Electrical Motors and Generators

7/28/2019 Electrical Motors and Generators

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" A t a n i n d u s t r i a l e x h i b it i o n in Vi e n n a , i n 1 8 7 3 , a n u m b e r o f G r a m m e m a c h i n e[dynamos , o r gene rato r s] w ere be ing p laced in pos i ti on . . . . I n m ak i ng the e l ect ri ca l connec t ions to one o f these ma ch ines wh ich ha d no t a s ye t been be l ted to the eng ine -sha[d r iven by a s t eam-eng ine ] , a ca rel e ss wor km an a t t ach ed to i t by mis t a ke a pa i r o f w i rew h i c h w e r e a l re a d y c o n n e c te d w i th a n o t h e r d y n a m o - m a c h i n e w h i c h w a s i n r a p i d m o t i oTo th e a m a z e m e n t o f t h is w o r t h y a r t is a n t h e se c o n d m a c h i n e c o m m e n c e d t o r ev o lv e wgreat rapidi ty in a reverse d i rec t ion . . . . "

" G r a m m e . . . a t o nc e p e rc e iv e dthat the s e c o n d m a c h i n e w a s p e r f o r m i n g t h e f u n c t i o n

o f a m o t o r, a n d t h a t w h a t w a s t a k i n g p l a c e w a s a n a c t u a l t r a ns fe r en c e o f m e c h a n i c ap o w e r t h ro u g h t h e m e d i u m o f e l e c t ri c i ty. .. . [ U p t o t h a ttime]a lm os t t he on ly p rac t i ca l u seto wh ich the e l ec t ri c mo to r had been app l i ed was in the ope ra t ion o f den ta l a ppa ra tus . . . .

~F. L. Pope,Past President of the Am erican Ins titute of Electrical Engineers,Electricity n Daily Life(1890)

C h a p t e r 1 3

Mechanical Impl ica t ionsof Faraday ' s Law: Motorsa n d G e n e r a t o r s

Chapte r Overv iew

The previous chapter repeatedly considered the steady-state motion of a loop pulledinto a region of m agnet ic fie ld , f inding that such mo tion produces in the loop an emfand an electric cu rrent. S ection 13.1 discusses ho w such emfs and currents becam e apart of m ode rn civilization's daily life. Section 13.2 discusses the break through s tha tal lowed motor and generator eff ic iency to become suff ic ient ly high that e lectromag-netic induc tion could be taken seriously as a practical so urce of p ower, rather thanas a me re novel ty. Sect ion 1 3.3 presents some general considerations abou tm o t o r s(which convert electrical energy into mechanical energy) and aboutgenera to rs (whichconvert mechanical energy into electrical energy). Both of these are calledelectricmachines . To be specific, Section 1 3.3 introduces a simple m odel of al inear machine,wh ich is the subject of analysis throu gh Sec tion 13.7. The linear machine, w hich canserve either as a mo tor o r as a generator, is an exam ple of the use and gene ration of

dc ~d irec t current~e lectr ic i ty. (By the use of mode rn pow er electronics i t is possible todrive ac~al ternat ing current~motors with dc electr ic i ty, and to dr ive dc motors withac electricity.) Sec tion 13.4 derives the equa tions th at describe the electrical behav ior(Ohm's law) and the mechanical behavior (Newton's law of motion) for this system.Section 13.5 solves these equa tions for the initial and the steady-state response. Forthe linear generator aback force develops (for a rotational generator aback to rquedevelops). For a m oto r ab a ck e m fdevelops. Se ction 13.6 discusses the different typesof m echanical and electrical load s to wh ich a moto r or gene rator can be subjected.The discussion of motors and generators c loses with Sect ion 13 .7, which considers

5 5 9


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5 6 0 Chapter 13 ~ Motors and Generators

th e transients tha t occur w he n a c i rcu i t is f i r s t tu rne d on or off . Sec t ion 13 .8 cons idersa s i t ua t i on whe re eddy cu r r en t s , s o unwa n te d i n t h e ca se o f mo t o r s an d g ene r a to r s ,a r e u se fu l - -m agn e t i c l ev i ta t i on a n d m agne t i c d r ag . m

13.1~nt ional

H o w E l e c t r i c i t y B e c a m e P a r t o f D a i l y L i f e

The poss ib i li ti es for us ing the mot ive pow er of e lec t r ic i ty (hencemotor)we r erea l ized ear ly on . In 1821 , Faraday deve loped two s imple motors us ing a ma gnet

and a cur ren t -car ry ing wire in apool of mercury. (Mercu ry i s agood electr ical conductor; herei t provides a path for electr ic cur-rent . ) See Figure 13.1. Electr icalenergy f rom a source of em f (no tshown) i s used to sus ta in mo-t ion (against viscous drag) in thetwo m otors . O n the lef t, a mag-ne t in a ba th of l iqu id mercuryturns about th e ax is o f the e lec-tr ic current , which is vert ical .In o ther words , the N magnet icpole tu rns in the d i rec t ion ofthe magnetic f ield. On the r ight ,

Figure 13.1 Faraday's two types of motors. In the wir e tu rns ab ou t th e axisthe motor on the left, the magnet moves about of the m agn et, wh ic h is also ver-the axis of the current. In the motor on the tical. (You are inv ited to verifyright, the current moves about the axis of the tha t the sen se of rot atio n is cor-magnet , rect in each case.) Des pite their

common cu r r en t , t h e se mo to r sare independent o f each o ther.

Bui lding on Bar low's 1826 invent ion of the e le c t ro m agn et~ Ar ago ' s ear l iermagn et iz ing of a magn et ic needle wi th a he l ica l wi re d id no t inc lude the idea ofa n e l ec t r om a gn e t~ He nry improved t he l i ft ing s t r eng th f rom 9 to 2300 pounds .Henry then used such an e lec t romagnet to bu i ld a motor in 1831 . These motors

It has been surmised that Faraday developed mercury poisoning from his use of mercury.

Even in 1821, at the age of 30, Faraday complained abou t his fading mem ory. In 1828,he remarked upon "nervous hea daches and weakness." From 1839 to 1844, he wasplagued by almost co nstant and severe headaches. Somehow, he managed to do hisstudies with wh at w e n ow call the Faraday ice-pail during 1843. Despite the headaches,he remained in robust physical health, hiking s om e 30 miles a day even in his mid-5Os.By 1845, the giddiness and malaise were gone, but he continued to suffer from aperm anent decrease in his powers of mem ory. By 1855, he was nearly unable to performresearch, and in 1861 he had to disco ntinue his famous Christmas Lectures for children.He died in 1867.


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13.1 H ow Electricity Became P art of Da ily Life561

V

t ~

V,.,

t

(a) Co)

Figure 13 .2 (a) O scillating voltage. (b) Rectified oscillating voltage.

required that the current be produced by a battery, which was very expensive(zinc-copper batteries we re used, in whic h the zinc was consumed): in the 1840s,Joule showed that , for the same am oun t of energy productio n, zinc cost 120 t imesas much as coal.

