847 tance R,capacitive reactance XC,and inductive reactance XL. The circuit’s resistance is the Samplegiven resistance Problem R.Its capacitive Calculation: Solving Eq. 31-65 for f leads to reactance is due to the given capacitance and, from an earlier Current amplitude, impedance, and phase constant XL Ϫ XC 86.7 ⍀Ϫ177 ⍀ X ϭ ⍀ ␾ ϭ tanϪ1 ϭ tanϪ1 sample problem, C 177 .Its inductive reactance is due R 200 ⍀ to theR ϭ given⍀ inductanceC ϭ m and,L ϭ from another sample problem, In Fig. 31-7, let 200 , 15.0 F, 230 mH, We then find ϭϪ ЊϭϪ f ϭ 60.0 Hz, andX Ᏹ ϭϭ 36.0 V.⍀ (These parameters are those 24.3 0.424 rad. (Answer) d L m 86.7 .Thus,the circuit’s impedance is Ᏹ 36.0 V used in the earlier sample problems above.) I ϭ m ϭ ϭ 0.164 A. (Answer) Z ϭ R2 ϩ (X Ϫ X )2 Z 219 ⍀ The negative phase constant is consistent with the fact that I L C (a) What is the current amplitude ? the load is mainly capacitive; that is, XC Ͼ XL.In the com- ϭ 2(200 ⍀)2 ϩ (86.7 ⍀Ϫ(b) What177 ⍀ is) 2 the phase constant f of the current in the KEY IDEA circuit relative to the driving emf?mon mnemonic for driven series RLC circuits, this circuit is ϭ 2192 ⍀. an ICE circuit—the current leads the driving emf. The current amplitude I depends on the amplitude Ᏹm of the KEY IDEA driving emf and on the impedance Z of the circuit, accord- The phase constant depends on the inductive reactance, the ing to Eq. 31-62 (I ϭ Ᏹ /Z). Additional examples, video, and practice available at WileyPLUS m capacitive reactance, and the resistance of the circuit, Calculations: So, we need to find Z,which depends on resis- according to Eq. 31-65. tance R,capacitive reactance XC,and inductive reactance XL. The circuit’s resistance is the given resistance R.Its capacitive Calculation: Solving Eq. 31-65 for f leads to reactance is due to the given capacitance and, from an earlier XL Ϫ XC 86.7 ⍀Ϫ177 ⍀ 31-10X ϭ Power⍀ in Alternating-Current␾ ϭ tan CircuitsϪ1 ϭ tanϪ1 sample problem, C 177 .Its inductive reactance is due R 200 ⍀ to the given inductance and, from another sample problem, In the RLC circuit of Fig. 31-7, the source ofϭϪ energy24.3ЊϭϪ is the0.424 alternating-current rad. (Answer) XL ϭ 86.7 ⍀.Thus,thegenerator. circuit’s Some impedance of the is energy that it provides is stored in the electric field in the 2 2 The negative phase constant is consistent with the fact that Z ϭ R ϩ (XL Ϫ XC) capacitor, some is stored in the magneticthe loadfield is inmainly the capacitive;inductor, that and is, someX Ͼ X is.In dis- the com- 2 2 2 C L ϭsipated(200 ⍀as) thermalϩ (86.7 ⍀Ϫ energy177 ⍀ )in the resistor.mon Inmnemonic steady-state for driven operation, series RLC thecircuits, average this circuit is sin θ ϭstored2192 ⍀ .energy remains constant.The netan transfer ICE circuit—the of energy current is thus leads fromthe driving the gener- emf. +1 ator to the resistor,Additional where examples, energy video, is dissipated. and practice available at WileyPLUS The instantaneous rate at which energy is dissipated in the resistor can be 0 θ written, with the help of Eqs. 26-27 and 31-29, as 0 π 2 π 3π 2 2 2 2 31-10 Power in Alternating-CurrentP ϭ i R ϭ [ ICircuitssin(vdt Ϫ f)] R ϭ I R sin (vdt Ϫ f). (31-68) –1 In the RLC circuitThe of average Fig. 31-7,rate the atsource which of energyenergy is isthe dissipated alternating-current in the resistor,In however, an alternating is the current aver- supply, the voltage and current vary generator. Some of the energy that it provides is stored in the electric field in the sinusoidally with time: u (a) capacitor, someage is stored of Eq. in the 31-68 magnetic over fieldtime. in Overthe inductor, one complete and some is cycle, dis- the average value of sin , 2 1 sipated as thermalwhere energy u isin anythe resistor. variable, In steady-state is zero (Fig. operation, 31-17 thea) butaverage the average value of sin u is 2 sin 2 stored energy remains constant.The net transfer of energy is thus from the gener- θ sin θ (Fig. 31-17b). (Note in Fig. 31-17b how the shaded areas under+1 the curve but ator to the resistor,above where the energy horizontal is dissipated. line marked ϩ1 exactly fill in the unshaded spaces below +1 The instantaneous rate at which energy is dissipated in the2 resistor can be + –1 that line.) Thus, we can write, from Eq. 31-68, 0 θ 2 written, with the help of Eqs. 26-27 and 31-29, as 0 π 2 π 3π 2 2 2 2 2 2 0 θ P ϭ i R ϭ [I sin(vdt Ϫ f)] R ϭ I R sin (vdt ϪI Rf).I (31-68) 0 π 2 π 3π Pavg ϭ ϭ R. –1 (31-69) The average rate at which energy is dissipated in the resistor, however,2 ΂ is the2 aver-΃ (a) (b) age of Eq. 31-68 over time. Over one complete cycle, the average value1 of sin u, i = i0 sin(ωt) where u is any Thevariable, quantity is zero (Fig.I/ 31-172 is acalled) but the the average root-mean-square, value of sin2 u is 1or rms, value of the current i: 2 2 (Fig. 31-17b). (Note in Fig. 31-17b how the shaded areas under the curve but Thesin power2 θ delivered to a loadFig. fluctuates 31-17 as P(a=) AP plot0 sin of(ω sint) u. versus u. 11 +1 above the horizontal line marked ϩ2 exactly fill in the unshadedI spaces below The average value over one cycle is zero. (b) –1 2 that line.) Thus, we can write, from Eq. 31-68, Irms ϭ (rms current). + 2 (31-70) A plot of sin u versus u.The average value 2 1 2 2 0 θ I R I 0 π 2 π 3π over one cycle is 2 . Pavg ϭ ϭ R. 1 (31-69) 2 ΂ 2 ΃ (b) The quantity I/ 2 is called the root-mean-square,1 or rms, value of the current i: u u 1 Fig. 31-17 (a) A plot of sin versus . I The average value over one cycle is zero. (b) Irms ϭ (rms current). (31-70) A plot of sin2 u versus u.The average value 2 over one cycle is 1 . 1 2

