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Sample_Test_3 - Version on — Third Exam — Hofhnarm...

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Unformatted text preview: Version on] — Third Exam — Hofhnarm — {some} 1 This print—out should have 15 questions. Mifltiple—choice questions may continue on the next column or page — find all choices before answering. lUahie [15 [I131 11].!) points The figure below shows a cylindrical coaxial cable of radii as h, and c in which Equal, uni— formly tiistl'ihutevtl1 but. antiparallel currents ii exist in the two conductors. {I Which expression gives the magnitude 3&3} at D of the magnetic field in the re gion ii <1 r3 4: a? in; 1. em} = iii-m2 imilo2+r§ — 2s?) 2 air-r3 [n2 — {12} i 3. em} = Jig—3 2. sea} = -i1".2—h2 _ mi [a2 - bi) 5' 3”“ — m 6.3{1-3}: ““1“ 2st:2 - 2 2 7_ B = M {r3} 2171'3 [a2 _ :52] correct _ theirs E. B{r3] — 2M? 9. Eli-3] = U no. em.) = E or T3 Explanation: Ampere’s Law states that the line inte- gral if E - (iii arouud any closed path equals pal, where I is the total steady current poser ing through any surface hounded by the closed path. Considering the symmetry of this problem, we choose a circular path, so Ampere‘a Law simplifies to B{21TT3)=,UOI€", where 1"; is the radius of the circle and Im is the current enclosed. Aer]. _ Tr {Ta — b2} Aeyfinder " all? —b2}’ r3 a: o for the cylinder. Since when b s: B: Mien 2am .airi-bii] ”'3' [12—irrfc2—h2} Earn; 1: G2 '- TE #0 '32 _ 52 2TH"; _ imitate-i) _' 21rr3{a2—i12} I Rectangular Loop and Long Wire ["12 mo points Consider a long wire and a rectangular current loop. '—I1‘—3" H——£’—HC Determine the magnitude and direction of the net magnetic force exerted on the rectan- gular current loop due to the current I1 in the long straight wire shove the loop. 1|I.-I’ersion [lill — Third Exam — Hoffman]: — {:5de5] ‘2. 1.F=M( a ).down 251' n+1!) 2.s=i‘222;22{e-t),iee —-_nul'1l'2t‘ _ . 3'F_—2w {o b},rrght —« #011125? tr 4_ =— — F 21w [n{n+h] ,upcorrect 5,F=M{a+b]rup 2w —- mflfzd. ELF: tu- 2WE [ ajlup T.F‘=‘uflflh£ob,down ‘lrr 3 F=lm1112(flh),down 2w o+b Explanation: To compute the net force on the loop. we Editionsider—the forces on segments E. BC, CD, and BA. The net force on the loop is the vector sum of the forces on the pieces of the loop. The magnetic force on E due to the straight wire can be calculated by using ._ B _. Fgfi=hf {$37K B. A In order to use this, we need to know the magnitude and direction of the magnetic field at each point on the wire loop. We can apply the Biot-Savart Law. The result of this is that the magnitude of the magnetic field due to the straight wire is itch B: 2n!" and the direction of the magnetic field is giwn by the right hand rule; the field curls around the straight wire with the field coming out of the page above the wire and the field going into the page below the wire. We can not.r find the force on the segment E; applying the right hand rule to find the direction of the cross product, ds'x B. we see that the force will be in the up flection. Since the wire along the segment ALB is straight and always at a right angle to B, the cross product simplifies to Bds. Since the magnitude of the—magnetic field is constant along segment AB, it can come out of the integral which simplifies to give us the result, FAB = I2 E 31 = 1'25 (so 11) . 