W ith Faraday's 1831 discovery of electromag netic induction, i t beca me pos-sible to think in terms of an al ternative source of electr ical power for motors.In what may have been the f irst use of the com mu tato r ( i.e. , the sl ip r ing),at Arnpere 's suggestion, Pixii in 1832 rectif ied current produced by Faraday'slaw: a handle turned a U-shaped magnet , whose poles passed beneath coppersolenoids f il led with iron core. The cu rrent in duced in the solenoids wen t thro ugha comm uta tor and was rectif ied. Figure 13.2(a) shows the unrectif ied signal , andFigure 13.2(b) shows the rectified signal. These early generators of electricity,driven by mechanical power, were calleddynamos .

Two major forces driving the use of electr ic power were public l ightingand electr ic trains, both of which used batteries for their electr ici ty. In 1808,Hum phr ey Davy discovered the carbon arc lamp (dr iven by a la rge ba t tery of

voltaic cells), which became used in public facilities to provide lighting at night.Around 1851, i t was found that electr ic power could be transmitted effectivelythrough train rai ls so that a locomotive using an electr ic motor did not have tocarry its own batteries.

The first practical electr ical generator of the Pixii type was used in 1858to power a carbon arc lamp in a l ighthouse. (Presumably, the generator wasdriven by a coal-powered steam engine.) As indicated in the quotation at thechapter head, by 1873 there was considerable interest in electr ical generators,primarily for arc l ighting, but i t seems to have been forgotten th at the electr ici tythey generated could also power motors, unti l the fortuitous rediscovery by a"worthy art isan." (As early as 1842, i t was realized that motors and generatorswere inverse to one another.)

Nevertheless, the generators were not very good. An 1876 study for theFrankl in Ins t i tu te , by then h igh school teacher El ihu Thomson~la ter inventorof the w at tme ter and of the jumping r ing tha t wi l l be d iscussed in the nextchap ter- -sh owe d tha t the Brush company 's genera tor, a t only 38%, was the m ostefficient then available. The low efficiency was largely due to losses within thesedevices: as discussed in the next section, unwanted eddy currents were inducedbecause of the high conduc tivity of iron, and mag netic energy losses were causedwhile cycling the iron magnetization back and forth. Once recognized, the eddycurrent problem was quickly solved, and efficiencies rose to nearly 80%. By


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562 Cha pter 13 M Motors and Generators

1878, arc lamps had begu n to appear on the streets of bot h P hiladelphia andBoston. The hysteresis problems took somewhat longer to solve, but by 1890motors and generators were operating near 90% efficiency. Modern motors andgenerators aren' t much better than this , but they are much l ighter, cheaper, andmore reliable.

In the mid-19th century, industr ial machines were driven by the turningpow er of steam engines or nearby rivers. However, by the early 1880s, i t beca mepossible for smaller companies to operate electr ical motors with electr ici ty pur-chased from a power company. Electr ici ty-powered streetcars (with overheadlines for electr ic power produced by generators driven by coal-powered steamengines) became practical: indeed, they cost only one-tenth as much as horse-drawn streetcars. There was a rapid conversion to electr ical power of streetcarsand elevated trains, with a resultant skyrocketing in the rate of equine unem-ployment .

I t was quickly realized that electr ic power could be transported over largedistances, so that power from the Niagara Falls could be sent to New York (or

power from the Columbia River could be sent to Seatt le) . Again, from the samesource as the quotation at the chapter head:

Electricity, in its important applications to machinery.., is merely a convenientand easily manageable a ge nc y.., by which m echanical power m ay be transferredfrom an ordinary prim e m otor, as a steam engine or a water w heel, to a secondarymo tor-- it may be at a great distance--whic h is employed to do the work.

Once electr ical power became more commonly available, new electr ical de-vices, such as the c arbon fi lament lamp, developed aro und 1880, made th e worlda different place. The electr ic powe r transformation, whic h inclu ded labor-savingdevices such as the washing machine, was more noticeable than even the com-pute r transformation occurring today. The following two ch apters touch upo n i tseffect in the area of communication s: signal generation, transmission, detection,and manipulat ion. The present chapter concentrates on electr ic power.

13 2 B r e a k t h r o u g h s i n E f f i c i e n c y o f M o t o r sa n d G e n e r a t o r s

I t is of interest to indicate how the inefficient pre-1870's motors and genera-tors evolved in to the eff ic ient~ albei t bulk y~an ces to rs of mode rn motors andgenerators. As indicated earl ier, there w ere tw o major advances.

1. Elimination of eddy curr ent losses. The original mo tor and generator designshad large amount s of condu ctor in their metal casing (for structural s trength)and in their i ron cores ( to confine the magnetic f ield). W he n subjected tot ime-varying magnetic f ields, these conducting materials had emfs inducedin them, by Faraday's law. Hence, by Oh m's law, currents we re induced.Such currents were not needed in the casing and the core, so the result ingJoule heating consti tuted a huge waste of energy. By the late 1870s, engi-neers were aware of this problem, and impro ved designs mini mize d i t . Themo st obvious solution was to use less solid metal in the casing, leading to a"squirrel-cage" frame. See Figure 13.3(a). Such eddy currents, following


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13.2 Breakthroughs n Efficiency563

Figure 13.3 (a) Squirrel-cage frame for mo tor casing. Being mostly hollow, theinduced currents and Joule heating losses are small compared to a solid frame.(b) Laminated transformer core. The induced currents circulate in the samedirection as the curre nt in the prima ry; the laminating gives the in ducedcurrents a high-resistance path and thus a small amplitude.

Lenz's law, circulated ab out the direction of the ma gnetic f ield. In the iron

core, by using laminations or a packe d set of mu ch thinn er iron r ings, the pathof the eddy curren ts cou ld be given a mu ch smaller cross-sectional area, thusincreasing the electr ical resistance and decreasing the rate of Joule heating.See Figure 13.3(b).

Figure 13.4 illustrates a square, before and after it has been cut in four,and the associated eddy currents d ue to ma gnetic f ield change normal to thepage. Before cutting, let the effective resistance beRo , and le t the emf due toa t ime-varying ma gnetic f ield normal to t he page be Co. The total rate o f eddycurren t heating is the n Cg /R0. Figure 13 .4(b) shows tha t the shape o f the eddycurrent path is the same after cutt ing as i t was before. Each quarter squareis subject to an emf E0/4 and has resistance R = R0. (Although R =p l / A ofChapter 7 does not hold here, i t captures the correct ideas. For each quarter

square the eddy cur rent length in the plane decreases by a factor of tworelat ive to the original length, as does the len gth in the plane associated withthe current-carrying area. However, the current-carrying direction normal tothe page does no t change. Hence , effectively, bo th 1 and A halve, so the eddycurren t resistance is the same for bot h th e big square of Figure 13.4a and thequarter squares of Figure 13.4b.) The total heating rate of al l four quartersqua res is 4C 2 R - E2/4 R0, one-four th of the original rate.

2. Hysteresis losses. In ac motor s and generators, the iron in the electrom agnetcaused unnecessary heating because the applied magnetic f ield cycled the

Eddy currents

C)C)C )0

(a) (b)

Figure1 3 . 4 (a) Eddy currents for a square. (b) E ddycurrents for a square that has been broken up intosubsquares.


