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Exam 2 222 Fall 2009 solutions

# Exam 2 222 Fall 2009 solutions - 23 Two long ideal...

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Unformatted text preview: 23) Two long ideal solenoids {with radii 2t} nun and 3t] mm respectively} realtyI the same eurre'nt. The smaller solenoid is mounted inside of the larger one. along a common axis. it is observed that there is a zero magnetic ﬁeld within the inner solenoid. Therefore the inner solenoid must have X times as manyr turns per length as the outer solenoid, whereX in: mm m Wale, {Ja'ci 55-414. this—1101‘; is 213 @1112 %:/MuﬁT “it? at (\$110 {11:} 5&9?an a“ 3 E)?” M radium it gal-.31: he Lave- 31:43»; we nee; ego-L, 29) The diagram shows three arrangements of circular JODFE, centered on vertical axes and carrying identical cunents in the directions indicated. Rank the arrangements according to the magnitudes of the magnetic fields at the midpoints between the Icons on the central axes: smallest, intermediate, and iargesL ft “11111:; m1 1 2 a of. L __, 10:10 teatime; +22? ﬁes _. “a, 752%ng EV;” at m WNW ET' Liga- an) A beam efeIeetrens is sent heriecntalty dewn the axis cf- a tube to strike a ﬂucreeeehteereeﬂ at the end cftlte tube. 0n the way, the electrons eneeunter a magnetic ﬁeld directed 1v.ure1:'ti-'..tall].r dammed. The spat en the screen will therefcre be deﬂected: 1-1 A) upward q E, ﬂew _ _ (an ‘F D) tn the left, as seen Earn the electron source E) act at all 3: 3’38. l3 {3) 3 imme 91*”) 3!] A circular lee-p afraditte H} cm and three lcng straight wires carry currents eff. = tit} A. I; = 211} A, I; = ID A? and {4 =10 A reswctively as shown Each straight wire is 2G! cm From the center cf the leap. The Hemp-uncut of the resultant magnetic ﬁeld at the center cf the lee]: is cluseet tc: 1%.) +211} nT CED-1t a'r C) -15 [LT D} +15 141‘ E) +20 pT er I - it ”£3 - 3’4 3 13:3 2 Lo xte’5T =5L3—A-1l0a10q'l— ’ZJTKﬁ,_ ‘l‘ “2““ng Te: [0’25 “Kat-0.11m 2123.: we Mattie“: 32) At any:r point the-magnetic ﬁeld lines are in the direction of: n) the magnetic force on a moving positive charge, i: {d *‘W B] the magnetic force on a moving negative charge. C] the tttt'tlooitir ofa moving positive charge. the velocity ofa moving negative charge. . one: - @0111: of the above. E) 0.3011; Q11 Mm {RE} 23"" 331 Four 2!] ﬂ resistors are connected in series and the combination is connecteti to a EU V emf device. The current in an};r one of'Lite resistors is: ®oasa 11911 = 4(10R): 80.51.. B} Lon o} eon 33V _ mean 5: ““3331! ' 9:35p. E} toes 3c) The equivalent resistance between points 1 and 2 of the circuit shown is: (\$32.59 tit iﬂ H E”) .‘iﬂﬂ I o} eon o} 4.5 it 1 ﬂ 5“ It) son a so set E:- ii ‘5‘ d] l‘: 35) 1it‘iihich ufﬂte following graphs best represents the cment—vcttage- reiatichship for a device 11ml obeys ﬂhm‘s law? 1-” H 35} If the potential difference across a resistor is doubled: A calf,r the current is doubted Q crniy the current is halved V 7' IE C) only the resistance is doubled D] only the resistance is halt-red E} bath the current and rcsistancc are doubled 3?) Copper ccntains 3.4 K it)“ Free clectrensimj. A copper wire cf cross sectional area I mut-2 carries a current of l A. The electmn drift speed isappmximately: A) mama’s l _ E] Iclms j = T = “in“; 1 nuts @310" ms is“ m it‘d = l 33] A 1t:- m length ofwire carrying a current of It} A lies. on a horizontal table with a rectangular top of dimensions {L612} 111 X QED m. The ends of the wire are attached to opposite ends of a diagonal of tho rectangle. A vertical magnetic ﬁeld of G. l D T is present. What is the magnitude of the magnetic force that acts on thia segment of wire? (Q era N moon (1] 1D N D] Zero [-1) The force cannnthe determined without knowing the precise shape of the length of wire on top of the table. don—3ft curt ohm Sate.” CW, \$3: ﬂed-cc ‘__ A ...