2212_Test4_Key-1 - PHYS 2212 Test 4 Spring 2011 Name...

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Unformatted text preview: PHYS 2212 Test 4 Spring 2011 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. 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! 0 Show What goes into a calculation, not just the final number, e.g.: 7—3 = W = C 5x104 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 4th from 8:00 to 10:50am Locations TBD. The conflict final exam will be held during exam period ten Thursday May 5th from 8:00 to 10:50am in Howey N208. PHYS 2212 Do not write on this page! Problem 1 (25 Points) 0.05 m +L.—t+ To power $11 A bar made out of a conducting ma- terial is 0.21 m high with a rectangu- lar cross section 0.05 m wide and 0.004 m deep. The bar is part of a circuit and carries a steady current (the dia- gram shows only part of the circuit). 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). 0.21 m _Y_ To power B = 0.95 T supply Two voltmeters are connected along and across the bar and read steady voltages as shown (mV 2 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 voltage along the bar. There is only one kind of mobile charge in this material. (a 5pts) What is the sign of the mobile charges, and which way do they move? Explain in detail using diagrams to support your explanation. W “dot/{J 09‘ VDl'l'mb-l—fzrz 5".“5 us (REGHN pt Chm/vies have, callioi-M CM -HM. V‘f‘ll’d' side 09— +04. bar (364- jl M Hula/i3 Rom wll—welw 1 gives us wcgl— 4km COW/Wi‘latl cdvmvi+,f, films Mus—ch bur. Ii: we. - . r‘ kl- “55VW-e. + dag-(Qt Caulk/s] “Ha; Waowkcgfct 15 ’lb 'H": "I ’5 2x3 (b 5pts) What is the drift speed 1) of the mobile charges? Explain your reasoning. F 1 Z V ‘ . Alzlor skaJTS‘imk, Mars loam €Sfibligk¢rflj . So meil-e m viomcavdak ( Crass wt“) ant on Cup/{la 05246. - ' - dds/‘9. . a ' CW7“ '5 aw i “My “M '“ 5"“"7 ‘h‘ {’36 MW _ “us 00w“. 94w. 54/. l "V" m 4 ... / qwf‘tls f ' "—5" F"??” gr“; 6 ’mj W’ “W” I 5* amoeba) W“ '4" EL [ML/w (0.03140 “We, a; I) (a 5 Law—£- rfi-‘i m/s‘] amt r; M (c 5pts) What is the mobility u of the mobile charges? E V V V:,d( “ -—7 MsE“ : Mu/L (d 5pts) The current running through the bar was measured to be 0.55 ampere. If each mobile charge is singly charged = e), What is the mobile charge density n for this material? I "" I “W” ’7 = 137?? (e 5pts) What is the resistance in ohms of this length of bar? VSTK : Au -— R I gal” :y X K: 0.53% 4.71.}7, / Problem 2 (25 Points) A neutral copper bar of length L and resistance R is free to slide along perfectly conducting metal rails in a region in which there is a uniform magnetic field B directed into the page, as shown in the diagram. A power supply is connected across the metal rails and supplies a constant voltage emf. ® ® ® ® ® ® ® ® ® ® ® ® (a 5pts) As soon as the power supply is connected current run through the length of the bar. What is the magnitude and direction of the force on the bar due to this current. Label this force on the diagram. A‘+ vHM. ins-l-avvi’ "H'u. (owes-l1)“, {5 Mode / m CWNVVl’ in “ll/w (00M {S Sim‘aia em§/@_ 1‘4. ‘l’lom 44M lop Mil '40 “HM— lOO‘H‘OWL (ail (Snail. (b 5pts) As the bar slides down the rails, the motion of the bar causes it to polarize. Indicate this polarization of the bar by drawing pluses and minuses on the diagram. Note, we are not asking for the polarization created by the power supply. ‘ m Clmqes in “We bar an with 7L0 44w 0714+. a ’4’ ‘ M~1Mm apprcc on Wc char?