Alternating Current (AC)

Alternating current (AC) power supplies are the alternative to direct current (DC) power supplies. halliday_c31_826-860hr.qxd 11-12-2009 13:11 Page 847

PART 3 31-10 POWER IN ALTERNATING-CURRENT CIRCUITS 847

Sample Problem

Current amplitude, impedance, and phase constant

The current amplitude I depends on the amplitude Ᏹm of the KEY IDEA driving emf and on the impedance Z of the circuit, accord- The phase constant depends on the inductive reactance, the ing to Eq. 31-62 (I ϭ Ᏹ /Z). m capacitive reactance,PART and 3 the resistance of the circuit, Calculations: So, we need to31-10 find ZPOWER,which INdepends ALTERNATING-CURRENT on resis- according CIRCUITS to Eq. 31-65.847 tance R,capacitive reactance XC,and inductive reactance XL. The circuit’s resistance is the Samplegiven resistance Problem R.Its capacitive Calculation: Solving Eq. 31-65 for f leads to reactance is due to the given capacitance and, from an earlier Current amplitude, impedance, and phase constant XL Ϫ XC 86.7 ⍀Ϫ177 ⍀ X ϭ ⍀ ␾ ϭ tanϪ1 ϭ tanϪ1 sample problem, C 177 .Its inductive reactance is due R 200 ⍀ to theR ϭ given⍀ inductanceC ϭ m and,L ϭ from another sample problem, In Fig. 31-7, let 200 , 15.0 F, 230 mH, We then find ϭϪ ЊϭϪ f ϭ 60.0 Hz, andX Ᏹ ϭϭ 36.0 V.⍀ (These parameters are those 24.3 0.424 rad. (Answer) d L m 86.7 .Thus,the circuit’s impedance is Ᏹ 36.0 V used in the earlier sample problems above.) I ϭ m ϭ ϭ 0.164 A. (Answer) Z ϭ R2 ϩ (X Ϫ X )2 Z 219 ⍀ The negative phase constant is consistent with the fact that I L C (a) What is the current amplitude ? the load is mainly capacitive; that is, XC Ͼ XL.In the com- ϭ 2(200 ⍀)2 ϩ (86.7 ⍀Ϫ(b) What177 ⍀ is) 2 the phase constant f of the current in the KEY IDEA circuit relative to the driving emf?mon mnemonic for driven series RLC circuits, this circuit is ϭ 2192 ⍀. an ICE circuit—the current leads the driving emf. The current amplitude I depends on the amplitude Ᏹm of the KEY IDEA driving emf and on the impedance Z of the circuit, accord- The phase constant depends on the inductive reactance, the ing to Eq. 31-62 (I ϭ Ᏹ /Z). Additional examples, video, and practice available at WileyPLUS m capacitive reactance, and the resistance of the circuit, Calculations: So, we need to find Z,which depends on resis- according to Eq. 31-65. tance R,capacitive reactance XC,and inductive reactance XL. The circuit’s resistance is the given resistance R.Its capacitive Calculation: Solving Eq. 31-65 for f leads to reactance is due to the given capacitance and, from an earlier XL Ϫ XC 86.7 ⍀Ϫ177 ⍀ 31-10X ϭ Power⍀ in Alternating-Current␾ ϭ tan CircuitsϪ1 ϭ tanϪ1 sample problem, C 177 .Its inductive reactance is due R 200 ⍀ to the given inductance and, from another sample problem, In the RLC circuit of Fig. 31-7, the source ofϭϪ energy24.3ЊϭϪ is the0.424 alternating-current rad. (Answer) XL ϭ 86.7 ⍀.Thus,thegenerator. circuit’s Some impedance of the is energy that it provides is stored in the electric field in the 2 2 The negative phase constant is consistent with the fact that Z ϭ R ϩ (XL Ϫ XC) capacitor, some is stored in the magneticthe loadfield is inmainly the capacitive;inductor, that and is, someX Ͼ X is.In dis- the com- 2 2 2 C L ϭsipated(200 ⍀as) thermalϩ (86.7 ⍀Ϫ energy177 ⍀ )in the resistor.mon Inmnemonic steady-state for driven operation, series RLC thecircuits, average this circuit is sin θ ϭstored2192 ⍀ .energy remains constant.The netan transfer ICE circuit—the of energy current is thus leads fromthe driving the gener- emf. Alternating Current (AC) +1 ator to the resistor,Additional where examples, energy video, is dissipated. and practice available at WileyPLUS The instantaneous rate at which energy is dissipated in the resistor can be 0 θ written, with the help of Eqs. 26-27 and 31-29, as Alternating current (AC) power supplies0 are theπ alternative2 π to3π 2 2 2 2 direct current (DC) power supplies. 31-10 Power in Alternating-CurrentP ϭ i R ϭ [ ICircuitssin(vdt Ϫ f)] R ϭ I R sin (vdt Ϫ f). (31-68) –1 In the RLC circuitThe of average Fig. 31-7,rate the atsource which of energyenergy is isthe dissipated alternating-current in the resistor,In however, an alternating is the current aver- supply, the voltage and current vary generator. Some of the energy that it provides is stored in the electric field in the sinusoidally with time: u (a) capacitor, someage is stored of Eq. in the 31-68 magnetic over fieldtime. in Overthe inductor, one complete and some is cycle, dis- the average value of sin , 2 1 sipated as thermalwhere energy u isin anythe resistor. variable, In steady-state is zero (Fig. operation, 31-17 thea) butaverage the average value of sin u is 2 sin 2 stored energy remains constant.The net transfer of energy is thus from the gener- θ sin θ (Fig. 31-17b). (Note in Fig. 31-17b how the shaded areas under+1 the curve but ator to the resistor,above where the energy horizontal is dissipated. line marked ϩ1 exactly fill in the unshaded spaces below +1 The instantaneous rate at which energy is dissipated in the2 resistor can be + –1 that line.) Thus, we can write, from Eq. 31-68, 0 θ 2 written, with the help of Eqs. 26-27 and 31-29, as 0 π 2 π 3π 2 2 2 2 2 2 0 θ P ϭ i R ϭ [I sin(vdt Ϫ f)] R ϭ I R sin (vdt ϪI Rf).I (31-68) 0 π 2 π 3π Pavg ϭ ϭ R. –1 (31-69) The average rate at which energy is dissipated in the resistor, however,2 ΂ is the2 aver-΃ (a) (b) age of Eq. 31-68 over time. Over one complete cycle, the average value1 of sin u, i = i0 sin(ωt) where u is any Thevariable, quantity is zero (Fig.I/ 31-172 is acalled) but the the average root-mean-square, value of sin2 u is 1or rms, value of the current i: 2 2 (Fig. 31-17b). (Note in Fig. 31-17b how the shaded areas under the curve but Thesin power2 θ delivered to a loadFig. fluctuates 31-17 as P(a=) AP plot0 sin of(ω sint) u. versus u. 11 +1 above the horizontal line marked ϩ2 exactly fill in the unshadedI spaces below The average value over one cycle is zero. (b) –1 2 that line.) Thus, we can write, from Eq. 31-68, Irms ϭ (rms current). + 2 (31-70) A plot of sin u versus u.The average value 2 1 2 2 0 θ I R I 0 π 2 π 3π over one cycle is 2 . Pavg ϭ ϭ R. 1 (31-69) 2 ΂ 2 ΃ (b) The quantity I/ 2 is called the root-mean-square,1 or rms, value of the current i: u u 1 Fig. 31-17 (a) A plot of sin versus . I The average value over one cycle is zero. (b) Irms ϭ (rms current). (31-70) A plot of sin2 u versus u.The average value 2 over one cycle is 1 . 1 2 halliday_c31_826-860hr.qxd 11-12-2009 13:11 Page 850

