大学物理(下)东华大学物理教研室詹科利

1、标准文案第二十三章必做题：4, 10 , 12 ,19 , 22I Jie polential at Ilie siu face of a chained sphere is derived in Example 2-4.T P=4戌匕ta _ p1叫*_ 30 6W)佟虚5黑1UCN眉21Aiea4 77j4 斤%/=3.761 KlOc/ur（切10 :之 3.8 *10-lc/nrThe potential from the aurfaee of a charged sphere is also tietiv ed in Example 23-40 - 4%一匕 _ 瑞- T - e r

2、 / 4诏/r叽坦蜩 隔The poteuti3l at the coinei the sm】i of ihe polenlials due to each of The charges, using Eq. 23-5,j_L 山+_L+_LB.，却 t- t4笈/4tc0 f 4,T/ Pl V212 :We follow the development of Example 23-9. with Figure 2S-15 The chaige on a Thin ring of radius R nnddR / dq - adA - r(lRdR. Ue Eq. 23-6b ro find

3、 fho porfiKinl of ft conriinLoiKclmr distiihiition. b1 产 1 I b仁界火曲）19 :The elecTron tvrr ccelei ated ilnongli n KiTenrial di faience of 133 kV (moving fi oiu low potential to hi oh potcntml m siniiiijig 1.33 ker ofkinctic eneiy. The proton is accekiated diroush the npposti亡 poteiHial differenct? as

4、the(?lecTron. and Iihr The exact opposite charge.Thus Ihe proton gain、rhe same kineric energy, I 33 keV |iZ?) Budi 山t? piolou and the ehctiuu h e 山d 5ame KE. I_e 山M lo liud the lalio oi die speeds.ER% T1.67x10S_Tlie lighter electron is moving about 45 tmies taster than the heavier proton.22 :Let，rep

5、resent the distiiKe from The 卜什 charge tet point b. and let * represent tlic distnmc fron; the ri Jil ehdiMe to ponil lx Lei Q lepieutLie pown e chaiand kl /14? + 24J cui = 27.7S cmL；-t/ =7(i;-r) =-十- + 0 )1*0.14iu 0.2778m I lOIZiii 024j I I ( 1 1 1 0Jfi44m 0J778iii J10 12m 024ui /= 18.99xl0,Ntm7cJ)

6、(-L5xl0C)(33cl0fC)(-3.477ml)=L547J |1.5j|第二十四章必做题：5, 7, 9, 11 , 18We apply Eq. 27-5 to eacli circunisTance. and solve tor die infigiieKc field. LeT B = 5r i 4- SJ + 5.k卜.nil. LI lUI1U1；11L. l1 / - f iFb = /xB = (B.2A)2.Oqi 0一0=( -164AM11) ZfJ + (l6,4A-m , k = ( -2,5j) N t5 -0 : (-16- -2.5N n For Ili

7、e second ciicuiustance. / = 7=二Tv(6.6x10 -ka)二=|工石黑bqV tfB 2(1.60x10-19C)(O3J0T)9:The force 0n rhe eJectron is given by Eq. 27*5乱 maa, Iijkg 二行 x 在二一r 7。x 10* ni/s -6,0 k 04 ni/s 0 = -p(4 2 - 4.8) x IO1 T*ni/sk-0.80T0.60T0= -(l.60x 1O*0 -0,6 x IO4 T*ni/sk) = 9,6 x 10*N k 之 lx11 :The total, frrce ou

8、the uro ton is given by the Lorentz eqnahon. Eq 2-.I -M.-5,0x10* qsIJ=(E + vxB)= e(3 0i-42j)xtfv/m+ 6.0xl(fu5 3.0x10*0.451038T(1.60 xl0-19C)|(3.0i- 4.2j| + (lT9i- 225j + 0,95k )1 x 10 N/C二(L60汽 10 ”(4,9i -6.45j + U.95k)jx 10、N/C=(7,$4 x 10-u i -1.03 x 10-15 j +1.49 x 10-ltk) N7c=(Q.78i-1.0j + 0.15k)

