### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Principles of Electricity and Magnetism 1 PHYS 3310

GPA 3.96

### View Full Document

## 27

## 0

## Popular in Course

## Popular in Physics 2

This 110 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 3310 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/232115/phys-3310-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.

## Reviews for Principles of Electricity and Magnetism 1

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/30/15

COULOMB S LAW E FIELDS Coulomb39s law 131 1 on 2 k11qu Efiu 57 In the fig q1 and q2 are 2 m apart Which arrow can represent 5 2 12 BC gt D More than one or NONE ofthe above E You can39t decide until you know ifq1 and q2 are the same or opposite signed charges What is ilzquotfrom 1 to 2quot here r1X1vY1 5312 1 2 r1 A 8 IA 09 A X239X11Y239Y1 B X139X21 Y139Y2 C 239X1yY239Y1Jx2 in 02 m D X139X21Y139Y2IxZ 7102 y 7y E None of theseit dependsnot sure Two charges Q and Q are fixed a distance r apart The direction ofthe E field at A is AUp BDown CLeft DRight ESome other direction 39Q or E 0 or it s ambig depends on sign ofq you put at A Two charges Q and Q are fixed a distance r apart The direction ofthe force on a test charge q at A is AUp BDown CLeft DRight ESome other direction or F 0 5 charges q are arranged in a regular pentagon as shown What is the E field at the center C A Zero B Nonzero C Really need trig and a calculator to 39 1 of the 5 charges has been removed as shown What s the E field at the center B kqazj C 0 D Something entirely different EThis is a nasty problem which I need more time to solve There is a uniform electric field in a region of empty space shown at evenly spaced points Which of the following could be a depiction of the electric field in this region of space I I Aonly I I I l l l Ezl39o i39ly I DNeither 0 gt There is a uniform electric field in a region of empty space What distribution of charges could create such a field There is an electric field in a region of empty space shown below at selected points What is the electric field at point x A Up B Zero X C Something elsel To find the E field at P from a thin line uniform linear charge density 7 1 A E 8 t 111139 Y 471301 what is ER dl g I ER A x r X B y39 39 PXOO C 2 2 D 2 2 dl x Xy E Something completely different l gt A EG fA dl 80 Exltx00gt f 471 4718053 80 A ILSX y X d 39 g B I 0 ER x y lt0y0gt X o fdyigy 39 Px00 X d l l E Something else x y To find the E field at P from a thin ring radius R uniform linear charge density A ER E 7 42150 What is SB E NONE ofthe arrowsshown correctly represents ER To find the E field at P from a thin ring radius R uniform linear charge density A E 1 flJz Ml P00z 471750 ER What is ER A VRZJrzZ B R C lit2 D 2 E Something completely different To find E at P from a sphere radius R uniform volume charge density p using E 1f pdf ER 47150 What Is S given the small PXYYYZ volume element shown D None of these 1 1 A Ei i511 d 4nsnfm2 p 1 XYz 2 2 2 X716 Y7y 44272 Af l XYZ my er2Yry2Zz2 X7 y 2 CIpmyaz er Yry Alia Xixinva l zpdxdydz ENone ofthese er2Yy2Zl my Griffiths p 63 finds E a distance 2 from a line segment with charge density A a 1 2M A 0 E k 47rso Z lzz L2 L J What is the approx form for E if zltltL 2 47130 A 0 B 1 C 12 D 1zquot2 E None of these is remotely correct E An infinite rod has uniform charge density A What is the direction ofthe E field at he point P shown A 45 P 1 B Less than 45 1 gt C horizontal y I I Origin WM 5 m m a a pm mum wawi dimd WW U 39 m m 5m mum unmm m 0 mm Ehsumhm m an um Lhmmmbvm bmd emn vuss mm H u Lhmmmbvm andsamhruseshnd u z mmmmpmmpmmmcm A a Which ofthe following are vectors I Electric eld ll Electric ux lll Electric charge A I only B l and II only C l and Ill only D II and Ill only E I II and Ill You have an E field given by E c r Here 0 constant r spherical radius vector What is the charge density pr Ac Bcr 330 DScrquot2 E None of these is correct Given E c r c constant r spherical radius vector We just found pr 30 What is the total charge Q enclosed by an imaginary sphere centered on the origin of radius R Hint Can you find it two DIFFERENT ways A 43nc B4nc C 43 7 c Rquot3 D 4 n c Rquot3 E None ofthese is correct The space in and around a cubical box edge length L is filled with a constant uniform electric field E E What is the TOTAL electric flux g Eda through this closed surface Z AZero L BEL2 C2EL2 D6EL2 x EWe don39t know pr so can39t answer A Gaussian surface which is not a sphere has a single charge q inside it not at the center There are more charges outside What can we say about total electric flux through this surface gSEwl A It is qeO B We know what it is but it is NOT qaO C Need more infodetails to figure it out A point charge q is located at position R as shown What is pr the charge density in all s 7 A pltfgtq53ltfo B pltfq63ltf C pltfq53ltf D pltfq53lt f E None of these The flux through the pillbox surface IS cap area A E V 139 E l 1 I I I I l I 1 I radiusr AL c A EA B 2EA C 2ALnr2 E D Lrur2 E E None of these A spherical shell has a uniform positive charge density on its surface There are no other charges around What is the electric field inside the sphere A EO everywhere inside B E is nonzero everywhere in the sphere C EO only at the very center u nonzero elsewhere inside the sphere D Not enough info given A cubical nonconducting shell has a uniform positive charge density on its surface There are no other charges around What does Gauss law tell us about the E field inside the 7 A EO everywhere inside B E is nonzero everywhere in the cube C EO only at the very cent nonzero elsewhere inside the cube D Not enough info given 1O The surface of a thinwalled insulating cubical box is given a uniform surface charge What can be inferred about the E field everywhere INSIDE the box using Gauss Law A E everywhere is zero B E everywhere must be uniform but nonzero C Direction of E everywhere must be radially out from the center of the box D Direction of E everywhere must be perpendicular to one of the sides E None of the above Now we add a single extra charge Q just outside the sphere