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564 Chapter 13 ~ Motors and Generators

iron in a very lossy fashion, known ash y s t e r e s i s .(This word has i ts or iginsin a Greek word meaning "to lag," because the energy loss is due to themagnet iza t ion lagging behind the appl ied f ie ld . ) Cons iderab le energy waslost in demagnetizing, and i t was not recovered on remagnetizing. This effectwas f i r s t s tud ied and named by Ewing in 1881 , and much usefu l work inthis area was done by Steinmetz. Reducing hysteresis losses was a materialsproblem. In the early 1900s, grain-oriented s teel was found to be much lesslossy than ordinary i ron.

A mec hanica l e xample of hys te res i s. In o ld- fash ioned wind up c locks , en-ergy is ex t rac ted f rom a spr ing as the c lock mecha nism goes th rough a per iodof i ts motion. In addit ion to energy that is extracted by choice (useful en-ergy) , there i s energy was ted because the mec hanism has f r ic t ion (energy lossto heat) or is noisy (energy loss to sound). For example, i t would take moreforce F to s t re tch a c lock-spr ing f rom .09 m m to . 10 mm than to re lax it f rom910 m m to .09 mm . T he energy was ted dur ing a cyc le of the c lock mecha -nism would be cal led a hysteresis loss . The magnetic analogy would be that i t

would take a grea te r appl ied mag net ic f ield B to increase a magn et ' s m agne-t iza t ion M f rom 90 0 A/m to 1000 A/m tha n to decrease M f rom 1000 A/ mt o 900 A /m.

Let us now tur n to a s imple mode l tha t wi l l se rve to expla in the behavior of bo thdc motors and generators . The next chapter discusses ac motors and generators .

1 3 , 3 S i m p l e M o d e l f o r DC M o t o r a n d G e n e r a t o r - -T h e L i n e a r M a c h i n e

Consid er a circui t as in Figure 13.5, wh ere a cond ucting b ar of mass m can sl idealong a f ixed conductin g rai l a t the veloci ty v. (We p res ent a top view of thecircui t . ) A uniform magnetic f ield B points into the page. The bar 's posi t ioncan be l imited to a desirable range by the use of micro switches, whi ch causethe d i rec t ion of the cu r ren t to reverse , and thus cause the d i rec t ion of the forceon the bar to reverse. We include an external emf go, and let an external force

C O

X X

X X

x F ~ x -

T[ X . X

X X X X X X.....

X X X X X XB ~

x x x x " " - x

X X X X X X

X X - X X X X X X~ X "-" ' - --I~

Figure 13.5 Simple mode l for a dc motor and generator.A conducting rod can slide along a fixed conducting rail,to which a constant em f is attached.


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13.3 The Linear M achine 565

F0 act on the bar. For simplici ty, w e neglec t the self- inductance of the circuit .Howe ver, w e include the resis tance R of the circuit , and a fr ict ional resis tanceforce of the b ar against the rai l, which for simplici ty we assume to take theform - m f i / r f ( just as in Chap ter 7, for electrons in a wire) . Th e t im e Tf represents

a fr ict ional relaxat ion t ime. The shorter ther f ,

the greater the fr ict ion. A motorthat s l ides to rest in less t ime has more fr ict ion; oi l ing that motor decreases thefriction and lets it slide for a longer time. r~ 1, the inverse o fr f , has the units ofs -1 an d is a relaxa tion rate. (A co nsta nt sliding frictional force could b e w ritte nas Fsl~d~ -- --~2Fslide,with a direct ion -~ that opposes the veloci ty ~, but we wil lnot consider this case.)

O p e r a t i o n as a M o t o r

In running this l inear machine as a motor, we wil l consider that the externalpow er source, or input , is the constan t go. The ou tpu t is a mech anical load force.

The re are a nu mb er of possible mot or applicat ions, and thus different types ofmechanical load forces:

1. In lifting a we igh t W , the load is 1~0 and is due to W. The mass M -W / g ofthe object also mu st be included, along with the mass m of the bar.

2. In pull ing a boat th rou gh water, addit ion of the bo at increases the m ass mand increases the frictional resistance (so thatTf decreases). Such a velocity-dep end ent drag force leads to a load force that is not constant . F0 = 0here.

3. In compressing a spring (for energy storage), the load force F0 is propo rt ionalto the disp lacem ent x: F0 = -K x . Again, the load force is not constant .

~3~3~2 O p e r a t i o n as a G e n e r a to r

In running this l inear machine as a generator, we take the external po we r source,or input , to be the force/~0. As for the motor, for the generator there are anumber of applicat ions and load emfs.

1. If a bat tery of emf go and internal resis tance r is to be recharged, the electr i-cal load is both an increase in the electrical resistance by r and a constantemf g0.

2. If a lightb ulb of resistance RL is to be o pera ted, the electrical load is a resis-tanc e Rr~, wit h s = 0.

3. The electr ical load also could be a capacitor that is being charged, or aninductor that is given a current .

For both the generator and the motor, there are even more complex possibi l i -t ies , but their detai led study properly belongs to a special ized course, not to anin t roduct ion .


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5 6 6 Chapter 13 t Motors and Generators

13 .4 E q u a t i o n s D e s c r i b i n g t h e L i n e a r M a c h i n e

Ou r task is to solve for both th e mo tion of (i.e., the positionx ) , and the currentI through, the bar. The equations we employ are (1) Newton's law for motion,

subject to the load force and the magnetic force on the current-carrying bar;and (2) O hm 's law , with emfs produc ed by both th e dc source g0 and by themotional emf due to the bar. This yields two equations for the two unknowns,the c urre nt l th rou gh the circuit, a nd the positio n x of the bar. If it is true, asthe great inventor Edison is said to have stated in court, that Ohm's law is thebasis of electrical engineering, then surely it is just as true that Ne wto n's secondlaw of motio n is the basis of mechanical engineering.

New ton ' s L aw: The Bas is o f M ech anica l Engineer ing

Mechanical engineering is associated with the motion of machinery, describedby Newto n's second law of motion. W hat changes from problem to problem isthe nature of the forces that cause the mo tion.

Let us first consider the magnetic force. In Figure 13.5, the magnetic fieldpoints into the paper, and the battery discharges (l > 0) when its current flowsclockwise. Thus, for the bar a positive current flows downward. The magneticforce on the bar is given by

F B - I I x B . (13.1)

W ith [ point ing along $ (i.ea, -3)), and ~ poin ting along | [ x B points to theright, and is ofmagnitude I F B ] - I l l x B I - I I B s i n 9 0 ~ I l B .Hence

_.}

F B - - I l B Y c . (13.2)In addition, let there be an applied (or load) force/~L opposing this motion,

with

FL - - - FoYc . (13.3)

Finally, we will include a frictional force Ff proportional to the velocity, thatopposes the motion, in the form

-. _ _ _ d xF f - m V yc, v - . (13.4)

V f d t

This drag force is like that of (7.39).The net effect is that, including each o f the forces (13.2), (13.3), and (13.4),

Ne wto n's second law of motio n along )~ becom es

du my~ om - d - = ( F x ) n e t = l i b - F o r f (13.5)

If we t hink of F0 as the (leftward) applied force, thenI I B may be thought ofas a b a c k f o r c e because it opposes the applied force.