9、-|xl0-1-N18 :f) Tlie magnetic nionient of the coil is given by Eq. 27-10. Since Ihe current flows in the clockwise direcrioiL ric rihf linnd nile show$ thnr tIk magnetic iDonient is down, or in the negative r*direcrionP-f0 Am 乙 * ；ri口二= 1517 6 A归1-；(-kl 二7334kA皿-Y3kA*n? 1 /(b) We use Eq. 27-11(o fin

10、d th torque on the coil二E=(T.334 t A*m3)x(O,55i + 0.60j-0,65k)T =(2.6i-Z4j)mN(r) Wc use Eq, 27-12 to find the potcnrinl cncray of The coil.W= -p-B = -(-4334 k Amu2 |(0.55i + 0.0j-0 65k |T = (4334Ai/ 1(0.65 二毋1第二十五章必做题：6, 12 , 13 , 15 , 16 , 24Ilie magnetic Held at the loop duu to the Iguq wire 珞 int

11、o the page, aud can be calculated by Eq. 2K- l Tlie force on rhe segment of rhe loop closest to tlie wire is towards the wire, since rhe currents are in tlK same direction. The force on The segtncnf of the loop fnrrhesT from the wir ztiu. Tlni the tolal tbivc on thu letr and i idit euineiiis i* zero

12、, and so only the paialld eguieuts n?e(l to be consideieil iu The Ciikiihititui U货 Eq 2W.1 一嬴-I=15一 I m IO-* N, towards wire12 :Ww iise the results of Example 28-10 to find the maxi in inn and niiikiimun fields.u ni (4jrxl(T7 TWA)(687)(25.0 A)B=1A = L-= 12,7 mTg lr2工 0.270 m)4rxHJ 7 r*iii A)(6K7)| 2

13、5.0 A) =13.7 inT2Ko 250 iu)12.7uiT.5.B 7niT13 :Because ofche cylindncal syiumetiy, ihe injaietk tidds will be circular In cnch 国泗，口弋 can dHejininc ihe iDOgniic tild iMng Amperes L *i【h conceinric loops. The ciinem ilensides in llm wiiej, are given by the toial einrein divided by ihw cross-sctioual a

14、rea.1 _ 八 j _ Ai:皿工MA年）k/J Luidtf ihe uiDti 招 lie the enclosed icimeul is detenniiied b山胪 cniTenl density of the mneij m+杰.Jeoei 二为 JsiHR 4=%蜷”=镖gj Betweeu the wiies rbe citneut enclosed i the cuwein on the iniiei wue百疝为411d T取2丁五-p1T 8 -（ 1 Iiuk tlit i.uitei Ilie cimuir eneloseil 由 ih史 uiiijenr Ikn

15、jj iht 111111 ui nnd * potdou of Ihe cuneiu troni the oiitei wire.j A出=% =为4+ /皿工I * -抬）2jtR （居-R；）.r g川-硝 b0n=% 70-/0#1启-用）（rf） Outside the outer wire the net cuirent enclosed is zero.J B t/s = Wq/j = 0 t 别 = 0 t B = G（c 1 See Ilie .idjaceut gniph. TLespreikLslieel u+etl fui ihis pioblem fii be fou

16、nd on the Media Manaaer with fikniiincPSE4_ISM_C H2S XLS/1 on tab bProblem2y.31e15 :*舒+龄Silica rbu point C is along rhe line of the two straight sepnients of rhe ciinent, iliee segmenli* do not coutiibute lo the injuiietic lleld at C. Ve cjkulak lite tiifigiietic field by iiitegiatLiig Eq_ along the

17、 two curved segments. Along each in re jratio n rhe line segment is perpendicuhr to the radial vector and The radial distance is consul.二%*a。i二时叫1一4,16 :Since ihe cuifent in tlie two snaifht segment* flows radially toward and away ficm the center of the loop、tliey do nat contnbutc to the inasnielic

