2212_Test3_Key(3) - PHYS 2212 Test 3 Spring 2011 Name(print...

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Unformatted text preview: PHYS 2212 Test 3 Spring 2011 Name(print) K6 Instructions o Read all problems carefully before attempting to solve them. 0 Your work must be legible, and the organization must be clear. 0 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! a Show what goes into a calculation, not just the final number, e.g.: a—b — M51193) 2 ad — (2x10— )(4x10 ) 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 The Final Exam will be held during Exam Period Seven, Wednesday, May 4, from 8:00 to 10:50am. If you have a conflict, please contact Dr. Greco by Monday April 4th. PHYS 2212 Do not write on this page! Problem Problem 1 25 pts) | Problem 2 25 pts) Problem 3 25 pts) Problem 4 25 pts) AAAA Problem 1 (25 Points) A circuit contains one battery and three wires in series as shown in the diagram. The wires have different lengths and are made of different materials. Two of the wires are identical: they are each L1 = 7 cm long, and are made of material having 111 = 6 X 1025 mobile electrons per cubic meter. The central Wire is L2 = 5 cm long, and has n2 2 2.5 x 1025 mobile electrons per cubic meter. All wires have the cross—sectional area 4 X 10"6 m2 and the electron mobility of each material is 5 x 10—4 (m/s)/(V/m) The emf of the battery is 2.9 volts. In the steady state, the potential drop across the length L2 of wire 2 is 1.33 V. (a 15pts) In the steady state, what is the magnitude of the electric field in the left wire, of length L1? Show all steps in your work. / , ‘(0 M (C; —2 .. -7. .10 Ex , 2L, + a—‘LLZ (mm W 7’7» (b 5pts) How long would it take a single electron to travel from the negative end of the battery all the way around the circuit to the positive end of the battery? Show all steps in your work. ¢:‘tl2_+t +t\ :2t'+t7_ Ml L L l I :26 + ’3 V: ME ulelEl: T: l V; / (0-5 M/ §.(0_4MS >(H2V/m>:5§8 S ‘ [3M AWS) (IL/”2 > 2. '0’ \I;: E; : liq: (gm Wm)» H Mv as '_ ’2 no‘hcc V > I j vz , l 33 /o "4/5 [ ,‘+ shat/3! 1m —2 7 10"") (M : T: l (flirt/5) + (l«%3-/o"ZM/s) 28 015 Problem 2 (25 Points) A round light bulb is connected to two bat— teries, each with emf = 1.5 V, and a conven- tional current IO = 25mA runs in the circuit. An uncharged capacitor (C = 1 F) exactly like the one you used in class and lab is connected in series to the battery and the light bulb. 25 mA (a 5pts) Sketch a graph of conventional cur— rent vs. time in the new circuit (including the capacitor): 305 (b 12pts) The circuit with the capacitor is allowed to fully charge. On the diagram, at location A inside the wire, draw and label three arrows representing the: (1) net electric field Enet, (2) the electric field due to the capacitor plates E fringe, and (3) the electric field due to charge on the surface of the wires and in and on the batteries Esurf- If any fields are zero, explicitly state this. (0 8pts) Calculate the charge on the positive plate of the fully charged capacitor. [00? NILZ AVWO‘VL O 3],”? : 2em¥ «I but“ '5 ” Problem 3 (25 Points) A battery with an unknown emf is connected in series to an ohmic resistor with resistance R1 = 5 Q and a non—ohmic resistor. The resis- tance of the non—ohmic resistor is given by the formula Rn(T) = R0 +r0T, where R0 = 20 (2; 7’0 2 0.018 Q/kelvin, and T is the temperature of thenon—ohmic resistor in kelvin. non—ohmic resistor (a 15pts) When the circuit is first connected; the non—ohmic resistor is at a temperature of 273 kelvin, and the conventional current in the circuit is measured to be 3.00 A. Twenty minutes later, the non—ohmic resistor is at a temperature of 400 K. What is the conventional current in the circuit after the 20 minutes? Vise [004’ Mt Mtge: T:7_?5K : who? -— emhtfi. 