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Test 2 Fall 2008 - PHYS 2212 Test 2 Name(print<...

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Unformatted text preview: PHYS 2212, Test 2, October 15, 2008 Name (print)_________}<__% ______________________________________________________________ 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! ' . . . (1.1, _ 8x10'3 5x106 _ 0 Show what goes into a calculation, not Just the final number, e.g.: a: — WW — 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 (27 Points) A thin coil has 60 circular turns of wire of radius 4 cm. The current in the wire is 8 amperes. (a 9pts) What is the magnetic dipole moment of this coil? . ’ a r» ‘ h b\C ‘5’ “BAA/7:. Am, A ": "\TKUO‘lrL P :CLA> cg: otdtprmfld :— 60 '1 Y KCTT K0.Q‘17')X€D :: '.LH A . 7' {Ki/.3: (b 9pts) At a distance of 40 cm from the center of the coil, along the axis of the coil, what is the approximate magnitude of the magnetic field contributed by the coil? 2. ‘ as use l; 2 Bi 33—923 N _ ‘0.-~7<Q_1T in0‘l x8 69 “‘ serif :. LHTXeOMO’ -~e ' : Ztm x10“ :: 7-55me I . The coil is placed with its axis along the x axis; the direction of the current in the coil is show-W” figure (if you look at the coil from the location of the bar magnet, conventional current runs counterclockwise). A bar magnet whose magnetic dipole moment is the same as that of the coil is also placed on the x axis, the poles of the magnet are oriented as shown. / coil 6)] (c 3pts) At the origin (halfway between the coil and the bar magnet) draw the magnetic field contributed by the coil, and label this arrow chl- (d 3pts) At the origin, draw the magnetic field contributed by the bar magnet. Draw the arrows using the same scale (so if one vector has a larger magnitude than another, its arrow should be drawn longer than the other arrow). Label this arrow Bmagnet. (e 3pts) At the origin, draw the net magnetic field contributed by the coil and the bar magnet. Draw the arrows using the same scale (so if one vector has a larger magnitude than another, its arrow should be drawn longer than the other arrow). Label this arrow Bnet. Problem 2 (28 Points) y axis A small plastic ball of radius 0.01 m has a charge of -5X10_8 C spread uniformly over its surface. The center of the ball is located at (0.05, —0.04, 0) m, as shown. A thin plastic ring of radius 0.04 In whose center is located at (0, 0.04, O) In has a charge of 5x10‘8 C uniformly over its sur— face. (a 20pts) Calculate the net electric field at location A = (0, -0.04, 0) m (indicated by the dot in the diagram). Your answer should be a vector. Show all of your work. x axis ' : ‘\" ER ___, Efid e6 (‘6 “L, (Q ’_ Dub“ ’ D> ‘— <D ‘05)fioa 09,0) ‘2 (-0 .05: 03D) A 1 m < ‘ d~ -— j z, * * C‘xmfi X5XI5SXUDDY 'A En teacher/Mm ‘-"’> * WW ~ , : 503w; we. (—8) EM 3' \ 000 ~“D: .‘ N <80 , b 3ns,o> /<, c (b 3pts) On the diagram, draw an arrow representing the net electric field you calculated in (a). Make sure your arrow makes sense. (0 5pts) An electron that is moving through the air arrives at location (0, —0.04, O) In. What is the force on the electron at this instant? Your answer should be a vector. f; .; ammo“ x End" :3 Q 2.8%, 0.205 )D>m““ N fl/fl—d—A N Problem 3 (15 Points) This VPython program calculates the electric field at a single observation location due to a charged rod. from visual import * from _future._ import division ## constants oofpez = 9e9 =2.5 N=25 =—3e-8 scalefactor=le-3 ## assume this is correct deltay = L/N deltaQ = Q/N #initial values y = -L/2+deltay/2 while y < L/2: sphere (pos=vector (0, y, 0), radius=deltay/2, color=color.blue) y = y + deltay obslocation = vector (0. 3, 0, 0. 6) Enet = vector (0, 0, 0) y = -L/2+deltay/2 while y < L 2: sourcelocation = vector (0, y, 0) r = obslocation — sourcelocation mag = sqrt (r.x**2 + r.y**2 + r.z**2) rhat = r/rmag deltaE = oofpez * deltaQ * rhat/rmag“? Enet = Enet + deltaE print sourcelocation = y + deltay print Enet (a 3pts) How many pieces is the rod divided into for the purposes of this calculation? N z: 25 (b 3pts) What is the name of the variable representing the length of one piece of the rod? delta 6, (c 3pts) Circle the loop in which the electric field is calculated (that is, put a circle around all statements that are part of this loop). ((1 6pts) What will the second printed value of sourcelocation be? (Give numerical values for the components of the vector) Problem 4 (30 Points) A long thin metal Wire with