18、field at the centei. We calculate the magiieric field by iiitegi;itug Eq 2S-5 along ihe rxvo cune(l xettin?nts Motik e;ieli iutei；ition segment, ibe cmreDi ispeLpcndicuLii tu tlie rudial veckn atid tliu tadial clistaiictrcoilstanl By 山c imhNHnd-uik theiiiiiLue(ic field livtii the upper pen Hun 也 ill

19、 pouil into the pige ciud ihe iuhmuSiv field Ham 山e low er pom on will point 011t of rhe pace.出,k-R2，和k卜弟翁k(5啜k(0.35/0卜卜40A24 :(门) We set the magiieiic foice, usms Eq. 2S-2, equiil tQ the weight of the wjie and s(?hpe fbi the necessary cimeiir The cunei it must flow in The ame diiecticin as the ujip

20、er ciineiii fbr tke fbrci* to be upward.二 360 A, tightpgjrM (S900kg/mJ)( SOm/s2)1 (0.050x11)(1.00 xlO-3m)4/i0A 1 ( 4.t x IO_7 T-niyA K 4S.0 A)The lowei wiie i? in hmdcbk 廿qu】lihiimn|- imce if it isiaissd bahlly from emlibiniui. the magnetic fbrce would be iiicieawd. eauiug ilie wire io uiose tunliei

21、 liom e Idt. 11m i? i “iL* ui辿山口山 1：、u vertical displacement since li the wne is moved slightly off the eqiiilibniun pomt the magudic force will uicreaie or dcrae to push the ii# back io The eqmlibiiuni heizht.第二十六章必做题：2, 3, 15 , 16 2:W e chcwse np as the jositive diiechon. The average induced emf i

22、s given by Hie *diiiereiiceL x erion of Eq. 29-2a父 皿 AAB ,7(0.054mf(-0.25T-0.68T) |一W ：=1- - 5.3x10 M Ar0J6s3：(a) When the plane of the loop is peipendicular to the field lines, tlie flux is given by the nuutiniuin of Eq. 29-la事，=BA = Bh# =(0.50T)t 0.080 in T =|1 爪1代加(b) The augk is - 55(c) Ust Eq.

23、29-1 a, iji dii upvaul nrneikfields there will be a magnetic force to tfae left. To keep rhe rod moving, there must be an equal exTernal force to the nt*ht. given by Eq 27-1.F = NE = f L13Kxl0 |(U250m(U35T I = 3.621 x 1U *N =：|036niN第二十七章 必做题：3, 5, 10 , 153:We find the mutual iiiductaiice of the iim

24、er loop. If we assume rhe outer solenoid is carryiug cimetit,.卢一 一.Me.门 i/r tlien the magncTic field inside the outer solenoid is =L/. Thu magnetic flux ttirough each loop of the small coil i& Ike luagnetic field times the area peipeiidiculai to the field. The mutual inductaiic& k givwn by Eq. 30-1.

25、NI .小 A；A - 1v产，干1asM.M4疝18力,i = HAy sin8 =氏当 sin 8 :M = - = s-We draw the coil as two elements iu series* and pure 惟乐tance and a pure indiiclance There i、a vol忸gw drop due to tlie resisraiKt of the cod. given by Ohm 9 law. and an induced emf due to tbe ludiKitaute of the coiL gien b Eq 30-5. Smce I

26、 tie cmieut is iucleasing, tlie induciaiice will create a polential difleience to oppose the iticreasine cunem. and so liieie is a drop in rhe potenrinl due to (he inductance. The potential diffeience across the coil is (he $uiu of die two poteutial di op*.-IR + Z =(3.00 A 1(3.250) +(0 4411 J(3.60 A

27、/s) = 1113 V dt10 :( When coDtiectcd in 弓盯the drops 猛仃口/ ench inductor will add. while the cun cnts in cacli mdiicror art The snmeS 4 = .*-吟=-k 1聋=*T小岗，) 猛lieu coiuiecTetl in parallel The cuiieuts ininductor add io The equivalein culeujlI, while ibevoltage drop across ech inducloi is Ihe . the uquiv