2 a a. Following the same argument1 we see that the force on the segment CD is #1111 F =Ir — 22 2 l2nio+hi ’ and its direction is down. This is because the direction of the current is new in in the opposite direction along segment fir We can i the nithe same procedure for segments BC and DA. but because the mag— netic field decreases with distance from the straight wire. Ev“ is changing along these seg— ments. This means that the integrals are not as simple. Using the right hand rule, we see that the force on segment E is directed to- wards the right and the force on segment DA is directed towards the left. Because the two segments of wire are symmetrically placed, their magnitudes will he equal. Since these forces on the loop have equal magnitudes but opposite directions, theyr will cancel. Look— ing more closely at segments FE and m, we see Lhat for each small portion of the seg ment BC, we can find a small portion on the segment EYE such that the forces on these two portions are the same magnitude {The small portions are the same distance from the straight wire. II; because the currents are in op- posite directions, the forces on them will be in opposite directions. Their contributions to the net force cancel. The net force on the loop is then Fioop=FrABTF|CD #nIr .U'oir =1 E —I E’ — 2 (217a) 2 {seem} _pnr1.r2e t _ 2n n[n+h] I Version 881 — Third Exam — Hofi'mann — {55T45} 3 If B F A 23" léifl B '2 n ‘2® C I2 8 X .31 ‘8“ Fan Here, a positive force is in the up direction. and a negative force is in the down direction. So. the direction of the net force is up. Metal Ring 063 10.0 points No units are shown for the answer choices. The units are Amperes {A}. The magnetic flux threading a metal ring varies with time t according to e3=3afi—bfi. wM:i=EAs4nfiE mob: 5 s'2 r1112 -T. The resistance of the ring is 2.8 fl. Determine the maximum current induced in the ring during the interval from :1 = —2 s to t; = 2 s. 1. 8.12344 8.838889 8.488891 D.2TT?T8 [1.258518 [1.488111 0.15581 5.155999 . 0.233245 18. 8.182153 weeeeewm Correct answer: 0.15501 A. Explanation: From Faraday-’5 law, the induced emf should he dcpfl 2 g=__._.._.=_ t —25 in [9% ti. so the maximum 5 occurs when rif.‘ E ——l8o£+25=fl r—b “—8.: and the maximum emil.r is e 2a a _fi _ Ezflo' Thus the maximum current is I _mm_rfl m”_ R _8oR [5 s'E-m2 91‘]? :QmAs4-mLTH28m Conducting Bar and Wire 554 10.1] points No units are given for the answer choices. The units are milli‘v’oits [m‘ifl a conducting bar moves as shoum near a. long wire carrying a constant I = 57 a cur— rent. in" A e l—H—L—e a _l_._ I Ifu =15 mm1 L = T9 cm. ando = 25 mfs. what is the potential difiemnce. av E V4 — V3? . 25.48 . 18.3211 . 37.7143 88.1??4 57.2 183.3?6 414084 . 155.486 El. 18.8I 18. 82.3833 mummhmMn—t Correct answer: 155.455 mV. Version 1301 - Third Exam — Hofimann — {55745} Explanation: Given; _dfl dt =~EEBEri dt (it _“BEE =—Bi‘p. 5: From Ampere’s law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire IS I 3:”) 2 it a L Hence the potential difference is raV=BLe #01 = — L (2wa)l a] AP EM 1993 MC 63 v2 [105 11].!) points In the figure shown, the magnet is first with- drawn upward from the loop of wire, then moved downward toward the loop of wire. 11F then down Counter— Clockwise {7 ____ “\clockwme induced I :5- I induced eurren \ 5/, current As viewed from above, the induced current in the loop is 1. first counter—clockwise, then clockwiso. 