Test3_fall2009key - PHYS 2212, Test 3, October 28th, 2009...

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Unformatted text preview: PHYS 2212, Test 3, October 28th, 2009 Name (print) ___________________ “gig: ______________________________________________________________ __ 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. a 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—2 = 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 PHYS 2212 Do not Write on this page! Problem Problem 1 (15 pts) Problem 2 (30 pts) Problem 3 (30 pts) —_ Problem 4 (25 pts) Problem 1 (15 Points) The following VPython program is intended to calculate and display the magnetic field due to a single moving charged particle, at a single location. from visual import * from __future__ import division ##constants mu_over_4pi = 1e-7 L 2. scalefactor = 7e-9 deltat = 9e-20 #objects & initial conditions particle = sphere(pos=vector(0,0,-4e—10), radius=1e-11, color=color.red) velocity = vector(0,0,4e4) barrow=arrow(pos=vector(-8e-11,0,0), axis=(0,0,0), color=color.cyan) scene.autoscale = O #calculations while particle.y > -5e-10: particle.pos = particle.pos + velocity*deltat r = barrow.pos - particle.pos rmag = sqrt(r.x**2+r.y**2+r.z**2) rhat = r/rmag B = mu_over_4pi*q*cross(velocity,rhat)/rmag**2 barrow.axis = B*scalefactor (a 5pts) In what direction does the magnetic field point? ’5 a” .8?!» (Jiltch 0i:- Xckvmivqfi. ,b/Dbfir E, {5 in ,5 Aimvch’ou . ( r {ski Rama WIch +1 (b 5pts) Where is the particle located when the magnetic field is largest? Give the position as a vector and explain how you determined this. 4\ ‘9‘ Mo WV"? where, {s smut/54' lg act JAM WiUivL , (DID/a) (c 2pts) In order to change the particle to an alpha particle (a helium nucleus, containing two protons and two neutrons), what line or lines of code would you need to change? Circle the line(s) in the program, and write the changed line(s) here: 2: 2" Lee—l0! (d 3pts) After changing the particle to an alpha particle, you run the program. In What way will the display look different in this version? Give as precise and clear a description as you can. m 0W (bud mesew‘kwl “RN, «443th \MC‘kN v36“ be Lou/01¢ V V/(, 7) VS ahead—Cr but 5H“ T05A’i M ' Problem 2 (30 Points) A long wire is formed into a vertical rectangular coil in the my plane, as shown in the diagram, and connected to a battery (not shown). The height of the rectangle is h = 6 cm, and the width is w = 3 cm. A compass is placed a distance 6 = 4 mm beneath the bottom wires. When a current runs in the coil, the compass needle is deflected 20° East when viewed from the origin; Magnetic North is in the +m direction. (a 15pts) Determine the magnitude of the con- ventional current in the loop. Indicate the direction of this current on the diagram. The compass is close enough to the bottom wire that the magnetic field from the other three wires can be neglected when solving for the current in the loop. Bk) : Baal/£6 Ma L ' 3.. 4;- , we I: 0.;th (b 15pts) Determine the approximate magnetic field expressed as a vector at location (0,0, ~—0.7) In, due only to the current in the coil. Show every step in your work. If you make any approximations or assumptions, state them clearly. Appmxn'mqsk /00/ M Mfi7m/‘7‘c o/ipa/e, g: i“: 231 M:I/4=Izw ‘HT r3 / 2(0-/‘4(0H)(é-/013)(3-/o',:) M, 2:11“) ‘ ' -47”, ______/————— 5: If"? r3 ' (In) 7*.) (0«7’M)3 ’IO - , T ’7 ,, . ~’° * - /.§3/o 53(0/0/ Irilo 7/ Problem 3 (30 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 ma- terials. Two of the wires are identical: they are each L1 = 7 cm long, and are made of material having n1 = 6 x 1025 mobile electrons per cubic meter. The central wire is L2 = 5 cm long, and has 112 = 2.5 x 1025 mobile electrons per cubic meter. Both wires have the cross-sectional area 4 X 10’6 In2 and the electron mobility of each material is 5 X 10"5 (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 5pts) Determine the magnitude of the electric field in wire 2. Show all steps in your work. Av: .. 