“ {5 6 ‘ [56C (C 5pts) As the bar moves along the rails, the current decreases. Explain Why. "ML low academics 41, ~th New}. As {HS 8M {VICNVlw/S ‘l’W0 “Hiltvbfi “fir/mm: (’) Chap/yes ,‘h “71% 5 and ‘AI’Cfol 7‘6 ‘fl-e (MG dual [2) am e/(ch'c ar I —>) .50 4L1M‘5 QM FE ((41.1. 744/1 .‘n W 17v 7mm L4 Marin/(j. [ 7L“. fad/1L ' 1's #4 «#511le (Sew—er mzemzij'Cs It I‘ 1,; 7'0; éflr age ,4: 54% 0.9 p a f 4’ [’2 AM (‘5’! Off Jt’p. )' 7&4/5 34“ (Witt/UP 49“? Mp4 NM CC is Nod/cc 9?. (d 10pts) The bar quickly reaches a steady state (constant) velocity 1). Determine the magnitude and direction of this velocity. You must start only from fundamental equations or equations on the formula sheet to receive full credit. Hint: Do not attempt to use Faraday’s Law for this problem. [.F M bar is modi‘t] aL—l- Cows-laud- sececll I Problem 3 (25 Points) The electric field is measured at locations on every surface of a box—shaped surface oriented with its long side on the x—axis, as shown. The electric field at all locations on each surface of the box is found to have the value E = (—100, 500, O) V/m. For clarity, the electric field vector has been drawn only on four faces of the box in the diagram, but the electric field has the same value on the other faces (front and back). [New V rbc“ E f; a augiaué veal-or so 22:42:» (a 20pts) Calculate the net electric flux through the surface of the box. Be sure to show all of your work to receive full credit. I, a c . - , - : » $00 07- <—I 0,07 (0-03)(o-043 alrfi' I ¢‘ EA! <loO/ I I — 4>l:+. l1. ‘1‘“ (0.0?)(°«°”‘3 : 0 VM o.oH) :—.IZVM 2 (’(00) 503/07 .<+')ol 07(0'03)( .7 (0.0?)(00'4) v: 0"“ : <_loo FOP/07 ‘ 4010/, I (0.0?)(0'03) :fl'ogvm 45 :CHZVM)+(O) +(112Kn)+(o)+(1.0§vm)+(—(.oS‘\/.a) : 0v,k + , (b 5pts) Calculate the amount and sign of the charge enclosed inside this box—shaped surface. (FET: ZZ-CVK 6° :0 L5 Zwlctm / / Problem 4 (25 Points) Circular ring Ewe} VHW Power supply radius r2 6’ In the figure a very long solenoid of length d and radius 7'1 is tightly wound uniformly with N turns of Wire. A vari— able power supply forces current to run in the solenoid of amount I1 = p — kt, where p and k are positive constants. A circular metal ring of radius 7‘2 and re— resistance R sistance R is placed around the middle Very long solenoid: B 0f the SOIenmd' radius r1 N turns along length d 0...; an «Era ? mk- NKe, . (a lOpts) Calculate the magnitude and direction of the constant magnetic field inside the solenoid. Re— member that the field outside of a solenoid is zero. Consult.” “HM. aMp4n.M 1006) I __, -—’) —->;_ Z]: igimim law— “‘0 u. e . ‘HM; SARVlou‘oi Pond-s ‘l'b ‘HM rchlr (56¢ «ML 4.4"a jam was up «w. is (wole cams-ined- (sce «Ci-«‘3. {Wu/(Whine, else. ii— i”; 124w. 3 E,- L+0(w)+(o)L+(o)u.