fixing all the other charges exactly as they were What is the electric field inside the sphere A O everywhere inside B nonzero everywhere in the sphere C Not enough info given A point charge q sits outside a solid neutral copper sphere of radius A What is the magnitude of the Efield at the center of the sphere A E kqr2 q B E kqA2 C E kqltrAgt2 D E O E None of these 11 In which of the following can E anywhere be calculated easily from Gauss Law III A I only insulating sphere w B II only uniform volume charge Q I and H only ll insulating dumbbell w uniform volume charge D39 I and HI only lll same as II but only one B I II and Ill sphere charged For which of these Gaussian surfaces will Gauss law help us to calculate E at point A due to the sheet of charge PointA is at the top center of each Gaussian surface Sphere Cube co A Only the spheroeo B Only the cyl C Only the cylinder and the cube D Only the sphere and the cylinder E All surfaces will work iner A dipole sits near the origin We draw an imaginary Gaussian sphere radius r around it A Q Gauss law sayssffE39da Do we conclude that EO everywhere around that sphere AYes EO everywhere B No E is not 0 at all points on that sphere 12 We have a large copper plate with uniform surface charge density 0 Imagine the Gaussian surface drawn below Calculate the Efield a distance s above the conductor surface A E 050 B E 0250 C E 0450 D E 14rra0os2 E E 0 You have two hollow insulating spheres I 6Q distributed uniformly on its surface II 6 point charges Q arranged equidistant from one another on the sphere What is true about E INSIDE the spheres A E0 everywhere in both B nonzero everywhere in both C 0 in II but varies in I D 0 in II but uniform and nonzero in I E None ofthe above 6 point charges Q are put on a hollow insulating sphere Shown are a spherical and cubic Gaussian surface concentric with the sphere Which ofthe following are true CUDII Gwzmail Suvrm Sphei cal ecusm I 5mm Irsulatf39ig Sphere E constant on the cubic Gaussian surface I E constant on the spherical Gaussian surface II E radially outward on spherical Gaussian surface A I only B II only C III only D II III only E other Three thinwalled insulating objects have net charge Q uniformly distributed on their surfaces We can use Gauss law to find E at a point outside due to l 4 57mm Lmtalm Over 4 dew m Suilam mm mm mm Luilum MIME Jargn A cube only B sphere only G Sphere and cylinder only D Sphere and cube only E All three objects Which ofthe following are true I If E0 at every point on a Gaussian surface 135 through the surface must be zero ll lfthere is no charge enclosed inside a Gaussian surface E everywhere on the surface must be zero Hi If ltIgtEO through a Gaussian surface E everywhere on the surface must be zero A only B H only C and II only D and I only E II and I only 4 Gaussian surfaces are coaxial with an infinitely long line of charge with uniform A Choose all surfaces through whichCDE AL 0 rm mu WV 39 iii 7amp A only B and II only C and I only D I II and I only E All four emunsmces m Gamma n mm m m m Mum mm Change a sunaceswmch cquot be used mum anmMVusvaauss aw BNandH my c A my w and w my DJW H andH anW mum Apmm unavge o sneavatmn haHaW rsmawvg sum mdmsL w my myme msmbmed an Ms SAWS2 Whahstme Mann A mew t quot1477 mm a Mm am 5mm m mamam mama m Grif th s POTENTIAL I 1 pr395 The formula for Efield IS Er 4mfff W d1 With E71 r r39x x39y y39zZ39 However itturns out check that g Vi li l Where V 33i k6 6y 61 Question is the following mathematically ok 1 i g 4n5nf p l vimild quotVm w ism d A Yes B No C 7 Em Which of the following electric fields could exist in a finite region of space that contains no charges a Axyzd f b A2xy xzg c Ami xzj d A xyj x213 e Axy39 xzE Could this be a plot of Er Or Vr for SOME physical situation A Could be Er or Vr B Could be Er but can39t be Vr C Can39t be Er could be Vr D Can39t be either E 17 Given a sphere with uniform surface charge density 0 what can you say about the potential V inside this sphere Assume as usual Voo0 A V0 everywhere inside B V nonzero constant everywhere inside C V must vary with position but is zero at the center D None ofthese WhyistdZ0 inquot 9 a Because VXE 0 b Because E is a conservative field cBecause the potential between two points is Independent of the path d All of the above e NONE of the above it39s nottrue WORK amp ENERGY Three identical charges q sit on an equilateral triangle What would be the final KE 0 the top charge if you release keeping the other two fixed A 1 q B 1 21 43m 61 471380 3a 2 2 1 2L 1 E C 471380 61 D 4380 a E othernot sure Three identical charges q sit on an equilateral triangle What would be the final KE 0 the top charge if you release all three A 1 q B 1 21 43m 61 471380 3a 2 2 1 i 1 E C 47 a D 4W8 a E othernot sure CONDUCTORS 19 CAPACITORS Given a pair of very large flat conducting capacitor plates with surface charge densities c what is the E field in the region between the plates A 0220 B 020 C 2020 D 4020 E Something else not determined 20 H M A lonj rot RAJHJ a gnu UnIEwm Ifquot quotunev NRM I Uf NR m 2 JHCUHA Flag H 43904 B CV04 wkcff LKt Gaff x 61 w A I3 i IMaan 53 GMfk I Was Cu 2 Jla ma 1 50 am act 01M uni A equflt quot 7 f 1 ex at f O by Amfmt J lqw Mn an o Jj I g L C US I39Kt aqu fa11W I T J 39 coneh s F 3 r 5 a I exfen R Ni Lg fmnel 10 B Lecwsc free 105M A 3 a fqmmann yo M 0100 of cooIsa 0075112 ls vacudm 50 Mogul O A A n l K I M 13 5mm 1 3 B M I that 39amcd law no F 3003 10 be fmaHW394K n J i wxll 51 3930 3 0 7A W 1 Ice 1 Fl quotlooksquot M B 39 Q or rto mmrrto S 0 1 M U J 739 1f17ltou1l Diff5 71 quot L010 gut Aown JagINJ q J cn un aquot 657 VIl 0 ij novJ 7 H Rns 739 Jf ns 3 I A m s i f T39a39J izsj 1 3W 1 7MB 5 SM 63 Cr LfJ Jm or loam make no 3 camel u 2 i f some as Griff 40quot mag11gt ans WHO ma39itrlgi 3 A Q Omanc M30 to B Um OIJ WOH39 OI f K newue 105 quot7139quot 630 M TIH OI MJtrl lc a 3 3151ch M know Jlrcmm 6 M 1 Hnow nu Lg an B I46 Luquot we lt SMLK 4201007 Him uan AOL Al m jhhlfeS 6wquot 7C 10 1km Scan 3 1 lt11 qr3u va nawvf lj V1107 39 pl Ion KL lt gt 1km 1X3 w39 run Junn 1 52 0U7I lf WW on an Griff examale 06 