Chapter 31 Inducton and Magnetc Moment CHAPTER 31 INDUCTION AND MAG- NETIC MOMENT In ths chapter we dscuss several applcatons of Faraday s law and the Lorentz force law. The frst s to the nductor whch s a common electronc crcut element. We wll pay partcular attenton to a crcut contanng an nductor and a capactor, n whch an electrc current oscllates back and forth between the two. Measurements of the perod of the oscllaton and dmensons of the crcut elements allows us to predct the speed of lght wthout lookng at lght. Such a predcton leads to one of the basc questons faced by physcsts around the begnnng of the 20th century: who got to measure ths predcted speed? The answer was provded by Ensten and hs specal theory of relatvty. In the second part of ths chapter we wll dscuss the torque eerted by a magnetc feld on a current loop, and ntroduce the concept of a magnetc moment. Ths dscusson wll provde some nsght nto how the presence of ron greatly enhances the strength of the magnetc feld n an electromagnet. However the man reason for developng the concept of magnetc moment and the varous magnetc moment equatons s for our later dscusson of the behavor of atoms and elementary partcles n a magnetc feld. It s useful to clearly separate the classcal deas dscussed here from the quantum mechancal concepts to be developed later.
31-2 Inductors and Magnetc Moment THE INDUCTOR In our dscusson of Faraday s law and the betatron n Chapter 30, partcularly n Fgure (30-15), we saw that an ncreasng magnetc feld n the core of the betatron creates a crcular electrc feld around the core. Ths electrc feld was used to accelerate the electrons. A more common and accessble way to produce the same crcular electrc feld s by turnng up the current n a solenod as shown n Fgure (1). As we saw n our dscusson of Ampere s law n Chapter 29, a current n a long col of wre wth n turns per unt length, B B = µ n o produces a nearly unform magnetc feld nsde the col whose strength s gven by the formula B=µ 0 n (29-31) and whose drecton s gven by the rght hand rule as shown n the sde vew, Fgure (1a). If the col has a cross-sectonal area A, as seen n the top vew Fgure (1b), then the amount of magnetc flu Φ B flowng up through the col s gven by Φ B =BA=µ 0 na (1) And f we are ncreasng the current n the col, then the rate of ncrease of ths flu s (snce µ 0, n and A are constants) dφ B = µ dt 0 na d (2) dt It s the changng magnetc flu that creates the crcular electrc feld E shown n Fgure (1b). postve path E a) sde vew of col and magnetc feld area A of col E B up ncreasng rght hand rule for postve path B up ncreasng b) top vew showng the electrc feld surroundng the ncreasng magnetc flu Fgure 1 When we turn up the current n a solenod, we ncrease the magnetc feld and therefore the magnetc flu up through the col. Ths ncreasng magnetc flu s the source of the crcular electrc feld seen n the top vew. E d = dφ B dt (Faraday's Law) Fgure 2 Sgn conventons We start by defnng up, out of the paper, as the postve drecton. Then use the rght hand rule to defne a postvely orented path. As a result, counter clockwse s postve, clockwse s negatve. Wth these conventons, dφ B / dt s postve for an ncreasng upward drected magnetc flu. In calculatng the lne ntegral E d, we go around n a postve drecton, counter clockwse. Everythng s postve ecept the sgn n Faraday's law, thus the electrc feld goes around n a negatve drecton, clockwse as shown.
31-3 In Fgure (2) we have shown the top vew of the solenod n Fgure (1) and added n the crcular electrc feld we would get f we had an ncreasng magnetc flu up through the solenod. We have also drawn a crcular path of radus r around the solenod as shown. If we calculate the lne ntegral E d for ths closed path, we get by Faraday s law E d = dφ B dt E 2πr = µ 0 na d dt (30-17) (3) where the ntegral E d s smply E tmes the crcumference of the crcle, and we used Equaton (2) for dφ B /dt. The mnus sgn n Equaton (3) tells us that f we use a postve path as gven by the rght hand rule, and we are ncreasng the flu up through ths path, then E d must be negatve. I.e., the electrc feld must go clockwse, opposte to the postve path. (Do not worry too much about sgns n ths dscusson. We wll shortly fnd a smple, easly remembered, rule that tells us whch way the electrc feld ponts.) Drecton of the Electrc Feld In Fgure (2) and n the above eercse, we saw that an ncreasng magnetc flu n the col created a clockwse crcular electrc feld both nsde and outsde the wre as shown n Fgure (3). In partcular we have a crcular electrc feld at the wre, and ths crcular electrc feld wll act on the charges carryng the current n the wre. To mantan our sgn conventons, thnk of the current n the wre as beng carred by the flow of postve charge. The up drected magnetc feld of Fgure (3) wll be produced by a current flowng counterclockwse as shown (rght hand rule). In order to have an ncreasng flu, ths counterclockwse current must be ncreasng. We saw that the electrc feld s clockwse, opposte to the drecton of the current. We are turnng up the current to ncrease the magnetc feld, and the electrc feld s opposng the ncrease. If we already have a current n a solenod, already have an establshed B feld and try to decrease t, d/dt s negatve for ths operaton, and we get an etra mnus sgn n Equaton (3) that reverses the drecton of E. As Equaton (3) tells us that the strength E of the crcular feld s proportonal to the rate of change of current n the solenod, and drops off as 1/r f we are outsde the solenod. In the followng eercse, you are to show that we also have a crcular feld nsde the solenod, a feld that decreases lnearly to zero at the center. B up ncreasng E Eercse 1 Use Faraday s law to calculate the electrc feld nsde the solenod. Note that for a crcular path of radus r nsde the solenod, the flu Φ B through the path s proportonal to the area of the path and not the area A of the solenod. The calculaton of the crcular electrc feld nsde and outsde a solenod, when s changng, s a good eample of the use of both Ampere s law to calculate B and Faraday s law to calculate E. It should be saved n your collecton of good eamples. ncreasng current creates the ncreasng magnetc flu Fgure 3 If the sgn conventons descrbed n Fg. 2 seemed too arbtrary, here s a physcal way to determne the drecton of E. The rule s that the electrc feld E opposes any change n the current. In ths case, to create an ncreasng upward drected magnetc flu, the current must be flowng counter clockwse as shown, and be ncreasng. To oppose ths ncrease, the electrc feld must be clockwse.