850 CHAPTER 31 ELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT 31-11 Transformers Energy Transmission Requirements When an ac circuit has only a resistive load, the power factor in Eq. 31-76 is

cos 0° ϭ 1 and the applied rms emf Ᏹrms is equal to the rms voltage Vrms across the load. Thus, with an rms current Irms in the load, energy is supplied and dissipated at the average rate of Pavg ϭ ᏱI ϭ IV. (31-77) (In Eq. 31-77 and the rest of this section, we follow conventional practice and drop the subscripts identifying rms quantities. Engineers and scientists assume that all time-varying currents and voltages are reported as rms values; that is what the me- ters read.) Equation 31-77 tells us that, to satisfy a given power requirement, we have a range of choices for I and V, provided only that the product IV is as required. In electrical power distribution systems it is desirable for reasons of safety and for efﬁcient equipment design to deal with relatively low voltages at both the generating end (the electrical power plant) and the receiving end (the home or factory). Nobody wants an electric toaster or a child’s electric train to operate at, say, 10 kV. On the other hand, in the transmission of electrical energy from the generating plant to the consumer, we want the lowest practical current (hence the largest practical voltage) to minimize I2R losses (often called ohmic losses) in the transmission line. As an example, consider the 735 kV line used to transmit electrical energy from the La Grande 2 hydroelectric plant in Quebec to Montreal, 1000 km away. Suppose that the current is 500 A and the power factor is close to unity. Then from Eq. 31-77, energy is supplied at the average rate 5 Pavg ϭ ᏱI ϭ (7.35 ϫ 10 V)(500 A) ϭ 368 MW. The resistance of the transmission line is about 0.220 ⍀/km; thus, there is a total resistance of about 220 ⍀ for the 1000 km stretch. Energy is dissipated due to that resistance at a rate of about 2 2 Pavg ϭ I R ϭ (500 A) (220 ⍀) ϭ 55.0 MW, which is nearly 15% of the supply rate. Imagine what would happen if we doubled the current and halved the voltage. Energy would be supplied by the plant at the same average rate of 368 MW as previously, but now energy would be dissipated at the rate of about

2 2 Transformers Pavg ϭ I R ϭ (1000 A) (220 ⍀) ϭ 220 MW, Transformers change ∆Vrms and Irms simultaneously, while keepingwhich is almost 60% of the supply rate. Hence the general energy transmission the average power: rule:Transmit at the highest possible voltage and the lowest possible current. Pavg = Irms(∆Vrms) constant (conservation of energy). The Ideal Transformer ΦB S The transmission rule leads to a fundamental mismatch between the requirement for efﬁcient high-voltage transmission and the need for safe low-voltage generation Np Vp Vs R and consumption. We need a device with which we can raise (for transmission) and

Ns lower (for use) the ac voltage in a circuit, keeping the product current ϫ voltage es- sentially constant.The transformer is such a device. It has no moving parts, operates Primary Secondary by Faraday’s law of induction, and has no simple direct-current counterpart. The ideal transformer in Fig. 31-18 consists of two coils, with different num- This worksFig. via mutual31-18 inductance.An ideal transformer If the current (two in the ﬁrstbers coil of turns, wound around an iron core. (The coils are insulated from the core.) coils wound on an iron core) in a basic did not constantly change (AC) this would not work. In use, the primary winding, of N turns, is connected to an alternating-current transformer circuit.An ac generator pro- p duces current in the coil at theNs left (the pri- generator whose emf Ᏹ at any time t is given by ∆Vs = ∆Vp mary).The coil at the right N(thep secondary) Ᏹ ϭ Ᏹm sin vt. (31-78) is connected to the resistive load R when switch S is closed. The secondary winding, of Ns turns, is connected to load resistance R,but its Transformers

The reason for the voltage relation is that the iron core ideally contains all the magnetic ﬂux lines produced.

∆Φ Then the emf per turn Et = − ∆t is the same in both solenoids.

∆Φ ∆Φ ∆V = −N and ∆V = −N p p ∆t s s ∆t

Ns ∆Vs = ∆Vp Np Maxwell’s Laws

Amazingly, we can summarize the majority of the relations that we have talked about in this course in a set of just 4 equations.

These are together called Maxwell’s equations.

q E · dA = enc I B · dA = 0 I dΦ E · ds = − B dt I dΦ B · ds = µ E +µ i 0 0 dt 0 enc I Gauss’s Law for Magnetic Fields

The ﬁrst of Maxwell’s equations is Gauss’s Law for E-ﬁelds: q E · dA = enc I

The second is for Gauss’s Law for B-ﬁelds:

B · dA = 0 I But what about the reverse? A changing electric ﬁeld inducing a magnetic ﬁeld?

It does happen!

Maxwell’s Law of Induction

dΦ B · ds = µ E 0 0 dt I

Maxwell’s Law of Induction Faraday’s Law of Induction is the third of Maxwell’s laws. dΦ E · ds = − B dt I

This tells us that a changing magnetic ﬁeld will induce an electric ﬁeld. It does happen!

Maxwell’s Law of Induction

dΦ B · ds = µ E 0 0 dt I

Maxwell’s Law of Induction Faraday’s Law of Induction is the third of Maxwell’s laws. dΦ E · ds = − B dt I

This tells us that a changing magnetic ﬁeld will induce an electric ﬁeld.

But what about the reverse? A changing electric ﬁeld inducing a magnetic ﬁeld? Maxwell’s Law of Induction Faraday’s Law of Induction is the third of Maxwell’s laws. dΦ E · ds = − B dt I

This tells us that a changing magnetic ﬁeld will induce an electric ﬁeld.

But what about the reverse? A changing electric ﬁeld inducing a magnetic ﬁeld?

It does happen!

Maxwell’s Law of Induction

dΦ B · ds = µ E 0 0 dt I halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page 864

