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Sp06-Final-Smoot-Soln

Sp06-Final-Smoot-Soln - Department of Physics University of...

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Unformatted text preview: Department of Physics, University of California, Berkeley Final Examination Physics 713, Lecture 1, Prof. Smoot 8 AM - 11AM, 12 May 2006 Discussion Section: ______________________________ Name of TA: _____________________________________ Problem 1 Problem 5 There [‘3- no 5, Score: ___________________________________________ 2. [25 points] Resistor and Voltage Source Network 1» we on“ («Lifer-"3) uhoafic Hm, '9 dfmhhs ‘53 “Hm currents as shown LL. <————— (2 R2 42 16" 1'3th Figure 2: Resistor and Voltage Source Network. (a) [5 points] Write down the ap [1 through points P, (2) 12 resistor R6. propriate equations to determine the four currents: (1) through resistor R3, (3) 13 though point Q and (4) ‘14 through Junction Rules: ‘1‘" It: Izl+ls "B? I; 2 «[3413! C; I3+Iq:I‘ fl .' 15*1231'. ‘ ‘ m. I {I Not a” of Her: are .‘nJepeuJut. Ne w.” ell; 'k s a, fiance. tiny M m4 IaLe’eJ .A ‘H¢ fra en. :9 7) 1 egmtian m d Loop Rules) We need iéfee Independent ’00,; :0 we cu. Lou—e four quul‘fiw 41:1 SEMI Wilma/A," a—LK+&-L%'KL:° L25 -C13*L)Ru' 13R; * 83 + I‘R‘V £120 L/"" \(3 5011.] ugulhif (vi/(n (b) [5 points] Assume that the voltages E1 = 12 V , sources E2 and E3 are replaced with capacitors, and that all the resistors have the same value R1 = R2 = R3 = R4 = R5 = R6 = 100 Ohms. Solve for the current [1 and the voltage on each of the two capacitors E2 and E3. If 82‘; 83 0N- quadh’f § Hm Cn‘rcuul bah-lee is 1‘4 a stead; 511%, km (c) [5 points] Voltage sources E1 and E2 are removed and replaced by conductors and all the resistors have the same value R1 = R2 = R3 = R4 = R5 = R6 = 100 Ohms. Draw the . c v simplest equivalent circuit. Determine and label the equivalent resistance(s) and current I 1. ‘/ Br. “, FUZFIFL / "L 4 \l‘ R 5mm? K, 1 Rm‘Rz * ‘ ‘. .~ ¢‘ ,L ’ $9]!ch \\ F \ I PWQHI l K, > R. félz 1" g \l V R) 13 7 \ ‘ t R a. at: > a V a" : V0i+dly w n A {I 7; £35 901“} A! fbenfifrca E RI;2 RI! T0,] S=§=o (d) [5 points] Instead remove the right half of the circuit so that only the left hand loop circuit remains. The two remaining voltage sources are still E1 = 12 V and E2 = 6 V and that all the remaining resistors have the same value R1 = R2 = R3 = 100 Ohms. Again draw the simplest possible equivalent circuit with labels for the components and their values including the current 11. (e) [5 points] Keeping again only the left half of the circuit, at time t = O replace resistor R3 instanteously with an inductor With L = IOpH. What is the current 11 as a function of time? 8:2'421 1W 3R R P r; ‘ ‘ 8 £1 1 @ t<01 ind-4J7! ‘} “ ”pf,“ J I _ S L ‘> 8 L I: 3-; CHI/94’ (“1+ climy (Hr/Limeaql] JIU» W/M .Hdwfin, Io R Ice): 3—i- LOOP I‘M/9.: E‘aRI- L‘dl:0 d1. alt 9 ‘ : ~ :78 e g; “H. (L)I+Z- dt'V‘quB W30), > I( > <9 i + 3 4: 2 __._ -042) 8 - R ‘3 iifl-e m) __1; Mali ,0 2R : awn. : 0-5” 135111563? 5240“,; .—. 0.0544; 2W (L40? (0’0 (L400 %0 to 24W \lo CL 7? I “(FR b (PM : ”Impvmg: {oAxnoVQQVW ‘ “fl ’ 2R a G 2 ‘ . ‘ Q4m§3>hr9§ '3 «L‘ZQ = CPAAR “ Z4300; O 462 v “' (Lb/H04 \N "l CL‘Ef/szgyfim (7:50) “043] Q FW (2/ “Margo, USA leggeg N 4 G70 {30“} a W: PKGQW K th‘CACA. ZnAungncwf a flawed 0? kfitL ,Q N Am: (5.qu '2 B= WEI", @ww W 2 M73: “D? $1: -:> L: WWW T 4L = IBM/7W? 4} Cask COVE) a L, “(H5 N/A’L4 @UbleQT‘rQDQW’I 4x04» L“ (5.67 M“ «Fexmuflwhc Cove: #:4ovwo 3» L. = [WOOL 1”, 6.344 f 2). “" 06L: 377 %‘3 39mU2141LJL—VZL721; ~ / 45002L 4 “mm e. WLJWLM ms; WLQLQ a; COM}. W WW a WW0 WW 0 J WWWM/ We ngw moi/10mm BMWL CWW (9 WM wmtepx 0&0 Jet WWW WS/ W5 W07 W (L $55?” at ”0 EM: Ecuw) Twezggcm \: flzamgcfld’t « V? “%'a {ideas 3. 