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Test 4 Spring 2009 - PHYS 2212 Test 4 April 9th 2009...

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Unformatted text preview: PHYS 2212, Test 4, April 9th, 2009 Name (print)__ _________________ _.K§fi\__mw_fl___ww___m_ Instructions Read all problems carefully before attempting to solve them. Your work must be legible, and the organization must be clear. You must show all work, including correct vector notation. Correct answers without adequate explanation will be counted wrong. Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything you don’t want us to read! Make explanations correct but brief. You do not need to write a lot of prose. Include diagrams! Show what goes into a calculation, not just the final number, e.g.: 3+3 = 23:13::(223‘5) = 5 x 104 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 (25 pts) Problem 2 (25 pts) Problem 3 (25 pts) Problem 4 (25 pts) Problem 1 (25 Points) A metal rod of length 16 cm and mass 65 grams has metal loops at both ends, which go around two metal poles (as indicated in the figure). The rod is in good electrical contact With the poles but can slide freely up and down. A magnet sup- plies a large uniform magnetic field B in the region of the rod. When the rod is not connected to a battery, and no current runs through it, if it is released in the position shown in the diagram, it falls downward until it hits the table. However, when the metal poles are connected by wires to a battery, and an 8 ampere conventional current flows through the rod in the direction shown, the rod remains at rest when released in the position shown, and does not fall. In the diagram +x is to the right, +y is up, and +z is out of the page. (a 5pts) What is the direction of the magnetic field due to the large magnet? a €M M0 S'l” fbiw‘l‘ Ulo E CS {win 4&9. parody, l9/c. f“; Ing’h (b 5pts) What Is the change 1n momentum (acceleration) of the metal rod? Explain how you know this. AP 2‘ <o/olo> or K: <0 o o) b/c 4M9— Fbcllis Ml,— «Ami/{‘46 (c 10pts) Starting from a fundamental principle derive an expression for the magnetic field due to the large magnet? AP: <0 0/07 :— 0“ 2?: {0,0,07 2'— ~>2 Q; ’5 IAQE‘Mg B~ Q ”a EU. El" .2, , Chet/m (d 5pts) Calculate the magnetic field due to the large magnet. 33 (814)(lb‘(o' (55-10'3143N7'75M/fl m) Problem 2 (25 Points) (a 10pts) An electron travels with velocity < 3e5, 0, 0 > m/s. It enters a region in which there is a uniform magnetic field of < 0, 0.8, 0 > T. What is the magnetic force on the electron? A PM: fix? , (—r‘wo’mc) <s‘log,0,°>~/s “WW“ .—v = ('wwo‘me <0; 0) (3.(o§M/5)(0~‘Ti> ”i <0, oI—zwtmo'”) D (b 10pts) Despite the magnetic force, the electron continues to travel in a straight line at constant speed. You conclude that there must be another force acting on the electron. Calculate the additional force that is acting on the electron. ._.lq- FoQJIHk-ami: <O’O, 3.84'(0 >M q (c 5pts) What is the electric field in this region that is responsible for the electric force calculated in (b)? 2 w, FALWWJ f 25 ’5 . -/L/>/\/ 2 Faumm/ W ’ ______._’) .— Problem 3 (25 Points) A slab made of unknown material is connected to a power supply as shown in the accompanying fig— ure. There is a uniform magnetic field of 0.6 tesla pointing upward throughout this region (perpen- dicular to the horizontal slab). Two voltmeters are connected to the slab and read steady volt— ages as shown. (Remember that a voltmeter reads a positive number if its positive lead is connected to the higher potential location.) The connections across the slab are carefully placed directly across from each other. The distance w = 0.15 m. As- sume that there is only one kind of mobile charges in this material, but we don’t know whether they are positive or negative. (a 5pts) What is the Sign of the mobile charges and in what direction do these charges move inside the slab? For full credit, carefully explain your answer. ~Tint. vaH‘ wkr (w 4M Wansuuse Aivflc‘mk gkpws a was 46% \ aakw'l-wA *6 9%“ch firm“ 4M5 we Com C&ec&UL-& Wed— “.3an thwfleg \zuxw. accqulakA cu. m con/Lu: uckkai CAM/Mug"— chowe‘e CvarL-crs OLv-L ‘M k‘. SQAQ a; M$h€kbf W3 m rut/LS doakwxse“ L+.