864 CHAPTER 32 MAXWELL’S EQUATIONS; MAGNETISM OF MATTER

2 a changing electric flux will always induce a magnetic field whenever it occurs.) We + E – assume that the charge on our capacitor (Fig. 32-5a) is being increased at a steady rate by a constant current i in the connecting wires. Then the electric field magni- + – tude between the plates must also be increasing at a steady rate. 1 + – Figure 32-5b is a view of the right-hand plate of Fig. 32-5a from between the plates. The electric field is directed into the page. Let us consider a circular loop halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page i864 + – i through point 1 in Figs. 32-5a and b,a loop that is concentric with the capacitor plates + – and has a radius smaller than that of the plates. Because the electric field through the + – loop is changing, the electric flux through the loop must also be changing.According to – Eq. 32-3, this changing electric flux induces a magnetic field around the loop. Maxwell’s Law of Induction + : Experiment proves that a magnetic field B is indeed induced around such + – The changing of the aloop,directed as shown.This magnetic field has the same magnitude at every dΦ electric field between point around the loop and thus has circular symmetry about the central axis of 864 CHAPTER 32 MAXWELL’SB · ds = µ EQUATIONS;E (a) MAGNETISM OF MATTER 0 0 dt the plates creates a the capacitor plates (the axis extending from one plate center to the other). I 2 magnetic field. If we now consider a larger loop—say, through point 2 outside the plates 2 a changing electric flux will always inducein Figs. a magnetic 32-5a and field b—we whenever find thatit occurs.) a magnetic We field is induced around that loop E a + – assume that the charge on our capacitoras well. (Fig. Thus, 32-5 ) while is being the increased electric at field a steady is changing, magnetic fields are induced rate by a Bconstant current i in the connecting wires. Then the electric field magni- – E between the plates, both inside and outside the gap. When the electric field stops + tude between1 the plates must also be increasing at a steady rate. 1 changing, these induced magnetic fields disappear. + – b a Figure 32-5 is a view of the right-handAlthough plate ofEq. Fig. 32-3 32-5 is similarfrom between to Eq. the 32-2, the equations differ in two ways. plates. Ther electric field is directed into the page. Let us consider a circular loop i + – i B First, Eq. 32-3 has the two extra symbols m and ␧ ,but they appear only because through point 1 in Figs. 32-5a and b,a loop that is concentric with the capacitor plates0 0 + – we employ SI units. Second, Eq. 32-3 lacks the minus sign of Eq. 32-2, mean- and has a radius smaller than that of the plates. Because the electric field through: the : B ing that the induced electric field E and the induced magnetic field B have + – loop is changing, the electric flux through the loop must also be changing.According to R opposite directions when they are produced in otherwise similar situations. To – Eq. 32-3, this changing electric flux induces a magnetic field around the loop. : + see this opposition,: examine Fig. 32-6, in which an increasing magnetic field B , Experiment proves that a magnetic field B is indeed induced around such : : + – B directed into the page, induces an electric field E . The induced field E is counter- The changing of the aloop,directed as shown.This magnetic field has the same magnitude at every : electric field between clockwise, opposite the inducedcentral magnetic axis field B in Fig. 32-5b. (a) point (aroundb) the loop and thus has circular symmetry about the of the plates creates a the capacitor plates (the axis extending from one plate center to the other). Fig. 32-5 (a) A circular parallel-plate ca- magnetic field. If we now consider a larger loop—say, through point 2 outside the plates 2 pacitor, shown in side view, is being charged Ampere–Maxwell Law in Figs. 32-5a and b—we find that a magnetic field is induced around that loop : :by a constant current i.(b) A view from Now recall that the left side of Eq. 32-3, the integral of the dot product B ؒ ds within theas capacitor, well. Thus, looking while toward the the electric plate field is changing, magnetic fields are induced B : around a closed loop, appears in another equation—namely,Ampere’s law: E at the rightbetween in (a).The the electric plates, field both E inside is uni- and outside the gap. When the electric field stops 1 form, is directedchanging, into these the page induced (toward magnetic the fields disappear. : : (B ؒ ds ϭ ␮0ienc (Ampere’s law), (32-4 r plate), and growsAlthough in magnitude Eq. 32-3 as the is chargesimilar to Eq. 32-2, the equationsͶ differ in two ways. B m ␧ on the capacitorFirst, Eq. increases.The 32-3 has the magnetic two extra field symbolswhere 0iencandis the0,but current they appearencircled only by becausethe closed loop.Thus, our two equations that : : B inducedwe by employthis changing SI units. electric Second, field is Eq. 32-3specify lacks the the magnetic minus sign field of B Eq. produced 32-2, mean- by means other than a magnetic material B : : shown at ingfour thatpoints the on induceda circle with electric a radius field r (thatE and is, bythe a inducedcurrent and magnetic by a changing field B haveelectric field) give the field in exactly the R less than oppositethe plate radius directions R. when they are produced in otherwise similar situations. To same form.We can combine the two equations: into the single equation see this opposition, examine Fig. 32-6, in which an increasing magnetic field B , B : : directed into the page, induces an electric field E . The: induced fieldd ⌽E is counter- : : E (clockwise, opposite the induced magnetic field BͶ in BFig.ؒ ds 32-5ϭb␮. 0␧0 ϩ ␮0 ienc (Ampere–Maxwell law). (32-5 (b) dt Fig. 32-5 (a) A circular parallel-plate capacitor, shown in side view, is being charged Ampere–Maxwell Law :The induced E direction here is opposite the :by a constant current i.(b) A view from Now recall that the left side of Eq. 32-3, the integral of the dot product B ؒ ds within the capacitor, looking toward the plate induced B direction in the preceding figure. : around a closed loop, appears in another equation—namely,Ampere’s law: at the right in (a).The electric field E is uni-

form, is directed into the page (toward the : : B ؒ ds ϭ ␮ i (Ampere’s law), (32-4) E plate), and grows in magnitude as the charge Ͷ 0 enc B Fig. 32-6 A uniform magnetic on the capacitor increases.The magnetic field where i is the current encircled by the closed: loop.Thus, our two equations that : enc field B in a circular region.The field, B induced by this changing electric field is : specify the magnetic field B produceddirected by means into the other page, than is increasing a magnetic in material R shown at four points on a circle with a radius r : r E (that is, by a current and by a changingmagnitude.The electric field) electric give field the Efield in- in exactly the less than the plate radius R. same form.We can combine the two ducedequations by the into changing the single magnetic equation field is shown at four points on a circle E B : : d⌽E concentric with the circular region. (B ؒ ds ϭ ␮ ␧ ϩ ␮ i (Ampere–Maxwell law). (32-5 Ͷ 0 0 dt Compare0 enc this situation with that of Fig. 32-5b. E

The induced E direction here is opposite the induced B direction in the preceding figure.

E B Fig. 32-6 A uniform magnetic : field B in a circular region.The field, directed into the page, is increasing in R : r E magnitude.The electric field E induced by the changing magnetic field is shown at four points on a circle E B concentric with the circular region. Compare this situation with that of Fig. 32-5b. E halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page 865

PART 3 32-3 INDUCED MAGNETIC FIELDS 865

When there is a current butMaxwell’s no change in Law electric of flux Induction (such as with question a wire carrying a constant current), the firstThe term figure on showsthe right graphs side of of Eq. the 32-5 electric is zero, field and magnitude E versus so Eq. 32-5 reduces to Eq. 32-4, Ampere’stime t for law. four When uniform there electric is a change fields, in all electric contained within identical flux but no current (such as inside or outside the gap of a charging capacitor), the circular regions as in the circular-plate capacitor. Rank the E-fields second term on the right side of Eq. 32-5 is zero, and so Eq. 32-5 reduces to Eq. 32-3, Maxwell’s law of induction.according to the magnitudes of the magnetic fields they induce at the edge of the region, greatest first.

