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F08 - Exam 3 - Solutions(1)

F08 - Exam 3 - Solutions(1) - — Physics 116B —...

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Unformatted text preview: —_ Physics 116B — Exam #3 (Wednesday) — Fall 2008 i Only calculators and pens/pencils are allowed on your desk. No cell phones, music players, or additional scrap paper. You have 75 minutes to complete the exam. Name Section (Circle): 3 Helms TR 8:10-9:25 1 Helms TR 11:00-12:15 If you are in Prof. J ohns’s class, you’re in the wrong room! I pledge my honor that I have neither givennor received aid on this work. Signed MC Prob 9 Raw Score Percentage % MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question, Figure 29.1 {—— 20 cm —-> 10:) e5: 1) In Fig. 29.1, a wire and a 10—0 resistor are used to form a circuit in the shape of a square, 20 cm by 1) 20 cm. A uniform but nonsteady magnetic field is directed into the plane of the circuit. The magnitude of the magnetic field is decreased from 1.50 T to 0.50 T in a time intervai of 32 ms. The average induced current and its direction through the resistor, in this time intervai, are closest to: A) 75 mA, from a to b B) 75 mA, from b to a C) 190 mA, from a to b 30 mA, from b to a E) 130 mA, from a to b at: '7 £325; 2:: ngV r '8’" cf:- ‘f :1: r? 1051, is m aims QM vs aifrmms min in Page A/L‘E’“ iix/ Ewe]; law] Mailing] Curm’i @993th Chang,” if) (LL Pmaiuurs flA m3“; - Figure 29.8 a b 2) In Fig. 29.8, the inner loop carries a currentI that is increasing. The resistorR is in the outer loop. 2) The induced current through the resistorR is: ' I A) zero @from a to b C) from b to a hair“! mended“ Tbrooiuws a (plaid “mejr :3 {mowing Miro H4 [362%, TL OVAL—s (qrmeAr Mus—V produvc. A, Qtiflz‘ VYL‘R [5 govt o g ’HAL Fag/.27, A-2 Figure 30.1c 219 51H shown. Initially, the switch is open, and there is no magnetic flux in the inductor. At tirnet = f} s, the switch is closed. In Fig. 30.1c, when the resistor voltage is equal to the inductor voltage, the current in the circuit is closest to: @AA 8) 0.57A C) 0.86A D) 1.7A E) u A v 6V If 0 " Loo ewlt f 2‘}: 35: L g . = 60V ! i m : m CNN/vi} in. ‘i’LL féfiéL’léf .K‘S 2V (1, V "ls/LL CUFW‘!" m 444. LJLLoi-L cr‘raM-J: :e T O Figure 30.3 10.0 $1 30.0 (I! 3) An R—L circuit has a 60-V battery, a 51—H inductor, a 21—0 resistor, and a switch S, in series, as 3) H 4) In Fig. 30.3, the instant after closing the switch, the current through the 60.C—Q resistor is closest to: 4) pl @313 A B) 0.00 A C) 2.50 A D) 3.33 A s) 10.0 A ermine dongs 30$th 44¢ \‘nduclcrs lock le'iflf'opm gawk W‘HA {to CLUTet/bl’ 1‘“ 'HJLM. i :x 5;; et WM 1 (00¢ H (40“me A-3 Figure 30.5 75.9 D 290.0 V 35,0 HF 25.0 £1 5) In Fig. 30.5, the potential drop across the 70.0- pF capacitor after the switch has been closed for a very long time is closest to: A) 101 V 33 V pl'l) 5i long) alarm.) “’l‘dlflfi. Qa‘fcml looks ll l? e i 200v C) 200.0 V D) 0.00 V E) 80.0 V 25% Figure 31.121 X0 = 790 o _ R=500f1 XL: 270 o 6) The 60—Hz ac source of a series circuit has a voltage amplitude of 120 V. The capacitive and inductive reactances are 790 Q and 270 Q, respectively. The resistance is 500 Q. In Fig. 31.1a, the capacitance, in pF, is closest