2. for both cases clockwise with increasing magnitude. 3. for both cases clockwise with decreasing magnitude. 4. for both cases counterclockwise with in- creasing magnitude. 5. first clockwise, then counter-clockwise. correct E. for both cases counterclockwise with de— creasing magnitude. Explanation: When the north pole of the magnet is with— drawn npward from the loop, there is a de— crease in the downward magnetic flux at the loop. This means that the direction of the change in the magnetic flux is upward. Lane‘s law implies that the loop generates an induced magnetic field Em: to oppose this change, so Bind is points downward. The right hand rule implies that the indumd current is clock- wise as viewed fi'om above. As the magnet is moved downward toward the loop, follow— ing the same reasoning, the induoed current should now flow in the opposite direction. Falling Magnet [11 DDE 19.0 points a magnetic dipole is falling in a conducting metallic tube. Consider the induced current in an imaginaryr current loop when the magnet is moving advise,r from the upper loop and when the magnet is moving toward the lower loop. Version Dill — Third Exam — Hoffmann — {513745} 5 Determine the directions of the induced currents lame and Ibelom in an imaginary loop shown in the figure1 as viewed from ebot'e1 when the loop is above the falling magnet and when the loop is below the falling magnet. 1. lm : Wilmer—clockwise and Ibflm = counter—clockwise 2. no current flow 3. lame = clockwise and Imam = counter-clockwise correct 4. Lime = counter-clockwise end Imam = clockwise 5. fume = clockwise and Emma] = ClEIIEkWiSE Explanation: 1When the felling magnet is below the upper loop1 Hind must be down to attract the falling magnet and slow it down: i.e., clockwise as viewed from above. Before reaching the lower loop, Til-m must be up to oppose the falling magnet; ten, counter-clockwise as viewed from above. Inner Coaxial Coil [It] Short flfl‘i’ 10.0 points A long solenoid has a. tilted coil inside it made of fine wire1 as shown in the figure below. Inside coil has N turns Outside solenoid has it turns per meter What is the mutual inductance M between the solenoid and the coil? 1.M=manREcoslicorrect 1M =p43nNirrrzsinEl' 3-M=#DTl-NT 2 4..M =ngnN%ocsH ELM =hganR2sinfil 2 fi.M=_uunN% Version 001 — Third Exam — Hoffmann - [55745) E- 2 7.M=mflN%sinfl 3..M =pnnN1rr2coefl 0.M=nnnNirr2 10.M=nunN1rrR2 Explanation: Tho mutual inductance of jlifirc'gil satisfies MI— — N03. Thus M: 1 which yields, I M=ngnN1rflzoosH. Inductance of an LC Circuit 003 10.0 points No units are given for answer choices. The units are Henrg.r (H). Calculate the inductance of an LC circuit that oscillates at 130 Hz when the capacitance is 4.2 pF. 1. 0.425462 2. 1.15504 3. 0.151811 4. 0.11349? 5. 0.503137 5. 0.453459 7. 0.19480? 3. 0.159525 9. 0.312149 10. 0.215407 Correct answer: 0.312140 H. Explanation: Let: f=13lZIHz and C=4.2nF. The frequency of osnillations of an LC circuit is l f: 2 11th c“ ‘ Therefore, the inductance is 1 Linnea 1 _ [2u{139 H2312 (4.2 x 10-5 F} ELL-{S Current 04 000 10.0 points No units are given for answer choices. The units are Amperes (A). An AC voltage of peak 1ralue 85.7 V and ifetlljenltljgr 48.3 Hz is applied to a 24.5 HF capacitor. What is the rms current? 1. 0.342335 0.450557 0.303053 0.3Tfil3123 . 0.483197 . 0.403751 . 0.3712041 . 0. 35503 “£1 318857 [1. 0. 413353 _goon-qmcngap:5¢ Correct answer: 0.45050? 