31-4 Inductors and Magnetc Moment a result we get a counterclockwse electrc feld that eerts a force n the drecton of. Thus when we try to decrease the current, the electrc feld tres to mantan t. There s a general rule for determnng the drecton of the electrc feld. The electrc feld produced by the changng magnetc flu always opposes the change. If you have a counterclockwse current and ncrease t, you wll get a clockwse electrc feld that opposes the ncrease. If you have a counterclockwse current and decrease t you get a counterclockwse electrc feld that opposes the decrease. If you have a clockwse current and try to ncrease t, you get a counter clockwse electrc feld that opposes the ncrease, etc. There are many possbltes, but one rule the electrc feld always opposes the change. current n wre E B up ncreasng postvely orented path nsde wre for calculatng E d Fgure 4 The electrc feld penetrates the wre, opposng the change n the current. The lne ntegral E d around the col s just equal to the change n voltage V 1 around each turn of the col. Induced Voltage We have just seen that the changng magnetc flu n a solenod creates an electrc feld that acts on the current n the solenod to oppose the change n the current. From Equaton (3), we see that the formula for the lne ntegral of ths electrc feld around one loop of the col s gven by E d = µ 0 na d dt (4) where the path s at the wre as shown n Fgure (4). The n n Equaton (4), whch comes from the formula for the magnetc feld of a solenod, s the number of turns per unt length n the solenod. In our dscusson of the betatron, we saw that the crcular electrc feld accelerated electrons as they went around the evacuated donut. Each tme the electrons went around once, they ganed an amount of knetc energy whch, n electron volts, was equal to E d. In our dscusson of the electron gun, we saw that usng a battery of voltage V acc to accelerate the electrons, produced electrons whose knetc energy, n electron volts, was equal to V acc. In other words, the crcular electrc feld can act lke a battery of voltage V acc = E d. When actng on the electrons n one loop of wre, the crcular electrc feld produces a voltage change V 1 gven by V 1 = E d change n electrc voltage n one turn of the col (5a) If we have a col wth N turns as shown n Fgure (5), then the change n voltage V N across all N turns s N tmes as great, and we have V N = N E d change n electrc voltage n N turns of the col (5b)
31-5 Usng Equaton 4 for the E d for a solenod, we see that the voltage change V N across the entre solenod has a magntude V N = N E d = µ 0 NnA d dt (6) where N s the total number of turns, n = N/h s the number of turns per unt length, A s the cross-sectonal area of the solenod, and the current through t. To get the correct sgn of V N, to see whether we have a voltage rse or a voltage drop, we wll use the rule that the crcular electrc feld opposes any change n the current. Ths rule s much easer to use than tryng to keep track of all the mnus sgns n the equatons. In summary, Equaton (6) s tellng us that f you try to change the amount of current flowng n a solenod, f d dt s not zero, then a voltage wll appear across the ends of the solenod. The voltage has a magntude proportonal to the rate d dt that we are tryng to change the current, and a drecton that opposes the change. It s tradtonal to call ths voltage V N the nduced voltage. One says that the changng magnetc flu n the col nduces a voltage. Such a col of wre s often called an nductor. VΝ (voltage across col) E h col of length h n = N/h s the number of turns per unt length Inductance If you take a pece of nsulated wre, tangle t up n any way you want, and run a current through t, you wll get an nduced voltage V nduced that s proportonal to the rate of change of current d/dt, and drected n a way such that t opposes the change n the current. If we desgnate the proportonalty constant by the letter L, then the relatonshp between V nduced and d/dt can be wrtten V nduced =L d dt nducedvoltages are proportonal to d/dt (7) The constant L s called the nductance of the col or tangle of wre. In the MKS system, nductance has the dmenson of volt seconds/ampere, whch s called a henry. Comparng Equatons (6) and (7), we mmedately obtan the formula for the nductance of a solenod L=µ 0 NnA = µ 0 N2 A h nductanceof a solenod (8) where N s the number of turns n the solenod, A the cross-sectonal area and h the length. In the mddle term, n = N/h s the number of turns per unt length. N turns, area A Fgure 5 Our standard col wth N turns, area A and length h. If you try to ncrease the current n the col, you get an opposng voltage.
31-6 Inductors and Magnetc Moment Eample 1 The torodal Inductor The smplest solenod we can use s a torodal one, lke that shown n Fgure (6), where the magnetc feld s completely confned to the regon nsde the col. Essentally, the torod s an deal solenod (no end effects) of length h=2πr. To develop an ntutve feelng for nductance and the sze of a henry, let us calculate the nductance of the torodal solenod shown n the photograph of Fgure (6b). Ths solenod has 696 turns and a radus of R = 21.5 cm. Each col has a radus of r = 1.3 cm. Thus we have N = 696turns R = 21.5cm h=2πr =2π*.215 = 1.35m r = 1.3cm A=πr 2 = π.013 2 = 5.31 10 4 m 2 µ 0 = 1.26 10 6 henry/m Wth L= µ 0 N2 A h we get L= 1.26 10 6 696 2 5.31 10 4 1.35 = 2.40 10 4 henry We see that even a farly bg solenod lke the one shown n Fgure (6) has a small nductance at least when measured n henrys. At the end of the chapter we wll see that nsertng an ron core nto a solenod greatly ncreases the nductance. Inductances as large or larger than one henry are easly obtaned wth ron core nductors. 2r h = 2πR R Fgure 6b Photograph of the torodal solenod used n varous eperments. Although the col looks bg, the nductance s only 2.40 10 4 henry. (If you put ron nsde the col, you could greatly ncrease the nductance, but you would not be able to calculate ts value.) B Fgure 6a A torod s an deal solenod of length h = 2πR.