2 ”a? 7 SD 50/2. 7W8 Cemfv‘fimncmw 9W4 Mimi/1c! an 07) 6% SeUUnm‘d U4 W‘smmh "E” mm (m .. a) 13“ we We! W ox smhgcm% dxsfili b‘VL‘fim/x ) VX5163. Lab; CMC/C: 3V *4 “330 fi 1 +0 + ¢§r[fli)§ IQ cw 431) Wm M mm W Mod/Ly W3? b6 W 01 PW “(M5 t§ W W >2 mi X > 6. [25 points] Microwave Oven (a) [3 points] The operating wavelength of a microwave oven is /\ =2 12.24 cm. What is its operating frequency? 9 r V: 3 (10% W5 : 7 H946! ”’3; 7'2.“t€§@s‘r\ra x (D. \ l’LLi vw (b) [5 points] If the power of the microwave is about 1000 watts and the microwaves are generated in a device called a klystron and transmitted to the microwave cooking area through a wave guide which is one half wavelength wide and one quarter wavelength high. What is the average Poynting vector < [S[ > or Intensity I, and rms Electric ERMS and Magnetic B RMS field values in the waveguide, when it is operating? w: 2. 'L. H A 9 1 . T - <3» ; page”: L; 13922;? : 9.?‘4.“0 H. ('7’: r C)” (OAq’c-LLJ WW}?— W1 (3 mafia? E: 1‘93; “l :‘L r, (3sz ‘> 37‘ It. 46 “5938-8 “’0‘ W l K 2:5 gmg‘r \VLWAVAW PM‘ EM» .3 4777‘ (L (c) [5 points] The volume of the microwave cooking area is 1.4 cubic feet (3.9644 X 10”2m3), what is the energy density, and rms Electric and Magnetic field values in the microwave cooking area when it is operating? Assume that the Quality factor of the circuit - microwave cavity (an LC) and food (R) is such that there is about 10 times the energy stored than is being drawn out per cycle. 19" [’- £7 6.0;‘4'30 (Al/”‘4- : [??‘/O‘g< ~ 1 7. a raw-M " V93 1. (4: )(€E+§— f/LE(G(+<:- : é(€* Q :GE 2 y\ 2, (I), 7’ (J « 32;. (d) [2 points] Why is the microwave oven enclosed in a metal container? Note that even the window in the door is covered by a metal screen and the rest of the inside is fully enclosed by metal. m glen? I l‘r J3; m“, (W3 ‘ Wax / ’I {3&3 We. gasp/m is wdwm ' i ‘ to; w (e) [3 points] Why is it a bad idea to put a cup with a gold rim (e.g. 10 cm diameter circle) (9 in the microwave? How about food wrapped in tin foil? What about other metal utensils or Chinese food boxes with wire handles? How much power could be coupled into the circuit produced by such metal? \Quz WM “M9 06 W qflla’ilow (l-Mlow) MM LQ¥QMQJN\‘MM UJ‘VC WEN? 01%!)qu HMLU (mp3 M(m98;~23 \w % Ué-Qxxd‘\ mil/diss; QéEu KL Mummeggrfi W91 (f) [5 points] One warning buried in the microwave manual is that a clean smooth cup full of water can be superheated so that it has sufficient internal energy to boil but not yet started boiling because nothing has yet triggered the formation of steam bubbles. When picked up or disturbed, it will then errupt in a geyser of hot boiling water. For how long would one have to heat a container containing 100 grams of water starting at 20 C so that it was overheated and ready to errupt into boiling? Assume that the water’s internal energy is distributed uniformly. Rep}? mmfmqrt? wbwmfir \ 04%,? Egg,“ ‘33 W '00 WM SW10”? *0" 03“: my)»: 8000(5me @930 oak/re; : 330W): wt :2 3‘3; SQCQ‘Adé (OQOW “Ge (arm— LQM “(WM :363‘? 93%;»wa =5 53?,CBOC0ML34 7'2 gamma? 2217/9623: 21219699034ng /©QOW 21 7. [25 points] A coaxial cable has an inner conductor with radius r1 = 1 mm and an outer conductor with inside radius r2 = 2 mm and they are concentric. They are separated by an insulating dielectric With H = 5. 2 ‘——-————-—- L——-——> Figure 5: sketch of coaxial cable with dimensions labeled for problem 5. Treat as two concentric cylinders. (a) [5 points] What is the capacitance per unit length of the coaxial cable? To amt 4h mew‘wa , my“ tomb; a war 4am, in ,«A «44+ 05% 5" ~Ku lamr and owkr Cofldwc‘fiafi. M JW‘fi fiek‘ a“ 5L few”! wins Gauss‘ Law‘- {2: § ENE” =21”! E; = Q‘M/éu : '% my.