材料科学基础课后习题答案12.docx
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材料科学基础课后习题答案12
SolutionsforChapter12
1.Find:
Engineeringapplicationsofattractivemagneticforce.
Solution:
Applicationsinclude:
1)separatingiron(ferrousalloys)fromothercomponentsinasolidwastestream,2)Asanattachmentforcranesusedtopickupcarsinajunkyard,3)closingdevicesoncabinetsandrefrigerators,4)magneticstorageracksforholdingtools,and5)containmentfieldsforfusionreactors.
2.Find:
Engineeringapplicationsofmutualrepulsionoflikemagneticpoles.
Solution:
Applicationsinclude:
1)magneticallylevitatedtrains(andotherlevitationapplications),2)magneticbearings,and3)magneticstirringdevices.
3.FIND:
Whichclassesofmaterialsdoyouexpecttoshowsignificantmagneticbehavior?
GIVEN:
Magneticpropertiesresultfromthemotionofelectrons.
SOLUTION:
Metalshavehighelectronicmobility,soweexpectmetalstohaveinterestingmagneticproperties.Whenwerealizethatelectronicspinsinmoderate-to-highatomicnumbermaterialsarealsoimportant,thenmetalsagaincometomindaswellassomeceramics.Conductivepolymersmayormaynothaveinterestingmagneticproperties:
wejustcannotsayyet.
4.FIND:
DoesacompassneedlepointdueNorth?
SOLUTION:
Itcertainlydoesnot.Itpointstothenorthmagneticpole,whichisnotthesameastheNorthpole.
COMMENTS:
Compassmeasurementsneedtobecorrectedinordertoprovidetruereadings.Talktoyourlocalsurveyororsatellitedishinstaller.
5.Find:
Similaritiesbetween(B,H,and)and
(I,V,and).
Data:
Equation12.3-2:
B=oH
Equation10.2-1:
V=IR
Equation10.2-2:
=(A/L)R
Equation10.2-3:
=(1/)
Solution:
Combiningequations10.2-1through10.2-3yieldsV=I(L/A)(1/)whichcanberearrangedtoyield(I/A)=(V/L).Comparisonofthisequationwithequation12.3-2showsthatthemagneticfieldstrength,H,withunits(A/m),iscomparabletotheelectricfieldstrengthV/L,withunits(V/m).Botharemeasuresoftheexternaldrivingforce.ThemagneticinductionB,withunits(T=V-s/m2),isanalogoustothecurrentdensityI/A,withunits(A/m2).Botharemeasuresoftheresponseofthematerialtotheexternalfield.Finally,o(andinothermaterials)andarethematerialspropertiesthatdescribetheabilityofaspecificmaterialtorespondtotheexternalfield.
6.
Find:
Relationshipbetweenelectronmotionandmagneticproperties.
Solution:
A.Theorbitalmotionofanelectronaroundtheatomicnucleusalwaysresultsinadiamagneticcontributiontotheresponseofamaterialtoanexternalmagneticfield.
B.Thespinmotionofanelectronmayresultinparamagneticbehaviorifthereareunpairedelectronswithinunfilledinnerelectronshells.
7.Find:
Differencebetweenmagneticinductionandmagneticfieldstrength.
Solution:
Themagneticfieldstrength(H)isthedrivingforcewhilethemagneticinduction(B)isthesystemresponse.Thatis,inmostcasesHistheindependentandBthedependentvariableina
magneticexperiment.
8.Find:
Unitsforandcharacteristicsofmagneticpermeability,relativepermeability,andsusceptibility.
Solution:
A.Magneticpermeability,,hasunitsof(T-m/A)orusingthedefinitionofatesla(T=V-s/m2).hasunitsof((V-s)/(A-m)=(-s)/m).
B.Relativepermeabilityistheratior=/osoitisadimensionlessquantity.
C.Susceptibilityisalsoadimensionlessquantity.Becauserandaredimensionless,theyarethevaluesthataremorelikelytobefoundinhandbooks.Thisisbecausetheyalwayshavethesamevalueregardlessofthesystemofunitsbeingemployed.
9.FIND:
Estimatesforair,water,diamond,alumina,Pyrexglass,andepoxy.Comparetoaccuratevalues.
DATA:
UsethedatainTables11.3-1()and11.5-1(n)tocomputethevalues.Comparetonumberscomputedfromequation12.3-6(r=1+)andTable12.5-1().
SOLUTION:
Webeginbyusingequation12.3-8,whichcanbeusedtoprovideapproximatenumbers:
n=rr=n/.
Thenweuse12.3-6tocalculaterfromvaluesof.Thefollowingtablegivestherawnumbersandthecalculatedvalues:
nr=n/r=1+
air1.01.01.001.0
water1.3380.40.0165-12.9x10-61.0
diamond2.436.00.4-5.9x10-61.0
alumina1.769.00.2-37x10-61.0
Pyrexglass1.475.00.301.0
epoxy1.583.50.4501.0
NotethatIusedof0forair,Pyrexglass,andepoxy.Theyarediamagnetic,sothevaluesareessentiallyzero,asintheothercasesshown.
COMMENTS:
Thisisnotverygoodagreement!
10.Find:
Themagneticfieldstrengthtocreateaninduction
equaltothatoftheearthinMg.
Given:
(Mg)=13.1x10-6
Data:
(earth)=6x10-5T
Equation12.3-5:
=o(1+)H
o=4x10-7(T-m/A)
Solution:
Solvingequation12.3-5forthefieldstrengthHgives:
H=/[o(1+)].
Substitutingthevaluesintheproblemstatementyields:
H=(6x10-5T)/[(4x10-7T-m/A)(1+13.1x10-6)]
H=47.7A/m
11.Find:
RelativepermeabilityandpermeabilityofMg.