3104 Mjf 2 Q 00quot ant 7X14 JB calm x 71w ephan To Jf 14le IS OASISTBIH wnl mn7np115mj and offognc Il ix 901 390 comFuH 5 33 and quot3 wt quotcu M m silc am 1 lk ovn an r 73 7M mm my Known HWIJC 4quotan know LOW A ma7nhll 5 So meu ManorL s Hanj common mazeng s Maynnlle e refer mu 0 B g g 01 IE ELL o39urq n 1445 3 so XE E101 J39clmm WE ottflnt Km for Iliqgi Mujnf llc Manual 413 31 ax H no 1 his 071 quotSymmnib 47 cut Z Jlflel1r1 X5 15 dgflaCJ Icahn an E by for ma7nf7llt 3 g Mu39 rlu39s We Usc Hnn 3 gamut H is 10 vlw39 who you meanIf CdMIoJ n07 ifiJL COMETe 0 mouj C UCS 39 392 136m 7cm IS quotmalnotc fugiff14AIII7j UQIWIQSSJ IyrjcaHJ Shea 1 Q X is 1 or fammajnt l J M WHA H D 4 XM IS quot f0 4161quot47 1175 M of a u OfCOUISc 8 5O 0 lj 110 1314 704 50 I an 5 30311 m 7 5m Jlru39um Hf XMgtO and 33 if 7M40 as Ionj as n 439 595 I For nolmo39 hcntrl tlj xnhl 3 KC 10 7010 For Surgeonlu ats Xn i l Whtclx mega B 10 quot1517c 1014 S UDIJMJH Summary 7 3 X H 9 ftmug i39l ij 39 3 A A B oOXMH EHH gt 3 M v 39 g A Nowe ct a L I qr 9 Tm 5rd Q B 40 So 4 110 if 3 1 fUMguLnj of mtg Sraccl 645 5 Bark 10 our AL OJ Exaer x 3 T P311 muynnC We M Wquot vl L H 2 I A f n r Msquot 3 J 4 go A mSIJQ 2 7 l S T v59 small mm Cu an n 3 A Blanch A I Sj wch IS afWJ Mum 390 I v n WM W6 30 In iv5 wnkou Knowing bow quotquot47quotquot 75quot Sine 4 D4 Y HcMVU I gt70 A f a quotquot33 5quot Bunnie I3 quotMlmnlcal l K 71W fifjpmqynnmm A Off ourc has XMZlo39sj 6 B n le 5 1 So rCJVJ 027512 no of u quotpa114 4H Exam 1e 39 Twomslmsvs CmrL curt16W 1 L j l 2 5 701 m X J39nnwn x l 3M7 bonam M 39X tr 87 are saw CSqu 100 SUSCff7ILl 7j x 301W8 n 3531 15 quotImcar maVCrIrAl J 3 I wkqwtg B H M II S39QL Fits quot Ifquot A win quotNu aw 3 33 S7mmf u9 1 lrhou 39 a H of 001 any Mac 434 0912521 SlaLj w39 aquot 10 No 0mg unk HR 76 75393 54mg c QMfIC m flit Sloan NHL on JIQL LI 39 I So 933 Anteum LooP 31W NJL H L W v quot 2 1s a 1c K L 3 VA M l 390 J for so H 2 K3 In Lejwecn Loaf 3 0 CUISIOLC I o aw fx w mu Q 3 h 1 A B tonic quot1 H 1 k3 f40l7nKj eraMulrmm f MlMg Pre39ITj runcl 4m 5 In Srqct jug shill rah0m for Hwy 13 M m Ju 80 1quot COMB 90 KC quot VIW XOMfk o J Insult Vxn 0 Shut M IS Unlfun A 0 72 80005 33 Bu M3 39xnk Q on 700 70 5 k8 tmom 7ft 0quot by h u 5 arqllel 10 Kfue am very small1f 231gt0 3 Ant Jensc 7amp4quot 75 Mtc MISM 70 tnkqnlc B 10514 If Xngd 1b olenoses Kfzu ItoUrn 30 g Irnk Batneler Condht nj Slate 3ft Hymn IJ VN I quot q bounJarI S Ll HQ W139Lquot thow39L U or quot 7 WWW Stole H i m an An 129 fwdquot quot700MB WC 51 0le MUD obUrDC I g A 49 OVC gm Y 71 Can watt jDKIScf 11 jive 1711quotsz fa quot354cmquot thMft quc lX 3 quotGnu3952P H 3 a ibisrm so V 73 gt MmL 57 g7 alon kph I abut P39oJ HJ HJ quot ML 13 1 w vqnuh l R is can7l U0b SM A or Imo A A mmouqb and 34 IS Away commuow 67 V 13 0 J Mun Hi 5 aim 2 cominuou 007ka ii WArH XM Sudlcnlj await 14 of A MNPIh39 MM C 4 Iflce lodg 5fjg and 0 SM 6 6 quot53971 57Mmrwy In you gltUte I egglj a la AMftre s IQW Bu if 37mlnr uj 15 133571 Wolf If 1 0 WynAct you MIME MCvole 30 EvelynLacquot 1 Hok of q 1oj quotMinn 3057 Lacaux ix zo Q VHDhhng JUGS n mcqo Iquot 7 0 CV0 AVe voles 50 57mmc71j jumup can be anokCJ 9 Lb our fur ylce l OHM0k 63911 BID GIN4W quotNFCu 39Equ It 39 15 4 4 m U m 39quotquot quotquot 3 0 7L W7lwv Q 1L0 E015 1Lc 107039 30 full qusmj rah0 film quotOHM ant 7 70 1 im390n 1 H mqwrm39J 0 cIOOScX 5 X J 50 71mm manual has Mo 4 Bo Umfom l4 3 3 a BA A Ho LBO unlolmjg nJ Ho A39J MD cf u e 4 no MQTT39I W X 39331o 74 3 JIJ 0 quotncc c 1quotqu UMP an slamm MAN S 8quot LE 1393 7k lode Clearlj IT 0 In 7Aufl3915 q an 1k Cumf quotllwt are g tuneup 4 Nu LounJlu39j in I H u a so H But H below 0 quotn ne 01th 3 73 H 3 Ho 34 0384 3 1ch s A 4 801 7Mquot means 3 40H quot0H Z 05 7 lelm anm39 3quot quotMS 0 30quot a u u 0H0 B 0 392 8 3 f f f quotPXM So 8 as reduced In 7th9 2 71 gt0 quotIA incnuquot a bquot 70 Banal UHMU I In 3511quot 3 5 O quotM If Q Iquot 40 on wall J notMth mnk iiw l Picquot1 I7 I Like S0knul J9 50Pfgt quot Aml391q NLM F 0 am Clon as 513m knowMe my l quot 39quotquotJ quotV 7jfgtlcol 130quot InsgtrvMC1a B 7a 1 391 S Uf39r H 145quotquot 1 It Mk 8 Is 6am Wm 1M 15 as on C w39fhw Am 70 3 1 warn 70 eltmmqn x791an 3 quot15 f45lvdj 640 L05 Jo 1km for a Na er IIHC com7 7 quot 661 H 1613 quot 1 T n T ue B a 4 4 1113 7mm 1 can aljue BL39JWL Bi p39av 1 B 3 Bo Z a tncr 5 0 u wine 5 I quot31 maJlfch em m I 13 p 152 g s n HON Ivaum 40 co 4 39 A 0 5 Hanan uo B2 ofMo 40 for channel annual Nan 5 u Mm 4 1m quot Ncenm I adv 44 mmml c 951 1 AWN j f M SueJuly chairs 1 F i NM 11 KC we vc quotSureroyal o JuH K S u n J Baggy 7 w quot2139 If Juk has by ll hence I unc39lr njtx 1391 E PV exv flc 3 f m uruojul Sciatic14 quotWL 112 lmy 1 Q mule I h all FtrmqunN SM 2 WWW 5an a 70A o 71M on i39us 039 10mm coqrhcaud Micrdsc fl lli 939 me quot1479 1 have 0 V l vc North due 10 62M tummyquotH WWWMi 7 123 5010 quot71 171005 I 7 can can 5 wt SfMS Til 451C L fl v39l URI 10 39llyn bu I332 AVj Stun on 5ng WW J 15 run a MSSICI CCh L 39 quot 1 I mean M 39locks m jn my a rCSfansc 10 Bot 50 yam Ca mO I Jtmc M 4r A f fOUOMAJMLJI no 6 1mm quotMquot emu I q lacquot cc 170 1rd C 31 Jalauf l 5f 5 503 J r h M II am A See 61 m 6 f 15 in 70 I f ch you an q ln 15032 Jame05 In Drcchc 06 Eu Inclm may17 quotWhy 5 le gt33 Hie 41m 7 cinnamon W15 quot747790 nwuuj 756001 reldwq non MaltaI39M Jonman bulck lj g ch ucsn39w remain ma nnI quotnlnn jm H5 A WU jig11L gleammo I 6 all quot39 Fe CO an AOU7 7 inj mq rlqlr IRA 7A5 quot A 1175 7 cva fumancm m l75 w Wmff quotCurie 7quot IS 75C rht co raht 6avt Huck 73m 00 58 4l7m m17 79flca FL JOMMAS we N Mm a 45 quot quot Maynew qmw 1 4 40mm 15 1 IO A m gt Q Md 5399 35 13 9quotquot 8 3 f oM quot J 50 7 5 QLW w for 5flncal Amen 75 mow yow Ie MW 70 J0 flaw mmquot 71103 quot B 30 PrIOMq Rf Menl TC d 17n JOM IW 50 50m fen um raglan M39h e jn her 310 For ne mule Sa39tno J JR 7 if Ifftc we Mic m I V521 W4 3 m 2 3 3 A vere I N14551 1 n I h In Bur 1 Bmslic w 8 0 IM J can gt 7 m H mar 39 39 I Q Mao mwc 4071am 10mm afmcnuzj tut 0259C 700 QJ 3 g 43 uh bf Ben CIOM NJ lam M 1 Tflq xmwmc N If 7 an H C I quotI 7quot Avg meamj quotj K van rwnv In Dar Jan m a L L 1 