to: A) 20 .4 C) 9.7 D) 13 E) 6.5 ,L _ ML KfirCta.) F C218: “(9 “’1” c "M z 3.1%“) IF: 34w: SE : ISBV’ 6)B 7) If the voltage amplitude across an 8.50—pF capacitor is equal to 32.0 V when the current amplitude 7) through it is 3.33 mA, the frequency is closest to: A) 32.6 Hz .20 MHz C) 32.6 kHz D) 5.20 kHz E)32.6MHz \J':2 T: A“ : all... Cub gm“; 3.; $9 W‘SA Q sew 3%Egmmwww .zszmoHZ: 520%? F h 3 N (Kl r- C 9 8) An L—R-C series circuit consists of an 85.0—0 resistor, a 14.0- HF capacitor, a 1.50-mH inductor, and 8) a variable frequency ac source of voltage amplitude 13.25 V. The angular frequency at which the inductive reactance wili be 4.00 times as large as the capacitive reactance is closest to: A) 2.20 x 103 rad/s @38 x 104 rad/s C) 6.90 x 103 rad/s D) 3.45 x 103 rad/s E) 1.10 x 103 rad/s VL’JHXC Lm=q W CL.) A-5 _JE_ Physics 116B — Exam #3 (Thursday) — Fall 2008 Only calculators and pens/pencils are allowed on your desk. No cell phones, music players, or additional scrap paper. You have 75 minutes to complete the exam. Name Section (Circle): 3 Helms TR 8:10-9:25 1 Helms TR 11:00-12:15 If you are in Prof. Johns’s class, you’re in the wrong room! I pledge my honor that I have neither given nor received aid on this work. Signed MC Prob 9 Raw Total Percentage % MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Figure 29.9 1) In Fig. 29.9, a bar magnet moves away from the solenoid. The induced current through the resisto; 1) B R is: A) zero rorn a to b C) from b to a fls Masmi” i3 PwuiA gauche/J H €le Musi‘ [QM—36km}: a mat“ NAIL ‘R'filol ( ‘ a“, d§;;:::) niwwdt 9' 2) A small square coil is located inside an ideal solenoid at the center with its plane oriented 2) perpendicular to the axis of the solenoid. The resistance of this coil is 200 Q and each side is 4.00 I cm long. The solenoid has 125 windings per centimeter of length. If the current in the solenoid is increasing at a constant rate of 1.50 A/s, the current in the square coil is: A) decreasing at 1.50 A/s B) initially equal at 18.8 uA but is increasing ::::S:i:t81:: AIS gufl‘ {Q Vow Mud” “in C6! lmlaL-fi Ill : E)zero J .. EE’ZQGQ a- fl fies?“ré”sat an“) Ra‘s «L (Ll/16m 15 win g figfifigfivna Eisnflxgenn: y l ' anti-taper men I ‘flfi 5: :‘ET fl 5g ’5 C ” OF re W "at" ‘ ‘9 509 a? is (onsisci’) 5° :- “" (‘Olmvwmollgxgr X F > 25L 8 is. consdmwfl: so ’HILL aqua/A” l’f\ 2'1?.g*/LA gigxmg gas? :5 Kong‘lfifl'l": A-l D 1’: H»; mix) answer, Figure 30.1d 250 3) An R—L circuit has a 60—V battery, a 45—H inductor, a 25-0 resistor, and a switch S, in series, as 3) E shown. Initially, the switch is open, and there is no magnetic flux in the inductor. At timet = 0 s, the switch is closed. In Fig. 30.101, the rate of change of the current when the current is 1.5 A is closest to: A) 0.67 A/s B) 0.83 A/s C) 0.75 A/s D) 0.58 A/s @150 A/s I/é:) = 1:, 0 ~65”) I. SA 42W) vise/s) 3.4.3:... E, Jr): ; jf a; V 3; (99V ,3 LA 1.2:); see/rt 0:“ - ; a ? ER a» 7' I MAY —: \—€ i L, ng Z'Q [S ZAQ “mas/17$ ‘ r: = M ‘3‘; M f: . "t9 ._ J.