31-7 INDUCTOR AS A CIRCUIT ELEMENT Because a changng electrc current n a col of wre produces a voltage rse, small cols are often used as crcut elements. Such a devce s called an nductor, and the symbol used n crcut dagrams s a sketch of a solenod and usually desgnated by the symbol L. The voltage rse across the three crcut elements we have consdered so far are V R = R resstor (27-8) a) b) R C + V R = R V C = Q C V C = Q C capactor (27-31) V L =L d nductor (7) dt As shown n Fgure (7) the drecton of the rse s opposte to the current n a resstor, toward the postve charge n a capactor, and n a drecton to oppose a change n the current n an nductor. In (c), we are showng the drecton of the voltage rse for an ncreasng current. The voltage n the nductor s opposng an ncrease n the current, just as the voltage n the resstor (a) opposes the current tself. c) (ncreasng) Fgure 7 The resstor R, capactor C and nductor L as crcut elements. L V L = L d dt
31-8 Inductors and Magnetc Moment The LR Crcut We wll begn our dscusson of the nductor as a crcut element wth the LR crcut shown n Fgure (8). Although ths crcut s farly easy to analyze, t s a bt trcky to get the current started. One way to start the current s shown n Fgure (9) where we have a battery and another resstor R 1 attached as shown. When the swtch of Fgure (9) has been closed for a whle, we have a constant current 0 that flows down out of the battery, through the resstor R 1, around up through the nductor and back to the battery. Because the current s constant, d 0 /dt = 0 and there s no voltage across the nductor. When we have constant DC currents, nductors act lke short crcuts. That s why the current, gven the choce of gong up through the nductor L or the resstor R, all goes up through L. (To say ths another way, snce the voltage across L s zero, the voltage V R across R must also be zero, and the current R = V R /R = 0.) Snce R 1 s the only thng that lmts the current n Fgure (9), 0 s gven by 0 = V B R 1 (9) Equaton (9) tells us that we have a serous problem f we forget to nclude the current lmtng resstor R 1. L d dt L R R When we open the swtch of Fgure (9), the battery and resstor R 1 are mmedately dsconnected from the crcut, and we have the smple LR crcut shown n Fgure (8). Everythng changes nstantly ecept the current n the nductor. The nductor nstantly sets up a voltage V L to oppose any change n the current. Fgure (10) s a recordng of the voltage V L across the nductor, where the swtch n Fgure (9) s opened at tme t = 0. Before t = 0, we have a constant current 0 and no voltage V L. When the swtch s opened the voltage jumps up to V L = V 0 and then decays eponentally just as n the RC crcut. What we want to do s apply Krchoff s law to Fgure (8) and see f we can determne the tme constant for ths eponental decay. L 0 R swtch 0 0 R 1 Fgure 9 To get a current started n an LR crcut, we begn wth the etra battery and resstor attached as shown. Wth the swtch closed, n the steady state all the current 0 flows up through the nductor because t (theoretcally) has no resstance. When the swtch s opened, the battery s dsconnected, and we are left wth the RL crcut startng wth an ntal current 0. + V b Fgure 8 The LR crcut. If we have a decreasng current, the voltage n the nductor opposes the decrease and creates a voltage that contnues to push the current through the resstor. But, to label the voltages for Krchoff's law, t s easer to work wth postve quanttes. I.e. we label the crcut as f both and d/dt were postve. Wth a postve d/dt, the voltage on the nductor opposes the current, as shown.
31-9 If we walk around the crcut of Fgure (8) n the drecton of, and add up the voltage rses we encounter, and set the sum equal to zero (Krchoff s law), we get R L d dt =0 d dt + R =0 (10) L Equaton (10) s a smple frst order dfferental equaton for the current. We guess from our epermental results n Fgure (10) that should be gven by an eponental decay of the form = 0 e αt (11) Dfferentatng Equaton (11) to get d/dt, we have d dt = α 0e αt (12) and substtutng (11) and (12) n Equaton (10) gves α 0 e αt + R L 0 e αt =0 (13) In Equaton (13), 0 and e -α t cancel and we get α =R/L Equaton (11) for becomes t (L R) = 0 e ( R L)t = 0 e t (L R 0 e t T (14) We see from Equaton (14) that the tme constant T for the decay s T= L tme constant R = for the decay of an LR crcut (15) Everythng we sad about eponental decays and tme constants for RC crcuts at the end of Chapter (27) apples to the LR crcut, ecept that the tme constant s now L/R rather than RC. Fgure 10 Epermental recordng of the voltage n an RL crcut. We see that once the swtch of Fg. 9 s opened, the voltage across the nductor jumps from zero to V 0 = 0 R. Ths voltage on the nductor s tryng to mantan the current now that the battery s dsconnected. The voltage and the current then de wth an eponental decay. (For ths eperment, we used the torodal nductor of Fgure 6, wth R = 15Ω, R1 = 4Ω, and V b = 2.5 volts.) Eercse 2 The LR crcut that produced the epermental results shown n Fgure (10) had a resstor whose resstance R was 15 ohms. Quckly estmate the nductance L. (You should be able to make ths estmate accurate to wthn about 10% smply by sketchng a straght lne on the graph of Fgure 10.) Compare your result wth the nductance of the torodal solenod dscussed n Fgure (6) on page 6.
31-10 Inductors and Magnetc Moment THE LC CIRCUIT The net crcut we wsh to look at s the LC crcut shown n Fgure (11). All we have done s replace the resstor R n Fgure (8) wth a capactor C as shown. It does not seem lke much of a change, but the behavor of the crcut s very dfferent. The eponental decays we saw n our LR and RC crcuts occur because we are losng energy n the resstor R. In the LC crcut we have no resstor, no energy loss, and we wll not get an eponental decay. To see what we should get, we wll apply Krchoff s law to the LC crcut and see f we can guess the soluton to the resultng dfferental equaton. Walkng clockwse around the crcut n Fgure (11) and settng the sum of the voltage rses to zero, we get Q C L d dt =0 d dt + Q LC =0 (16) The problem we have wth Equaton (16) s that we have two varables, and Q, and one equaton. But we had ths problem before n our analyss of the RC crcut, and solved t by notcng that the charge Q on the capactor s related to the current flowng nto the capactor by = dq dt (17) If we dfferentate Equaton (16) once wth respect to tme to get d 2 dt 2 + 1 dq LC dt L d dt =0 L C Q C Fgure 11 The LC crcut. Ths s the same as the LR crcut of Fgure (8), ecept that the resstor has been replaced by a capactor. Fnally use Equaton (17) = dq/dt and we get the second order dfferental equaton d 2 dt 2 + 1 =0 (18) LC The fact that we get a second order dfferental equaton (wth a second dervatve of ) nstead of the frst order dfferental equatons we got for LR and RC crcuts, shows that we have a very dfferent knd of problem. If we try an eponental decay n Equaton (18), t wll not work. Eercse 3 Try the soluton = 0 e αt n Equaton (18) and see what goes wrong. We have prevously seen a second order dfferental equaton n just the form of Equaton (18) n our dscusson of smple harmonc moton. We epect a snusodal soluton of the form = 0 sn ωt (19) In order to try ths guess, Equaton (19), we dfferentate twce to get d dt = ω 0cos ωt d 2 dt 2 = ω2 0 sn ωt (20) and substtute Equaton (20) nto (18) to get ω 2 0 sn ωt+ 0 sn ωt =0 (21) LC The quantty 0 sn ωt cancels from Equaton (21) and we get ω 2 = 1 LC ; ω = 1 LC (22) We see that an oscllatng current s a soluton to Krchoff s law, and that the frequency ω of oscllaton s determned by the values of L and C.