l J 2 Y3 OJA 340 w H l quota M you com 51 j T Curlem m Salenmj ch wk 3175c 7 Hs Solemn tomI04 6413 Q cue47C quot Q quot 39 g 3 nu mm m lt lo denial A 0 we re WJC T IIAJ LIN0A C kqmemcn if you bank 07L 71w Curlew 3 Jams 6w some I 03911nmcm remains 7K1 I quothyymcsls39 Mcmorp 1 Tumoffl 539Ian We 7W f I end 70m 0 I 7 Prtmtm t m quot i r T a Chapter 3 concept questions LAPLACE S EQUATION AND UNIQUENESS Poisson s equation tells us that V2V 3 80 If the charge density throughout some volume is zero what else must be true throughout that volume A V0 B E0 C Both V and E must be zero D None of the above is necessarily true A region of space contains no Charges 39 What can I say about V in the interior A Not much there O are lots of p possibilities for Vr throughout in there this interior region B Vr 0 everywhere in the interior C Vrconstant everywhere in the interior A region of space contains no charges The boundary has VO everywhere What can I say about V in the interior A Not much there are lots of r V 0 possibilities for Vr in there 32 B Vr0 everywhere in the interior C Vrconstant everywhere in the interior 33 Two very strong big C ideal capacitors are well separated If they are connected by 2 thin conducting wires as shown is this electrostatic situation physically stable AYes BNo C Two very strong big C ideal capacitors are well separated What if they are connected by one thin conducting wire is this electrostatic situation physically stable AYes BNo C 34 35 If you put a test charge at the center of this cube of charges could it be in stable equilibrium A Yes B No C Earnshaw39s Theorem 36 What must be true for you to know that you ve found the potential in a region a It satis es V2V 80 b It satis es the boundary conditions c a and b METHOD OF IMAGES 37 A point charge Q sits above a very large grounded conducting slab What is Er for points above the slab A Simple Coulomp s law Q 5R a A Er4ngo W1th r dz B gomething more complicated l d 2 1w 38 A point charge Q sits above a very large grounded conducting slab What39s the electrzic force on Q A O B Q 2 downwards 4m02d 2 downwards C 4Jreod2 D Something more complicated l Q d z l wx lquot 38b A point Charge Q sits above a very large grounded conducting slab What39s the energy of this system 92 A 4mm B Something else 1 Q 39 Two oo grounded conducting slabs meet at right angles How many image charges are needed to solve for Vr A one B two three more than three Method of images won39t work here Vvvv C D E On paper don t forget your name in your own words by yourself What is the idea behind the method of images What does it accomplish What is its relation to the uniqueness theorem SEPARATION OF VARIABLES CARTESIAN 310 Suppose V1r and V2r are linearly independent functions which both solve Laplace39s equation VZV 0 Does aV1rbV2r also solve it with a and b constants A Yes The Laplacian is a linear operator B No The uniqueness theorem says this scenario is impossible there are never two independent solutions C It is a de nite yes or no but the reasons given above just aren39t right D It depends 339 Given the two diff eq39s LLX C 1 W X we 1 E3 2 where C1C2 0 Given the boundary conditions in the figure which coordinate should be assigned to the negative constant and thus the sinusoidal solutions Ax By yl 0 C C1 020 here V D It doesn t matter 33911 Given the two diff eq39s 2 2 idXC 1dY Xa xz 1 Ydyz C2 where C1C2 0 Which coordinate should be assigned to the negative constant and thus the sinusoidal solutions Ax By C C1 020 here V0 D It doesn t matter 311 The Xx equation in this problem involves the quotpositive constantquot solutions A sinhkx B coshkx What do the boundary conditions say about A and B A AO pure cosh B BO pure sinh C Neither you shou rewrite this in terms AemBeml D Othernot sure 33911 Given the two diff eq39s i de C 1 dZY x dxz 1 W 2 where C1C2 0 Which coordinate should be assigned to the negative constant and thus the sinusoidal solutions yi v0 Ax By C C1 020 here v0 D It doesn t matter 312 V2hat is the value of f sin2x sin3xdx A B C D E Zero J39E 2n 752 Something elsehow could I possibly know this VVVVV mmquot sinmty a x x quotx 439 39 39quot 39 j l Pk eQV 4 10 if 4b y raga 39 r x WE m k f39 vege Ha P1 3 I quot quot quot5 quot7 h r r kW 3 16 x I J K 4 ii In I l fkp ix e f 5 4quot 3 3 W Q fl r h a or u amp was t a quot1quot EE4 f39 2quot inn 4 c rquot ml it x d 15th w DJ 314 2 troughs 00 in 2 Le out of page have grounded sidewalls The base of each is held at V0 4V 0 1 33914 Vxy O 2 Slnmtxae n135 l How does Vxy compare 4 m above the middle of the base in the two troughs A Same in each B 4x bigger in 1 C 4x bigger in 2 D much bigger in 1 E much bigger in 2 mtya Discussion question Rectangular Pipe 2 is twice as wide as rectangular Pipe 1 and the shaded face is at a potential Vxy l Pipel Pipe2 How will the solutions to Laplace s Equation be qualitatively different or similar SEP OF VAR LEGENDRE POLYNOMIALS 315 Given VZV O in Cartesian coords we separated Vxyz XxYyZz Will this approach work in spherical coordinates ie can we separate Vr9cp RrP9gtFcP A Sure B Not quite the angular components cannot be isolated eg fr9cp RrY9cp C It won39t work at all because the spherical form of Laplace s Equation has cross terms in it see the front cover of Griffiths 316 The Rodrigues formula for generating the Legendre Polynomials is if 2 z Plltxgt 2ldxjltx 1 If the Legendre polynomials are orthogonal are the leading coefficients 1 necessary to maintain orthogonality 21 A Yes fmx must be properly scaled for it to be orthogonal to fnx B No the constants will only rescale the integral Given V09 EClPlcos6 0 The Pl39s are Legendre polynomials If we want to isolatedetermine the coefficients CI in that series first multiply both sides by A Pm9 B Pmcose C Pm9 sine D Pmcose sine E something entirely different 2 7r 1 fPlcosl9Pmc0sl9sinl9dl9 211 1 m 0 0 if 1 m 1 i if lm fPxPmxdx 211 1 0 if 1 m 318 00 B Vr6 2Alrl liljmcose 0 r Suppose V on a spherical shell is constant ie VR 6 V0 Which terms do you expect to appear when finding Voutside A Many A terms but no Bl39s B Many BI terms but no Al39s C Just A0 D Just B0 