28、filenf voltage drop.df /rf/普 苣格i i i由 dr1T4 A A4一There tbi e. mdiicTors in series and parallel dd the aame as resistors in series and parallel.15 :We create an Anipei eaii loop of radius r to calculate the mauelic field witlkiiL ihe wire usiiig Eq 283 Siuce the resuliuis masnetk field only depends o

29、u ladius, we use Eq. 30- br the uuurgy density in 山eohune d厂二 ZrWr 口nd iutegiate Hom zro tc Uie radius oi tbe wire,iR 疝=/%*T别3b仁闻:7君二名71 7T Jx 1/-TTjK第二十八章必做题：1 , 5, 8, 10Th* ciment in the wiis imit also bu ihw displacement cimnt in the capacitor; Use H】cdisplacement current to find the rate at whic

30、h the electric field is changing.dEHE卜U.S A)1.2xO15m*s1)If ve wiittf the aigiimeuT of 山u coMue fiiiietioii as kz加=叔: + cf we see thal the wave istiiiveling in the 一 二(lirggjjgiL or0 由小 5/ (S.S5 x io-12 c2/N un2) (o.o 160ni r E aud B me prpejj B luiE poiu( in Ilie uegtie: direction. B must point in t

31、he the - v diiecbcn. or |二 j| The mngnirnde of the niapnetic field is ibimd iroiii Eq. 51-11 as / = /c.8:(，)The general fbnn of a phne wave is ci ven in Eq. 31 -7. Foi this ivaveh = & sin(Az 一句0.077111-1-SI.60m i 82m二兰半山二3 66E0，H”际叫Note that A/ = (8L60m)(3.661xlOeHz)2.98xl0a xr.(Jm Hie magnitude of

32、rhe maanetic field is ci ven by 5,= E J c. Ihe wave is travelino! in thek diieciiou. and so 山已 ruiigtieic field be m ihe j diieciiun. since 山匕 diieetiou of Irivel is given by the direction of E x B=225X/in= 7 50 y 10-7T 7c 3.00 m 唠B- j|7,?0x 10 ?T)iiuj(0.077mJr-fZJxlOrad/slz10:The energy p&r unit ar

33、ea per unit time T gi en by the magnitude of the Poyntin9 vector. Let rcprescjK rhe energy that crosses orea A u。time yT.0J94W沁二(4 . X lOPm/A )(335 J)(1.00x 10 Jnf )(3 00 x 10!iil/$)(22.5 x 10 4T)=土77父0%| - 321 由户第二十九章必做题：5, 8, 12 , 14 , 16 , 26 , 27 , 31 , 33 , 36For consTmctive interference, the p

34、ath difference is a iniiltiple of the wave length, as si ven by Eq. 34- 2a The location on the screen is given by v = f hn a as seen in Fie. 3 1*7(c). For smnll 和里1亡f. we have stn h ii 0 = r/f. Second order m - J,t ,r ./阳fZmiAj献siu ff ma 日一mA v : I1= : : k,= e1(7- d建-4)力r(720-660)x 10011(2 H10m)dx =

35、#垦-Hj = = 7= l,76x lOni 空 0.2mmd(6* xlL(m)lhi& justifies usnisi tlie small appioHiiation, siuce m y 8:Foi constructintei饶tciicu. Hie pfltli diffeience F a multiple of the waveJength. given by Eq. 34- 2a. The.Il tlic croen is iivcn lv v 1 ivn，r wen m Ff ；4-葭：1 Fer 旬讥油 口口1”1veh.a a suit/ tEiu0 )%;F.v共

36、做(680，10山 ”)(2.6111) (7siii = mA d- - mA d=：=1.4 x lO-4!tr3孔0-712 :X*e equate tb R 二920nm : wH = 1 t A 640mnnf - 2 r /I = 3 84 nmThe epHLilion hem llic dmta iui I lie 5(_NJ mu light, hat expiebsiou is then used to hud the locatiou ct the maxiuia ibr llie 65 uiu light. Foi coustmctiv intrteieijce, th