3 343 E’ ,3 “New m4.» Av 1.33\/ - I?” a: t ’ “WM (b 7pts) In the steady state, calculate the magnitude of the electric field in the left wire, whose length is L1. Show all steps in your work. Moi: rule 3 AV 0 (00,9 t too r em; ~ 5": ' FLLI ' EIL’ :0 P V‘gfz AV 25L! : crvt-c ’ AV; £-Av afiv' 1.33V V “A 2 v , = //.2 / El: T ' l(¥~/ozm) m (c lOpts) 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. ’y _q ’ ‘ r ‘/ -: . ' m \Il 1 ME. — (5/0 Wmflna. IM) 5 bl (0 l5 (5 ' h ,3 v1.4a}: (9.10 fiwa/M) - 1.33/0 m/s Ad:va+ ) L! L; A In k LZ T: 7 + / +,— 1 «— _._ l VL V. V\ VZ 1(7 (0- W‘ (ilo’zm T'I +. _#——’— =231'N 5 (Cu (o'qW/s) (/3340 3.445) (d 8pts) How many electrons per second cross the junction between the two wires? Show your work. blzbl / .. -§N/5 ;,-,m,AAE.=(O'/°”7§-a)(4'/0 may” m)(”'2“/”) Silt/tall error b/L mumflivta_ ‘ a 1/1: “ZflMEL : '/0l e/S Problem 4 (25 Points) A metal slab of width 3 / 2 and area A is centered between the plates of a capacitor with a gap width 3 and area A. The plates are connected to a battery with emf by Nichrome wires and allowed to charge completely. Then the metal slab is removed quickly. (a 4pts) On the above diagrams above, show the electric field at location A inside the Nichrome wire, just before and just after removing the metal slab. If the electric field is zero state this explicitly. (b 9pts) Calculate the total charge on the right plate before the metal slab is removed? look) rule 3 Avloo‘) 7' (c 7pts) Just after removing the metal slab, will an electron current flow? If so, in what direction will the electron current flow at A? To receive full credit you must provide a clear explanation. T55 I Cow/Mr Clockwise -l-o Mac-c Achf W; W W H Um“ WWWQ W mum access ‘H’LL Cox‘oacflvr {s new l4?ij *fl/(avt {MC . (d 5pts) After the metal slab is removed and the circuit has reached equilibrium, calculate the total charge on the right plate. Avlm‘o : gym-9’ Q[_AVM(’ ’ g: 0 0 o _ an AVMP'vew/Li' AVMe : E3 ' éo 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‘dfi’ d5_ 4 dfiN r, a—dt EE— net dandarvmalfv<<c AUel=quV AV:—fi:Eodlm—Z(E$Ax+EyAy+EzAz) <I>el=onfidA <I>mag=fBofidA ffimczAzilgmfl fémdAzo 0 demag f E o elf: #0 Z Iinside path lemfl = fE—lNC Odi=‘ dt —-a —a d —» A f B . = M0 [inside path ‘l' 503% f E 0 Specific Results —» 1 2qs , —' 1 (18 . ‘Edipolefixis % (on 3X13, 7' >> 3) Edipoledl % 471.60 “7'3” (on J- a‘XIS: T >> 3) a 1 Q a E = —-———— _L f t l t ’ d' 1 t = 4: E - rod 47m) r T2 + (L/2)2 (7“ rom cen er) e ec I'lC 1pc e momen p qs, p or applied ~ 1 ZQ/L , .4 1 qz . Emd N R T (if 1“ < L) Ting = mm (2 along ax15) ~ Q/A Z . a Q/A Z Q/A . Edisk = 260 1 — m (2 along 3X18) Edisk ~ 260 — ~ 260 (1f 2 << A a A IEmpaa-W’ % Q/ (+62 and —Q disks) {Elm-We z 62/ just outside capacitor 50 60 2R » [AZ ‘ a a a AB = g X T (short Wire) AF = I Al x B 47r r2 —‘ _ _ [1,0 LI N [1,0 2] -¢ - _ .i lezre —— Em N E? (7‘ < L) iszre — ‘Bearth‘ tang —» _ [L0 ZIflRg N [1,0 21WR2 _ _ _ 2 .Bloop —EW AJE 23 (on ax1s, z>>R) ,u—IA—I7rR a 2 , a _ de-polemis % fill—73%: (on ems, 7' >> 3) leipoleaL‘ % fill—73% (on J. ems, 7" >> 8) ~ 1 —qc'i A A A _. Erad Erad : 4 2 J— ’U = rad X Brad Brad = 7m) 0 r c i=nA17 I=|qlnA17 77=uE I L = = —- : E 2 — 0 [611% J A a R 0A E - 1 1 Edielectm'c = applied AV = —-q— — — —— due to a point charge K 47reo rf n- IAVI I = R for an ohmic resistor (R independent of AV); power 2 I AV [AV] . . Q = C IAV[ Power = I AV I = R (ohmic res1stor) d-t -* 2 K m %mv2 if’u << 0 circular motion: d—i = liRl m LL]: Math Help a x '6: (am,ay,az) x (bx,by,bz) = (ay bz — a2 by)§: — (ax b2 — a2 bag + (am by — ay bm)2 / dm _1n(a+m)+c/ dry __ 1 +C/ d3: _ 1 +6 m+a_ (m+a)2— a+x (a+m)3_ 2(a+ar)2 ‘1 2 2 a 3 /adm=am+c /aacdm=§x +c /aa: dm=§$ +c Constant Symbol Approximate Value Speed of light 0 3 X 108 m/s Gravitational constant G 6.7 x 10—11 N - mZ/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 ~ mg/C2 0 Epsilon-zero 60 8.85 X 10—12 (N - m2/C2)_1 Magnetic constant 2%? 1 x 10—7 T - m/A Mu—zero ,uo 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 m 1 X 10—10 m Proton radius Rf, % 1 X 10—15 m E to ionize air Eiom-ze % 3 X 106 V/m BEmnth (horizontal component) BEarth % 2 x 10—5 T ...
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This note was uploaded on 04/10/2011 for the course PHYS 2212 taught by Professor Kindermann during the Spring '09 term at Georgia Institute of Technology.

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Test3_fall2009key - PHYS 2212, Test 3, October 28th, 2009...

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