CHECKPOINT 2 The figure shows graphs of the electric field magnitude E d versus time t for four uniform electric fields,all contained E within identical circular regions as in Fig. 32-5b.Rank the c fields according to the magnitudes of the magnetic fields a they induce at the edge of the region, greatest first. b t A a, b, c, d B a, c, b, d Sample Problem C d, b, c, a Magnetic field induced by changing electric field D d, c, a, b 1 A parallel-plate capacitor with circularHalliday, plates Resnick, of radius Walker, R is pageRight 865. side of Eq. 32-6: We assume that the electric field : being charged as in Fig. 32-5a. E is uniform between the capacitor plates and directed per- pendicular to the plates. Then the electric flux ⌽ through (a) Derive an expression for the magnetic field at radius r E the Amperian loop is EA, where A is the area encircled by for the case r Յ R. the loop within the electric field. Thus, the right side of Eq. 32-6 is m ␧ d(EA)/dt. KEY IDEAS 0 0 Combining results: Substituting our results for the left A magnetic field can be set up by a current and by induction and right sides into Eq. 32-6, we get due to a changing electric flux; both effects are included in Eq. 32-5.There is no current between the capacitor plates of d(EA) (B)(2␲r) ϭ ␮0␧0 . Fig. 32-5, but the electric flux there is changing. Thus, Eq. dt 32-5 reduces to Because A is a constant, we write d(EA) as A dE;so we have dE : : d⌽E (B ؒ ds ϭ ␮ ␧ . (32-6) (B)(2␲r) ϭ ␮0␧0 A . (32-7 Ͷ 0 0 dt dt The area A that is encircled by the Amperian loop within the We shall separately evaluate the left and right sides of this 2 equation. electric field is the full area pr of the loop because the loop’s radius r is less than (or equal to) the plate radius R.Sub- Left side of Eq. 32-6: We choose a circular Amperian loop stituting pr2 for A in Eq. 32-7 leads to, for r Յ R, with a radius r Յ R as shown in Fig. 32-5b because we want ␮0␧0r dE to evaluate the magnetic field for r Յ R—that is,inside the B ϭ . (Answer) (32-8) : 2 dt capacitor.The magnetic field B at all points along the loop is : This equation tells us that, inside the capacitor, B increases tangent to the loop, as is the path element ds: . Thus,B and linearly with increased radial distance r,from 0 at the cen- ds: are either parallel or antiparallel at each point of the tral axis to a maximum value at plate radius R. loop. For simplicity, assume they are parallel (the choice does not alter our outcome here).Then (b) Evaluate the field magnitude B for r ϭ R/5 ϭ 11.0 mm and dE/dt ϭ 1.50 ϫ 1012 V/m и s. : .B ؒ ds: ϭ B ds cos 0Њϭ B ds Ͷ Ͷ Ͷ Calculation: From the answer to (a), we have Due to the circular symmetry of the plates, we can also as- 1 dE : B ϭ ␮ ␧ r sume that B has the same magnitude at every point around 2 0 0 dt the loop.Thus, B can be taken outside the integral on the right 1 ϭ (4␲ ϫ 10Ϫ7 T и m/A)(8.85 ϫ 10Ϫ12 C2/N и m2) side of the above equation. The integral that remains is ds , 2 ͛ Ϫ3 12 which simply gives the circumference 2pr of the loop.The left ϫ (11.0 ϫ 10 m)(1.50 ϫ 10 V/m и s) side of Eq. 32-6 is then (B)(2pr). ϭ 9.18 ϫ 10Ϫ8 T. (Answer) halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page 865

PART 3 32-3 INDUCED MAGNETIC FIELDS 865

CHECKPOINT 2 The figure shows graphs of the electric field magnitude E d versus time t for four uniform electric fields,all contained E within identical circular regions as in Fig. 32-5b.Rank the c fields according to the magnitudes of the magnetic fields a they induce at the edge of the region, greatest first. b t A a, b, c, d B a, c, b, d ← Sample Problem C d, b, c, a Magnetic field induced by changing electric field D d, c, a, b 1 A parallel-plate capacitor with circularHalliday, plates Resnick, of radius Walker, R is pageRight 865. side of Eq. 32-6: We assume that the electric field : being charged as in Fig. 32-5a. E is uniform between the capacitor plates and directed per- pendicular to the plates. Then the electric flux ⌽ through (a) Derive an expression for the magnetic field at radius r E the Amperian loop is EA, where A is the area encircled by for the case r Յ R. the loop within the electric field. Thus, the right side of Eq. 32-6 is m ␧ d(EA)/dt. KEY IDEAS 0 0 Combining results: Substituting our results for the left A magnetic field can be set up by a current and by induction and right sides into Eq. 32-6, we get due to a changing electric flux; both effects are included in Eq. 32-5.There is no current between the capacitor plates of d(EA) (B)(2␲r) ϭ ␮0␧0 . Fig. 32-5, but the electric flux there is changing. Thus, Eq. dt 32-5 reduces to Because A is a constant, we write d(EA) as A dE;so we have dE : : d⌽E (B ؒ ds ϭ ␮ ␧ . (32-6) (B)(2␲r) ϭ ␮0␧0 A . (32-7 Ͷ 0 0 dt dt The area A that is encircled by the Amperian loop within the We shall separately evaluate the left and right sides of this 2 equation. electric field is the full area pr of the loop because the loop’s radius r is less than (or equal to) the plate radius R.Sub- Left side of Eq. 32-6: We choose a circular Amperian loop stituting pr2 for A in Eq. 32-7 leads to, for r Յ R, with a radius r Յ R as shown in Fig. 32-5b because we want ␮0␧0r dE to evaluate the magnetic field for r Յ R—that is,inside the B ϭ . (Answer) (32-8) : 2 dt capacitor.The magnetic field B at all points along the loop is : This equation tells us that, inside the capacitor, B increases tangent to the loop, as is the path element ds: . Thus,B and linearly with increased radial distance r,from 0 at the cen- ds: are either parallel or antiparallel at each point of the tral axis to a maximum value at plate radius R. loop. For simplicity, assume they are parallel (the choice does not alter our outcome here).Then (b) Evaluate the field magnitude B for r ϭ R/5 ϭ 11.0 mm and dE/dt ϭ 1.50 ϫ 1012 V/m и s. : .B ؒ ds: ϭ B ds cos 0Њϭ B ds Ͷ Ͷ Ͷ Calculation: From the answer to (a), we have Due to the circular symmetry of the plates, we can also as- 1 dE : B ϭ ␮ ␧ r sume that B has the same magnitude at every point around 2 0 0 dt the loop.Thus, B can be taken outside the integral on the right 1 ϭ (4␲ ϫ 10Ϫ7 T и m/A)(8.85 ϫ 10Ϫ12 C2/N и m2) side of the above equation. The integral that remains is ds , 2 ͛ Ϫ3 12 which simply gives the circumference 2pr of the loop.The left ϫ (11.0 ϫ 10 m)(1.50 ϫ 10 V/m и s) side of Eq. 32-6 is then (B)(2pr). ϭ 9.18 ϫ 10Ϫ8 T. (Answer) Since we could have a situation with both a changing E-field and a current, we can express it more generally with the Ampere-Maxwell Law: dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

