ES3F10 - .019.“ Physics 227 — Final Monday, December...

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Unformatted text preview: .019.“ Physics 227 — Final Monday, December 20, 2010, 4:00 PM - 7:00 PM Colleage Ave Gym / Gym Annex Your name sticker .—.> S o L L2 “H 0 MS with exam code Use a #2 pencil to make entries on the answer sheet. Enter the following z'd information now, before the exam starts. In the section labelled NAME (Last, First, MI.) enter your last name, then fill in the empty circle for a blank, then enter your first name, another blank, and finally your middle initial. Under STUDENT # enter your 9-digit RUID Number. Under CODE enter the exam code given above. Enter 227 under COURSE. You do not need to write anything else on the answer sheet for now, but you may continue to read the instructions. During the exam, you may use a calculator and are allowed three 8.5 x 11 inch sheets of paper with whatever you want written on it. For each of the 32 multiple choice questions mark only one answer on the answer sheet. There is no deduction of points for an in— correct answer, so if you cannot work out the answer, you should make an educated guess. If you have questions or problems during the exam, raise your hand and a proctor will assist you. We will provide values of physical constants that are needed but not given, but rye/will not provide equations. //// A proctor will check your name stickeraand/ymf student ID some— time during the exam. Please have/them ready. You are not allowed to give help to any other student, ask for help from anyone but a proctor, or change your seat without permission from a proctor. Doing so will result in a zero score for the exam. When you are done with the exam, hand in only this cover sheet and your answer sheet. ' Please sign above by the name sticker to indicate that you have read and understood these instructions. Possibly useful constants: 60 = 8.85 x 10-12 C2/N-m2, k = 1/47reo = 8.99 X 109 N-m2/C2, c = speed of light 2 3.00 X 108 m/s, _Qelectron = Qproton = X 10—19 07 melectmn = electron mass = 9.11 X 10‘31 kg, mpmtm = proton mass = 1.67 X 10‘27 kg, no = 47? X 10‘7 Tm = 12.57 X 10'7 Tm, and 1 eV = 1.602 X 10—19 J. Some metric prefixes: f = femto = 10—15, p = pico = 10—12, H = nano = 10‘9, ,u 2 micro = 10‘s, m = milli = 10"3, k = kilo = 103, M = mega = 106, G = giga = 109 A square loop and a circular loop of equal area lie in a plane per— pendicular to a uniform magnetic field. Each loop then rotates at the same angular frequency w about an axis in its plane pass ing through its center. In which loop will the root—mean—square average induced emf be greater? a) the square b) the circle c) Depends upon the orientation of the x x x axis in the square. d Depends on w. ‘ X fix X e) The induced emf ’s are the same X X X SWVQ: ‘ ‘ wBA sinuot. A 200-turn coil of wire is wrapped on a square frame with 18.0—cm sides. The total resistance of the wire is 2.00 ohms. A uniform magnetic field perpendicular to the plane of the coil is turned on and it changes linearly from 0.00 to 0.500 T in 0.800 seconds. What is the magnitude of the induced EMF in the coil while the field is changing? b) 0.0202v \w‘ ST. L 810V = 2... (ll 5.06V (.35) ©ng 200 e) 3.24V 3 ‘405 V The current through a long solenoid of radius 3 cm is increasing linearly, so that the rate of change of the magnetic flux (BA, where A is the area of the solenoid) is 4.75 X 10‘6T—m2/s. What is the magnitude of the induced E—field at a point midway along the solenoid and 5 cm from its axis? a) 9.2 x 10‘6 V/rn b) 2.5 x 10"5 V/m c) zero d) 4.2 x 10-5 V/m 5 £2,313 :. -0592 mo? db 211‘ (o .05 =- "f: seal = afCBA usth ='m fig ’I;:-d}o l 2M9 it: dtSBcl thU am- E ‘—'- “JS'XW‘E Tw‘ L *s‘ - WI — M 5. A square wire loop of area A = 0.16 In2 is in the any plane moving at a constant velocity 2.0 m/ s 3,7. For y < 0, the magnetic field vanishes, but for y 2 0 the magnetic field is E = 0.4093 + 0.202 T. What is the magnitude of the emf around the loop while it is crossing the y = 0 plane? a) 0V 2mg; «<29 A.“ gin? *5 at c 0.32V [gwi -_~_ ‘ dz 2 d) 0.36 V r 37*: 4 71—» y e) 0032 V -'-' 5-90“? 3- Z'Omx (it :(OQI) (DJ-kn.) (2.0 84:15) .. odév The electric field in a circular parallel plate capacitor increases at a rate of 72 V/m in 10"6 s. The diameter of the capacitor is 0.4 m. What is the magnitude of the magnetic field at the outer rim of the capacitor that results from this changing electric field? J :- a) 7.2x107T $80.2 : ill??? b) 6.4x10-5T dine c) 6.4x10-10 T 931'”; ‘-‘- “Lg—5‘ a} (.3 8.0x10-11T I e) 90T B= gaé~a= M’- 1——-—;1V"' ' g g“ a: moms) to“ s 4: \Ho“ 1‘ Two coils are close to each other, with one connected to a current source which produces a current of [1(t) = 3t2 + 5 amps, Where t is expressed in seconds. If the mutual inductance of the two coils is 10 mH, what is the voltage V2 across the second inductor at time t = 3 seconds? 