15L. Explanation: Let: V... = saw, f = 48-3 Hz. and c = 24.5 “F = 2.45 x 10—5 F. The capacitive ['eflflta-IICE is 1 1 XC=E=esfd so the rms current is = (E) 21:1!me (Via 211185.?” a) (43.3 Hz] x {2.45 x 1D'5 F} = 0.4505575. . ll Version Ufll — Third Exam _. Hoffmann — [5517145) "I rFransformer fllfl 10.0 points a transformer has input 1roltage and current of 14 V and 2 a respectively. and an output current of ISLE a. If there are 1325 turns turns on the see ondary side of the transformer. how many turns are on the primary.r side? 1. 397.5 . 224.0 . 129.2 _ 1T3.fl5 . 223.1] . 1T9.85 T. 144.114 3. 213.03 9. 184.2 10. 145.529 menarche Correct answer: 32?.5 turns. Explanation: Let 1 at, = 1325 turns. IF = 2 A. and I3 = 0.6 A. Energyr is conserved, so Pp: Pf In V = I V; V: __ I V _ I For the transformer V cc n a _ E a m‘n‘a I_ [LB a n: itsfi— — {1325 turns] 2 A = _- Maximum Displacement Current 1311 10.1] points No units are given for answer choices. The units are Amperes {A}. A sinusoidal voltage is applied directl},r acrom a 16.29 nF capacitor. The frequency of the source is 2.85} kHz. and the voltage amplitude is 59.4 V. Find the maximum 1u'alue of the displace— ment current in the capacitor. 1. 15.223 2. 3.73459 3. IDJSSEI 4. 2|].8364 5. 6.13293 5. 45.241 ’4". 3fl.33fi4 8. 12.5705 9. 8.111555 1|]. 13.1241 Correct answer: 1T.5Tflfi A. Explanation: Let: C = 15.22 11F = 1.329 X 13—5 F, f = 2.39 kHz = 2890 Hz, and EM: = 59.4 V. The angular frequencyr of the source is w = 2 1r 1" = 2 it {2890 Ha) = 13153.4 3“. The voltage across the capacitor as a function of t is V = V...” sinfiwt} = {59.4 if) sin [(131534 s’ljnt] . Now we can find the charge on the capacitor, Q = CV. and to calculate the displacement current L11 by differentiating Q with respect to t: :1 Q d d V Id— 5— E[CV}— C— = to‘ C V7,“; oos[w t] . Thus. the maximum value of the displacement current in the capacitor is Jrthermal: = ”J0 me = {13153.1 s_1][1.fi2‘3} x 111—5 F)[59.4 V] Version flfll — Third Exam — Hoifrnann — [56T45j S Alternative eolut ion: V = IXC 1 1 X = = — ‘3 211' for no V I = — = C V XE: a: Id‘mz = we Vim: = {13153.4 s'ljtlfiflfi >< 10-5 F]{59.4 in RC and LC Circuits [III 012 lflfl points Consider the following circuit. After leaving the S‘W'ltch at the position “a,” for a long time... more the switch from ”c” to ”b". There will be current oscillations. The maximum current Twill be given by lufmmfig'lflLC IIIL 2IIm1=£ E 3. Ln” = 81.9% correct 5 L 4. In“ = \/g E 5 511mm:— H E fl-Imar=—R“VLC E {'0 7.1mmz—F — R L at ~5H’i 'TTMII_ LC Explanation: l l 2 2 EL-Imax Ecqrrmm 1 2 E05 C Ifllflm‘ — EE- Electron in Varying E Field v 313 10.1] points An electron under the influence of some central force moves at speed tr; in a counter— clockwise circular orbit of radius R. A uni— form magnetic field B perpendicular to the plane of the orbit is turned on [see figure}. B increasing ._..--':‘" circular path of electron Suppose that the magnitude of the field d B changes at a given rate E' What is the magnitude of the electric field induced at the radius of the electron orbit? 1.r«:=2«.rr3rrE er 2.E=%%correct cs .13: 2— 3 R or are 4 E: is?