31-11 Eercse 4 In Fgure (12) we have an LC crcut consstng of a torodal col shown n Fgure (6) (on page 31-6), and the parallel plate capactor made of two alumnum plates wth small glass spacers. The voltage n Fgure (12c) s oscllatng at the natural frequency of the crcut. a) What s the capactance of the capactor? b) The alumnum plates have a radus of 11 cm. Assumng that we can use the parallel plate capactor formula C = ε 0 A C d where A C s the area of the plates, estmate the thckness d of the glass spacers used n ths eperment. (The measured value was 1.56 mllmeters. You should get an answer closer to 1 mm. Errors could arse from frngng felds, effect of the glass, and non-unformty of the surface of the plates.) L a) The LC crcut C b) Inductor and capactor used n the eperment c) Oscllatng voltage at the resonant frequency. Fgure 12 Oscllatng current n an LC crcut consstng of the torodal nductor of Fg. 6 and a parallel plate capactor. We wll dscuss shortly how we got the current oscllatng and measured the voltage.
31-12 Inductors and Magnetc Moment Intutve Pcture of the LC Oscllaton The rather strkng behavor of the LC crcut deserves an attempt at an ntutve eplanaton. The key to understandng why the current oscllates les n understandng the behavor of the nductor. As we have mentoned, the voltage rse on an nductor s always n a drecton to oppose a change n current. The closest analogy s the concept of nerta. If you have a massve object, a large force s requred to accelerate t. But once you have the massve object movng, a large force s requred to stop t. An nductor effectvely supples nerta to the current flowng through t. If you have a large nductor, a lot of work s requred to get the current started. But once the current s establshed, a lot of effort s requred to stop t. In our LR crcut of Fgure (10), once we got a current gong through the nductor L, the current contnued to flow, even though there was no battery n the crcut, because of the nerta suppled to the current by the nductor. Let us now see why an LC crcut oscllates. One cycle of an oscllaton s shown n Fgure (13) where we begn n (a) wth a current flowng up through the nductor and over to the capactor. The capactor already has some postve charge on the upper plate and the current s supplyng more. The capactor voltage V C s opposng the flow of the current, but the nerta suppled to the current by the nductor keeps the current flowng. In the net stage, (b), so much charge has bult up n the capactor, V C has become so large, that the current stops flowng. Now we have a charged up capactor whch n (c) begns to dscharge. The current starts to flow back down through tn nductor. The current contnues to flow out of the capactor untl we reach (d) where the capactor s fnally dscharged. The mportant pont n (d) s that, although the capactor s empty, we stll have a current and the nductor gves the current nerta. The current wll contnue to flow even though t s no longer beng pushed by the capactor. Now n (e), the contnung current starts to charge the capactor up the other way. The capactor voltage s tryng to slow the current down but the nductor voltage keeps t gong. Fnally, n (f), enough postve charge has bult up on the bottom of the capactor to stop the flow of the current. In (g) the current reverses and the capactor begns to dscharge. The nductor supples the nerta to keep ths reversed current gong untl the capactor s charged the other way n (). But ths s the same pcture as (a), and the cycle begns agan. Ths ntutve pcture allows you to make a rough estmate of how the frequency of the oscllaton should depend upon the sze of the nductance L and capactance C. If the nductance L s large, the current has more nerta, t wll charge up the capactor more, and should take longer. If the capactance C s larger, t should take longer to fll up. In other words, the perod should be longer, the frequency ω lower, f ether L or C are ncreased. Ths s consstent wth the result ω = 1/ LC we saw n Equaton (22). Before leavng Fgure (13) go back over the ndvdual sketches and check two thngs. Frst, verfy that Krchoff s law works for each stage;.e., that the sum of the voltage rses around the crcut s zero for each stage. Then note that whenever there s a voltage V L on the nductor, the drecton of V L always opposes the change n current. a) f) V L V L = 0 b) g) c) V + + L V C h) d) L C ) e) V L d dt = 0; Q = 0 + + V C V L + + V C + + V C V L + + V C + + V C Fgure 13 The varous stages n the oscllaton of the electrc charge n an LC crcut. L = 0 d dt = 0; Q = 0 V + + L C V C
31-13 The LC Crcut Eperment The oscllaton of the LC crcut n Fgure (13) s a resonance phenomena and the frequency ω 0 =1/ LC s the resonance frequency of the crcut. If we drve the crcut, force the current to oscllate, t wll do so at any frequency, but the response s bggest when we drve the crcut at the resonant frequency ω 0. There turns out to be a very close analogy between the LC crcut and a mass hangng on a sprng as shown n Fgure (14). The ampltude of the current n the crcut s analogous to the ampltude of the moton of the mass. If we oscllate the upper end of the sprng at a low frequency ω much less than the resonant frequency ω 0, the mass just moves up and down wth our hand. If ω s much hgher than ω 0, the mass vbrates at a small ampltude and ts moton s out of phase wth the moton of our hand. I.e., when our hand comes down, the sprng comes up, and vce versa. But when we oscllate our hand at the resonant frequency, the ampltude of vbraton ncreases untl ether the mass jumps ω off the sprng or some form of dampenng or energy loss comes nto play. It s clear from Fgure (14) how to drve the moton of a mass on a sprng; just oscllate our hand up and down. But how do we drve the LC crcut? It turns out that for the parallel plate capactor and ar core torodal nductor we are usng, the resonance s so delcate that f we nsert somethng nto the crcut to drve t, we kll the resonance. We need a way to drve t from the outsde, and an effectve way to do that s shown n Fgure (15). scope oscllator = sn ω t 0 L Run a wre from an oscllator around the col and back to the oscllator. Do the same for the scope. C m Fgure 14 To get the mass on the end of a sprng oscllatng at some frequency ω you move your hand up and down at the frequency ω. If ω s the resonant frequency of the mass and sprng system, the oscllatons become qute large. B magnetc feld created by the current from the oscllator Fgure 15 Drvng the LC crcut. The turns of wre from the oscllator produce an oscllatng magnetc feld nsde the col. Ths n turn produces an electrc feld at the col wres whch also oscllates and drves the current n the col. (A second wre wrapped around the col s used to detect the voltage. The alternatng magnetc feld n the col produces a voltage n the scope wre.)
