Final Exam Solutions Physics 6C

# Final Exam Solutions Physics 6C - 1 Pond-S TRMTIQE FOR FRU...

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Unformatted text preview: 1% Pond-S TRMTIQE FOR FRU- loo‘ me. Wmu__.wm w , -—-. PHYSICS 6C FINAL EXAM Name Sb 11370“ s FALL 2001 December 11, 2001 Signature 100 POINTS TOTAL Student ID # __ SECTION A: Multiple Choice 2 points each (no penalg for wrong answers} A B 1) Consider two identical metal spheres. ® @ Sphere A has a charge of +4 microcoul, Sphere B —12 microcoul. They are touched together, then separated. Then @q 4’ . Sphere B .is grounded, They next are touched together again and ﬁnally @ separated. The ﬁnal charge on each is: El +4 microcoul El -1 microcoul Cl +2 [a -2 U +1 D -4 El 0 El none of above 2) For a negative source charge qs, the electric ﬁeld E points: atoward qs "E“ z E! away from qs * a; El concentric around qs [:1 none of above 3) For two points A and B separated by 10 cm in an electric ﬁeld E = 100 Me (as shown), VA —— V3 is: E +--O-—-—---—r——-—) El+1000 volts [:1 + 1 volt E] -100 volts h B D +100 1:; - 1 [3-1000 e—l‘M—‘a 04, 7 El +10 1::l- 10 tjnone ofabove AV AV: ‘0“ ,, E: if Avxrox/ 4) In the previous question, the work done by me (an, external force) to move a charge of+ 0.01 coul from A to B is: bodes {mulls} 'ou.) 13+ 10 volts uwoﬁoku 1:1 -1ng El +1 [3- 0.01 E; ~10 El+0.1 12:1 -0.1 -[:[.none of above 10) For a magnetic dipole moment (“mdm”): If the current is Edoubled If the area is Eldonbled doubled,the mdm is: Dhaived doubled, the mhalved [:1 unchanged mdm is: [:1 unchanged [3 none of above mnone of above 32 11) Two slabs of Aluminum (which has a non-zero nuclear magnetic A dipole moment) are being studied with NMR as shown. 3*: m T 316\$ A The primary ﬁeld in the z—direction changes such that 8; ‘ 1.7-1. 5 5‘ t, B Slab A has B1 of 1.0 Tesla on it and Slab B has 1.2 Tesla on it. The ratio of the 13x oscillation frequency to excite _ mm the nuclei in'SIah B to that for Slab A_ is: [I 1.44 [I 0.83 El 1.20 [:1 0.69 £11.00 [:1 none of above 12) For the diffusion situation as shown, If the cross~sectional area of the tube is doubled, the current is: Edoubled ﬂhaived munchanged gnone of above If the length of the tube is doubled, the current is: Ddoubled Eﬂhalved {Junchanged [31mm of above 13) Heat transfer by conduction in air applies in cases of : 13 moving air Emmi-moving air Heat trait. --'-fer by convection in air applies in cases of: 13 moving air I] non—moving air [9 Points B-2) Consider an inﬁnitely long straight wire of uniform 5 ——-""" charge per unit length .\ = 10A-6 coulfm. We will £_ _ - _ 5'" aim to establish the radial electric ﬁeld a distance of {6- oc- }\= O'btl'm 5.8 cm from the wire. a) State Gauss’s Law precisely (either mathematically ‘ (‘91) or in words). g email = Lurk Qwiml b) To pick Gaussian surfaces which {I pt) make the integration of Gauss’s law ,- easy (for some particular source charge conﬁguration), what criterion or criteria must the surfaces satisfy? c) What shape Gaussian surface would (I 9") work well to establish the radial electric ﬁeld near the wire? watt) 5 .3 u (I) Find the algebraic expression for the @E‘ R (l A '-' 0+ 0 ‘l' El?) Inn L - q“ k L k (3 pt.) radial electric ﬁeld near an inﬁnitely long uniformly charged wire. Find the numerical value of the radial electric field at a point 5.0 cm from a wire with charge per unit length of A = 10A-6 coal/m. =7 EM= Mot/R r6 _ '1thqu y lo 5’ : , 1 l ‘05, “um t/c i '\‘ (Fit L‘lli-‘UO Vb: ‘9. U SECTION C: “Medium Length” to p.