E Something elsell 318 b Vr6EAlrl B l rl1 l0 Plcos 6 Suppose V on a spherical shell is constant ie VR 6 V0 Which terms do you expect to appear when finding Vinside A Many A terms but no Bl39s B Many BI terms but no Al39s C Just A0 D Just B0 E Something else 319a P0 cos 6 1 Plcos 6 cosB P2cos6 200526 l P3cos6 cos36 cos6 2 2 2 2 Can you write the function V01 cos2 6 as a sum of Legendre Polynomials V01 cos2 6 Q 2 CP 0059 0 A No it cannot be done B ltwould require an infinite sum of terms C It would only involve P2 D It would involve all three of P0 P1 AND P2 E Something elsenone of the above 319 Vr6 514 B l rl1 l0 Plcos 6 Suppose V on a spherical shell is VR6 V01 cos2 6 Which terms do you expect to appear when finding Vinside A Many A terms but no Bl39s B Many BI terms but no Al39s C Just A0 and A2 D Just B0 and 82 E Something else 319 Vr6 514 B l rl1 l0 Plcos 6 Suppose V on a spherical shell is VR6 V01 cos2 6 Which terms do you expect to appear when finding Voutside A Many A terms but no Bl39s B Many BI terms but no Al39s C Just A0 and A2 D Just B0 and 82 E Something else 320 How many boundary condItIons on the potential V do you use to find V inside the spherical plastic shell A 1 B 2 C 3 D 4 E It depends on Vow V009 33921 How many boundary conditions on the potential V do you use to find V inside the thin plastic spherical shell 709 A 1 B 2 C 3 D 4 E depends on 00 321 Does the previous answer change at all if you re asked for V outside the sphere a yes b no Since the electric field is zero inside this conducting sphere and v f E dl is vo inside as well a Yes b No MULTIPOLE EXPANSION 322 a A small dipole dipole moment pqd points in the z direction We have derived V z E 4 80 r3 Which ofthe following is correct and quotcoordinate freequot A 39 A 1 13 A V 47180 r2 B V 47180 If A 1 F A 1 x C Vlquot 2 Vlquot 2 47180 r 47180 E None of these 22 An ideal dipole tiny dipole moment pqd points in the z direction We have derived 1737 p 3200s6f sine 4neor Sketch this E field What would change if the dipole separation d was not so tiny 322 C You have a physical dipole q and q a finite distance d apart When can you use the expression 1 13 f 2 w 47rso r A This is an exact expression everywhere B lt39s valid for large r C lt39s valid for small r D 322 d You have a physical dipole q and q a finite distance d apart When can you use the expression 1 q V r l 47180 Xi A This is an exact expression everywhere B It39s valid for large r C It39s valid for small r D 323 Griffiths argues that the force on a dipole in an E field is F 13 VE lfthe dipole p points in the z direction what direction is the force AAlso in the z direction B perpendicular to 2 C it could point in any direction 324 Griffiths argues that the force on a dipole in an E field is F f o VE If the dipole p points in the z direction what can you say about E if I tell you the force is in the x direction A E simply points in the x direction B Ez must depend on x C Ez must depend on 2 D Ex must depend on x E Ex must depend on 2 325 Which charge distributions below produce a potential which looks like Cr2 when you are far away 2q 2q W O O O C O 2q q 2q B D E None of these or more than one of these Note for any which you did not select how DO they behave at large r 326 Which charge distributions below produce a potential which looks like Cr2 when you are far away 2q 2q O O O C 0 2q 1 q B D E None of these or more than one of these Note for any which you did not select how DO they behave at large r 327 What is the magnitude of the dipole moment of this charge distribution A qd B2qd C 3qd D 4qd E It39s not determined To think about How does Vr behave as r gets large 327 What is the magnitude of the dipole moment of this charge distribution A qd B2qd C 3qd D 4qd E It39s not determined To think about How does Vr behave as r gets large 328 In which situation is the dipole term the leading nonzero contribution to the potential A B C 00m coo noom q q q 2q q 2q q A A and C a kCOS6 B B and D C only E D A and E E E Some other combo 329 In terms of the multipole expansion Vr Vmono Vdip Vquad the following charge distribution has the form COO 0000 oq oq A Vr Vmono Vdip higher order terms B Vr Vdip higher order terms C Vr Vdip D Vronly higher order terms than dipole E No higher terms Vr0 for this one 330 a bf c 00 What is the direction of the dipole moment of the blue sphere 0 k sin6 e the dipole moment is zero or is ill defined Week 12 Vector Potential magnetic dipoles MD12 1 A toroid has average radius R winding diameter AR a total of N windings with current I We quotidealizequot this as a surface current running around the surface What is K A IR B I2 11 R o NIR D NI2 11 R E Something else MD12 2 Atorus has N windings and current Answer using Ampere39s Law The magnitude ofthe magnetic field inside a torus is Aconstant independent of position within the torus B nonconstant depends on position within torus 519 The vector potential in a certain region is given by AXy Cy C is a positive constant Considerthe imaginary loop shown What can you say about the magnetic eld in this region Bbmm B is nonzero parallel to zaxis y B is nonzero parallel to yaxis A B is nonzero parallel to Xaxis Dowgt VZA 1qu In Cartesian coordinates this means Vzle Z ILIOJx etc39 Does it also mean in spherical coordinates that VZAV Iu0Jr A Yes B No 0 5 0M Can you calculate that integral using spherical coordinates A Yes no problem B Yes r39 can be in spherical but J still needs to be in Cartesian components C No MD12 3 The vector potential A due to a long straight wire with current I along the zaXis is in the direction parallel to A 2 B p azimuthal C radial MD12 4ab A circular wire carries current I in the xy plane What can you say about the vector potential A at the points shown At point a the vector potential A is A Zero B Parallel to Xaxis C Parallel to yaxis D Parallel to zaxis a g y At point b the vector potential A is x A Zero Parallel to Xaxis B C Parallel to yaxis D Parallel to zaxis Efield around electric dipole Bfield around magnetic dipole current loop From Purcell Electricity and Magnetism MD12 5 Two magnetic dipoles m1 and