We add up the contributions from the changing electric ﬂux and the current.

Ampere-Maxwell Law However, a changing electric ﬁeld is not the only cause of a magnetic ﬁeld.

We know from Ampere’s Law:

B · ds = µ0ienc I that a moving charge (current) causes a magnetic field also. Ampere-Maxwell Law However, a changing electric field is not the only cause of a magnetic field.

We know from Ampere’s Law:

B · ds = µ0ienc I that a moving charge (current) causes a magnetic ﬁeld also.

Since we could have a situation with both a changing E-ﬁeld and a current, we can express it more generally with the Ampere-Maxwell Law: dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

We add up the contributions from the changing electric ﬂux and the current. Ampere-Maxwell Law

The Ampere-Maxwell Law is the fourth and last of Maxwell’s laws.

Ampere-Maxwell Law

dΦ B · ds = µ E +µ i 0 0 dt 0 enc I Maxwell’s Equations

q E · dA = enc I B · dA = 0 I dΦ E · ds = − B dt I dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

1Strictly, these are Maxwell’s equations in a vacuum. dΦE This works because the units of 0 dt are Amps.

displacement “current”

dΦ i = E d 0 dt

Note: The displacement “current” is not a current and has nothing to do with displacement.

Ampere-Maxwell Law and Displacement “Current” Ampere-Maxwell Law: dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

It can be convenient to imagine that even the part of the B-ﬁeld that is produced by a changing E-ﬁeld is actually produced by some kind of virtual current. Note: The displacement “current” is not a current and has nothing to do with displacement.

Ampere-Maxwell Law and Displacement “Current” Ampere-Maxwell Law: dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

It can be convenient to imagine that even the part of the B-ﬁeld that is produced by a changing E-ﬁeld is actually produced by some kind of virtual current.

dΦE This works because the units of 0 dt are Amps.

displacement “current”

dΦ i = E d 0 dt Ampere-Maxwell Law and Displacement “Current” Ampere-Maxwell Law: dΦ B · ds = µ E +µ i 0 0 dt 0 enc I

It can be convenient to imagine that even the part of the B-ﬁeld that is produced by a changing E-ﬁeld is actually produced by some kind of virtual current.

dΦE This works because the units of 0 dt are Amps.

displacement “current”

dΦ i = E d 0 dt

Note: The displacement “current” is not a current and has nothing to do with displacement. Ampere-Maxwell Law and Displacement “Current”

This lets us rewrite the Ampere-Maxwell law as:

B · ds = µ0id + µ0ienc I

Looking at it this way can give us some insights. halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page 867

PART 3 32-4 DISPLACEMENT CURRENT 867 B-ﬁeld around a charging capacitor Suppose a capacitor is being charged with a constant current, i. A During charging, magnetic field is created by both Before charging, there the real and fictional currents. is no magnetic field. id

i i (a) (b) + –

BBB

During charging, the How does i relate to id the “current” from the E-field between the plates? right-hand rule works for both the real and fictional currents. After charging, there Fig. 32-7 (a) Before and (d) after is no magnetic field. the plates are charged, there is no magnetic field. (b) During the charging, magnetic field is created by both the real i i (c) current and the (fictional) displacement + – (d) + – current. (c) The same right-hand rule works for both currents to give the direction of the magnetic field. B B B

becomes d⌽ d(EA) dE i ϭ␧ E ϭ␧ ϭ␧A . (32-14) d 0 dt 0 dt 0 dt Comparing Eqs. 32-13 and 32-14, we see that the real current i charging the capacitor and the ﬁctitious displacement current id between the plates have the same magnitude: id ϭ i (displacement current in a capacitor). (32-15)

Thus, we can consider the fictitious displacement current id to be simply a con- tinuation of the real current i from one plate, across the capacitor gap, to the other plate. Because the electric field is uniformly spread over the plates, the same is true of this fictitious displacement current id,as suggested by the spread of current arrows in Fig. 32-7b.Although no charge actually moves across the gap between the plates, the idea of the fictitious current id can help us to quickly find the direction and magnitude of an induced magnetic field, as follows.