31-14 Inductors and Magnetc Moment In that fgure we have taken a wre lead, wrapped t around the torodal col a couple of tmes, and plugged the ends nto an oscllator as shown. (Some oscllators mght not behave very well f you short them out ths way. You may have to nclude a seres resstor wth the wre that goes to the oscllator.) When we turn on the oscllator, we get a current osc = 0 snωt n the wre, where the frequency ω s determned by the oscllator settng. The mportant part of ths setup s shown n Fgure (15b) where the wre lead wraps a few tmes around the torod. Snce the wre lead tself forms a small col and snce t carres a current osc, t wll create a magnetc feld B osc as shown. Part of the feld B osc wll le nsde the torod and create magnetc flu Φ osc down the torod. Snce the current producng B osc s oscllatng at a frequency ω, the feld and the flu wll also oscllate at the same frequency. As a result we have an oscllatng magnetc flu n the torod, whch by Faraday s law creates an electrc feld of magntude E d = dφ osc dt around the turns of the solenod. Ths electrc feld nduces a voltage n the torod 0.75 ampltude n volts ampltude peaks at ω = 0 3.6 10 6 rad/sec whch drves the current n the LC crcut. We can change the drvng frequency smply by adjustng the oscllator. To detect the oscllatng current n the col, we wrap another wre around the col, and plug that nto an osclloscope. The changng magnetc flu n the col nduces a voltage n the wre, a voltage that s detected by the scope. In Fgure (16a) we carred out the eperment shown n Fgure (15), and recorded the ampltude V C of the capactor voltage as we changed the frequency ω on the oscllator. We see that the ampltude s very small untl we get to a narrow band of frequences centered on ω 0, n what s a typcal resonance curve. The heght of the peak at ω = ω 0 s lmted by resdual resstance n the LC crcut. Theory predcts that f there were no resstance, the ampltude at ω = ω 0 would go to nfnty, but the wres n the torod would melt frst. In general, however, the less resstance n the crcut, the narrower the peak n Fgure (16a), and the sharper the resonance. N turns torrodal radus R T capactor plate area A c R T 0.50 d 0.25 cross-sectonal area A of the turns T capactor plate separaton 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 10 6 ω resonant 0 radans/sec frequency Fgure 16a As we tune the oscllator frequency through the resonant frequency, the ampltude of the LC voltage goes through a peak. Fgure 16b The LC apparatus.
31-15 MEASURING THE SPEED OF LIGHT The man reason we have focused on the LC resonance eperment shown n Fgure (15) s that ths apparatus can be used to measure the speed of lght. We wll frst show how, and then dscuss the phlosophcal mplcatons of such a measurement. The calculaton s straghtforward but a bt messy. We start wth Equaton (22) for the resonance frequency ω 0 ω0 = 1 LC (22) and then use Equaton (8) for the nductance L of a solenod L=µ 0 N 2 A/h (8) and Equaton (27-32) for the capactance of a parallel plate capactor C = ε 0 A C /d (27-32) For the apparatus shown n Fgure (16b), the length of the torodal solenod s h =2πR T, and the crosssectonal area s A =A T, so that Equaton (8) becomes L torod = µ 0 N 2 A T /2πR T (23) For the capactor, A C s the area of the plates, d ther separaton, and we can use Equaton (27-32) as t stands. If we square Equaton (22) to remove the square root ω 2 0 = 1 LC (22a) and use Equatons (23) and (27-32) for L and C, we get ω 0 2 = = 2πR T µ 0 N 2 A T d ε 0 A C 1 2πR T d (24) µ 0 ε 0 N 2 A T A C The mportant pont s that the product µ 0 ε 0 appears n Equaton (24), and we can solve for 1/µ 0 ε 0 to get Fnally, recall n our early dscusson of magnetsm, that µ 0 ε 0 was related to the speed of lght c by c 2 1 = µ 0 ε (27-18) 0 Usng Equaton (25) n (27-18), and takng the square root gves c= 1 µ 0 ε 0 = ω 0 N A T A C 2πR T d (26) Eercse 5 Show that c n Equaton (26) has the dmensons of a velocty. (Radans are really dmensonless.) At frst sght Equaton (26) appears comple. But look at the quanttes nvolved. ω 0 = the measured resonant frequency N = the number of turns n the solenod A T = cross sectonal area of the torod A C = area of capactor plates R T = radus of torod d = separaton of capactor plates Although t s a lot of stuff, everythng can be counted, measured wth a ruler, or n the case of ω 0, determned from the osclloscope trace. And the result s the speed of lght c. We have determned the speed of lght from a table top eperment that does not nvolve lght. Eercse 6 The resonant curve n Fgure (16a) was measured usng the apparatus shown n Fgure (16b). For an nductor, we used the torod descrbed n Fgure (6). The parallel plates have a radus of 11 cm, and a separaton d = 1.56mm. Use the epermental results of Fgure (16a), along wth the measured parameters of the torod and parallel plates to predct the speed of lght. (The result s about 20% low due to problems determnng the capactance, as we dscussed n Eercse 3.) 1 µ 0 ε 0 = ω 0 2 N 2 A T A C 2πR T d (25)
31-16 Inductors and Magnetc Moment In our ntal dscusson of the specal theory of relatvty n Chapter 1, we ponted out that accordng to Mawell s theory of lght, the speed of lght c could be predcted from a table top eperment that dd not nvolve lght. Ths theory, developed n 1860, predcted that lght should travel at a speed c = 1/ µ 0 ε 0, and Mawell knew that the product µ 0 ε 0 could be determned from an eperment lke the one we just descrbed. (Dfferent notaton was used n 1860, but the deas were the same.) Ths rased the fundamental queston: f you went out and actually measured the speed of a pulse of lght as t passed by, would you get the predcted answer 1/ µ 0 ε 0? If you dd, that would be evdence that you were at rest. If you dd not, then you could use the dfference between the observed speed of the pulse and 1/ µ 0 ε 0 as a measurement of your speed through space. Ths was the bass for the seres of eperments performed by Mchaelson and Morley to detect the moton of the earth. It was the bass for the rather frm convcton durng the last half of the 19th century that the prncple of relatvty was wrong. It was not untl 1905 that Ensten resolved the problem by assumng that anyone who measured the speed of a pulse of lght movng past them would get the answer c = 1/ µ 0 ε 0 = 3 10 8 m/s, no matter