m2 are oriented in three different ways quot11 quot12 Which ways produce a dipole 1 field at large distances A None of these 2 A7 B All three C 1 only 3 E D 1 and 2 only E 1 and 3 only 529 This is the formula from Griffiths for a magnetic dipole at the origin is hmxf 47 r2 Is this the exact vector potential for a at ring of current with mIa or is it approximate A6 A It39s exact B lt39s exact if r gt radius of the ring C lt39s approximate valid for large r D lt39s approximate valid for small r Whatis lidi A The current density J B The magnetic field B C The magnetic flux qDB D lt39s none ofthe above but is something simple and concrete E It has no particular physical interpretation at all MD12 6 In the plane ofa magnetic dipole with magnetic moment m out the vector potential A looks like kinda like this with A 1r2 At point X which way does curlB point A Right BLe Cln DOut ECurl is zero 528b In general which of the following are continuous as you move past a boundary A A B Not all of A just A C Not all of A just A Perl3 D Nothing is guaranteed to be continuous regarding A 53927 Suppose A is azimuthal given by A5 s What can you say about ourlA A ourlAO everywhere B curl A urlA is nonzero everywhere C D O everywhere except at s0 o 777 MD127 The force on a segment of wire L is F I I x B A currentcarrying wire loop is in a constant magnetic eld B B zhat as shown What is the direction ofthe torque on the loop A Zero B X C y D z E None of these 2 t B Iout I B G y 59 Iin Griffiths argues that the torque on a magnetic dipole in a B field is iszB How will a small current loop line up if the B field points uniformly up the page ll 62 Griffiths argues that the force on a magnetic dipole in a B field is F yarnfa lfthe dipole m points in the z direction what can you say about B if I tell you the force is in the X direction B simply points in the X direction Bz must depend on X B2 must depend on 2 BX must depend on X A B C D E BX must depend on 2 6 01 21 memo Crrnlv5 g 1 1 an 619 ITCAJJ Mame quote s 1er Minerquot 1 W617 one junfkfoj wwf 7A9 0515 as so I lam elerjfons C I le mqner Bug m t 01 01 m n we 59M m 614 f 3quot is NOW 1M3 1 3 w TentQ To Ime u 3 I 7 I T w z SJ 9 C 8390 339M quotNos 7116 8 seiaoln armnu wee5 73gt 707 57W fwm I 10195 vf W 360 Em A F msohnes of bu aw make an n7ei 0i E ted QfEOSL weaHenmj E 8K J am 70 Ben 7k 8160 55 Ce farami 71 m 71 gmt 5 0 CD HUE 71m 011m 7 73m C A ffen 75 39 am loam 1216 0711C Way cgl amgvrIcafb 009056 70 838571 7L1 25 DMMAC NUJs 39 73fthle WEqPEer 111cm Foamy H 39 OESQnCJ 17 f gt x p fafqmaj 0575 I q seI IJ Ban 4 5 e a 19 300 m65 J m mime 31meu 3 1 f6 o bk E24quot 7dr of 3186710115 whicA quot96nd 75 quot117 ql7h 760 a 69 Whart 090 cammaj he39 jm com rum 7 Zen amwer 13 CMsycA Exf qnmlan sacgem 17m their max Sense So 00516 l 39 a W 7 0 66 V svah j WhLov 3956ij 4quot 700 l7ef afjj J L W Onslnh r 8 in 0343 6Q QHR f V So I E215 3 ev flan 7 a R L Meanwkll mame disrolc quot6sz W I r39 q Elf K 3V2 JDK 9 AWE A j UJGU39 mOan Wm L 2 My R my mans of 6quotquch mquot Maj meme17 N e 4 on 901qu 06 elusive m L 017 Jsremoq a t So o q my Momemum Covnrilaurj 70 r of Eff ian 71 005 150 m om 5m 3 Q l 9quot ism NAM 3 mccx A 4 V0 61 L F 6 awevo K mm 2 R Now Siqto on B mic m 2 JVQOIW v 2 5707 50mg 66 130 V CAIH1295 76 V A 1 1 new F it evB UDEQIL 63 As B 701M on oQBjf to 50 vee 5 0m In llCch E Wino Fal39aoaolj f Law Funnmj mc uclcs L77 c401n7c5 K6 479605 IUms 0117 10 be quotva7 5039 7m K I undue174 517 V CZ qn g 6L P 1 eV39B 3 Mevn hnfofl39 39 quot I I L J R So vokgt CV 8 p 1Inn0L 39R If B Isnvkuje VnLVo L 1 VIA3 VJVa A SV 1233 5v 131 a B SO gv a aestz arm 1 39k 8V7 remamlntf W EVIL 50 gm Z 66 all 8 L 3 amQ LIMQ 1 E Air 5 m t Sfeejs of 9 m 5 LJJJU pun rug 1 000051 ilfx on due 70 39 50 D 6 39llus u me nanqjjsmj M B 70 r Wm WM my C1 1 Was lnvzl 7gtslow down m 7875 smaller gm t5 V19 511 m vE olucmwn 6 W 80710 t 1 MCI a 161 DQthnglU Jgfdlq MOW 75 I 60 96a if of gleam J aim71 oyhpr MedanSm me fuss Cgt 7hr Daramajnevmm Jommavtj quot alVl ma nnxC Q C mmmo WWer 71 7C 7 J 1 m 397 1 quotmammal IS mayIMIH J 1 MED 4w 7Z0 VolumL M11 hm en 5 pmaneoUSI 0139 Raqusg ELK 7 Pf f 7 Ban 7 MM majerla 5 It Solemn 6U ngnQ39 7116 me BInJuceo quot0 056 3911 regal Mm 5 I 1 fquot S y mmerla I v T T f 7 A 00 me m1 1 mar BmeJ H 77 l I farale 1 175 an quotno 89 ememb 5 B 8531 94ch Cut my 3mg 71 5 i121 quotH ckcv MQJHC39J HJIHI 6g 5309 I 7 1 Fer0 may 6715 A ISO re 439 lm p c eo It 395 NavalI 4 6 S J E Wm 3057 Q5 WQ 00190 N 11 L OFA mmj Clecquc JQajf39 3 Wt com Mug 8 10m quotMAJ quot7quotnC NC OilWk A09 Joy at we 0561 Vehaje 70 J0 7W5 In 44H 1421 wequot use A Acre 3 gt 1 A mz cnbeilc 1 4 m X 02 2 quotLqu I quotra39e 02 gt quota 39 g0 ACf fir39 5 J C in oak 1n frlncirle qu s hquot 3w beargwe C010 mq55q jg VHIS 70 7t 0x VhUcL MWQ Qv woof 56c Cran FREQ amp 4m7unvc VCSuhi A 2 30 Kbowi Joquot j wnalr 39t an F R Surf ldlw c 7L9 mag FO39RWH J due 10 0213 mqj W009 foxmlq Surchc currem alle 7 OffIAan Currem Jens my Q 3 39 Wm WL MXD j JBZ VXB gce jnmg 3 3942 0 lose 0 7 F 3 J f8 f 6 ltnw39 39 analogues 6ng I 339 H 1 A 09 o So where an es mqjtnmJ or cffccnvg ow Cuneij 0x majne cqnj fofarifeJ o jtd ail RL 7 he oremm Cancun flowmj m 397 71mm Wk 41 J1 leqraea 60quot g frctl CIFISMj Ian1 SUCL Comm5 76 V3 35017 8 fwd Le v s WAR 419 2k quot1840121 0 95 0 711cm 761015 Consnler 53 4 Shaun Cian 0 6 m476rlu punk 1 9 m I C XJyotl a 1 Sq F loomns In g dummy 91 j 3 as WOULJ 23 100 corfe q7 IdaIX 5 really 50mg TL O and of L ngg WQH NOW MZI AFQQ amp39Qlgtama390n VB Ube Q4J Wta opxjj Nit who 50 VLHJS A K Dirtolan 06 LWH an ffC WE maag mm 1235 1 X h Q g I 1 Is consmmjvken 37 can be Nam71x7 06 or Ell IE ar v mj J35 Mm fwfq Curler Cfmcmor Lumen canal I I 50 301 3 mm 0 ch 15 M33 lms 4 WW0 Fvnnj Conan wme 4 VaL ldvmt mushP 64 3 xxhm If R 5 101 0133361 COMM1h j M loom15 m 1 Lu 813sz m 700 at 7AM In bo mcd m Now limeMI Cartemx Jo nu39 canal 73 L5 a J TI 3 n 1amp3 C055 IlnTPrIor 33 4A I v 00 you 36 d3 00 6329 My Buy We br juPrer IlaMfr J6 USC 12 K 1609 IJE MjS 2 Qg M 2JZ L L 39 OJ M 5342 S A j v nk M Mel M2 2quot SO 2 m ns Cast1 L7 53 r of Loure if 1094 ons doc g we d 797 33103 a g 3 A q 1 BUj