-. h’ts 0-1) A container has 2.0 liters of salt}’ seawater with 60 gm m of NaCl in it (as shown) and also 20 gm of K" ions. ’— The atomic weight of NaCl is 58 and of k" is 39. M A9 for NaCl is 12.8 and for Kl is 7.4. The seawater has a hollow insulating cylinder of length ”tn“: "‘3‘?“ 10.0 cm and radius 1.0 cm immersed in it (and filled with seawater). A 12.0 volt battery is connected to metal plates (Area: It r1) on each end of the cylinder. [Note: (fr-'- ha‘. (2'. ] l?» ,1.) a) "Talculate the conductivity of the electrolyte. ‘ I i I G‘Tot‘ 0"" u'l' 01+ ': A ll‘lﬂ-Ql) Quelsl 'l” As it." H C mu = 6:31.53— "' snhole/liicr P. k.“ = 3‘-7-:..—-= It“. Male/tier O. 1‘.) b) Write the ormu a ior {lie registiwty 1n " P terms of the conductivity and calculate ' the resistivity. =) (Egg lmmsn asmtts ‘ESIA—m (a .12) 0) Write the ________fprrnula for the resistance between the ends of the cylinder in terms of the resistivity, etc, and calculate the resistance. (5. t.) d) W through the cylinder. ’ P I =V/R-7 ”/3732.“ 0.31191me 2. {Q t) e) Calculate the power diSSipated into the E) 51151 '5 0,311 '3! 37.3.. "3 3. 8 l: ”Ml-t 'P .. electrolyte within the cylinder. [7:512:10 ‘3 Po‘: n1. (39*) Us 93-) H at) misﬁt-"N SECTION D: “Long” Del) A mass spectrograph was one of the very arduous (but viable) ways during WWII in which ﬁssionable U235 (0.71% abundance) was separated from U238 (99.28% abundance). 'In the process, Uranium atoms (92 protons; 143 neutrons for I U55, 146 for U233) were ionized, typically with 7 electrons V = no.0!) a removed. They were then accelerated across 40,000 volts and entered a 0.3 Tesla B—ﬁeld at right angles and traversed a semi-circle . (as shown). The U235 has a slightly smaller orbit and was k, ion source. e? collected in a different slot from the U238._ ‘ t?:§§,::£3§‘:uf _ _ _ 7 7 _ _ _ 'nn‘vud (hmovttn . a) Derive the algebraic expressron'rfor the speed of the‘Uramurn 10115 when they entered. the B-ﬁeld and then the expression for D (the distance of the slot from the entry point) in terms of the Uranium mass, the net ionized charge of the Uranium ion, the accelerating voltage and the size 01th:: B-fie d. - V . m..— s- 07’ GbV=€w~v1 ear: m 2m-w¢s 1-2. 1. 1. "- b m m ovguﬁm‘té’: “33:: 9:23 a D ._-—— t V _ _ :1th 1m 1 V _ 3:» EL- '753'1.;75"3b-g\imrei 1, [-—--————'—‘—', - [-16.35th way across \ hate my "I 3;. b) Find numerically: D235, D233, and their difference (it’s not very bi gt). You may use the mass of protons and neutrons of 1.67 x IDA-27 kg. 0) What is the fractional difference between F _ __ ) D235 and D238? freshen“) btuercucﬂ. " (Note: If need be, this can be answered without = 0 .0953 = G - L32: doing part (b).] [a Poi ntS (int) it (11ft) (ii-t) SECTION B: “Short” [ r—ir 3-1) For the circuit as shown: a) Find Im the total current ﬂowing out of the battery. b) Find the potential difference VB .. V... c. 5 o g 0' c) Find 150 (the current in "' 3-2: ' 0J5”? the 50 ohm resistor) and Im (the current in the 200 ohm resistor). un-I’ U; V ‘VL . E" _ l1“ £5: "an— 6‘0““? ENott '- T\$e+11ou= To] (1) Find Fm (the power dissipated in the 200 ohm resistor). Rm? has“: noo= 0.32. We ll loam. 5) A parallel plate capacitor has an area of 0.001 mg, a separation of 2.0 mm, and a dielectric constant of 5.0. Its capacitance is: A‘iO-ahn‘ J i=1 M [213.5 x IDA—9 farad [max 10A—11 farad “‘F—Fs ‘r "‘ Elaax ion—11 C14.4x10"—11 JR A [34.4 x10A-11 1:] 8.8 x 10"-13 C : Ha [I 3.3 x IOA-ll [:1 none of above q “ K 6) For a sphere of uniform charge qs, the electrical equipotential lines are: ,. 