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13 .4 Equa t ions Desc r ib ing the L inea r M ach ine567

I3~4o2 Ohm's L aw: The Bas i s o f E lec tr ical E ngineer ing

Electr ical engineering is associated with the motion of electr ic current , describedb y O h m ' s la w. Wh a t c hanges f rom p rob l em to p rob l em is t he na tu r e o f t he emfstha t cause the m ot ion . Le t us now cons ider the emfs tha t ac t in the presen t case.

In F igure 13 .5 , the ba t te ry em f tends to dr ive cur ren t c lockwise . Fur thermo re ,ther e is an indu ced emf. By Lenz 's law, i f the bar m oves to t he r ight , i t gains f luxin to the paper. He nce th e induc ed e mf m us t p r oduce a f lux tha t is ou t o f thepaper, so the ind uced em f mus t be counterc lockwise . Le t us comp ute i t explic it ly.Con s ider t h e m o t ion a l em f

E m o t - - f f i X B .d-~, (13.6)

wh ere d~ circulates in a clockwise sense to pro duc e a posi t ive emf. This is non zeroonly for the par t of the circui t that includes the bar. Since i t mov es to the

right, and the field is into the paper, 5 x B points along 3). This corresponds to acounterclockwise circulat ion, as obtained by Lenz 's law. Moreover, s ince 5 andB are normal to each other, I~ x BI -v B . Wi t h d ~ - d y e , (13 .6) becomes

f o foE m o ~ - - v B S , . d y ~ - v B d y = - v B l . (13.7)

In agreeme nt wi th our qua l i ta tive d i scussion, the mot iona l em f tends to dr ivecur ren t counterc lockwise to oppose the increase of f lux f rom the r igh twa rdmot ion .

W i th bo th t he b a t t e ry g0 and t he mo t iona l emf , Ohm ' s l aw become s

g. go - vB lI = - = . ( 1 3 . 8 )

R R

If se l f - induc tance were inc luded , then- L d I / d t would be added t o t he nu m er-ator of (13.8) .

If we thin k of g as the (clockwise) app lied emf, thenv B l m a y b e t h o u g h t o fas a b a c k e m fbecause i t opposes the appl ied emf .

~ Backemf theory

Find the back em f for v - 1m / s , B - 0.1 T, and l -- O. 1 m.

Solution: Equation (13.7) givesv Bl - 0 .01 V.

E ~ ~ ~ Backem f measurement

Often the back em f can be measured. Let a rotat ional motor have a 6 A currentpass through it when a dc emf of 12 V is switched on, but only a 2 A currentpasses through it when it is in steady operation. Consider thatL / R is soshort that the current im mediately reaches the dc value before the mo tor canstart to turn. Find the resistance and back emf.


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5 6 8 Chapte r 13 u Motors and Generators

Solution: At t = 0 + there is no motion , so there is no back ems Th en I = C0/Ryields R = 12 V /6 A = 2 fa. W hen the m otor is in steady operation, use of I =(go - Eb ack)/Rgives Eback= gO -- I R = 12 V - (2 A)( 2 S2) = 8 V.

13 .5 Solv ing the Equ a tions

Ou r goa l now i s to so lve (13 .5 ) and (13 .8 ) s imul tan eous ly fo r v and I . Subs t i -tu t ing (13 .8 ) in to (13 .5 ) y ie lds

( ) ( 1 B212 )d v B l g o F o - r n v + . (13.9)

m--d-[ - R ;f m R

W e def ine the ne t r e laxa t ion t im e r and the magne t ic re laxa t ion t ime rB v ia theinverse o f the ne t r e laxa t ion ra te

1 1 1 mR= ~ , rB = (13 10)

r r f rB B2 l 2 ' "

and the e ffec t ive fo rceF 4f via

BlgoFeff - R Fo. (13.1])

Clear ly the ne t r e lax a t ion ra te o f (13 .1 O) i s the sum of a mec han ic a l r e laxa t ionra te and an e lec t romechan ica l r e laxa t ion ra te . L ikewise , the to ta l fo rce cons i s t s

o f the app l ied fo rce F0 and the m agne t icb a c k f o rc e ,in this case given byB l C o / R .

M a g n e t i c r e la x a t io n t i m e a n d m a g n e t ic b a c k f o rc eFor m = 0.0l kg, R = l0 -2 S2, B = 0.l T, a n d / = 0.l m, find rs and themagnet ic back force.

Solution: Equa tion (13.10) gives rB = 1 s. Thus m agnetic drag can be do min antover mechanical drag because, for a low-friction surface, rf can be hundreds ofseconds. For go = 10 V, by (13 .11) t he m agnetic bac k force isB I C o / R = 1 0 N .

By (13 .11) , the cons ta n t pa r t o f the cur ren t , induc ed by the c ons tan t emf , p ro-d u c e s a c o n s t a n t r i g h t w a r d m a g n e t i c f o r ce o fB I s Using (13 .10) and (13 .11) ,E q u a t i o n ( 1 3 . 9) c a n b e w r i t t e n i n t h e m o r e c o m p a c t f o r m

dz) mT)- - - - . ( ] 3 . ] 2 )m - d [ F e f f T

Once (13 .12) i s so lved fo r v, (13 .8 ) wi l l y ie ld l .W e f i r st cons ider the in i t ia l r esponse o f the motor, and then i ts s t eady s ta te .

Sec t ion 13 .6 cons iders how i t a t t a ins tha t s t eady-s ta te behav ior.


11/22

13.5 Solving the Equations 569

13~5,1

13q:~

In i ti al Res ponse o f Mo tor

Initially, the current I and velocity v are taken to be zero. However, in theabsence o f self-inductance, whi ch is a mea sure of the mag netic field "inertia"(and therefore of the current , to w hich the magnetic f ield is proport ional) , thecurrent can build up very quickly. O n th e other hand, the bar 's mass inert ia wil lprevent the velocity from building up immediately. Hence (13.8) and (13.2)yield an ini t ial current and mag netic force

80 Eol BI o- --~, FB (t - O ) - IolB - ----R--" (13.13)

This can be a large current and a large force. For startup, a large force is oftendesirable. However, a large current cannot be sustained for a long time becausethe Joule heating (12 R) can cause the insulat ion on the windings to melt . On theother hand, the larger the current , the larger the startup force, and the shorterthe t ime i t takes for the motor to start operating.

S teady S tate o f M oto r

After a long t ime, the velocity reaches i ts ma xim um value, the termina l velocityv~, where dv/dt - O.Equation (13.12), with its left-hand side set to zero, yieldsv~ - F4fr /m.Equation (13.11) then yields

v o ~ - F ~ f f r - ( B IC ~ F o ) r- m - R m (13.14)

In (13.14 ), increasing all types of load tends to decrease vo~: for a more massive

load, the mass m in the den om ina tor increases; a larger load force F0 is subtracte d;and for a larger drag force, the relaxation time r in the numerator decreases.Equation (13.14) also has some interesting consequences as a function of F0:

1. If the load F0 increases, the velo city v and theback emf,in this case givenby vBl , decrease. Then, by Ohm's law, as in (13.8), the current increases.W hen we fix v = 0, meaning that the mot or is held in place, the full s tartupcurrent wil l pass throu gh t he motor, causing perma nently largeI2R heating.This explains why, when motors "freeze up," they stop working: too muchheat is produced, so the insulation on the coils melts or burns up, and themotor shorts out .