— :: R 26:51 lgs e [@7/ 2:4 .345 t-gfi 6" -td 1': . E=fi4(37s) O'SC‘A ) so : warmtssflfims Figure30.3 10.0 0 30.0 0 r 15.0 1000‘)? “‘H 4) In Fig. 30.3, after the switch has been closed and left closed for a very long time, the potential drop 4) A across the 600-!) resistor is ciosest to: @000 V B) 85.7 V C) 66.7 V D) 100.0 V E) 90.0 v Witr a lens 05M, 44¢ LiamH {noiuc or acid hire a sitar—F “£5955 “i'kt, (60.31. rescsvtor. Wage, 0° Curr'evfjr Mow; ibvmsh J/LA» crazier and W Pcicvriccd Gimp amass P3? )5 grav- Figure 30.5 75.0 {2 200.0V . 35,0 PF 25.0 D 5) In Fig. 30.5, the potenfia drop across the 15.0—mH inductor just after closing the switch is closest to: A) 0.00 V 50.0 V C) 50.0 V D) 200.0 V E) 100.0 V [mien/)9} m cqpqaiflfj $0.4" Séar'zs‘ arm/ Ihdwca/vrfi /ooé 01:93“ Cr‘rem'JS. W-..W..WM/ww V“WV“ Wit % «r» ; LWJMAM V ffi J|M§+ 5;"‘59 Citrbfifi 4154... P6§P¢g¥af , \J 20:20 “:Ww. . ‘2: 2 A 594% Mac.) r71. A-8 5) .3. Figure 31.1b XC = 710 o R=4900 6) The 60-Hz ac source of a series circuit has a voltage amplitude of 120 V. The capacitive and inductive reactances are 710 Q and 320 Q, respectively. The resistance is 490 D. In Fig. 31.1b, the inductance, in mH, is closest to: A) 1900 C) 3200 D) 4000 s) 5300 XufiLW “My”? “page 01;?st 7) The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequency of 1400 kHz. What ithhe value of the inductance? H pk. A) 1.97 ;H B) 1.43 «1H C) 4.42 mH D) 9.20 saH E) 1.36 pH Wm; if LL00 6) 7)_______ 8) A series L—R—C circuit consists of a 175—0 resistor, a 10.0—mI-I inductor, a 2.50—uF capacitor, and an 8) ac source of amplitude 15.0 V and variable frequency. If the voltage source is set to 3.00 times the resonance frequency, the impedance of this circuit is closest to: @243 Q B) 386 Q C) 525 o D) 344 o E) 175 o (c) Calculate the magnetic force on the slide wire. W Faxes: Saw “7%” 2 2 (d) From your last answer, show that the velocity of the slide wire as a function of time is given by v(t) 2 v0 e“B_mii“. (4) 22 iflaz‘z i FEWmdfiwqgge"E LV flAw" E>Lf Git Q MR _ 1 aien'i 44%, \I'” H. “The infra; (v: V” a“ % 4:) 1% la gFggfii’itr éfitfééfLIG-Q (A5; V. Qh/x gig: at: “ .22? V V t‘ 9“! I TL? RT “‘2 “‘ Es. ate: «it? m g (e) Calculate the total distance traveled by the slide wire from the point where it was released. a4 a9 = _. “E11 (4) Ax [we -V”§€ m at 0 _M€ flgfll .2» V; at e I I I l ! FREE RESPONSE. Write your final answer for each part in the box provided. Partiai credit will only be given if you show your work and it is clear. 9) A rectangular loop with width L and a slide wire with mass m are shown in the figure below. A uniform magnetic field B is directed perpendicular to the plane of the loop, into the page. The slide wire is given an initial speed 00 and then released. The resistance of the loop is negligible and the resistance of the slide wire is R. There is no friction between the loop and the siide wire. X X X X X X X X X X X X X away.) 9: an abs-sex Em he. wait? it new. Womb (a) Calculate the magnitude of the emf induced in the loop enclosed by the fixed wire and the slide wire. (Your answer should depend on v). i 013: lgli Egg: £84): BEE-:31“ ELV 5(v)= BLV (4) (b) Calculate the current in the loop and indicate its direction. f“ E: “a BLV R “ a“ flank; “Emit?! {35>th this Hrs and nil/LL Max is immersing) Lit-42:3 law Stu/s ‘I’LL t‘hdhrefip ganged?“ @roduws a W we deem w ems; e. Riel M oil HM FQSMQ. ‘ ...
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