40mm) 50/ 571+ ZofL—k o‘Olfér—E (£43K) affl/ It: Io KtaKnHOO) : 34 SJUZo/L-POD‘SMJ‘EHW‘) l": 2.23:4 < Io (b 10pts) Considering the microscopic view of resistance, what material or geometric property of the non- ohmic resistor likely changed during the 20 minutes? As best you can7 describe how temperature affected this property. m MENU \/Lo.s decmqsecgi 24$ vH/tL resielzsr 6 MS Vank; Wow-e, With U? [m whim/18c o \ (ollc’J/KD MS Problem 4 (25 Points) At a particular instant a proton is moving with velocity (5.5 x 107, 0, 0) m/s, and an electron is moving with velocity <0, —7 X 107, 0) m/s. The electron is located 1.5 x 10’“6 m below the proton (in the y direction). V p T @_> 33 I ' y {7‘3 ‘ ‘. E : ‘_>’ 9! ’3 (I’__» <1) t "‘- : EMSLT ei "‘ I 2<-" . (9 x _l_ 59$ _:.) z (out of page) (a 4pts) On the diagram draw an arrow representing the electric field at the location of the electron7 due to the proton. Label it Ep. (b 4pts) On the diagram draw an arrow representing the magnetic field at the location of the electron, due to the proton. Label it Bp. (c 4pts) On the diagram draw an arrow representing the electric force on the electron at this instant. Label it Fez- (d 4pts) On the diagram draw an arrow representing the magnetic force on the electron at this instant. Label it Fmag. (e 9pts) Calculate the magnetic force on the electron at this instant. Your answer should be a vector. W wr. v? A Sho all 0; J a __ 7g : £9 gp VP xr (Ma -/0"'7C) {gr/07,0,» W5 x<o, «i, 0) ——>> ~¥TM //'T——_—’— BP:(/'/0 7(- (Lg-IO'6M) 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 1 d6 er a d" azg V £23161; Handfiwmaifv<<c Ayelzqév AV:—f1/:Eodl~—Z(EmAa:+EyAy+EzAz) <I>el=onfidA (PmangBofidA fE.fidA:;M ffiofidAzO 60 demag |emf| = fENC 0 df: dt f g . dz: #0 [2 [inside path + 60d dtf E . ndA] Specific Results 7» 1 2 s . —» 1 s . |Edipole,aa:is 7“) 47T€0 TQ3_ (on 3X13: 7’ >> 3) 'Edipole,_L[ z 47r€0 (:13 (on J- aXIS, T >> 3) a 1 Q a E = —— J. f t l t ‘ d' 1 t = ”2 E ~ Tod 47T€0 7‘ r2 + (L/2)2 (7“ rom cen er) e ec r1c 1p0 e momen p qs, p a applied —' 1 2Q/L _ —. 1 qz _ Erod N 47"60 7" (1f 7' << L) ring :04’IT6 W (2 along axrs) a Q/A z . Q/A Q/A . [Edisk = —2 60 1 — ———(z2 + R2)1/2 (z along ax13) 'Edwkz 260 [1 — El ~ 260 (if z << R) Ecapacitor ~ Q—e/ (+62 and —-Q disks) [Efringe N Q/ (2—S‘R) just outside capacitor 0 6 ~ I AZ _. _. _, AB: ”—9 J T (short Wire) AF = [Al x B 47r 7' ‘4 M0 [11%de 2—1 { —¢ _. B ~ = ———————— L B . = I t a ”LU’LTE 471’ 7” 7'2 + (L/2)2 WEI—7T 7’ (7' << ) wwe Bearth an no 217TR2 N [10 2171'R2 . 2 ZZWWNE Z3 (on ax15,z>>R) [.LZIAZI’ITR #0 2M . -» ,uo , m E? (on ax1s, 7‘ >> 8) Bdipole,.l.‘ ’5 Eff),- (on —L 3X15: 7" >> 3) ~ E —. 1 — a A A A _. Tad Erad : Q2 J— 7} : rad X Brad Brad 47T€0 c 7’ c iznAr“) I=[q|nA?7 v—uE I L 0 IQ| W A a 0A E ~ 1 1 Edielecm'c = applied AV 2 q —— — — due to a point charge K 47reo rf 7‘2- A 121w for an ohmic resistor (R independent of AV); power = I AV R |AV| ‘ _ Q = C IAV| Power = I AV 1 = — (ohmic I‘eSlStOI') d” -' 2 K % %mv2 if 11 << 0 circular motion: d_1:1 = [—17%] I]?! % -mRi Math Help a x 5: (am, ay,az) >< (bmby, bz) = (ay bz — az by)a": — (ax bz - az bag + (am by — ay b$)2 / dm 1 ( + ) + / dm 1 + c / da: ' 1 + c = n a a: c ———— = — = ____ x+a (96+a)2 a+m (a+a:)3 2(a+a:)2 a 2 2 a 3 /adx:a$+c /amdm=§x +c fax dngm +c 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 e 9 X 109 N - rri2/C2 7F 0 Epsilon—zero 50 8.85 X 10“12 (N - m2/C2)“1 Magnetic constant g9 1 X 10“7 T - In/ A 7T Mu—zero H0 47r x 10“7 T - m/A Proton charge e 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 Ra "NV 1 x 10—10 In Proton radius R1, m 1 X 10“15 In E to ionize air Emma z 3 x 106 V/m BEarth (horizontal component) 7 BEaTth m 2 x 10—5 T ...
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