2. By (13.14), if the load force F0 >BICo/R, the moto r wil l run backwards 1.(In other words, it will run as a generator.)

Equations (13.14) and (l 3.8) yield the c urrent I~ at long t imes. Using (13.10)for rB , we obtain

I ~ - 8 ~ - R ( BISOR F ~ r R

+~ ~ +R -mR R r f + TB

BITm-~ Fo. (13.15)


12/22

5 7 0 Chapter 13 9 Motors and Generators

Clearly, the larger the lo adFo, the greater the cur rent tha t mu st f low. In addition,the smaller the fr ict ional t imer f , the greater the current .

13 ,6

!3,6.I

!3~6~2

E ff i c i e n c y a n d L o a d

A given device, ei ther electrical or mechanical , provides po wer to al oad . Thereare numerous types of load.

M otors : Types of L oad

For a motor, the inpu t energy is provided by the em f Co, and the electr ical powe rinpu t is the rate of discharge of that emf:

~)input -- ~0 I~ . (mo tor input) (13.16)

In pulling a boat th roug h water, the load is the drag forcemv~/'rf ,s o we mayset F0 = 0. Then the useful mechanical power is

~ u se f~ l m v ~ m y 2- - ~ v~ = ~ . (drag load) (13.17)rf Tf

The efficiency is given by (13.17 ) divided by (13.16).For a lifting device, the lo ad m ay be ta ken to be F0 ~ 0. In that case, the

useful mechanical power is

~se f~ l = FO V~. (lifting load) (13.18)

The efficiency is given by (13.18) divided by (13.16).

G enerators : Types of L oad

For a generator, the inpu t energy is provided by the forceF o, and the mechanica lpower input is the rate of work by that force:

~)input ~ F o v ~ . (gener ator inpu t) (13.19)

We often consider generators for the case when the power source causessteady motion. In that case, (13.15) provides the current . We then interpret F0as coming fro m the pow er source, and Co as being due to a batter y tha t is beingrecharged.

For a generator used to recharge an emf, the useful electr ical power is

~useful - - ~0 I~ , (recharging em f) (13.20)

and the efficiency is given by (13.20) divided by (13.19). Here the rechargingem f is the load.

For a generator used to l ight a bulb, the useful electrical powe r is

72~seful - 1 2 R, (pow ering lightbu lb) (13.21 )

and the efficiency is given by (13.21) divided by (13.19). Here the resistor isthe load.


13/22

13.7 Trans ien ts 571

(

J _ _ 2

Figure13 .6 Genecon generator. Turning the handleproduce s a voltage difference across the leads. Likewise,connecting the leads to an applied voltage differencecauses the handle to turn.

O ne impl ica t ion o f the re laxa t ion t im e TB can be seen i f we "shor t ou t" agenera to r (E0 - 0 ) by conne c t ing i t s ex te rna l l eads to each o ther. The n the re -s is tance is due only to i ts own wires and is re la t ively small . By (13.10) , this

leads to a very short t im e TB, so T ~ TB, and the bac k force is ap pro xim ate lyg iven by m v / T B , whic h can be surpr i s ing ly l a rge . A ro ta t iona l ana log can be seenin the inexpens ive an d comm on ly used Gen eco n genera to r. See F igure 13 .6 .C o m p a r i s o n b e t w e e n t h e e f f o r t n e e d e d t o t u r n t h e h a n d l e w h e n i t is d i s co n -n e c t e d ( R = ~ ) , w h e n i t is c o n n e c t e d t o a f la s h li g h t b u l b ( m o d e r a t e R ) , a n dwhen i t i s shor ted ( smal l R) revea l s tha t the l as t case requ i res by fa r the l a rges teffor t .

By connec t ing one G ene con to ano ther, we can tu rn the hand le on one ( so i ti s a genera to r ) , and tha t causes the hand le on the o ther to tu rn ( so it is a mo tor ) .T h i s is j u s t w h a t t h e w o r k m a n d i d in t h e q u o t a t i o n a t t h e c h a p t e r h e a d .

~ Force to a gen era torrive linear

A l inear generator is used to provide the 2 A current req uired to run a l ightbulbwith a res is tance of 20 ~. Assuming that the generator is perfect ly eff ic ient ,wh at force mus t be provided to the generator i f v~ = 15 m/s?

Solution: By (13.21), 7 " ~ u s e f u l =I ~ R - ( 2 ) 2 ( 2 0 ) = 80 W. Since all the powerinput to the generator is converted to useful power,72useful -- 72input.By (13.19),72input = F ov ~ , so F0 = 72input/V~= 80/15 = 5.3 N. Note that for a rotationalgenerator to generate this pow er a t angular velocity 2 turns/s, or co = 12.56radians/s, the torque r would be given by~input = rOco~,which leads to a torqueof 80/12.56 = 6.37 N-m.

13o7 T r a n s i e n t s

W h e n a n e le c t ri c a l o r m e c h a n i c a l d e v i ce s t a rt s u p o r s h u ts d o w n , o r w h e n t h e r ei s a pow er surge , the sys tem is sub jec t tot r a n s i e n t s .

M otor S tartup

On s ta r tup , o r on a change in the load , the sys tem mus t ad jus t . We cons ider thecase where the re i s a loadF o, a n d w e s t a r t u p t h e m o t o r. T h u s w e m u s t s o l v e


14/22

57 2 Chap ter 13 ~ Motors and Generators

( 1 3 . 1 2 ) , w h i c h c a n b e r e w r i t t e n a s

dudt

= a - - , ( 1 3 . 2 2 )

w h e r e

a - ~ , (13.23)m

an d Feffs given in (13.11) and r is given in (13.10) .Equ a t ion (13 .22) i s ve ry like our o ld f r i end , the charg e-up of the capac i to r in

a n R C c i rc u i t w i t h a n e m f g . C h a p t e r 8 s h o w e d t h a t t h e c h a rg e Q o n a c a p a c i t o rsa ti s fi es the equ a t ion

d Q g Qdt = R RC" (13.24)

W h e n s t a r t e d f r o m Q - 0 a t t - O, t h e s o l u ti o n w a s f o u n d t o b e g i v en b y

Q - c c [ 1 - e x p ( - t / r R c ) ] , r R c - R C . ( 13 .2 5 )

W i t h (Q, C/R, R C) of (13 .24) -+ (v, a , r ) o f (13 .22) , an d no t ing tha t CE -(C/R) (RC) , so CE -+ a t , the so lu t ion to (13 .24) w i th v - 0 a t t - 0 i s

v - - a r [ 1 - e x p ( - t / r ) ] . (13.26)

No te tha t a r i s the sam e as v~ of (13 .14) . See F igure 13 .7 fo r a ske tch o f (13 .26) .From (13 .26) and (13 .8 ) , w e can f ind I i f des i red . Thu s the charac te r i s t i c re lax-a t ion t ime i s r o f (13 .10) , a comb ina t i on o f the f r i c t iona l t im e r f a ssoc ia ted wi th

t 1 .0 -v__ 0 .9-aT 0.8-

0 .7 -0 .6 -0 .5 -0 .4 -

0 .3 -0 .2 -0 .1 -

00

I I I I I I

1 2 3 4 5 6

tT

Figure 13.7 Velocity versus time for the linear motor ofFigure 13.5, with the constant forceFeffof (13.11).