— ” alt 5E=nR2d—B di RdB as 7E: a— ” dt 8E=rrERd—B Version {1131 — Third Exam — Hofi‘mann — {551'45} 9 dB ‘3. E = R — (it 1o. E = 211' R E dt Explanation: Basic concepts: Faraday's lat-tr states dtIi 5=—a- Magnetic flux, lJrhital magnetic moment of an electron, and Newton‘s second law mag,r he germane. Solution: To determine the induced elec— tric field apply.r Faraday‘s lat-tT .. dilig- E-e“=e~—. jet- 5 ext The line integral around the circular electric path reduces to Efds=E2rrR_. and the magnetic file-c through the surface enclosed by the electron’s path is ee=eafla Since we are interested in the magnitude of E . we will drop the "‘—” sign in Faraday’s law. Thus 2 E2IrR= tittfl‘R B}! dt RdB "3- e a The induced electric field at the electron gen— erates an induced magnetic field which op— poses the increase of the magnetic flux. Thus the induced magnetic field points downward. correspondingly the induced electronic field pointing clockwise as viewed from above. Induced EMF [IT (114 10.0 points No units are given for the answer choices The units are niilli‘tl'olts {miter}. The current in a solenoid is increasing at a rate of 4.16 life. The cross-sectional area of the solenoid is e cm2, and there are 435 turns on its 7.35 cm length. What is the induced emf oppodng the in. creasing current“? The permeability of free space is 4 it x 15‘? T . m/A. 1. 5.52341 2. 0.630823 3. 4.22235 4. 4.59288 5. 4.89291 5. 2.51539 7. 11.9202 3. 1.3325? 9. 1.15538 15'. 2.36141 Correct answer: 4.22235 mV. Explanation: d I Let : — = 4.16 A [it is 1‘ A = 1r ern2 . n — 435 , and L = 1'35 em = 5.5735 m and on =4rr x ltl—TTamfA. The magnetic field induced by the current is B _ MN I the solenoid. , so applying Faraday‘s Law for n_ E__ @1324} 5‘ th_ N tit _ maria er _ L E _ (4e 1-: in-TT.mm}{435}2 _ notes m x [If cm2}{4.lfi Afsj( 1m )2 Hit] em 15GB mi." 1 V Inductor in a Circuit [12 ms 1o.o points |5| X Version [I'D] — Third Exam — Hofl‘n'iann — {SENS} . No units are given for answer choices. The units are Henr},r {H} An inductor that has a resistance of 1m} H! is connected to an ideal |:Ir.I.’E.terj..r of 184 V. E14 n15 seconds after the switch is thrown the current in the circuit is 11112 not. Calculate the inductance. 1. 25.9495 2. 193.453 3. 139.304 4. 443.303 5. 24LUE‘2 5. 145.117 7. 50fl934 8. 133475 9. 91.3231 1|]. 113123 Correct answer: 5313934 H. Explanation: Let: R=1neko=1x105n. . 8:184V. 11 = 11312 mi. = Dflfllfllfl a, and t1 = 0.4 me = cairn s. The magnitude of the current in the circuit as a function of time is in; = E. (1—3-3415) R -Rth=1_E E 5 an L: 1 LR 111(1— T) —{c.ccees}{1 x 105 a} 1n (1 H [11130112112 A} {1 x 105 oi) 184i!r L: J; CL; f?- fig Mm -- ,1; 2 (3:1 9 t m ML .“fih‘ “ha—32L I. 51' T I'M. Z. .. I i :55- ff — 1'. f L at I *L “£5“ W"..- (‘5 3 (2. Es )( 6 q. ) Ll . __ .1 _. NUQHJ. Fame, .1 “i . $94-$31"? __ ___ _ _____ _ _ F w __ fl ____ mi #3.»; .- __..__._______ ________ ___ _ .. _— .. .5 WfluémLL‘ 3.- __._*_._ __%_— ________________ _ ..-_.___..____‘__-,.. . ___—___ ___ ________ ___ :I ‘12-'- Cflhfiififlwgt * m ,__ §I L a of __ _=_.____C__1_;_l j; £¥§u$(WEJ———- “___—Tm __.. _ _ ___-___“-.. . _._ ___ .. .-;_-_fl_____.___m :; b = '2.” L Q ~_'_-'-____2_.r_[ E's-h: we. Egvffiflfifi'ji 3 ___“:TH—_ “5-3—- “_ - % = C 8 = is; 419223;) €§E __§__; "F {Dc—313— E1, __ . off Cahsmwwi-t) . _ ..... WC? 3’7? tummy 12:51”; *- CZ“ fif‘g" —_ “(mafiacm ")(<ré.. a»). .(sz...(z. 89m.) A.. ”L 4252:,4 _/ p 4-3 H _.L_.= hour-tuna C __A5_alqn1 in than: . LEI-f: (he. - -E L" £:'£-_Q.Lt K ”L -Hzt Jail?! = -—Q L‘ (1 Afr-4: r - Q— at- = I“ E. c E. ...
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