how they were movng. And f everyone always got the same answer for c, then a measurement of the speed of lght could not be used as a way of detectng one s own moton and volatng the prncple of relatvty. The mportance of the LC resonance eperment, of the determnaton of the speed of lght wthout lookng at lght, s that t focuses attenton on the fundamental questons that lead to Ensten s specal theory of relatvty. In the net chapter we wll dscuss Mawell s equatons whch are the grand fnale of electrcty theory. It was the soluton of these equatons that led Mawell to hs theory of lght and all the nterestng problems that were rased concernng the prncple of relatvty. Eercse 7 In Fgure (17a) we have an LRC seres crcut drven by a snusodal oscllator at a frequency ω radans/sec. The voltage V R s gven by the equaton V R =V R0 cos (ωt) as shown n the upper sketch of Fgure (17b). Knowng V R, fnd the formulas and sketch the voltages for V L and V C. Determne the formulas for the ampltudes V L0 and V C0 n terms of V R0 and ω. Fgure 17a An LRC crcut drven by a snusodal oscllator. The voltage V R across the resstor s shown n Fgure (17b). Fgure 17b Knowng V R, fnd the formulas and sketch the voltages for V L and V C. V R V L V C C R V R= V R0cos(ωt) V = V... L L0 L V V R L V C V = V... C C0
The second half of ths chapter whch dscusses the concept of magnetc moment, provdes addtonal laboratory orented applcatons of Faraday s law and the Lorentz force law. Ths topc contans essental background materal for our later dscusson of the behavor of atoms and elementary partcles n a magnetc feld, but s not requred for the dscusson of Mawell s equatons n the net chapter. You may wsh to read through the magnetc moment dscusson to get the general dea now, and worry about the detals when you need them later. 31-17
31-18 Inductors and Magnetc Moment MAGNETIC MOMENT We wll see, usng the Lorentz force law, that when a current loop (a loop of wre wth a current flowng n t) s placed n a magnetc feld, the feld can eert a torque on the loop. Ths has an mmedate practcal applcaton n the desgn of electrc motors. But t also has an mpact on an atomc scale. For eample, ron atoms act lke current loops that can be algned by a magnetc feld. Ths algnment tself produces a magnetc feld and helps eplan the magnetc propertes of ron. On a stll smaller scale elementary partcles lke the electron, proton, and neutron behave somewhat lke a current loop n that a magnetc feld can eert a torque on them. The phenomena related to ths torque, although occurrng on a subatomc scale, are surprsngly well descrbed by the so called classcal theory we wll dscuss here. Magnetc Force on a Current Before we consder a current loop, we wll begn wth a dervaton of the force eerted by a straght wre carryng a current as shown n Fgure (18a). In that fgure we have a postve current flowng to the rght and a unform magnetc feld B drected down nto the paper. In order to calculate the force eerted by B on, we wll use our model of a current as consstng of rods of charge movng past each other as shown n Fgure (18b). The rods have equal and opposte charge denstes Q, and the postve rod s movng at a speed v to represent a postve current. The current s the amount of charge per second carred past any crosssectonal area of the wre. Ths s the amount of charge per meter, Q, tmes the number of meters per second, v, passng the cross-sectonal area. Thus = Q coulombs meter = Q v coulombs second v meter second (27) In Fgure (18b) we see that the downward magnetc feld B acts on the movng postve charges to produce a force F Q of magntude F Q =Qv B =QvB whch ponts toward the top of the page. The force f on a unt length of the wre s equal to the force on one charge Q tmes the number of charges per unt length, whch s 1. Thus a) current n a magnetc feld B (down) force on a unt length of wre f =F Q 1 = QvB (28) F B F B + + + + + + + Fgure 18 Magnetc force on the movng charges when a current s placed n a magnetc feld. b) model of the neutral current as two rods of charge, the postve rod movng n the drecton of the current v B (down) B (down) Fgure 19 Sdeways magnetc force on a current n a magnetc feld. The force per unt length f s related to the charge per unt length λ by f = λv B. Snce λv s the current, we get f = B.
31-19 Usng Equaton (27) to replace Qv/ by the current n (28), and notng that f ponts n the drecton of B, as shown n Fgure (19), we get f = B force per unt length eerted by a magnetc feldb on a current (29) where s a vector of magntude pontng n the drecton of the postve current. Eample 2 Calculate the magnetc force between two straght parallel wres separated by a dstance r, carryng parallel postve currents 1 and 2 as shown n Fgure (21). Soluton The current 1 produces a magnetc feld B 1, whch acts on 2 as shown n Fgure (20) (and vce versa). Snce B 1 s the feld of a straght wre, t has a magntude gven by Equaton (28-18) as B 1 = µ 0 1 2πr (29-18) The resultng force per unt length on B 2 s f = 2 B 1 whch s drected n toward 1 and has a magntude f = 2 B 1 = µ 0 1 2 2πr (30) Equaton (30) s used n the MKS defnton of the ampere and the coulomb. In 1946 the followng defnton of the ampere was adopted: The ampere s the constant current whch, f mantaned n two straght parallel conductors of nfnte length, of neglgble crcular cross secton, and placed 1 meter apart n a vacuum, would produce on each of these conductors a force equal to 2 10 7 newtons per meter of length. Applyng ths defnton to Equaton (30), we set 1 = 2 =1 to represent one ampere currents, r = 1 to represent the one meter separaton, and f = 2 10 7 as the force per meter of length. We get 2 10 7 = µ 0 2π r 1 From ths we see that µ 0 s now a defned constant wth the eact value a) top vew 2 µ 0 =4π 10 7 by defnton (31) 1 (up) F B B 1 2 (up) b) end vew showng the magnetc feld of current 1 eertng a magnetc force on current 2 The force between parallel currents s attractve. Wth the above defnton of the ampere, the coulomb s offcally defned by the amount of charge carred by a one ampere current, per second, past a cross-sectonal area of a wre. Lookng back over our dervaton of the formula f = B, and then the above MKS defntons, we see that t s the magnetc force law F =Qv B whch now underles the offcal defntons of charge and current. Fgure 20 Force between two currents.
