2212_Test4_Key - PHYS 2212 Test 4 Dec 1st 2010 Name (print)...

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Unformatted text preview: PHYS 2212 Test 4 Dec 1st 2010 Name (print) 6/ Instructions 0 Read all problems carefully before attempting to solve them. 0 Your work must be legible, and the organization must be clear. a You must show all work, including correct vector notation. 0 Correct answers without adequate explanation will be counted wrong. 0 Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything you don’t want us to read! 0 Make explanations correct but brief. You do not need to write a lot of prose. 0 Include diagrams! 0 Show what goes into a calculation, not just the final number, e.g.: 3—17 = éfigffiifig‘ig = 5 X 104 0 Give standard SI units with your results. Unless specifically asked to derive a result, you may start from the formulas given on the formula sheet, including equations corresponding to the fundamental concepts. If a formula you need is not given, you must derive it. If you cannot do some portion of a problem, invent a symbol for the quantity you can’t calculate (explain that you are doing this), and use it to do the rest of the problem. Honor Pledge “In accordance with the Georgia Tech Honor Code, I have neither given nor received unauthorized aid on this test.” Sign your name on the line above PHYS 2212 Do not write on this page! Problem 1 (25 lots) — A Problem 2 25 pts) Problem 3 25 pts) Problem 4 25 pts) Problem 1 (25 Points) to power supply A bar made out of a conducting ma- terial is 0.21 m high with a rectan- gular cross section 0.05 m Wide and 0.004 m deep. The bar is part of a cir— cuit and carries a steady current (the diagram shows only part of the cir— cuit). A uniform magnetic field of 0.95 tesla is applied perpendicular to the bar, going into the page (using some coils that are not shown). Two volt- meters are connected along and across the bar and read steady voltages as shown (mV 2 millivolt = 10“3 volt). The connections across the bar were carefully placed directly across from each other to eliminate false readings corresponding to the much larger volt— age along the bar. Assume that there is only one kind of mobile charge in this material. to power supply (a 5pts) What is the sign of the mobile charges, and which way do they move? To receive full credit, explain your answer with a supporting diagram. . 4 1 Post-HM Char?” “(CI/n4!va on. “HA-b W7“; S‘J‘: "f 3%" baa (val Wk“ ) WW (vaumw 2) v En PMth &9“j"‘ ‘H’u’ (VJ Pond—s down 9%") If ,Chfl¢~ CM“! e3 A QCW 0W Max/h!» [>675 7, ; 5./\£V\, «not flan/L13 7‘!) M 5:5 “flu MW“th 0k“/'65 4/4 Pasnhw—e and! Mdv-L Jan/916m,“ (b 5pts) What is the drift speed 1; of the mobile charges? Explain your reasoning. :‘L 551641752Lq¢o 7%, Mob/c char/g) mge oil/o 3.4 7€fc¢ W 04.111506 dc’MD‘I/I‘Dfl/ (c 5pts) What is the mobility u of the mobile charges? \/: ME“ 2: VLAV“/9~ Vs “55"(0'4 “4% 0‘2"“) ’ / /_ \A’ AV“ ’ (2‘1V) ,5 M m: 50840 V/m (d 5pts) The current running through the bar was measured to be 0.55 ampere. If each mobile charge is singly charged = 6), how many mobile charges are there in 1 m3 of this material? ’5 jigv’lAV :17 m: “J; i/iv’ 0.5571 AM ._ __ «4 Vi : 0‘9‘(a‘iqc)<o.05m)(0.004m>(("5 3'“) "4/5) 25' ’3 ': 21(34'0 W‘ (e 5pts) What is the resistance in ohms of this length of bar? D ~_< |\ 7D I H)? Problem 2 (25 Points) A neutral rectangular copper bar with length L = 0.55 m, height = depth = w = 0.02 m, is pulled to the right by a rope (which is not shown), at constant speed 11 = 0.3 m/s., through a region in which there is a uniform magnetic field of 0.43 tesla7 directed into the page. The copper bar slides along metal rails, with negligible friction. (You may not need to use all the