\ VLALtS h [L M MQSNA-‘Le 'mecc 4L.»j {KP-{VLQMg-Q \5 *0 We. \t‘C-i Eh" 874KB (b 5pts) What is the drift speed of the mobile charges? Afi-130r‘icg (sitcfi38kkk) .2, —)2 '—’? O 1:. F+FE*1 fmc ohm-lo MVMSWV“ E Ml Wt AWN" __/’__________._. ’ (0.4T) (c 5pts) What is the mobility of the mobile charges? AV \Io\ :MEH ’5 ,u [J Moi E+ la/Q \Ao\es WCOVL u (A) +5 slag-log (0%} Vol K) (€,b-(0-3M/$) (O-lg-W‘X d\’wec_-\—\ov\0 M: _...__. : __/ = [.2-{0‘3___ ILWul ( 043v) v5 -3 M/S : /.2~/0 —— (d 5pts) The current running through the slab was measured to be 0.3 ampere. If each mobile charge is singly charged (qu = 6), how many mobile charges are there in 1 m3 of this material? .—-. A .L I=zm V4 Vt: ZAVOK 0. 3A \A: ~l‘1 _ _ 4. (l.(:'(b (3)0110 zm\(%-to 1%)(8'4910 Mt/s) until: (tunic: Vt = 3710??) M/m3 (e 5pts) What is the resistance in ohms of a 0.15 m length of this slab? ___L_-____ L. IQ: 074' y ZM’MA 017m ’— K : Now—'76) [3'7” “aw/'2'” 1% (z : We ”BJZ Problem 4 (25 Points) A neutral metal rod of length 0.65 m slides hori— zontally at a constant speed of 7 m/s on friction- less insulating rails through a region of uniform magnetic field of magnitude 0.4 tesla, directed into the page as shown in the diagram. (a 5pts) On the diagram draw the charge distribution on the metal rod and indicate the direction of the Coulomb electric field inside the rod. (b 5pts) After the initial transient, What is the magnitude of the net force on a mobile electron inside the rod? ka 7: O W _>, «>3 FM+¥GZB C\£cxlfv\‘<_ Cielg Wm." Aowdofs olv-L ”‘1: 4i»... turtle $Qe0~focuiiM fiemmks am’ aoQalH-x‘md Came. ow {w ‘HJ-fi- V0<§~ \HaUL eie c-l—vous (c 5pts) What is the magnitude of the electric force on a mobile electron inside the rod? 32%;: ’3 FEM: led 31$th '5 E=VE E: {film/53(0Li'r): 2.?HJ/c (d 5pts) What is the magnitude of the magnetic force on a mobile electron inside the rod? PM: 6V3 ;_ ("b‘IOF‘qC3ii’v-4/5XOA (e 5pts) What is the magnitude of the potential difference across the rod? [NE EL = (2.2«I/c)(o.evm) /(82\/ 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* d ~ 0!" 6:25 d115,: net andfisziifv<<c Ave; = {AV AV— _ —ff E. dl~ —z (Ema: + EyAy + EzAz) $81: IE . fidA Lmag _fB . TLdA fBofidA=;—q:‘idg fBondA=0 o a J ch> J J lemfi = f ENC 0 C” = l (gag f 3 ° C” = #0 thide path f B . dl: #0 [Z Iinside path. + 60d— df E . fidA] Specific Results " 1 2 3 , —¢ 1 s IEdipole,aa:is N 47reo {—3- (on 3x15, 1" >> 3) ‘EdZ-WBA z 47reo 3—3 (on J. axis, 1' >> 3) E = ——-——— J. f t l t ‘ d' 1 = "‘2 ~ rod 47F€0 r ,._______7‘2 + (L/2)2 (7" rom cen er) e ec r1c 1po e moment p qs, p a Eapphed —» 1 262/ L , ~ 1 qz rod 471'60 r (1 r < ) rmg 47T€o (22 + R2)3/2 (2 along ax1s) ~ _ _ Q/A z . ~ ~ Q/A Q/A lEdzsk — 260 1 — (22 + R2)1/2 (2 along ans) Edzsk ~ 260 [1 — Bl ~ 260 (if 2 < R) ~ A J A Ecapacitor N Q—/ (+62 and-— -—Q disks) Efringe~ Q/ (3- —~) just outside capacitor 60 60 2R J I " * a J J AB = _H_0 AZ X 7" (short Wire) AF = [Al X B 47r r2 -* #0 LI p0 2I l .. _. B I = ———————— ~ —"’_ L B ' = B I wzre 471' ’r’ 7'2 + ( L / 2)2 471' 7‘ (7" << ) ID’L’I‘C earth tang p0 2MB2 #0 21m2 . _ - 2 Blow: 47W a 23 (0“ WWI?) “—IA-IWR I311 ole axis “ 13% (on axis 7" >> 5) lédz‘ leJ. ~ Nfl— (on J. axis 7" >> 3) p ’ 471’ 7'3 ’ p0 ’ 41r7‘ 3 ’ d 1 _ a A A A —. grad Erad = 4 qz J— ’U = rad X Brad Brad = 7m) 0 r c i=nA17 IlelnAfi 17=uE I L 0 IQI W A a R 0A E - 1 1 Edielecmc = apphed AV = q — —- — due to a point charge K 47r60 rf n- IAVI I = R for an ohmic resistor (R independent of AV); power 2 I AV IAVI . . Q = C |AVI Power = I AV I = (ohmic res15tor) d4 -* 2 K m émv2 if 1) < c circular motion: 31;; = kjl-ll- lfl z m?v Math Help a x (3’: (aw,ay,az) x (bx, by, bz) = (ay bz — az by):i: — (am bz — (1: bag) + (am by —— ay bz)2 / d3: —1n(a+x)+c/ d2: —- 1 +6] da: __ 1 + x+a_ (x+a)2— n+3: (a+x)3_ 2(a+ar)2 c a 2 2 a 3 /ada:=aa:+c faxdx=—2-a: +c fax dx=§x +c Constant Symbol Approximate Value Speed of light c 3 X 108 m/s Gravitational constant G 6.7 X 10‘11 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 47% 9 X 109 N - m2/C2 0 Epsilon—zero 60 8.85 X 10—12 (N - mz/CZ)_1 Magnetic constant g 5—: 1 X 10‘7 T - m/A Mu—zero M0 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 z 1 X 10’15 m E to ionize air Eimize % 3 X 106 V/m BEmth (horizontal component) BEarth z 2 X 10“5 T ...
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