VKI 2 L L K 911 5 quot all I3 3 9 a3 90X 0 o M A 339 A 0 61 We re yng nj 383va 275T R 3 x 1 3m Surface cumin 0 Moan OLDHole 3 EH 3 M NEFL we see new Wren 8W 1 39J 00 3f If as e 7 J A M In sh exquk 1 mac ev jwr amppuwm Look m3 T T 73x 33 Jojo4 Jown 9x 1 Jce 7km WU l i fax 1013 cur 0 56 DAJC39 J linJUJum fomlj Mayhngcqu 01ale x quot Insult 5 7 O Umff lm 50 0 Iowan Curler M MSG x 5 on 709bo7 N 3 4 A 8039 0n WaHS K 3 ng umfom ICerjmffg gufchg Curmnm 5 W115 15 11K6 frma Solenoid W 6397 so 1 U 0x fUmdner mo 1 JamCL JUy I R L 0 finnt Sd enoll Insult n I and om chh lml 7er Is no Lune T1 Wl q currcn Wfanmw NOLMJ 1x C9 C9 BUML 06 my 4 quot9 C9 lt9 C n5fnrmj Examng 1 Umfmmlj m 47hmael 53Wc39 ajmn J13 O 7 Bu K2 xgz MFMf395me 3 QrIHDLS oJGJ for CL 0 fowl kh m K 4150 MJ mm 5 393X a A He dung 2 was unifol 340 E M 5 171 DR3 of cu5r 3 3007 Is a fe fe cw 413113 CIJWJ7L m x J BM HT 1 10 1 jemqi rhside my mmf riab you AWE gcunen ii 194510113 W Wej Unnuj 501711 7 20 quot fol am 3005 and Jag 4 an J g Edi Pram WinCL fw er fojd ze 7216 Inqut rm ale93919 V 1L9 curfew5 MJrzwL m 7010 4170 lm Mal won mare A H 70337156 m e ughfoqum gt A A 7173 3 IS 196 f aj j J 3 3 IM 439 JlooUnJ 1 2 7f 7 7 C 3096 0 3 figquot W wkm 7w wt le mmm Amfem j few JLyf fdf I t3 MJP C r0n 1 wh ix quot quot 33 gt 3 0 0 am 39 a f 0 C6f1m m 4jhf39vosmud 71 5 mferc j NW Set new F i 39 ILe Hawk 3701 2 Q W95 1quot 51mfa so V x Z ion1770 D3 J r0 3 3 p i 3 A 50 we mL H w 8 5 E j 7quot mam 16 7A V 7 count In w nts nOJmaUy J W171 x 0r Ifa lxroui1 a w l BELLE 6 H arc A gi Jam 753053 N07 reg 5w Wk j 0 59 1 It m I 33quot5 Jusa H g 1 65H Exqm k 39 Chapter 5 Magnetostatics LORENTZ FORCE A proton qe 5 re eased from rest m a umform E e d and a umform B e d The E new poms up and the B new poms me the page Wmch of the paths M the proton f Mom A posmve y charged pamde movmg u Ward Wm a 5 eed v e t pamde at the moment u enters the regwon No net force E Not enough nformauon A proton with speed v enters a region with a uniform magnetic field B The velocity ofthe proton makes an angle theta with the B field Which choice best describes the path ofthe proton after it enters the region AHelical motion BContinues along straight line H CCircular motion perpendicular a W to the plane ofthe page The plane ofthe circle is perpendicular to B DCircular motion perpendicular to the plane ofthe page The plane of the circle is at an angle theta to B EThis situation is impossible The velocity of the proton should always be perpendicular to B gt B 54 A wire loop in a B field has a current I The mass is quotlevitatedquot by the magnetic force FmaglLb If you increase the current does the magnetic force do positive work on the mass B into page uniform AYes klt B No I V CURRENTS amp CHARGE CONTINUITY A student argues that the current through a wire flows throughout its volume you A Agree resistance is inversely proportional to cross sectional area not circumference B Disagree it must flow only on the surface of the wire because the negative charges repel each other C Agree for different reasons D Disagree for different reasons 55 Positive ions flow right through a liquid negative ions flow left The spatial density and speed of both ions types are identical Is there a net current through the liquid A Yes to the right B Yes to the left C No D Not enough information given 57 Current flows down a wire length L with a square cross section side a If it is uniformly distributed over the entire wire area what is the magnitude of the volume current density J A JIa2 B JIa C K J4a D J IaZL E None of the above 56 Current flows down a wire length L with a square cross section side a If it is uniformly distributed over the outer surfaces only what is the magnitude of the surface current density K A KIa2 B KIa C KI4a D KIa2L E None of the above 58 A quotribbonquot width a of surface current flows with surface current density K Right next to it is a second identical ribbon of current Viewed collectively what is the new total surface current density 53 K 58 A quotribbonquot width a of surface current flows with surface current density K Right next to it is a second identical ribbon of current Viewed collectively what is the new total surface current density B 2K C K2 D Something else 510 Which of the following is a statement of charge conservation ampp aP AE fJ dl B at ffJ dA 6p o 6p CE fffv Jdr DE V J E Not surecan39t remember Discussion Why does B follow the right hand rule Is it contained in Ampere s Law When you find the B field for a point in space near a long current carrying wire what could B depend on Given the form of BiotSavart law what would you GUESS BIOT SAVART LAW Earth s Magnetosphere E r r y quot quot quotl quot7 x I Eu lN 39 x 2 7K l N oti ce th at the Ea rth s B f ieldx is nearlyEparallel and discontinuous on the rig htxxside the field lines from the top are magneticsOuth the bottom ones are magnetic north Is the Earth s magnetic field driving a sheet of current in outer space If so where does it go 511 To find the magnetic field B at P due to a currentcarrying wire we use the Biot Savart law d xix Br lf R2 4 In the figure with dl shown what is 9 541 To find the magnetic field B at P due to a currentcarrying wire we use the Biot Savart law d 32 a M0 X B I r 4 f 2amp2 In the figure with dl shown which purple vector best reresents I 512 To find the magnetic field B at P due to a currentcarrying wire we use the Biot Savart law d 32 a M0 X B r I 47 f 29x2 What is the direction of the infinitesimal contribution dBP created by current in dl A Up the page B Directly away 39 in the plane 0 C Into the page D Out of the pag E Some other direction 513 To find the magnetic field B due to a currentcarrying wire below we use the BiotSavart law Em lf 0391 X3 What is the magni R2 a dlsinH b SR2 233 C dlcosH d 6110059 e i 2R2 m3 2R2 And what39s ER here 514 What is B at the point shown J39IIS A 1 E is 3 M0 1 277 S I C MO I 47 S D 1 I 8 S What direction E None Of these V does it point 516 What do you expect for direction of BP com How about direction of dBP generated JUST by the segment of current dl in red A Bp in p ane 0 page I 0 or y red