--’E! J. —HI "I .4 ...-'- 1: : IIJQKB; 16%)}! : .LLJ‘E .4" .J eh Mr- E H mm} = i 13‘9”“ ll (1.51; of: MY ‘1 It?“ 3'?) A mass spectrometer is operated with deuteron. : Q" 1 N . which have a charge of+e and a mass of3.34 I III” kg. I 9 39 :1 EB T Deuterona emerge from the source, which is groundﬁd. I G i ' with anegligihle velocity. The deuterons are accelerated gage-moan I by the potential difference between the source end the " " m x 1051“}?! accelerator grid, reaching a velocity ofli.t} 1: It}5 route as 1 they pass through the grid iteelf. A uniform magnetic __ ﬁeld of magnitude ﬂit} T, directed out of the page, in ‘ 1"“: present at the right of thcgtid. Once the deoterona are on I orator grid the right of the grid they more in e circular orbit in'the magnetic ﬁeld. The radius of the orbit and the initial score of deﬂection are cloeect to: 13] 'Ft] mm, upward “fol-LL a“ w cut ‘ T— “ ("BF-B ( EH 9.) C] 15D mm, downward ' 1 t ' Til mm, downward Wm” mm MW E) m, downward £1- " 15' E -r 1'“ -- El 1-- . -m 5 ) . , l0 m3 "“5" 5 (%%“‘° “'3‘ an“ ( = Orﬁ'ﬁcﬂdm t“: 915 ("lento“lqﬁl‘ﬂ G.WT\ iii) In the circuit shown the newer dissipated in the 2 £1 resistance in the circuit is closest to: MEDW Ea“ ~_t’..[.+ l : 6,11: mssw .L._.,L. C“; saw ‘3: G D 16"!” ‘ [1V sums # ___. ._ II‘ 6.3. 9‘“ '1... J1. u ,, "I = 91:15 gin \$231. ={1-5L)(tgﬁj-r new 4]) A multilcctp circuit is shown. The emfc. is cluscst to which ef the cheiccs heicw? [f the emf is in opposite direction of that shown in the ﬁgure. it hues negative sign. Hete: It is net neeessurjpr tc solve the entire circuit to snsuter this questiurn. {is -5't~' E}5V | cuss! n) en: 51 51v Kid-L- .ﬁot ken? We: ._ g1_(6ﬁ3(5ﬂ\+ﬁo\’ : o a: RD..QS:-—§V 42} A matal rad has a 4 mm1 cross section and'is lﬂ cm long and has a misunce if 1 1119. If 1111.: rod is drawn (rmhapedj inn: 3 uniform'wire that is U] m lung, whal is the: rasiatance of the wire? L..| E. A} EL] 1.1.0 at :. D _- E1 -: B] lﬂuﬂ IL A! in A1 C1 lmﬂ l 65‘“:an IE1, L1 AH L2. E an “- '- "‘— i "‘ } 111 L1 P11. LL Cummkqﬂum; “V: AIL-'7 {5&ng 7- _i__{}m 1;“; 111 5 Q.(L1=\_(1m§1) .51.— L4 0‘ l m 43] lfthecircuitshawn hasarealhatteryasits _. — TEF l" Bmfsuume wiﬂl §= IU Wand-an internal resistant: 1: ' ' x':‘__ of4 £2,th is ﬁrecurrcntlhmugh 111135 E: 1' rcsislﬂr'? :5 .- 5; ‘31": A} 0.19 A 3;; B RESA I... .33 A [1.513 A . E} {135 H. L._._._—-':.L:[-.Ij‘1‘.'-\.-'\ﬁ.:_.‘x til-.9..- lﬂ” "“' ” '1 I: WWI- lﬁ) .' 11g 1 I3 ”5.11 IUJL 44} A long solenoid has 3W tomﬂom and Barrios: a current. I. An electron moves within the solenoid in: circle of radius Lt] om perpendicular to tho solonoid axis.- Tho opted of me eleeooo is 3 u no“ nos. Find the current in the solenoid. mm - E ream ox. Eb- em. 3) 1M — and“ 5% armed 5‘ E‘ CJI}.45A . o- ozm r: _"'“__ .G‘lﬁﬂ qB {dimes B 1 ”and: E W - {W» E mm I = - _ gram” (touodqqaooo‘i 2%”. (NEW '){0101m) 45} A In: of length L = 0.5 m is ﬁoo to slide, without ﬁ'iction, on horizontal rails, as shown in the sketch below. Thorn is a uniformroogootio field, B = 3.11} T directod into the plane of the: ﬁgure. At one and ofthe rails there: is a Im'ttor},Ir with emf - T V. The bar has a mass of Lt] kg and a realm of 121} :1. (All other msismnoos in thc circuit can be ignored.) What is the maﬁa] speed of the bar? A.) 11.33 mo o) 1.611115 1:12.: mo 3.? mere 1 1.; '[' 5 @ﬁ ““5 if: 45) A 5 H inductor carries a-cun'ent of 2 A How can a self-induced emf of in V be made to appear across the inductor? @ Change the current at a rate of It} Ms. Change the cunentto it} A. C] Break the circuit instantaneously. D) Change the current uniformly to zero over 20 s. E) None ofthesc. I _._.... —- = UL- -'- L" d1 Gk L Elli" 41'} The units of induced emf are sometimes written as: A) We. a) Vccrc l E] _; iii C} Vﬂ“ at“ D Tfs '1 @T'mi’s v - 1'" S 43) A ﬂat. square loop [Ill “1 on a side} is in a region of space that has a magnetic field that points into the page; Between t= {i s and r= 5 s the magnetic ﬁeld is decreasing, at a constant late, from H] T to l] T. The loop has a resistance of 4 it. At r = 3 5, what is- the size and direction of the current ﬂowing in the loop? A) 5 x at2 n,c1cclcwicc a} 5.x “1‘1 n, cccctcrctcckwisc C} 5 x to" n, clcckwise - . 5 1‘ o s x to" a. counterclockwise : I one ol'the above i - Hume. . i | a -r . 49’ O ‘0 5,975- __-_.--I- Etc. it; i— i a ‘5‘ c (may???) c. mow — 49) The circuit show below is rustle ofs real battery with emf It] V'and internal resistance r = 5.0 ﬂ, and four resistors R1=Rz= R3 = 2.!) £1st R4 =ﬁ.t'.l \$1 Atr=ﬂ, 3 4t] mF capacitor is inserted hemeen points a and b in the ﬁgure below. 1What is the charge in the capacitor Eﬂﬂ ms later? it) ItitirnC (F3 teem: E} 21411113 I 13335th tacit -— BEL-t H) Home 51‘ (535%): ‘Qiiiﬂ .2 @(QﬂﬂmsD _-_ (40mF\ (mg) (i “I (E. E FEE. O’C- ﬁﬂ] A RHL circuit has a 6-1] V battery, s41 H inductor. 3125 Q r/5 resistor and a switch, 5. all in series, as shown. Initially. the 6D '53" switch is open and there is no magnetic ﬂux in the inductor. At time t‘ = II}, the switch is closed. ‘Wiieri the time is r= LB 51 the current in the circuit is ciosest to: 41H 25:: A} uses _§f_c LIA ‘1 : iiﬂ—ﬂ 1.3a o) 1.9a E E3144 L - 1' Is: “It “2: 60 (M‘JL){IS{‘I "LL-Esme) a J l _. «(2 out : i219: 5-1] A simple generator is made ﬂora a flat coil that has. dimensions of3'ﬂ cm by It] cm and has T5 loops of wire wound aretmd it. If the coil 1s rotating in a uniform magnetic ﬁeld oft}. 2 T with what frequency noes' :t need to spin so as to generate a maximum emf of 1213 V assuming that at some point in the rotation that the B-ﬁe'lcl is perpendicular to the plane of the coil? 31125313511}; @: eseemfw-t) E] 3121:: e: - — 5%- .. r meme! wt?!“ gm g MEAN ._. 11mm". 5”“ ﬂ : mil-He :' .———'_—— ‘5' HE. P1 9:1? an (011T)(0.310.1)NL 52) An tit] nil-I inductor. a 2.1:! ﬂ resistor. and a switch. S. are wired in series to a EU V battery. The switch is initiallyr open and there is no current in the circuiL The switch is eloseri at t = 0. After a long time, the current in the resistor and the current in the inductor are: 1110.0 we 11. tamer e We entrance HFRJG“ [MES ’ ”L c; lﬁA,2.SA m £299 Lens 91:: (missions 1 E)_ll]A.tﬂA . ”I = ﬂ -. tor-‘1 €131 50’fo EXTRA CREDIT I'RDB LEMS 53] “men the switch. 3, in the circuit shown below is ciceed, the time constant fer the gmwth cf the current in R; is: A UR. I t I B HR; 5 C} ”(RI +31} 13'] LERt +Rz)f{R.Rz) R: E) LI[2_{RI+R2)] 54} A rectangular Iccp ufwirc is placed midway between him, lung, straight, pereiiel wires as ehcwnt below. The wiree cert-y currents i1 and E; as shcwn. If i. is increasing and i1 ie constant, then the induced current in the [cup is: A) zero _.... clockwise L_ C) cuuntcrclcckwise i . 1.1 '13] depende null-11 1 E} [tune cf the abuve “LL WW F—F’ WMEG mew girﬁLQ/{é “=13 HID I-nlgljﬂuﬁh b—b ' . . rt. e; LhuﬁcuﬂﬂB =3 3 r3 rnmg 1“ {nap til-TEA ﬂ =5. 3‘.“ {a {3) 6-;- ‘Ir-h is. {IN ...
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