“ I", — ‘ 5* NHL 'Vbh;( f: 63 ML ”‘7 E" 2:6), 4“ e / C /I<. ’ .‘3 ’ R ._ A WWW” “KW“ 13 A“: E“ "Vt—fa.“ =52. V2379 (42‘5" /‘ F/M c/ - 21165 I: , 21:: (ESTHER) 5' _ “(div 2 q .. ,/ ’ m . Ln. “/21 In .2 . . (b) [5 pornts] What IS the energy stored per umt length of coax1al cable, when 1t contains a charge q?? it wok ow?- (Lt/5K LN, 40.2474 1 , flan 1 L “V M3 :: 1. 3% : L...“ J: 32‘ LdMll- lujhn Z It a S" 10*“: 22 (c) [5 points] What is the inductance per unit length of the coaxial cable? first Rut 'HUL 3 mu 3* M wt 35 I - («,1 am Aqun [00? oQ my“; r, W AmPu—Q Um say;- §il3$ =2urB 2f”: =5 13* [IE Z‘Rr el$> New “2% «u. too? on W «x My: Late, m «Cam MA 3 v---‘Zl/’:(4&\Z"“é \aof- v‘s ( Kl 2&4 $541337} 7- SOJ" SK ArIBl flux. Bis J. 4- Ha Santana. "~— = ( : pr: lo ‘K fifth. ”if “a; ‘4 fig. MMthMeuM 53 §=LL s. L‘~ if: fink/2‘ xo'?‘ . IZJUW _’ Z: 0“% = if; M— : ”“32 2“ a 7.1:. (d) [5 points] What is the energy stored per unit length of coaxial cable, when it carries a current I? =1. 2 ; £1221 2 1w— z u 13%. = war? 1‘ 2 23 (e) [5 points] ‘Nhat is the impedance of the coaxial cable per unit length for angular frequency w? Evaluate for w 3: 108 Hz. What is the average stored energy per unit length of coaxial cable, if it carries an. EMS current I ? ‘HAI \‘MF‘AR-"lu Op "Hi-4L I‘m_ (:5 21;:RJL harm J m “re—3m :5 .2641. m unmi- Mode/f Jim" 'H'U— CAIN-e i5 W W t I r __ £— TLU. inflame 0F Ms WJMSftG-fi be; :5 2- I}: (You Hut, 1405- «Mk4 {15, PM“; m5) ' " (a. M .‘A 4‘, LC “cited”, 14 W 1%»? m flu; $33 I “lbw-Crap] YOK m j 71“” «1:!th sfomé Low/3'3 ilk 4&4. defau‘fiwfs 17w] 4? #1 QV'Crigic gimraci ‘M‘j‘j i4 {bu winch“. ’(Lu. 5“or-L([. Lu'jj/bij :4. Ru. 1‘me :5 ”'1. 30 1m we 9‘1”“ W3“) (‘5 m (1,“) + <qc§ : Z<ML\ = LIT»... 24 8. [35 points] Tethered Satellite for electrodynamjc power generated from orbit consists of a major satellite and a secondary satellite lowered on a conducting cable. This concept was tested on a Shuttle flight in a near equitorial orbit around the Earth. The smaller satellite was lowered 20 km below the Shuttle. (a) [5 points] If the orbital speed is 7’ km/s and the mean Earth’s magnetic field is 0.5 x 10‘4 T, what is the voltage difference generated between the Shuttle and secondary satellite? ® Th aemml learn/Lula i944- gwfie é: [Olen €5,302 5d2(€4 Pal-34) A m Aguilar?) VLE) we MW 5: E E V 1- o..¢x/o“"-r. 20Kr03wrx Wham/J z: S¥OOC3V (b) [5 points] What charge must be on the secondary (lower) satellite for this voltage Ap- proximate the satellite as a conducting sphere of radius r— — 2 m with the Shuttle very (nearly infinitely) far away? First calculate its capacitance. Then calculate the charge. a: (2,111 ® V: {/T—é‘ é? bdmu Q Wed 6O t: ”trial/6L: VWWW/U’qumx 7‘“ 55: 15¢ ></o”‘c 25 (c) [5 points] Approximate both the Shuttle and secondary satellite as tWO conducting spheres of radii 3 m and 2 In respectively connected by an electrical cable 20 km long and supporting the voltage difference you found in (a). Sketch the field lines for this configuration and label. Explain what is different than one would have naively thought for the electrostatic case. 1, at declmicl’hc case, 1% ’le We Comdad ta 6: am “He? wok! loud, 'to 515.9%“? "H1. 