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1 3 .8 E d d y C u r re n ts a n d M A G L E V573

the bar or load, and the magnetic relaxat ion t ime rB associated with dissipat ionof induc ed cu rrents in the resis tance. The shorter of these t imes do minate s r. A nefficient motor will have rB


16/22

57 4 Chapter 13 a Motors and Generators

component and a l if t component. (For the loop pulled through the f ield (13.9)wit h Co = 0, F0 = 0 and no mec han ical drag yields a ma gne tic d rag force tha tis proport ional to the velocity.) The drag component tr ies to bring the magnetto rest; the l if t comp one nt tr ies to push the mag net away. Both of these serveto decrease the rate of change of the magnetic f lux seen by the conductor, inagreement with Lenz's law.

Ther e is a very simple way to think of the effect of moving a magnet. Since anymagnet can be thought of as a l inear combination of monopoles, for simplici tylet us consider the magnet to be a monopole. After understanding a monopole,we can consider mo re general types of magnet. Let us also restr ict ourselves tothe case of a conduc tor with a f lat surface.

First consider the response on the sudden creation of a monopole above theconductor, as appears to happen when the monopole is moving very quickly.(First the monopole is not above a given point on the surface, and then al lof a sudden, it is there.) In that case, eddy currents are set up in the con-ductor, and these succeed in opposing the change in magnetic f lux due to themonopo le. To an observer above the cond uctor, i t is as if there were an imagemon opol e bene ath the conductor. See Figure 13.8(a) for the image mono pole andFigure 13.8(b) for the eddy currents. This is precisely the situation discussedin Section 11.11, where Figure 11.25(a) depicts a monopole above a perfectdiamagnet, and Figure 11.27(a) depicts the corresponding surface currents. (InSection 11.11, however, the surface currents do not die off with t im e becauseperfect diamagnets are superconductors.) The force on the monopole due to i tsimage is

El km(qm)2- ~ . ( co n du c to r w it h fla t s urf ac e) (13.30)

(2s) 2

Thus, in the limit of high velocity v (where "high" is relative to a velocity v0to be defined short ly) , the force on the monopole is al i f tforce Fr, and

lim Fc = F I. (co ndu ctor wit h flat surface) (13.31)V/72 0 ~ OO

Let us now give more careful cons idera t ion to what happens when themonopo le moves . As noted by Maxwel l, mot ion of a monopo le f rom A to Bcan be thought of as the s imul taneous superpos i t ion of an ant i -monopole a t A(thus canceling the exist ing monopole at A) and a new monopole at B. (This is

Figure13 .8 (a) Image and source. (b) Monopole appearing above aperfect diamagnet, where above the perfect diamagnet the field can bedescribed as being due to the source and image of part (a).


17/22

13.8 Edd y Currents and M AG LE V5 7 5

1 3 o 8 ~ 2

Figure 13.9 Surface currents caused bymoving a monopole to the right. Thechange is equivalent to creating amonopole-antimonopo le pair, themonopole at the new position and theanti-monopole (in dark) at the old position.

very much l ike the "teleportat ion" that occurs in the television showStarTr e k ~ p r o m p t i n gCaptain Kirk 's immortal words "Beam me up, Scotty!") The

eddy current response to this motio nis obtained by superposing the eddycurrent response to the f ield changefrom adding both the ant i -monopoleat A and the monopole at B. SeeFigure 13.9. This gives a qualitativesense of the nature of the eddy cur-rents that will be set up. Eddy cur-rents set up at a previous time, ofcourse, die down, so the most recenteddy currents dominate.

M axwell's R eceding Image Construction

W e are now going to write down a remarka ble result . As Maxwell did in his f irstpaper on the subject , we wil l present i t wit hou t proof. This is not the preferredappro ach in physics courses, which ten d to derive eve rythin g from first principles.However, because the result is so simple, and the proof is so complex, Maxw elldid not follow normal proc edure whe n he f irst published th e result, and neithershall we. [For details, see theAm erican Journal ofPhysics,Vol. 60, p. 693 (1992).]

The result is t rue only for thin, f lat sheets of nonm agnetic conduc tor ( l ikealuminum or copper, but not iron). Herethin means th at the distance of themag net from the sheet is mu ch greater than th e thickness d of the sheet , whichis taken to have con ductivity ~. Here is the result .

The eddy currents set up by the image poles create magnetic f ields whosefuture behavior is the same as the magnetic f ields set up by image poles thatmove away from the sheet with what we call theM axw ell recession velocity

1v 0 - 2Jrkmcrd" (13.32)

Thus, i f a mono pole suddenly appears, so does an image monopole, with th eimage mon opole moving away at velocity v0. See Figure 13.10(a) for the si tuationas viewed from above, and Figure 13.10(b) for the situation as viewed frombelow. (Note that the conductor is now a thin sheet , as opposed to the moregeneral case depicted in the previous f igures.) We call the theory based upon(13.32) M axw ell 's receding image construction.

It is a pity that suc h an interesting literary device as teleporta tion violates some ofthe m ost fundam ental laws of physics. When an object dematerializes in the telepor-tation device, the energy associated with that object cannot suddenly disappear, butmust smoothly get transported from the interior of the spaceship to the desired place.How such a huge amo unt of energy can pass throu gh the hull of the spaceship is notaddressed.


18/22

576 Chapter 13 9 Motors and Generators

o o

17 . I I

VO

(a) (b)

Figure13.10 Maxwell's receding image construction on the suddenappearance of a monopole above the cond ucting sheet: (a) as seen fromabove, (b) as seen from below.

When the source monopole moves at velocity v paral lel to the conductor 'ssurface, as in Figure 13.9 the change at each step may be thought of as creationof a mon opole at the n ew posit ion of the source monopole, and as creat ion ofan antimon opole at the old posi t ion of the source monopole. In this way, thenew si tuat ion is a superposit ion of the original si tuat ion and the change. Th eresponse of the co nducting sh eet to the change is thus to produ ce a mon opoleand an antirnonopole beneath the conducting sheet , with a horizontal separat ionthat is proport ional to v.

As the velocity v grows, the most recent image becomes more domi-nant because the horizontal separat ion between images increases. CompareFigures 13.11 (a) and 13.11 (b). Fro m Figure 13.11 (b), it sho uld b e clear thatfor high velocities (i.e., v >> v0), the force on th e mo nop ole is repulsive, w ithmagnitude given by the image force of (13.30); it is thus a lift force.

In fact, (13.31) applies to any magnet moving at velocity v parallel to thesheet , with the appropriate image force. Of course, the image force depends onthe magnet, and for a monopole only is given by (13.30). In this limit, the self-inductanc e of the sheet , which wants to opp ose al l changes in f lux, dominates.