information given in this problem.) In the following questions, if any quantity is zero, state this explicitly. :3 wka . A 3 J) (Adah/‘05 (F:% J )(8 ) (a 5pts) On the diagram, draw ”+’s” and ”—’s” indicating the charge distribution n and/ or on the bar. Explain your answer. Full credit will only be awarded to a complete explanation. “NH/u— t/f‘Uw’of (hitch cm vita, Cltlotféle SCEOJu-clts _ (b 5pts) On the diagram, draw an arrow indicating the direction of conventional current through t e resistor, and label it ”conventional current”. (c 10pts) What is the absolute value of the potential difference between the ends of the bar? To receive full credit you must start from fundamental principles (do not start with a formula that is not on the formula sheet). Show all steps in your work. If vii/w, know Is (JUKJJ a} (oquwi’ Speedl/ 44m. CUW‘LVL‘i’ :5 akawi“ Ami ‘Hu, Wolm‘i’fc. "Q/CL 330m “HA4 altarxlés iv. “HAL bow is )Uéi’ («my Waqui/u, éieo‘l’hc f‘ofci Aida 41. “We 9~€J>AKKM oil/WWTC. lfmi'l' iii :7 $1132 glib” Dr {W 2 «B :> WWI: EWL :VBL 3(0‘3m8)(0‘%7)(m‘m) (Moat: 0.07)\/ (d 5pts) Assuming a 109 resistor, calculate the power that is necessary to move the bar at the constant speed 2)? CM». ofo “HMVJ +000 WW Si 1"“ W» «Li ’ : o VJ D P J when, j P: Q [(911 {.0 /0 Problem 3 (25 Points) / \ \ / \ \ \ Two point charges, one positive and \ one negative, are brought close to an / uncharged metallic bar magnet. As I shown in the Figure, the bar mag- net becomes polarized. Consider an arbitrarily shaped Gaussian surface around the bar magnet. The two point charges remain outside the Gaussian / surface. ' / K \ \ sw) (a 6pts) Determine the net Electric flux through this Gaussian surface? Be sure to show your work to receive full credit. (b 6pts) Determine the net Magnetic flux through this Gaussian surface? Be sure to show your work to receive full credit. (PM : @fi-MA : 0 Via Mir MKYMHC flux“ (c 6pts) A new Gaussian surface is shown in the Figure on the right [This surface includes one of the point charges]. In this situation, does. your answer to part (a) or (b) change (circle one)? If so, you must explain why? Electric Flux is now: @@ Same Magnetic Flux is now: Different éucloseoa malt Cl/totvje 220m): ' a $13: 16:: figwfi Wt M 3 0 film? ((1 7 pts) A second uncharged bar magnet is brought inside the Gaussian surface shown in Figure 1 (see Figure 3). In this situation, does your answer to part (a) or (b) change? / \ \ . . / (Cerle one) If so, you must explaln wh? \ \ Electric Flux is now: Different \ Magnetic Flux is now: Different @ I I , We enclosecfl M'lV’CQ/L‘C/‘fifi so flame-<0 ' / ¢Z§E.RJLA:Q l \ E \ @E’M/P :0 @430 Problem 4 (25 Points) A coil of wire is connected to a power supply, and a current runs in the coil. A single loop of wire is located near the coil, with its axis on the same line as the axis of the coil. The radius of the loop is 4 cm. At time t1 the mag- netic field at the center of the loop, due to the coil, is 0.5 T, in the direc- tion shown in the diagram; the current in the coil is constant. trail (a 3pts) What is the absolute value of the magnetic flux through the loop at time t1? ®M=\%'hd¥l 2: BA 3) 2: 313,: {osflfiXaollMZ : 2.;i-(o'3TmZ M (b 3pts) What approximations or assumptions did you make in calculating your answer to part (a)? VW Mac‘vxxln‘z. e\ «l (s: appmkamkb M ‘ Sam WWJW swan ‘HA—L «AER we” bog -/ m :JdM O‘HOW M; L ls (Ammas WWqu “(but £5“ £332 (0 3pts) When Viewed from the coil, What is the direction of the ” curly” electric field inside the wire of the loop at time t1? (the current in the coil is constant.) EQCRUDC “HA1 OWN—(4+ l‘s (Kassie/wt} “liar £2162 £065 woi' Cl/MVLZC-wm" ‘kmb that“) °Huzm :5 Mo Co’s/[L7 eléolm‘c «Mal a} t,. (d 3pts) At a later time t2, the current in the coil begins to decrease. When viewed from the coil, what is the direction of the ”curly” electric field in the loop at this time? Indicate this on the diagram with arrows. see/litM “L” “likl‘ CM 9 imp b§~ i; 4? 