> =33sz C M L. 14011;“ =- MON: 3,;- W rhea-Ho” c4: cunncf' pguiwfi‘nj *“M Sdf‘cuL-e . (b 5pts) Determine the flux through the circular ring of radius r2 as a function of time? .a -? n JloNrt ‘ 42M: BHA ’13—- d -: Thu-kip) A: TT‘QZ l7/C. 3:0 outside— _ 4L“ gal-euoicil J4 NTW‘I’L 7520 (c 10pts) What is the magnitude of the induced current in the circular ring of radius 7’2? Nike). / go /5 N1‘4?! Makes Jtfls-e l7/c, ‘H/u. CVVNVLf‘th dean-(“5037 / (d 5pts) Extra Credit: On the diagram, indicate the direction of the current you calculated in part Briefly explain how you determined this. f (see a l7d fe m4 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 _, _ d1? d1? _ —o d}? N _, , a—dt dt— net fianddt~ma1fv<<c AUel = qAV AV = — ff E . dl % — Z (Exm + EyAy + EzAz) @elzffiofidA <I>mag=f§ofidA fE-fidA=Z—q:”sfli find/1:0 0 r 4 do a r r lemfl : fENC . = ' 6;: g ° : M0 ZIinside path r r d r A f3 . = M0 Iinsidepath + 60% f E . Specific Results 4 1 2qs , -» N 1 qs _ Edipole,azzis R: arm—7:3— (On 3X13; 7‘ >> 3) IEdipoleA.’ mg (on J— a‘XlS) 7“ >> 5) _. 1 Q _ _ _. ET = ——————-———— 'r' .L from center electric d1 ole moment = s, “’2 oz Ea 1-6 0d 4mm 7. T2 + (L/2)2 ( ) p p q 17 ppl d a 1 2Q/L _ a _ 1 qz . Emd N F60 T (1f 1' < L) Em”g —— m (Z2 + R2)3/2 (2 along axrs) _. A Q/A z _ ‘ a Q/A Z Q/A . ‘ = — —— -s z — — m f R ’Edlsk 260 1 (Z2 +R2)1/2 (2 along axrs) Ed1 k 260 [1 R] 260 (1 z < ) ~ Q/A . e Q/A s . . . Emma-tor % 60 (+Q and ——Q dlsks) ‘Efm-nge z 60 Just outsrde capamtor AB; = flIAE X T (short Wire) AF 2 IAfX E 471' r2 " #0 LI #0 2I _. .. Bwire = —————— % ———- 7‘ < L }Bwire = B a, tan6 l 47M1 /7~2 + (L/2)2 47r 7" ( ) 6 M —o _ ,LLo 2I7TR2 N no 2.771'R2 , _ __ 2 ‘Bloop — E (22 + R2)3/2 ~ 4” Z3 (on axrs, 2 > R) p, —— IA — I7rR ~ N #0 2M . ~ N #0 H - leipole,aIis ~ E773 (on mm, 7" >> 3) Edi-pole; ~ E773 (on 1. 3X18, 7" >> 3) a E -+ 1 —an_ A A A _. rad = —— = Ta. Bra B'Nl : Erad 4WD 027, v d X d d c i=nAfi I=|q|nA17 27=uE I L __—_- = — = R : —“ a |q| nu J A (TE 0A Eapplied q 1 1 . . . = = _ _ _ t Edwlecmc K AV 4WD Tf n due to a pom charge _ [AVI I R for an ohmic resistor (R independent of AV); power = I AV IAV| , _ Q = C |AVI Power = I AV I = R (ohmic reastor) d-i —' 2 K m 5—va if 1) << 0 circular motion: fii : % x L]: Math Help a x 13': (amayflz) x (bm,by,bz) = (ay bz — az byfi: — (asc bZ — a2 bag? + (ax by — ay bmfi / dac __1n(a+m)+c/ dm _ 1 +0/ dm _ 1 + x+a— (:c—l—a)?‘ (1+3: (a+a:)3_ 2(a+:1:)2 c a 2 2 a 3 /adm=ax+c /axdx=—2-x +0 fact dnga: +c Constant Symbol Approximate Value Speed of light 0 3 X 108 m/s Gravitational constant G 6.7 X 10—11 N - n12/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 EE— 9 X 109 N - m2/C2 0 Epsilon—zero 60 8.85 X 10‘12 (N - m2/C2)_1 Magnetic constant 52 1 X 10‘7 T v m/A Mu—zero no 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 Ra % 1 X 10—10 m Proton radius RP % 1 X 10—15 m E to ionize air Eiom'ze % 3 X 106 V/m B Earth (horizontal component) BEarth % 2 X 10—5 T ...
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This note was uploaded on 05/23/2011 for the course PHYISCS 2212 taught by Professor Shatz during the Spring '10 term at Georgia Institute of Technology.

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2212_Test4_Key-1 - PHYS 2212 Test 4 Spring 2011 Name...

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