Given:
(Mg)=13.1x10-6
Data:
Eqn.12.3-5:
=o(1+)
Eqn.12.3-6:
r=/o=(1+)
o=4x10-7(T-m/A)
Solution:
A.r=(1+)=1+13.1x10-6=1.0000131
B.=o(1+)=(4x10-7T-m/A)(1.0000131)
=1.257x10-6(T-m/A)
12.Find:
MagnetizationandinductanceofMginafieldof
105A/m.
Given:
=13.1x10-6
Data:
Eqn.12.3-1:
M=H
Eqn.12.3-3:
=o(H+M)
o=4x10-7(T-m/A)
Solution:
Usingeqn.12.3-1,themagnetizationis
M=13.1x10-6(105A/m)=1.31A/m.
Usingeqn.12.3-3,theinductanceis
=o(H+M)
=(4x10-7T-m/A)(105+1.31A/m)
=0.1257T
13.FIND:
Howcanyoupickoutasteelwith<1?
GIVEN:
Moststainlesssteelshavethisproperty.
SOLUTION:
Holdamagnetuptothemetals.Anon-magneticsteelisthistypeofstainless.Aliseasytodetectbecauseofitslowdensity.Brassisashadeofyellow.
COMMENTS:
Simpleandeffective
14.FIND:
Wouldagiantferromagneticsinglecrystalincreaseinlengthwhenplacedinamagneticfield?
SOLUTION:
WhileIhaveneverseenanequationthatdescribesthegrowthorshrinkageofamaterialplacedinamagneticfiled,Iexpecttheremightbeachangesindimensionsasthedipolesalign.Theeffectwouldbeverysmall.Sincealargesinglecrystalisnothuge,thechangeinlengthmightbedifficulttomeasure.
15.Find:
Relationshipbetweenvectors,H,andMindia-andparamagneticmaterials.
Solution:
Theexternalmagneticfieldvector,H,andtheinductionvector,B,alwayspointinthesamedirection.Themagnetizationvector,M,alsopointsinthesamedirectionasHinparamagneticmaterialsbutisantiparalleltoHindiamagneticmaterials.
16.Find:
Whydia-andparamagneticfindfewintersectingmagneticapplications.
Solution:
Bothdia-andparamagneticmaterialshavesusceptibilityvaluesverynearzero.Therefore,theirresponsetoamagneticfieldisnearlyindistinguishablefromthatofavacuum(orair)andtheirmagneticpropertiesarenotveryinteresting.
17.Find:
ExplainwhyNaCl,LiF,Si,Ge,Cu,andAgarealldiamagnetic.
Solution:
NaClandLiFareioniccompounds.Afterelectrontransferfromthecationtotheanion,allionshavefilledelectronshells.Thus,therearenounpairedelectronsandbothcompoundsarediamagnetic.SiandGeareGroupIVelements.Bothelementscrystallizeinthediamondcubicstructurebyformingfourcovalentbonds.Assuch,theirelectronshellsarefilled,therearenounpairedelectrons,andSiandGearediamagnetic.CuandAgaremetalsandhavesomeunpairedelectrons.Theirunpairedelectrons,however,areinthevalenceshellanddonotcontributetoparamagneticbehavior.Sincetherearenounpairedelectronsininnerelectronshells,CuandAgarediamagnetic.
18.
Find:
ShowelectronconfigurationforSrthroughCdandpredictmagneticbehavior.
Data:
ElectronconfigurationsinAppendixB.
Solution:
ELEMENTELECTRONCONFIGURATION#OFUNPAIREDMAGNETIC(INNER)e-sCLASS
Sr[Kr]5s20DIA
Y[Kr]4d15s21PARA
Zr[Kr]4d25s22PARA
Nb[Kr]4d45s14PARA
Mo[Kr]4d55s15PARA
Tc[Kr]4d55s25PARA
Ru[Kr]4d75s13PARA
Rh[Kr]4d85s12PARA
Pd[Kr]4d100DIA
Ag[Kr]4d105s10DIA
Cd[Kr]4d105s20DIA
19.
Find:
Can"tin"cansbeseparatedfromawastestreamusingamagnet?
Data:
FromTable12.5-1,Sniseitherdia-orparamagnetic
dependingonthecrystalstructure.
Solution:
Sncan'tbeseparatedfromawastestreamusingamagnet."Sn"cans,however,aremadeofFewithaSnlining,andtheferromagneticFeiseasilyseparatedusingamagnet.
20.
Find:
Contributionofunpairedvalenceelectronstoparamagneticbehavior.
Solution:
Withinasolidthevalenceelectronsofindividualatomsareeithersharedortransferred.Ineithercase,therearenounpairedvalenceelectronswhenthesolidisconsideredasawhole.Thus,thereisnoparamagneticcontributionfromthevalenceelectrons.
21.
Find:
Whyaremostoxidesdiamagneticandarethereanyexceptionstothetrend?
Data:
FromTable12.5-1,weseethatonlyafewoxidessuchasCeO2,CsO2,andCr2O3areparamagnetic.
Solution:
Mostoxidesareeitherionic,covalent,oramixtureofthetwo.Assuchtheywillexhibitfilledvalenceshellsandnounpairedelectrons.Thus,theyarelikelytobediamagnetic.Iftheoxidecontainsatransitionmetalwithanunfilledinnerelectronshell,itmayexhibitparamagneticcharacteristics.
22.
Find:
Mechanismsbywhichanexternalfieldincreasesthemagnetizationofaferromagneticmaterial.
Solution:
InaferromagneticmaterialanexternalfiledcanincreaseMeitherbydomaingrowthorbydipoler