Finding the Induced Magnetic Field In Chapter 29 we found the direction of the magnetic field produced by a real current i by using the right-hand rule of Fig. 29-4. We can apply the same rule to find the direction of an induced magnetic field produced by a fictitious displacement current id,as is shown in the center of Fig.32-7c for a capacitor. We can also use id to find the magnitude of the magnetic field induced by acharging capacitor with parallel circular plates of radius R.We simply consider the space between the plates to be an imaginary circular wire of radius R carrying the imaginary current id.Then,from Eq.29-20,the magnitude of the magnetic halliday_c32_861-888hr.qxd 11-12-2009 13:14 Page 867

i i (a) (b) + –

BBB

During charging, the How does i relate to id the “current” from the E-field between the right-hand rule works for bothplates? the real and fictional currents. After charging, there Fig. 32-7 (a) Before and (d) after is no magnetic field. the plates are charged, there is no magnetic field. (b) During the charging, magnetic field is created by both the real i i (c) current and the (fictional) displacement + – (d) + – current. (c) The same right-hand rule works for both currents to give the direction of the magnetic field. B B B

PART 3 32-4 DISPLACEMENT CURRENT 867

A During charging, magnetic field is created by both Before charging, there the real and fictional currents. is no magnetic field. id

i i (a) (b) + – B-ﬁeld around a charging capacitor Suppose a capacitorBB is being charged with a constantB current, i.

During charging, the right-hand rule works for both the real and fictional currents. After charging, there Fig. 32-7 (a) Before and (d) after is no magnetic field. the plates are charged, there is no magnetic field. (b) During the charging, magnetic field is created by both the real i i (c) current and the (fictional) displacement + – (d) + – current. (c) The same right-hand rule works for both currents to give the direction of the magnetic field. B B B

How does i relate to id the “current” from the E-ﬁeld between the plates? becomes d⌽ d(EA) dE i ϭ␧ E ϭ␧ ϭ␧A . (32-14) d 0 dt 0 dt 0 dt Comparing Eqs. 32-13 and 32-14, we see that the real current i charging the capacitor and the ﬁctitious displacement current id between the plates have the same magnitude: id ϭ i (displacement current in a capacitor). (32-15)

dΦ dE i = E = A d 0 dt 0 dt

Gauss’s law allows us to relate q, the charge on the capacitor to the ﬂux: q = E · dA = EA 0 I

The current is the rate of ﬂow of charge: dq dE i = = A dt 0 dt

So, id = i !

The B-ﬁeld outside a circular capacitor looks the same as the B-ﬁeld around the wire leading up to the capacitor. Outside the capacitor at radius r from the center: µ i B = 0 d 2πr

Inside the capacitor (plates have radius R) at radius r from the center: µ i B = 0 d r 2πR2

But remember: id is not a current. No current ﬂows across the gap between the plates.

Magnetic Field around a circular capacitor

Can be calculated just like the ﬁeld around a wire! Magnetic Field around a circular capacitor

Can be calculated just like the ﬁeld around a wire!

Outside the capacitor at radius r from the center: µ i B = 0 d 2πr

Inside the capacitor (plates have radius R) at radius r from the center: µ i B = 0 d r 2πR2

But remember: id is not a current. No current ﬂows across the gap between the plates. Another Implication of Maxwell’s Equations

Using all 4 equations (in their diﬀerential form) it is possible to reach a pair of wave equations for the electric and magnetic ﬁelds:

1 ∂2E ∇2E = c2 ∂t2 1 ∂2B ∇2B = c2 ∂t2 with wave solutions:

E = E0 sin(k · r − ωt)

B = B0 sin(k · r − ωt) Another Implication of Maxwell’s Equations

1 ∂2E ∇2E = c2 ∂t2

The constant c appears as the wave speed and 1 c = √ µ00

c = 3.00 × 108 m/s, is the speed of light.

The values of 0 and µ0 together predict the speed of light! Relation between Electric and Magnetic Fields

These oscillating electric and magnetic ﬁelds make up light.

Faraday’s Law of Induction A changing magnetic ﬁeld gives rise to an electric ﬁeld.

Ampere-Maxwell Law of Induction A changing electric ﬁeld gives rise to an magnetic ﬁeld. Light

Faraday’s Law ⇒ a changing magnetic ﬁeld causes an electric ﬁeld.

Maxwell’s Law ⇒ a changing electric ﬁeld causes a magnetic ﬁeld. Light (Electromagnetic Radiation)

All light waves in a vacuum travel at the same speed, the speed of light, c = 3.00 × 108 m s−1.

Maxwell’s equations possess the ‘wrong’ symmetry for Gallilean transformations between observers; they are Lorentz-invariant. This gave Einstein an important idea.

All observers, no matter how they move relative to one another all agree that any light wave travels at that same speed.

Since light travels at this ﬁxed speed and c = v = f λ, if the frequency of the light is given, you also know the wavelength, and vice versa. c c λ = ; f = f λ Electromagnetic spectrum Electromagnetic spectrum Summary

• applications of Faraday’s law • inductance • self-induction • RL Circuits

Homework Halliday, Resnick, Walker: • NEW: Ch 30, onward from page 816. Problems: 41, 45, 61, 63, 67, 69, 73 • NEW: Ch 31 onward from page 858. Problems: 62, 63 • NEW: Ch 32, onward from page 883. Questions: 1, 3; Problems: 1, 5, 13