31-20 Inductors and Magnetc Moment Torque on a Current Loop In an easly performed eperment, we place a square loop of wre of sdes ( ) and (w) as shown n Fgure (21a), nto a unform magnetc feld as shown n (21b). The loop s allowed to rotate around the as and s now orentated at an angle θ as seen n (21c). If we now turn on a current, we get an upward magnetc force proportonal to B n the secton from pont (1) to pont (2), and a downward magnetc force proportonal to B n the secton from pont (3) to pont (4). These two forces eert a torque about the as of the loop, a torque that s tryng to ncrease the angle θ. (Ths torque s what turns the armature of an electrc motor.) Followng our earler rght hand conventons, we wll defne the area a of the loop as a vector whose magntude s the area (a = w) of the loop, and whose drecton s gven by a rght hand rule for the current n the loop. Curl the fngers of your rght hand n the drecton of the postve current and your thumb ponts n the drecton of a as shown n Fgure (22). Wth ths conventon, the loop area A ponts toward the upper left part of the page n Fgure (21b) as shown. And we see that the torque caused by the magnetc forces, s tryng to orent the loop so that the loop area a s parallel to B. Ths s a key result we wll use often. To calculate the magntude of the magnetc torque, we note that the magntude of the force on sde (1)-(2) or sde (3)-(4) s the force per unt length f = B tmes the length of the sde F 1,2 = F 3,4 = f =B When the loop s orentated at an angle θ as shown, then the lever arm for these forces s lever arm = w 2 sn θ Snce both forces are tryng to turn the same way, the total torque s twce the torque produced by one force, and we have torque = 2 w 2 snθ B 2 tmes leverarm tmes force w as end of loop torque = B w sn θ (32) a) a wre loop carryng a current, free to turn on the as end vew of loop F = B 1,2 w (3) F = B 1,2 (2) as (1) B θ F 1,2 (4) B F3,4 F 3,4 = B F 3,4 ( w sn θ) 2 w = wdth of loop b) magnetc force actng on horzontal loop Fgure 21 Analyss of the forces on a current loop n a magnetc feld. c) magnetc force actng on tlted loop
31-21 The fnal step s to convert Equaton (32) nto a vector equaton. Frst recall that the vector torque τ s defned as τ = r F where r and F 1,2 are shown n Fgure (23). In the fgure we see that r F and therefore τ ponts up out of the paper. Net we note that n Equaton (32), θ s the angle between the magnetc feld B and the loop area a. In addton, the vector cross product a B has a magntude a B = absn θ = w Bsn θ and ponts up, n the same drecton as the torque τ. Thus Equaton (32) mmedately converts to the vector equaton τ = a B (33) where s the current n the loop, and a s the vector area defned by the rght hand conventon of Fgure (22). a Magnetc Moment When you put a current loop n a magnetc feld, there s no net force on the loop ( F 1,2 =-F 3,4 n Fgure 19b), but we do get a torque. Thus magnetc felds do not accelerate current loops, but they do turn them. In the study of the behavor of current loops, t s the torque that s mportant, and the torque s gven by the smple formula of Equaton (33). Ths result can be wrtten n an even more compact form f we defne the magnetc moment µ of a current loop as the current tmes the vector area a of the loop µ a defnton of magnetc moment (34) Wth ths defnton, the formula for the torque on a current loop reduces to τ = µ B (35) Although we derved Equatons (34) and (35) for a square loop, they also apply to other shapes such as round loops. F 1,2 end vew of loop r Fgure 22 Rght hand conventon for the loop area A. τ 1,2 = r F 1,2 τ 1,2 ponts up, out of the paper Fgure 23 The torque τ 1,2 eerted by the force F 1,2 actng on the sde of the current loop. The vector r s the lever arm of F 1,2 about the as of the col. You can see that r F 1,2 ponts up out of the paper.
31-22 Inductors and Magnetc Moment Magnetc Energy In Fgure (24) we start wth a current loop wth ts magnetc moment µ algned wth the magnetc feld as shown n (24a). We saw n Fgure (21b) that ths s the orentaton towards whch the magnetc force s tryng to turn the loop. (a) µ θ = 0 If we grab the loop and rotate t around as shown n (24b) untl µ s fnally orentated opposte B as n (24c) we have to do work on the loop. We can calculate the amount of work we do rotatng the loop from an angle θ = 0 to θ = π usng the angular analogy for the formula for work. The lnear formula for work s (b) B τ = µ B 2 W= F d (10-19) θ µ τ = µ B sn θ 1 Replacng the lnear force F by the angular force τ, and the lnear dstances d, 1,and 2 by the angular dstances dθ, 0 and π, we get B W= τdθ 0 π (36) If we let go of the loop, the magnetc force wll try to reorent the loop back n the θ = 0 poston shown n Fgure (24a). We can thnk of the loop as fallng back down to the θ = 0 poston releasng all the energy we stored n t by the work we dd. In the θ = π poston of Fgure (24c) the current loop has a potental energy equal to the work we dd n rotatng the loop from θ = 0 to θ = π. (c) B µ θ = π Fgure 24 The restng, low energy poston of a current loop s wth µ parallel to B as shown n (a). To turn the loop the other way, we have to do work aganst the restorng torque τ = µ B as shown n (b). The total work we do to get the loop nto ts hgh energy poston (c) s 2µB. We can thnk of ths as magnetc potental energy that would be released f we let the loop flp back down agan. We choose the zero of ths potental energy half way between so that the magnetc potental energy ranges from +µb n (c) to µb n (a).
