Test 4 Fall 2008

Test 4 Fall 2008 - PHYS 2212 Test 4 Name(print_l_...

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Unformatted text preview: PHYS 2212, Test 4, November 19, 2008 Name (print) _______ __l§_§_._ ______________________________________ .. 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! . . . a4, 8x10-3 5x106 _ a Show what goes into a calculation, not Just the final number, e.g.: m = W — 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 PHYS 2212 FINAL EXAM will be: Dec 9th from 11:30am - 2:20pm Please read the announcement on WebAssign. PHYS 2212 Do not write on this page! Problem Problem 1 (30 pts) Problem 2 (25 pts) Problem 3 (20 pts) Problem 4 (25 pts) Problem 1 (30 Points) A bar (length L = 18 cm, height h = 5.2 cm, and thickness d = 2.1 mm) made of a new conducting material is connected in series (as shown in the diagram), to a power supply with emf = +115 volts. The bar is oriented along the x-axis. A voltmeter is attached vertically across the bar, with the leads directly opposite each other, as shown and reads +2.8 X 10‘5 V. Large coils not shown in the diagram create a magnetic field of 0.7 tesla in the +2 direction, as shown. Voltmeter Remember, that a voltmeter gives a positive reading if the negative lead (COM) is connected to the lower potential location, and the positive lead is connected to the higher potential location. (a 3pts) What is the direction of E”, the electric field inside the bar due to charges on the surface of the wire, and in and/or on the power supply? in is Eda eS‘VCxLUslmcl g W bad-Om ’ flxmgsli '“U‘O (b 3pts) What is the sign of the mobile charges-’iMW ' A ‘ h ‘ I “move. 1A ,VC Lek . (Tm mem 6&3) > “"6” ashram cm “X @330 - (c 3pts) What is the direction of the drift velocity of the mobile charges? “Fur —-Ve £935 wobble. Mao Q cw “ Mod #15 'm +— x sanctum . Com“? En D (d 3pts) What is the direction of the magnetic force on the mobile charges? Ema -——_ (e 3pts) What is the direction of El, the transverse (or perpendicular) electric field due to the magnetic “mowe 90104360 as Swab”) in . Tm; AJS’MBTR‘lah \W '\ 1A a $3424. ‘1’" , (problem continued on next page) pblarization of the bar? E M (f 6pts) What is the drift speed of the mobile charges? Show all steps in your work. fling?" 3’ 0 53> fit limzj in a—slvedim flELzfiCV XE) «K E)? VB —' > V2,“: - V?) Va ) El” Mia fl -' . Veiec'xfafl/ 3 mil/‘9) ’6 :; 1: 7&3 54700—2 )k'OT? . (g 9pts) There are 4 x 1024 mobile charges per cubic meter of this material. The absolute value of the charge of one mobile charge is +e. What is the value of the conventional current in the circuit? Show all steps in your work. : \n 3:; neAv : caulk)! A ’A d - Ls : Wool“ x Lexm’“ X 2.\X\D”3 xsambz x 7.1 x10 —- —2 2: 00538 Amps -; fifekmo ANS. W W. Problem 2 (25 Points) At a certain instant a horizontal metal bar is 14 cm above a table and is being pulled upward with speed 7 m/s. There is good electrical contact along the two vertical metal rods that are 24 cm apart. Throughout this region there is a uniform horizontal magnetic field out of the plane of the page, with magnitude 0.8 T, made by large coils that are not shown. it voltmeter Boo? \‘fimk ' Analyze this situation to determine the sign of the reading on the voltmeter. The mobile carriers are (positive! holes. (a 5pts) On the diagram, draw an arrow representing the magnetic force acting on a mobile charge in the d) fl 6f. horizontal bar, and label 't F . » V r ’ v Q a 7 MO V“ 1’ 1 mag {30: 4mg Mac, 591an 7? , x E 6 gave. (b 5pts) On t diagram, indicate the charge buildup that occurs due to the magnetic force. WWO 'l'VL X --V£ Chair ‘50 gLouows in "* . (c 5pts) On thg‘fiagam, draw an arrow representin the electric field in the horizo tal bar produced by the charge buildup, and label it E. TM b034,} U60 W183 mam an 80%: M (£1 5pts) On the diagram, draw an arrow representing the electric force on a mobile charge, and label it Felec- “PM We clam/Bea 47k {jam 'wfb \w W W: p ”E:or”-”. 