radius 7‘ and length L is surrounded by a con- centric long, narrow metal tube of ra- dius R. Insulating spokes (not shown in the diagram) hold the wire in the center of the tube and prevent electri— cal contact between the wire and the tube. It can be shown that the magnitude of the electric field in the air between the inner wire and the outer l 2Q/ L 47T€0 x the inner wire, —'Q 1s the charge on the outer tube, and m is the radial distance from the center of the inner wire (1' < :1: < R). tube (not too near the ends of the long, narrow tube) is given by , where +Q is the charge on (a 5pts) On the diagram, draw an arrow indicating the direction of the electric field E) in the region between the wire and the tube. (b 15pts) In terms of the charge Q, length L, inner radius r, and outer radius R, what is the potential difference Vmbe — Vwm between the inner wire and the outer tube? Explain, be sure to checks the units and Sign of your answer. AV 243;, _, i 'IKQJ. ex ,1 ”2% ‘52,,m\:;+2® Qul‘il<o _. , .1 , c NM - Nm \ (0K) sore QMQ) 13 was that, 2e ._ ”1.)“ M \Ims , ~ o ;,;_.,._l;\_, ems/Wm, gvm‘ (Lack l ~—- {1:3 4 O EX (AX wort: \DCAA’)‘ if 1A Sou/3A5 44m mpg (c 10pts) The charge Q is slowly increased until you see a glow 1n the air very near the inner wire. NeT'P—z‘ Q“( E) 40 Calculate Vmbe — Vwm (give a numerical value), and explain your calculation. The length L = 80 cm, the inner radius 7" =: 0.7 mm, and the outer radius R = 3 cm. ,.._. Gig-u; é EU) ‘; 3 thfi $6 ( KN lent-2.54 fl“ Lfih1;%7‘f20.7mwj 31 EC?) 9 “-3 M J. Of :GXIngfl‘K W 9 Lm‘ib ‘15 iii—Ye; ML: AV 2' fibz%1h(§) : 3xto‘ Y 114th .‘W/ .. 7 " 3 9’91) they - 2mm 1,, ,5 7,. 3?? x103 VATS This page is for extra work, if needed. Things you must know Relationship between electric field and electric force Electric field of a point charge Conservation of charge The Superposition Principle Relationship between magnetic field and magnetic force Magnetic field of a moving point charge Other Fundamental Concepts _, ‘_ d1? (1—1): —1 d]; N _, . a—a d—t‘ net and dt~ma1fv<<c Ave, = qAV AV— — —f.f E . dl~ —z (EmAx + EyAy + EzAz) ©6l=onfidA cbmangBondA ffimdA=qumde ffiofidA=0 0 _. —» d<I> _. _. lemfl = fENC . dl : (17:09 f B . dl = [JO Zhnsz‘de path -’ d<I> .. _. |emf| = f ENC 0 dl = (gag f B o dl= no [2 Imside path + 60— df E o ndA Specific Results s 1 2 s . a 1 q_3 ‘Edipolme z 47T€0 7‘13 (on 32:15, 7‘ >> 3) ‘Edipole,_l.‘~ 471'60 7'3 3(on J. axis, 7" >> 5) s 1 Q a .. E = —— .L fr t l t ' d' l t = s, = aE - rod 47T€o r ,—_———_T2 + (L fly (7" om cen er) e ec r1c 1po e momen p q p applied E — —-—-1—————Q— (7' J. from center) "’d 47reo n/fl + (L/2)2 _. 1 2Q/L —' 1 qz . Emd z m 7‘ (If 7" < L) ring 2 47FEO W (2 along 3X15) ~ Q/A Z . ‘ ~ Q/A Q/A - . = _ _— - m — —— ~ f R Edzsk 260 1 (22 + R2)1/2 (2 along anus) Edwk 250 [1 R] 260 (1 2 << ) —1 A Q A lepamm ~ ~% (+62 and —Q disks) ~ 6/0 ( R)?— just outside capacitor A3 = gIAqlfiEo 7“ (short Wire) AP" = IAl'X E —» . __ #0 LI ~ No 21 _. . __ ~ | |Bunre — Em ~ E? (r < L) ‘sz're — ‘Bearth tan0 a _ I10 2MB? _0,1 21%:2 _ _ 2 l-Bloop —E(22—+‘1{2—)3/§~ 4—,” 23 (on 3X18, Z>>R) fl—IA—IWR E ~ #02” ' é - ~M-—0 J. 7' >> 3) l dipole,aa:is N E? (on 3X13, 7' >> 5) dzpole,_l_ ~21-7TT—3 (on aXisa _. 1 — a A A —« “1 Erad = 471'60 chTJ— ’U = Er ad X Brad Brad 2 C i=nA1') I=|qlnA17 17=uE I L Uzlqlnu J=Z=UE R_H E ~ 1 1 Edielectric = applied AV 2 q —— — —— due to a point charge K 471'60 'rf n IAVI . . . I = R for an ohmlc res1stor (R Independent of AV); power = I AV Q = C IAVl |AV| . . Q = C IAVI Power = I AV I : (ohmlc reSIStor) ~ ~ 2 K z %mv2 if 1) < c circular motion: %1 : ILRI Ifl % @RL Math Help {i x 3: (a$,ay,az> X (bm,by,bz) = (ay bz — az by)§: — (at; bz — at; bay)? + (am by — ay b$)2 / dm —ln(a+x)+c/ dx —~ 1 +c dm ————1—+c sc+a_ (m+a)2_ a+m (a+x)3_ 2(a+:r)2 a 2 2 a 3 /adm=ax+c [amdxzizc +c /ax dx=§x +c Constant Symbol Approximate Value Speed of light 0 3 x 108 m/s Gravitational constant G 6.7 x 10‘1:l N - m2/kg2 Approx. grav field near Earth’s surface g 9.8 N/ kg Electron mass me 9 x 10‘31 kg Proton mass mp 1.7 X 10—27 kg Neutron mass mm 1.7 X 10—27 kg 1 Electric constant 4 9 x 109 N - In2/C2 7r60 Epsilon-zero 60 8.85 x 10‘12 (N - m2/CZ)_3l Magnetic constant 5—0 1 x 10‘7 T - m/A 7r Mu—zero p0 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, z 1 X 10—10 In Proton radius R], m 1 x 10~15 m E to ionize air Eiom-ze z 3 x 106 V/m BEarth (horizontal component) BEaTth z 2 X 10"5 T ...
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