B Bp into page dBP by red into pa e C Bp into page dBP by red out of page D Bp complicated has mult component not i or H to page ditto for dBP by red E Something else 515 To find the magnetic field B due to a currentcarrying loop we usAe the Biot Savart law gmamp1f dl x29 4 R2 What is the magnitude of x R2 z2 Z C dlsinqb D zdl 2 z2 a2 Z a E Something quite different Which colored arrow ism r r 515 To find the magnetic field B due to a currentcarrying loop we use the Biot e A u dl x2 Savart law BU lf R2 What is the dBZ the contribution to the vertical component of B from this dl segment dl a d A Z2a2 Z2a2 Z2a2 611 Z D dlc Z2az lz2a2 z2 E Something quite different C z r g BiotSavart Br OJJO X J lha mz39 39 422 912 What does Jr X Rhat even mean A It is the swirling of the B field at r B It is the swirling of the B field at r due to the current at r C It is the direction and magnitude of the B field at r D It is the direction and magnitude of the B field at r due to the current at r DIVERGENCE amp CURL OF B STOKES THEOREM 53916 Rank order 39 1 over blue surfaces where J is uniform going left to right Aiiigtivgtiigti Biiigtigtiigtiv Cigtiigtiiigtiv D Something else E Not enough info givenll 517 a If the arrows represent a B field note that B is the same everywhere is there a nonzero J perpendicular to the pae in the dashed reion A Ye B No C Need more information to decide SWIf the arrows represent a B field note that B is the same everywhere is there a nonzero J perpendicular to the pae in the dashed reion A Ye B No C Need more information to decide Swlf the arrows represent a B field is there a J perpendicular to the page in the dashed region A Yes B No C Need more information to decide AMPERE S LAW 518 Pick a sketch showing B field Hnesthat violate one of Maxwell s equa ons within the region r bounded by E a dashed lines What currents would be needed to generate the others 522 What is gigdl around this purple dashed Amperian loop quot 1 I IV A AM0II2I1 B MOII2I39II1I C M00 I2 I 13m9gt DM0 Izl39 I 13m9gt E Something else 519 The magnetic field in a certain region is given by Bx y Ayfc A is a positive constant Consider the imaginary loop shown What can you say about the electric current passing through the loop A must be zero B must be nonzero C Not enough info 53920A solenoid has a total of N windings over a distance of L meters We quotidealizequot by treating this as a surface current running around the surface What is K A B NI C L D NL E Something else 521 a A thin toroid has average radius R and a total of N windings with current I We quotidealizequot this as a surface current running around the surface What is K approximatel A R B 2 n R o NIR D NI2 n R E Something totally different AR is involve 521 What direction do you expect the B field to point A Azimuthally B Radially C In the z direction perp to page D Mix of the abovedepends on where E zero everywhere 521 C A Large azimuthal loo B Small loop in region II C Smallish loop from region II to outside where BO D Like A but perp to page What Amperian loop would you draw to find B inside the Torus region ll E Something entirely different 521 Two long coaxial solenoids each carry current I but in opposite directions The inner solenoid radius a has n1 turns per unit length and the outer one radius b has n2 Find B i inside the solenoid ii between them and iii outside both 523 mp2 l l In the case of the infinite solenoid we used loop 1 to argue that the Bfield outside is zero Then we used loop 2 to find the B field inside What would loop 3 show a The Bfield inside is zero b It does not tell us anything about the B field c Something else MAGNETIC VECTOR POTENTIAL 524 If the arrows represent the vector potential A note that A is the same everywhere is there a nonzero B in the dashed reion A Ye B No C Need more information to decide Compare the magnetostatic triangle p240 with the electrostatic triangle pg 87 How is the potential similardifferent to the vector potential 2 gt gt V A MOJ In Cartesian coordinates this means VzAx lu 0Jx 2 etc Does it also mean in spherical coordinates that VZA MOJ r r A Yes B No 225 a a 0 39 39 Ar fffodt Can you calculate that integral using spherical coordinates A Yes no problem B Yes r39 can be in spherical but J still needs to be in Cartesian components C No What is di A The current density J B The magnetic field B C The magnetic flux DB D lt39s none of the above but is something simple and concrete E It has no particular physical interpretation at all 527 Suppose A is azimuthal given by AE S What can you say about curlA A curlAO everywhere A O everywhere except at sO urlA is nonzero everywhere C c 7 B cur D On paper don t forget your name in your own words by yourself What is the idea behind the magnetic vector potential What does it accomplish In what ways is it like or NOT like the electric potential BOUNDARY CONDITIONS Choose all of the following statements that are implied by 53 damp 0 for any closed surface you like I Bejbove Bszalow ll only lll only I and II only I and Ill only A B C D E All of the above In general which of the following are continuous as you move past a boundary A A B Not all of A justA C Not all of A just A Perp D Nothing is guaranteed to be continuous regarding A DIPOLES MULTIPOLES Griffiths derives a B field gt c A A B r 3200s6 r s1n6 6 Sketch this what does it look like This is the formula for an ideal magnetic dipole c A A B r 3200s6 r s1n6 6 What is different in a sketch of a real physical magnetic dipole like a small current loop 529 This is the formula from Griffiths for a magnetic dipole s msin6 A AU ET 4717 r Is this the exact vector potential for a flat ring of current with mla or is it approximate A It39s exact B It39s exact if M gt radius of the ring Clt39s approximate valid for large r Dlt39s approximate valid for small r The leading term in the vector potential multipole expansion involves g div What is the magnitude of this integral A R B 2 n R C 0 D Something entirely differentit depends See Chapter 6 concept tests for force and torque on dipole questions

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.