8pm. Here) llxe, dam? Chili’s/9M2 is CL/ Miriam} Lyra tie, 17‘le for-u 1L +LQ / \ use. (d) [5 points] Power was taken from the system by letting a current flow and power experi- ments on the Shuttle and tethered spacecraft. If the current is I and the voltage is V what is the electrical power provided? How much force IS exerted on the cable? Evaluate for a cur— rent of 1 ampere Does it speed up or slow down the Shuttle—cable—satellite system? (Hint: Does it change the altitude? Which way is the force?) What makes the circuit or closed path? The basically stationary Earth’s upper atmosphere, the iOIlOSpl’leI‘e; is conductive and completes the circuit. F~=IV = 320603. 1:; IQ l3 7» j'ZOK/03/t .SK/O’q: 1 M, ad, at is ferrufiilztwlcr '59 Ha mm. 8% (guild [QUI‘HQ "force- “ Q C(MS {er-(12., GUI—I It \S‘AaU/c/ dag/gait ~H’L Q/Ttl‘tudq. 7Q 'romlahre acts as Q (fiSFQJ‘SJCSA Kidd 1723’“ 71c cQo/mM. 26 Qxd “Tle Vlflikl "Karen 6%»! So we Ural ““6 (e) [5 points] Could you run the experiment backwards? That is could the Space Station use solar cells or fuel cells to produce electrical power and run current down a tether cable to a secondary satellite and produce a force to match or overcome the residual atmospheric drag at its altitude and keep its orbit stable? How much electrical power would be required to keep the Space Station of mass 183,283 kg in its orbit, if its altitude is 353 km and loses about 2 km per year. If the tether cable is 5 km long, what is the required current and voltage? The OrE‘rls ll: £va— J r; l1» lowg milq MM, “4. “while {om Charm o4 6». All; lflflSfl—Utrl: —_ frT-"é‘.é% “0/616“? (5332] 5??!" ‘ c€ 2w] ,. Tie Wife 04 Power find I! “'—"=“"‘~ "‘42 [003500’l70cKLi/L 77%? J/(c/ r~. )I H O (A) & Q-B-v Ext/03x5xKo‘qTJcVMkésfiax/om 27 (f) [5 points] At the Space Shuttle altitude of 400 km the typical electron density in the ionosphere is around 105c'nf3 or 10117713. The typical temperature (thermal energy) is around 1200 K. Near the second satellite (2—m radius moving 7000 m/s) what is the necessary drift velocity of the ionized electrons away from the secondary satellite to produce a 1 Ampere current? How does that compare to their thermal velocity? How much higher would the electron density need to be around the secondary satellite to make the drift velocity reasonable and comparable to their thermal velocity? Chg dare‘lu ix. ‘0le At l! Ao:o i H dN‘z nAvdt so VA‘L N :. nAv= {0”M‘3 m a 7X tMoo Ml: : (RM/0’5 ”(L Cymbal a at}: firm/0'5- Mar/0”": .001 A“? Th). 1 an»? Curried mam/lac c509. to Invert cl) (for? E9;— fld, wed had dwells; {000 {as Mater. (g) [5 points] The deployment was almost complete when the unexpected happened: the tether suddenly broke and its end whipped away into space in great wavy wiggles. The failure was not caused by excessive tension, but rather that an electric current had melted the tether. Unwinding of the reel uncovered pinholes in the insulation and escaping air converted to plasma. The instruments aboard the tether satelite showed that this plasma diverted through the pinhole about 1 ampere at 3500 volts to the metal of the shuttle and from there to the ionospheric return circuit. That current was enough to melt the cable. Why is the current higher at the pinhole point where residual air plasma is escaping rather than at the surface of the Shuttle? Is the electrical power enough to melt the copper braid which is the conducting (not strength) portion of the cable? The heat capacity of copper is about 0.09 Cal/ gm/ K and its melting point is 1084.62 C. How large a piece of copper would be melted, if the heat were concentrated? [Tu Why (“in t busts at It. Orclu— +0 datum was at _ “Paint. will melt , Lei Mod #9 Mada/"fit clad/m Ami/{2m 6m! ac, its Muck POW it tars“ oeu we we em, fir Warm! adding 04 1Q; 0”ng $0 to (5&ch [é ”Adi” fie Carrel?“ (or) {car #4 twitter) tie ram) 258"“ otmf‘afe fle W. ...
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