At low velocities, self-inductance is less important than resistance. Considera related examp le. If in (13.9) w e setd v / d t = O, E0 - O, an d 1 /r f = O , then the

V V

L .. . . ! I I

o , 19 0 ~o 9 0 ~o

9 9 09 0 90 0

O

(~) Co)

F i g u r e 13.11 Maxwell's receding image construction for a monopole abovethe conducting sheet, moving rightward. The monopoles and antimonopoles(in dark) move downw ard at velocity v0. (a) A slowly moving monopole.(b) A quickly moving monopole.


19/22

1 3.8 E d d y C u rr en ts a n d M A G L E V577

magnetic force FB cancels the applied force Fo. This situation corresponds to thehand providing power, so it is a generator. Explicitly, the magnetic force is thengiven by ( B 2 1 2 / R ) v, which is the same as obtained in the previous chapter for acircuit pulled across a magnetic field. This is a drag force, and for thin sheets theproportionality to v holds in general at low velocities. Since F1 and vo are thenatural units of force and velocity, at low velocities we thus expec t tha t

Vlim F D - o t F i - - , (l 3.33)

V/vo~O VO

wh ere ~ is some con stant that d epends u pon th e details of the m agnet.

13o8,~3 L ift-Dr ag R elations hip

We now derive a result, based on the receding image construction, that willenable us to relate the lift force FL and the drag forceF D when a magnet movesat any given velocity parallel to the sheet. Then, from our kn owledge of the highvelocity l imit of FL~ given by (13.31 )~ w e can determine the high velocity l imitof FD. Moreover, from our know ledge of the low velocity l imit of FD ~give n by(1 3.3 3)~ w e can determine the low velocity l imit of FL. This will permit us toextrapolate th e behavior of both FL andF D for all velocities, and thus give us afeeling for the behavior of eddy current MA GL EV systems.

First, note that for a magnet to move at a constant velocity parallel to thesheet, it must be subject to zero net force. Hence, in addition to the lift and dragforces, some equal and opposite forces must be acting. Let them be providedby your hand. The rate at which your hand does work against the drag force isgiven by

"]-')hand~--- FDV. (13.34)

By energy conservation, this must equal the power that goes into eddy currentsin the conducting sheet, which in turn must be the same as the power going intopushing the image charges away from the surface. Since the image charges allmove with velocityvo, and since, by action and reaction, the total force on themmust be equal and opposite to the lift force, we obtain

T~ im ages-- ~ ] i ( F i v o ) - - ( ~ ] i F i ) v o - F L vO . (]3.35)

Equating (13.35) and (13.34) yields the important result that

F o v = F L v o. (13.36)

13 .8 .4 Theory of Eddy Current MA GL EV

Combining (13.36) and (13.31) (for high velocity) yields

V0l im FD - F ;~ . ( I3 .37)

v / v o ~ V


20/22

5 7 8 Chapter 13 ~ Motors and Generators

F i

l /F

W -

I \Liftoff V V o

Figure13 .12 Force versus velocity for a magnetmoving along a conducting sheet. Force is in units ofthe image force FI, and velocity is in units of theMaxw ell recession velocity v0 of (13.32).

Hence the d rag fo rce decreases as v(13.33) ( for low veloci ty) yields

- 1 a t h igh ve loc i ti e s . Co mb in ing (13 .36) and

lim FD -- ~F I , (13.38)v/vo--*O

so tha t the l i f t force var ies as v 2 a t low veloci t ies .F igure 13 .12 ske tches th e beh av ior o f the l i ft fo rce F r as a func t ion o f ve loc i ty

v. A con sta nt force W (for the we igh t of a t ra in) is given. At low veloci t ies , thel i f t fo rce Fr i s inadequa te to suppor t W, bu t a t h igher ve loc i t i e s , FL can suppor tW, an d l i f t -off can occur. At eve n highe r velocit ies , the drag s tar ts to decrease,and the l i ft sa tu ra tes a t the image fo rce . ( W hen l i f t -o ff occurs , the l if t fo rcewil l tend to decrease, due to the increasing separat ion, so this f igure, based on acons ta n t sep ara t ion , i s a b i t mis lead ing . O nce l i f t -o ff occurs , the he igh t ad jus t su n t i l W = F t . ) Yo u a r e n o w a n e x p e r t i n e d d y c u r r e n t M A G L E V .

A n d s o t h e c h a p t e r e n d s. T h e n e x t c h a p t e r w i l l w o r k o u t m o r e o f t h e i m -p l ica t ions o f the l aws of e lec t rom agne t i s m. Bes ides s tudy ing ac genera to rs andt rans formers , and ac power, i t a l so cons iders how eddy cur ren t s a re p roducedwh en a un i fo rm magn e t ic f i e ld osc i ll a tes wi th i ts d i rec t ion in the p lane o f aninf in i t e shee t o f conduc tor. In th i s way, we wi l l l ea rn how eddy cur ren t s l eadto e lec t romagne t ic sh ie ld ing . (We wi l l use the t e rmino logyelectrostatic screeningfo r the s ta t i c sc reen ing of e lec t ri c f i elds wi th i n a conduc tor, andelectromagneticsh ie ld ingfo r the d ynam ic sc reen ing of bo th e lec t r i c and magne t ic f ie lds wi th in aconduc tor. )

P r o b l e m s

1 3 - 1 . 1 Ve rify from the magnetic fields and themagnetic forces that Figure 13 .1 give s the cor-rect direction for the motion of the magnet of themotor on the left and for the wire of the m otor onthe right.

13 -1 .2 For both ofth e motors in Figure 13.1, thevelocity is proportional to the force acting on them ,and the ir a cceleratio n is negligible. This is similar

to what happens for electrons carrying current in awire. (a) Discuss why the velocity is proportionalto the force. (b) W hat role, besides being a goodelectrical conductor, does the liquid mercury play?

1 3 - 2 . 1 Which would make the best transformercore and why: solid plastic, plastic rods, solid iron,iron rods?


21/22

Problems 579

13-2.2 (a) Expla in why the eddy cur ren t hea t ingfor Figure 13.4(1:)) should be o ne-fou rth that forFigure 13.4(a) . (b) If the loop o f Figure 13.4(a)ini t ial ly has an eddy cu rrent h eat ing rate of 0.45 Wand i t is now broken into nine equivalent subloops,

what i s the n ew eddy cur ren t hea t ing ra te?

13 -3 .1 I f a moto r charges up a genera tor, whatis considered to be the external power source, andwhat i s cons idered to u t i li ze the p ower?

1 3 - 3 . 2 If a generator drives a motor, w hat is con-s idered to be the ex te rna l power source , and whatis considered to ut i l ize the power?

13-4.1 Durin g startup of a motor, at 120 V emf,the cur ren t i s 12 A. Once the motor ge ts moving ,the current is a s teady 2 A. Find (a) the resistanceand (b) the back emf.

13 -4 .2 For a l inear moto r wi th B = 0 .035 T andl= 24 cm, the back force is 0.84 N. Find thecurrent .