15g mm: “M (allot 5.53.); l 2 ‘ 30 Mi ’45 Powds fig: H~UA€¢ girl/UL (WM/V! élffilVl‘C/ field Pm‘vtts {lean/41V clockwis€ WW“ view/(LO “(l/ow» % Carl. (e 9pts) At time t2 the rate of change of the magnetic field at the center of the loop, due to the coil, is —0.25 T/s. At this time, what is the magnitude of the electric field at location P, which is inside the wire? (f 4pts) Now the Wire loop is removed. Everything else remains as it was at time t2; the magnetic field is still changing at the same rate. What is the magnitude of the electric field at location P? Explain how you know this to receive full credit. . I" «311/ Saw. £5145 / tug :53“ /C w 4M {5 CNcULCA? wit” “Hams, 4/00/ M Vlo C'Ulfv‘ev‘t‘i 14/09“ [7/6 {/Lo (90?. This page is for extra work, if needed. Things you must know Relationship between electric field and electric force Conservation of charge Electric field of a point charge The Superposition Principle Relationship between magnetic field and magnetic force Magnetic field of a moving point charge Other Fundamental Concepts d" d" _, d“ 6:67: d—f: net and—d—fzmé’ifv<<c AUel = qnv AV 2 — Lilli" o dfw — 2 (Ean: + EyAy + EzAz) (EelszofidA @mangBofidA ffiofidAzw féofidA=0 0 lemfl = fENC' . dl : ’ dygag fB . Z #0 ZIimside path —' —’ d —a A f B . : :u'O Iinside path + 60% f E . Specific Results —» 1 2 s . a 1 8 . Edipole,awis % 47T€0 "‘53— (On 3X15, 7' >> 3) Edipole,i‘ % 47r60 % (on J— aXISa T >> 8) —; 1 —; Emd 2 47m) :3ng (r J. from center) electric dipole moment p = qs, 16’ = a Eapph-ed .. 1 262/ L . -’ 1 qz . Erod ~ 471-60 r (If T < E1"ng 2 (Z along aXlS) a Q/A z _ a N Q/A z Q/A , ‘Edisk = 260 1 — m (2 along aX18) ‘Edisk ~ 260 [1 — ~ 260 (1f 2 < R) Emma-to," % Q/ (+Q and —Q disks) ‘Efm-nge x Q/ just outside capacitor 60 60 2R 6 [AZ A a a 1 AB 2 £9 2X T (short wire) AF = [Al x B 7r 7" r 0 LI 2] a _, 'Bwire = W 7' < L) Bwire = ‘Bearth tang _. _ ,uO 2I7rR2 N no 2]7rR2 , _ _ 2 Bloop —EW~E Z3 (on ax15,z>>R) ,u—IA—I7rR ~ #0 2 . a . Bdipole,ams % (on aXISa T >> 8) ‘Bdipole,.l_‘ % (on J— aXlSa T >> 5) 4 1 —q[i_L A A A _. Erad Erad : 47T60 Czr '0 : rad X Brad Brad : c i=nA17 I=Iq|nA17 17=uE I L or Iql nu A a 0A E - 1 1 Edielectric = algzw AV 2 4 q [— — —] due to a point charge 71' 60 T‘f Ti IAVl I = R for an ohmic resistor (R independent of AV); power 2 I AV [AVI . . Q = 0 [AV] Power = I AV I = R (ohmic res1stor) dfl -' 2 K m @7202 if 1) << 0 circular motion: d—IZL = 11R, m % Math Help (ix 5: (az,ay,az) >< (bx,by,bz) = (ay bz — a2 byfi —- (am bz — az bx)? + (am by — ay (7&2 /d:c 1(+m)+ / dx 1 +c/ dx 1 +0 2 n a c —— = — —— = ——-— x—l—a (510+a)2 (1+9: (a+a:)3 2(a—l—ac)2 a 2 2 a 3 /adm=ax+c faxdmzix +c /aa: dxz-gac +0 Constant Symbol Approximate Value Speed of light 0 3 X 108 m/s Gravitational constant G 6.7 x 10—11 N . m2/kg2 Approx. grav field near Earth’s surface 9 9.8 N / kg Electron mass me 9 x 10—31 kg Proton mass mp 1.7 X 10—27 kg Neutron mass mn 1.7 x 10“27 kg 1 Electric constant 4% 9 x 109 N - mZ/C2 0 Epsilon—zero 60 8.85 X 10—12 (N - InZ/C2)_1 Magnetic constant 5—0 1 x 10—7 T - m/ A 7r Mu—Zero M0 47r x 10’7 T - m/A Proton charge 6 1.6 X 10“19 C Electron volt 1 eV 1.6 X 10—19 J Avogadro’s number N A 6.02 x 1023 molecules/mole Atomic radius R, w 1 x 10—10 m Proton radius R1, m 1 x 10‘15 m E to ionize air Eiom'ze rd 3 x 106 V/m BEarth (horizontal component) BEarth % 2 X 10—5 T ...
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2212_Test4_Key - PHYS 2212 Test 4 Dec 1st 2010 Name (print)...

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