4:, on, WC” Qae/lel. (e 5pts) On the voltmeter in the diagram, indicate the meter reading as thou rtw, 0% W is (kw 3‘, Problem 3 (20 Points) The electric field is measured on a cylindri- cal surface of length 20 cm and radius 3 cm (See Figure). On the curved part of the sur- face, the electric field is perpendicular to the surface, and is found to have a constant mag— nitude of 3 X 103 V/m. On the flat end caps, the electric field is everywhere parallel to the surface, but varies in magnitude. On the end caps, the magnitude of the electric field is 4.5 X 103 V/m at the radius shown (1.5 cm). What is the magnitude and sign of the charge inside the cylindrical surface? Carefully and explicitly show all steps in your reasoning. Wm: 31.: 0 mt 6» 1 r“ x 3M6?) " C} '7‘ 6° *1": A] : 6~7< 3X\Q3 x(2W .4 "at ’ ‘v—m— XBXZOXZTTXKD , m a “W a WX‘WMD : \“C. [7/41 Because if \5 \‘n [:6 ‘ [ta Wee 0‘1 ‘3 \fi (2 gflfia‘rm 30 W 03% SD A? f 6 3{ E 3A demahfiu E a :21“ ‘5 cm 0 Problem 4 (25 Points) You measure magnetic field all around a triangular path and find the pattern shown here. ——> —> Going around the path counterclockwise, find f B - dl for: (a 6pts) The bottom side. Show your work. ’ me, We, a \e Mme—E)" E V 2. 8:5. g; a; Mm sdr » me'sxjtmifilx (p.959 ” (b 6pts) The sloping side. Show your work. ’- ‘ 09 Fe: stools?) sac Mat: \Demmzn E 2. on s r- ‘ —5 h ,— .._ ‘ n—-" I u 8 .Cu .1. muosx §iléxmso — 2 2 ‘0 Tm oo (o 6pts) The left side. Show your work. 3;“— Ls?r slate. await Mmem E2& :36 3 02,390“ :— Q .' , $3 .43 :2 Q (d 7pts) What is the magnitude of conventional current passing through the triangle? Show your work. Tm Mixamfiie Dre W cm lac, omneA mus «be Amfe/réo low-IQ , Le, fuel: 6M 5-? +62 LBW loof~ 31> : l-oq +0 + 224))«xo’5 : 1.17xm”5 3 z. ‘47?“55: H7ND'"5 m W :. .glA 3 P“ or“ x167 3‘ $2) 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 4_ d1? d5 _ a d1? a. a—dt 2-73-— net dande—Jt—~ma1fv<<c AUel=qAV AV=— .ondlz— EmAx+EAy+EzAz) cel=onadA (Dmag=fBofidA onfidA=M3 ffiofidA=0 60 —o 4 d<I> —. a iemfl : fENC . = (17:09 f B . = “'0 z Iinside path s a do s s d a A lemfl = fENC 0 dl = (gag f B 0 dl = [1,0 Iinside path + 60E); f E O ndA] Specific Results _. 1 2 s , ~ 1 3 . iEdipolemis “ 4WD "rig (0n 3X15, T >> S) ‘Edipole,J_‘ z 4ND Z—3 (on J. ax1s, 7' >> s) _. 1 d 4 Emd = (r _L from center) electric dipole moment 1) = qs, p = a Eapphed 0 73/?" E — 1 ————Q— (1" _L from center) "’d 4m m/r2 + (L/2—)2 _. 1 2Q/L . ~ 1 (Iz . End ~ 47m) T (1f 7' < L) Emmy = mm (2 along ax1s) ~ Q/A Z . ~ Q/A Z Q/A . . = _ - z — — m f Edzsk 260 1 (Z2 + R2)1/2 (2 along ems) ‘Edzsk 260 [1 R] 260 (1 2 < R) —» A . ~ A s . . . IE0”,th % 62:0 (+62 and —Q disks) IEfrmge z 9% Just out31de capac1tor a IA? “ a a # AB = 5—0 r: r (short wire) AF = IAl x B " #0 LI [to 2] —+ _. ‘ere = am 3 E7 (7“ < L) {Bwire = ‘Bearth tang a _ M0 2I7rR2 N no 2I7rR2 , _ _ 2 Bloop —EW~E 23 (on ax15,z>>R) M—IA—IWR 4 2 . -* 0 ,u . 'Bdipole,awis z (011 axxs, 1" >> 3) leipole,_L z (on J_ mus7 1- >> 3) _. E d —. 1 ~ (1 A A A —» “1 Erad 2 4W6!) :2; v = rad X Brad Brad 2 c z'znA'U I=|qlnA17 17=uE I L E . Edielectric = applied AV = q i — E— due to a point charge K 47r60 rf n IAVI . . . I = R for an ohmlc resrstor (R 1ndependent of AV); power = I AV Q = C IAVI IAVI . . Q = C |AV| Power = I AV I = R (ohrmc resrstor) a a 2 K % %mv2 if 1) << 0 circular motion: 3—:i — ERI z % Math Help Eix 6: (am,ay,az) >< (bz,by,bz) = (ay ()2 — az by):% ~ (any bz — az bay) + (am by — ay bx)2 / dx —ln(a+x)+c/ dm —— 1 +0] dx —-——1—-——~+c w+a_ (m+a)2_ a+x (a+m)3— 2(0L+:v)2 ‘1 2 2 a 3 /adac=ax+c /amdx=§x +c fax dx=§a3 +c Constant Symbol Approximate Value Speed of light 0 3 x 108 m/s Gravitational constant G 6.7 X 10‘:[1 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 m” 1.7 x 10—27 kg Electric constant 4 1 9 X 109 N ‘ mZ/C2 7T€0 Epsilon-zero 60 8.85 x 10‘12 (N - m2/C2)_1 Magnetic constant 5—0 1 x 10‘7 T - m/A 7r Mu—zero no 477 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 z 1 X 10‘10 m Proton radius R1, z 1 x 10'15 m E to ionize air Eiom-ze z 3 x 106 V/m BEarth (horizontal component) BEarth m 2 x 10‘5 T ...
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This note was uploaded on 09/12/2011 for the course PHYS 2212 taught by Professor Kindermann during the Spring '09 term at Georgia Tech.

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Test 4 Fall 2008 - PHYS 2212 Test 4 Name(print_l_...

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