31-23 We can calculate the magnetc potental energy by evaluatng the ntegral n Equaton (36). From Equaton (35) we have so that τ = µ B = µbsn θ W = 0 π µbsn θdθ = µbcos θ π 0 =2µB (36a) Thus the current loop n the θ = π poston of (24c) has an energy 2µΒ greater than the energy n the θ = 0 poston of (24a). It s very reasonable to defne the zero of magnetc potental energy for the poston θ = π/2, half way between the low and hgh energy postons. Then the magnetc potental energy s + µb n the hgh energy poston and µb n the low energy poston. We mmedately guess that a more general formula for magnetc potental energy of the current loop s magnetc potental energy of a current loop E mag = µ B (37) Ths gves E mag = +µb when the loop s n the hgh energy poston wth µ opposte B, and E mag = - µb n the low energy poston where µ and B are parallel. At an arbtrary angle θ, Equaton (37) gves E mag = - µbcos θ, a result you can obtan from equaton (36a) f you ntegrate from θ = 0 to θ = θ, and adjust the zero of potental energy to be at θ = π/2.
31-24 Inductors and Magnetc Moment Summary of Magnetc Moment Equatons Snce we wll be usng the magnetc moment equatons n later dscussons, t wll be convenent to summarze them n one place. They are a short set of surprsngly compact equatons. A = area of current loop µ A (34) τ = µ B B (35) E mag = µ B (37) Charge q n a Crcular Orbt Most applcatons of the concept of magnetc moment are to atoms and elementary partcles. In the case of atoms, we can often pcture the magnetc moment as resultng from an electron travelng n a crcular orbt lke that shown n Fgure (25). In that fgure we show a charge q travelng at a speed v n a crcular orbt of radus r. Snce charge s beng carred around ths loop, ths s a current loop, where the current s the amount of charge per second beng carred past a pont on the orbt. In one second the charge q goes around v/2πr tmes, therefore =qcoulombs =q v 2πr v meter / sec 2πr meter v q A Snce the area of the loop s πr 2, we get as the formula for the magnetc moment µ =A=q v 2πr * πr2 µ = qvr 2 magnetc momentof a charge q travelng at a speed v n a crcularorbt of radus r (38) We can make a further refnement of Equaton (38) by notng that the angular momentum J (we have already used L for nductance) of a partcle of mass m travelng at a speed v n a crcle of radus r has a magntude J = mvr More mportantly, J ponts perpendcular to the plane of the orbt n a rght handed sense as shown n Fgure (26a). Ths s the same drecton as the magnetc moment µ seen n (26b), thus f we wrte Equaton (38) n the form µ = q 2m mvr (38a) and use J for mvr, we can wrte (38a) as the vector equaton µ = q 2m J relatonbetween the angular momentum J and magnetc momentµ for a partcle travelng n a crcularorbt (39) Equaton (39) s as far as we want to go n developng magnetc moment formulas usng strctly classcal physcs. We wll come back to these equatons when we study the behavor of atoms n a magnetc feld. r Fgure 25 A charged partcle n a crcular orbt acts lke a current loop. Its magnetc moment turns out to be µ = qvr/2.
31-25 (angular momentum) J J = mvr m v r (a) angular momentum of a partcle n a crcular orbt (magnetc moment) µ q µ = (mvr) 2m q v r (b) magnetc moment of a charged partcle n a crcular orbt Fgure 26 Comparng the magnetc moment µ and angular momentum J of a partcle n a crcular orbt, we see that µ = q 2m J
31-26 Inductors and Magnetc Moment IRON MAGNETS In ron and many other elements, the atoms have a net magnetc moment due to the moton of the electrons about the nucleus. The classcal pcture s a small current loop consstng of a charged partcle movng n a crcular orbt as shown prevously n Fgures (25) and (26). If a materal where the atoms have a net magnetc moment s placed n an eternal magnetc feld B et the torque eerted by the magnetc feld tends to lne up the magnetc moments parallel to B et as llustrated n Fgure (27). Ths pcture, where we show all the atomc magnetc moments algned wth B et s an eaggeraton. In most cases the thermal moton of the atoms serously dsrupts the algnment. Only at temperatures of the order of one degree above absolute zero and n eternal felds of the order of one tesla do we get a nearly complete algnment. Iron and a few other elements are an ecepton. A small eternal feld, of the order of 10 gauss (.001 tesla) or less, can algn the magnetc moments at room temperature. Ths happens because neghborng atoms nteract wth each other to preserve the algnment n an effect called ferromagnetsm. The theory of how ths nteracton takes place, and why t suddenly dsappears at a certan temperature (at 1043 K for ron) has been and stll s one of the challengng problems of theoretcal physcs. (The problem was solved by Lars Onsager for a two-dmensonal array of ron atoms, but no one has yet succeeded n workng out the theory for a threedmensonal array.) The behavor of ron or other ferromagnetc materals depends very much on how the substance was physcally prepared,.e., on how t was cooled from the molten mture, what mpurtes are present, etc. In one etreme, t takes a farly strong eternal feld to algn the magnetc moments, but once algned they stay there. Ths preparaton, called magnetcally hard ron, s used for permanent magnets. In the other etreme, a small eternal feld of a few gauss causes a major algnment whch dsappears when the eternal feld s removed. Ths preparaton called magnetcally soft ron s used for electromagnets. Our purpose n ths dscusson of ron magnets s not to go over the detals of how magnetc moments are algned, what keeps them algned or what dsrupts the algnment. We wll consder only the more fundamental queston what s the effect of lnng up the magnetc moments n a sample of matter. What happens f we lne them all up as shown n Fgure (27)? A current loop has ts own magnetc feld whch we saw n our orgnal dscusson of magnetc feld patterns and whch we have reproduced here n Fgure (28). Ths s a farly comple feld shape. (Out from the loop at dstances of several loop rad, the feld has the shape of what s called a dpole magnetc feld. In certan regons earth s magnetc feld has ths dpole magnetc feld shape.) B et B et Fgure 27 Ferromagnetsm. When you apply an eternal magnetc feld to a pece of "magnetcally soft" ron (lke a nal), the eternal magnetc feld algns all of the magnetc moments of the ron atoms nsde the ron. The magnetc feld of the current loops can be enormous compared to the eternal feld lnng them up. As a result a small eternal feld produced say by a col of wre, can create a strong feld n the ron and we have an electromagnet. Ths phenomena s called ferromagnetsm.