37、e path difference is # iiuiltipk of the waveleiigdi. as ci ven by Eq. 3小2 孔 The lot alien on die screen is given by工二 a ns $en in Fig 34-79). For small triples, we have sintf 之片1】小黑/F.I - d &, .T . Atut Aw/.f力成dsmb- friA T d- - mA rf = f 1= |65。力向口 （5001也）tt x,d= 10,4nlm 离 111 mm (2sig.fgj16 :To chn

38、nge rtie center pon)T from consnuctive interference to deTmcttve interference. Tlie pliase shift produced by the lutroducticu of rhe plastic uiit be eqnivaleut to half a wavelengtli. Tlie wa vekuglb of the? lialit is shaiter in the plastic thn m th air. so the iinnbci of WAx Eeii/ha m the phslic mus

39、t be 1 2 Ln eaiei than the nutubei iu the same thickness of air. The nunkbei of wavelenThs in the distance equal to the thkknc of the plate is the thickness of the plate divided by tbe apprupnate wavelengthA _ 6801II】斗仁1广口比6。-1）26 :An luviideut wave that reiki；ts ixoni the top surface of thr coating

40、 lias a phase change of 机=开. An incident wave thatreflects from the glass 1 日餐 I，)at the bottom surface of thecoating has a phase charge dne to both Tlie addirional path length find a phase change of on reflection, so2& = 2尸 + 丁 For ccinsmictive interference with atninjimun non-zero Thckn电皓 of coaim

41、g. thi? net pha&e chmiae nnut be 2才一Tbt? leiis leflecb I he jdqsI fbi - 570iun The nmhiuuui uou-zeiu tliKkue。occurs (hr 解二 1:A t 57仆mil I2(125)228 mni 4lnSince rhe middle of the pectmii is being selecrively reflected the IrnEisiinited hgbT will be strojiccr in tbe iel and bine uortions of ilie visib

42、le siiectrum.27 :(a) When illuminated from above at A, a light ray reflected from the air-oil interface undergoes a pha百色 shift ofA rayat thu oil-wat interface uiidraces nnstmcriTe nifeiferinee. the net phase change must be a imilriple of 2t./ 江七二食一的三 丁 工花一开二川口工)T z = 4(m+1) = 1(w+4) From the diagia

43、m. we see thni point B is the second rluckness that yields consmicrive interfereiice fot 5SO mn. and so we use i J. (Tlie first locatiou that yields coustnieTi e niterference would be for m = 0.),1 /111 5 5X0iuu290njnl*W + )一一1 +力531 :With respect tc the uicitlein wave, Hie ivave that retlecTs tioni

44、 die air at the top suriate of the air layer has n phase chauge of %=0一 Wiih respetl to ihe iiicideut wave, the wave that(t)2leflecTi fi。口i the 乂括登 at ihe bonom $mface oi ihe aa layer tias 片 phsc elmnuc iluc io both the addirionL paih lenglh and1 fl toii5ltik：ne iiKeilVieuce.the uet phase change mus

45、t be an evt-u non-zer o mtejer unUtiple of t.限二网一二122a+总-0 = 2脚k，二打抑一力九小二】、2.，JIlie iinnimntii liickiieiS lb wiih ffi - 1.,皿=i(450iiniH I - I) = I 】3mliFor de弓tractive inrerfetetKC the net phase change iniisr be an odd-intaet iuukile of t 取11fl =4-4=(2h + .t -0 = (2f4 + l)/T F=*川人=0. L 工The uimiinum noivzero tlnckiiess is = 4( ：5Qnin H 1) = 225nm.33:参见课件相应容36 :We use the equation derivedi口Problem 33. wherek the radius ofihe lens (1.7 cm) to solve tor rhe radius of cm alurc Sukc il