13-4.3 A linear motor with mass 0.58 kg, length14 cm, and resistance 0.65 ~, in a f ield of mag-nitude 0.46 T, is driven by a 60 V emf. Insteady-state operat ion, i t has veloci ty 2.3 m/s.(a) Determine i ts ini t ial current . (b) Determinethe ini t ial magnetic force and the correspondingacce le ra tion . (c ) Determine the back em f whe ni t i s opera t ing . (d) Determine the cur ren t whenit is operat ing. (e) D eter min e T . (Neglect th eforce F0.)

13-4.4 A linear generator with mass 0.58 kg,length 14 cm, and resistance 0.65 ~2, in a field ofmagnitude 0.46 T, is driven by a 24 N force. Thereis no constant emf E0. In steady-state operat ion, i thas a cur ren t o f magni tude 36 A. (a) Determ ine i tssteady-state veloci ty. (b) Dete rmin e T .

13 -5 .1 Cons ider a c i rcu i t wi th a moving a rm of

mass 46 g and length 12 cm, in a f ield of mag nitud e0.08 T. (a) If the resistance is 0.86 ~2, det erm ine themagn etic relaxation t ime. (b) If a 6.3 V battery isadded to the circuit , f ind the ini t ial current and theini t ial mag netic force.

13-5.2 Consider a c i rcu i t wi th a moving a rm ofmass 125 g and length 16 cm, in a field of mag-nit ude 0.08 T. (a) If the resistanc e is 2.4 ~a, deter-mine the m agnetic relaxation t ime. (b) If i t is drivenby a 4.6 V emf, and there is no driving force, find

its steady-state velocity for z'f- - " 120 s. (c) Find itssteady-state current .

1 3- 5 .3 Cons ider a c i rcu i t wi th a moving a rm ofmass 68 g and length 14 cm, in a field of magni-

tud e 0.08 T. (a) If the resistance is 0.42 ~, deter-mine the m agnetic relaxation t ime. (b) If i t is drivenby a 25 N force, and there is no driving emf, findits s teady-state veloci ty for T " - 120 s. (c) Find itssteady-state current .

13-5.4 Consider a conduct ing rod of mass m andlength l, ini t ial ly at rest , but free to sl ide do wn twoguide wires m aking an angle ~) to the horizontal .The guide wi res a re connec ted a t the bo t tom, sothe rod and the guide wires form a complete cir-cuit . There is a vert ical f ield/~. See Figure 13.13.In the l imi t where se l f - induc tance domina tes overresistance (i.e., retainLdI /d t but neglec t I R in theemf equa t ion) , der ive the equa t ion of mot ion forthe p osi t ion z along the guide wires and for the cur-rent . Solve for the posi t ion a nd curre nt as a functio nof t ime , assuming tha t the pos i t ion and cur ren t in i-tially are zero. Take L to be constant.

>Figure13.13 P r o b l e m 1 3 - 6 . 4 .

13-6.1 A blender motor dr iven by a 120 V emfuses a 12 A current . If it operat es at 60% efficiency,with a mo me nt of inert ia 27 = 0.00 42 Kg-m 2, at14 .5 Hz , de te rmine the ro ta t iona l d rag t ime r f .

13 -6 .2 A l i ft ing moto r driven by a 250 V emfu sesa 16 A current. If it operate s at 80% efficiency, andlif ts a weight of 417 N, at what veloci ty does i t l i f tthe weight?

13 -6 .3 A linear mot ion genera tor i s d r iven by a12 N force at veloci ty 2.3 m/s. I t provides a curren tof 3.6 A at 6.8 V. Find the efficiency.

1 3 - 6 . 4 A rotat ional motion generator is driven bya 2 N- m torq ue at angular veloci ty 4.5 s -1 . If it is70% efficient , f ind the current i t provides at outputvoltage 5.6 V.


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58 0 C hap t e r 13 ~ Moto r s and Gene ra to r s

1 3 - 7 . 1 A circuit with a moving arm of mass48 g and length 15 cm is in a field of magni-tude 0.12 T. The resistance is 0.54 ~. (a) Deter-mine the magnetic relaxation t ime. (b) If thereis a 2.4 V em f in the circuit , determ ine the ini-tial force and acceleration. (c) If r f = 86 s, de-termine the steady-state velocity and the velocityafter 1.2 s.

1 3 - 7 . 2 A circuit with a moving arm of mass 212 gand length 24 cm , is in a f ield of magnitud e 0.086 T.The resistance is 0.94 ~. Let rf= 78 s, and let th erebe a 4.5 V em f in the circuit . Deter min e (a) themagnetic relaxation t ime, (b) the steady-state ve-locity, (c) the velocity 3.4 s after the e m f has beenremoved.

13 -8 .1 Com pute v0 of (13 .32) for Cu wi th d =l 0 nm, 100 nm, 100 / lm, and 1 mm.

1 3 - 8 . 2 The lift and drag forces at a velocity of0.5 m/s are 2.4 N and 15.8 N. The lift force at a ve-locity of 45 m/s is 38 N. Find v0 and the drag forceat velocity 45 m/s.

13 -8 .3 Using Figure 13 .11(a), appropr ia te forv (( v0, show that FD ~ v forv ( ( Vo . ( H i n t : Thereis no drag force for the nearest image, but for allpairs of images at the same horizon tal posit ion, par-tial cancellation of the force along the surface oc-curs. Show that for each pair this leads to a forceproport ional to v.)

1 3 -8 .4 De t e rmine the cu r ren t densi ty K gene r-ated if a monop ole su ddenly material izes at a heighth above a n y flat conductor. (b) For a thin conduct-ing sheet , use the receding image construction todetermine how K varies with t ime.H i n t : Refer toSection 11.11.

13 -8 .5 Treat a conduct ing shee t as i f i t were aperfect diamagnet. Use Oersted's r ight-hand rule inwhat follows. (a) If a current-carrying wire is paral-lel to the sheet , how should the image curr ent f low?

See Figure 13.14. (b) If wire carries curr ent into th e

sheet , perpendicularly, how should the image cur-rent f low?

Figure13 .14 Prob l em 13 -9 .5 .

13 -G .1 Consider a box containing a f luid. I t hastwo holes cut in i t through which the f luid canempty. (a) Is the rate o f relaxation bigger whe n b othare open than when only one is open? (b) Shouldthe relaxation rates be added, or should the relax-at ion t imes be added?

1 3 - G . 2 (N. Gauthier.) Consider a cylinder of ra-dius a and len gth l >> a. It is coated w ith a fixedcharge density a on i ts round outside and has mo-me nt of inertia 27. W rap ped around i ts outside is amassless str ing, one end at tached to the outer sur-face of the cylinder, and the other end at tached toa mass M. See Figure 13.15. (a) Find the accelera-tion of the mass for a - 0. (b) Find the accelerationof the mass for finitea . H i n t : A time-varying rota-t ion of the charged cylinder causes a t ime-varyingmagnetic field along the axis.

Figure 13 .15 Prob l em 13 -G .2 .

1 3 - G . 3 Using electronic switches, capacitors areoften added in paral lel to motor circuits duringstartup, but are short ly removed. Explain in quali-tat ive terms why they are added, why they are re-

moved, and how they operate.

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Electrical Motors and Generators