1r E] toward qs concentric around qs D away from L}, a , 53 none of above 7) For the circuit shown, the time constant to charge the capacitor is: . [110“+8 sec [:1 10A+2 see a 10m sec vb: 10 v = W” 1310A+6 [:1 10A+0 u IDA-6 I , _ EllO"+4 [:110A -2 a long L— to ' 1:] none of above 8) For an electron moving it? a magnetic ﬁeld as shown, the force iii [:1 down E Into the board Elleft C] out of the board 9) For an elaqtron moving in the opposite direction from a proton, the mforces between them are: _, e on the electron is: [1 up n U right gnome of above ; it —-—-—9 E attractive a zero grepulsive 1:] none of above A: f t l? Petrol-S 8-2) At t=O, the switch is closed in the circuit as shown. (a: 0.1;»; V3: 10V [I pt) a) Calculate the current tltmnghtheresismLanQ—\ _ V}. to \l. “WOW-ﬂ 42- F0: 0. i can)? (591) b) Shm; til—atlas t1 approaches inﬁnity: My: 0:, all ultimate“ in“). ctllgg‘l‘giéé‘iﬁew; ‘ I * V0 ‘3 {lav-ought R. :2 = 3 32:: f. V| 5 it“! “t“ Qt: VI CI: 16‘1”“; ‘ 10MB I 3 = f . . ‘- .J, __ l, _.3— = J’ ‘ .. Ql =20 m1crocoul cz‘lts “‘“ﬂ‘é' " Cat—“why" :1" C3 Elsi: 5:4 " «31;;th Q2 = 4 microcoul S‘ V A! a to . - , Q; = 4microcoul name 3 c i Q‘r‘v“: Q1‘ Q3 2 VCva: lm-‘t‘l' Lime (3,91) 0) After all capacitors are charged up ® . 91mm ® _ (in part (b) above), at t’ = 0, capacitor \$_ —-— "H C2 is shorted out (as shown). [80, all \I U Q C p? of +Q2 on the top of C2 flows and V51 10v "—3. - ' {- neutralizes all of —Q2 on the bottom of C2.) . - Q3 '1“ At t’=0,whatare: V -V andV -—-V ‘2 Vk'v c: l G V (1W (3 P10 e) For t’ > 0, write Kirchhoff” s Second Law for the sum of the voltage changes around thencircuitcohkotwemg 'lzlu. Inﬂu- + (.3. outﬁ'h ’ﬁ ghoul: (at A 3 (.30 °l°El<w\5u 3: it'll R - ﬁctive. +Vh=0 ,0 ’3‘) 1') What is the numerical value of the ti - constant as C, charges up ? l T = RC3: 1063114,.“ 1 g 1695““ l (3 t) g) At t’ = infinity, what are the numerical ? values of the current through the resistor and the charge on each capacitor? itt'm) =0 g Q :L|V551MFﬁlOI=3XlQ L \ Q3 =c3vb= 1»? m) =1x 155;, (Q3 [0 Pet-“ts C-Z) In the “classical” Bohr model of the atom, in a Hydrogen atom the electron orbits the proton in circles of radius é‘i‘ﬁ'm pain“ A 0.529 x 10A-10 m and at a speed of 2.19 x 106 rnfs. IE :uJJu s +——. B (-1 Pt) a) Show that the period is 1.5 x IDA-16 sec and the current is 1.05 x IDA—3 amp. ”.mm “to an; r : e.smxlo , _ T n; W: x [641; I: inbixig ”’SCC- 1.; 1/1,, Lexis” l.§'>.x|o"‘ = Le s’xto‘3 (.1 pt) b) Calculate the magnetic dipole moment M =IA. W A: n w‘:[o,§1qxlo"°:l ﬁ= ‘E‘J‘ix loll»: M: I A '- Logxto'z x seq 7: [0-14 2 61,13 k 30"“ WW"? (3 Pt) 0) If it is in a magnetic ﬁeld of B: 0.1 Tesla, calculate the potential energy difference between the maximum potential energy state and the minimum. Compare this with thermal energy of ~ 6 x IDA-21 joule. Will this B ﬁeld be able to keep the atoms polarized with the magnetic dipole moment along the B field? APE. E PEmkx'PE'Q‘“ whim-(i FE:P££.E = nam<>°mawl°° e lust-s 1:1“..13): lib-H xOrl = 1353‘ [0'12“ 30min. m Em,“ ~LMH\$WK ”33—2-di w‘,“ 53: Keep «items olcétimd ii («at “‘03" zﬁ‘km‘ l—m ' 0 1mm: =M3Wq‘3 5» —1 = 9.33M?" m =11; no n-m d) Show the direction of the magnetic dipole moment (relative to the direction of the B ﬁeld) for the maximum torque on the magnetic dipole moment. Calculate this maximum torque. l3 9'1) 7200 !2 Po‘mts 3-3) Consider the expression for the potential V from a electric dipole oriented along the +x—hat axis, as shown. a) Re-write the expression for V in terms of x and y (ht) (rather than 1' and theta). b) Show that this yields the correct expression for Ex, U! pt) the x—component of the electric ﬁeld. “ﬁlthy A T a A ——-"—"—‘>+x 34 me T?- ...
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