5;: -M“C‘/dl= 11(t) é, —> a) 0.32V _ - «aw-m lel ‘ M MW 1 c) 0.23 V ; t . d) 0.36 V M g (“3 V2 e) 7.2 V = lOM‘H' ($515 = {BDMV 7. The current in a solenoid increases steadily from 0 to 10 mA in 11. A 0.2pF capacitor is connected across a 110 V (rms) 60~HZ volt~ 40 ms. If the induced emf across the solenoid is 0.8 mV, what is age source. What is the ma)de current flowing through the the solenoid’s inductance? E 3 .- L. d: I db capacitor? /K“f&{)vrm$ #221159 a) 0.2mH 2) L ‘ ‘2; (am) a) 829m vain. (ro new: 9: t) m " b) 584 mA ‘ aw . :l m I C 1.32 M L =— c 7: by. 10 1)((i)£“°‘) (2.17 LO) d) 0.8 mil : 3.1m H d 11.7 mA ‘ V " she-Tr ‘03 e) Cannot be determined from the information given e) 145 mA LMA: :. @* m-‘Z)(¢fl(no)fizjr Go) :1 un «A 8. An inductor carrying a current of 1mA is designed to store 1m] 12‘ f Wh d 7 A voltage from an AC power supply with w = 103 rad/s is applied 0 energy. at is its in uctance. across a resistor with R 2: 100 Q in series with a capacitor with @ E 2. Ji— L, I?- C' = 0.2 ,uF. The current given by 2' = Imam cos(wt) results from b) 1 H an applied voltage ’U = Vme cos(wt + ¢) with a phase angle given 0) ZmH L. = 15/17- bytan¢= d; i In; : ‘2 ‘8. \O'SJ /(lD- 3:432 a) 50 b ~50 c) 2X10"6 d) ~2><10‘6 e # = 2 H any: wL .4er ___ o— #9090lequ __50 9. A 2.0 volt battery is suddenly connected across a series LR com~ R “3° bination of a 1.0 Q resistor and a 10 mil inductor. How long does 13‘ jAn RLC drum; is driven by 8“ Steady_100'HZ Source With 3‘ Ufa” it take for the current to build up to 1‘5 A? imurn EMF of 100 V. The current in the system has maanma ® VL :. L- A‘IA“ (amplitude) of 0.50 A. The average power dissipated in the resis~ a . ms :3 .UM .5 “a + I tor is 15 W. The phase angle between the current in the circuit b) 2.88ms Vac/ gr. . VJ P ) 1 39 C at) and the applied EMF is C . m8 = = _. e"- a e) 6.02 mg as h goo , c -> Lousy, vL—w => Em»: = la b) 0.30 rad .. was , Tb“: L’R ='> 13‘th (t- 4M") % “5:10”: 2W»: ‘ C054) : ZPave- {33/ i (91 ‘53/Qooxoé) ; 0.4 10. The resonant frequency of a certain LC circuit is 105 Hz. If the Q = q 8) 0.42 rad capacitance and inductance each increase by a factor of 6, what Loot 33g will be the new resonant frequency? ‘ _ '3 1m 5 9 0 a; op —_ “ “ a 3 ‘ V‘q a) (1/36) x 105 Hz U3 ’- rL-E Cf " 5- ” ' ' ‘ c) 1 x 10 Hz Lo! ____ A .. .. ll ._\ lLL-éc é fit-aw d) 6 x 105 Hz e) 36 x 105 Hz 14. 15. 16. A resistor of 8 Q is connected to the 60 turn secondary Winding of 17. an ideal transformer. The primary has 900 turns and is connected to an AC EMF of 110 V (rms) and 60 Hz. What is the rms current in the resistor? a) 13.75 A (rms) N1 :900 b) 206 A (rms) vm-s=110v c 917 mA rm , f=60Hz R=89 a 15 A (rms) 18 648 mA N ‘60 ' e) (HHS) +'““5'$”'Me.v- : W. N: 2) V?— : u '2 V‘ ~ fi[ rms Cmszvvms = A radio wave of frequency 107 Hz hits a perfectly absorbing planar surface that is perpendicular to the direction of propagation of the wave. The average power per unit area of the wave at the position Where it is absorbed is 10 W/mz. Calculate the radiation pressure on the surface. 19. \ d Wlm" a (3.3 x 10—8) N/m2 3 I: =~ 53:: gimgmfs b) (18.6 x 10‘) N/m2 c) (3.3 x 10-9) N/m2 : g3 “0-? MIN: d) (6.7 X 10‘8) N/m2 e) 18.6 N /m2 In the United States, accepted standard for the safe maximum level of continuous Whole—body exposure to microwave radiation is 10 milliwatts/cm2. For this energy flux, What is the corresponding maximum amplitude of the electric field? 0 a) 388 V/m E M1,,“ __._ \DmV‘fx G’Tfm) b) 194 V/m Scum. = 5“ CV“ "uc ‘- c) 137 V/m ,1 W 2" e) 8 v In em» 3 QMo (’— saw-3 2E1 \fL‘u-‘dO-‘z \C 37403 X :: 1"” WM Taking c 2: 3 X 108 m / s, the velocity of an electromagnetic wave: a) depends on its frequency. b) depends on its amplitude. c) is alwa s c. V WW ‘“ “1 e) is KeKm . $0rmd‘a 9‘832A‘ An electromagnetic wave has a magnetic field of form 3 2 BO sin(wt + What direction is the Poynting vector? a) +y"dlrecfion\ 9 Move; it; ~1~d "BCJHEn‘ b) —~y—d1rection H u ‘ ‘ J c) +z—direction E M“ “(W1 ewuflj move; “ " l' e) +m—direction “ r r u Want-313 Vacbf The electric field in an electromagnetic wave is given by E 2 E02' sin(k;z — wt). The associated magnetic field is given by B: A J ' | a) B02; cos(k;z — wt) 6 m 1" Oo.\v‘c.+m b) Bolf cos(k:c — wt) n A" . c) B02 cos(kz — wt) emufifl “‘0'” “1 2' ‘Nchc” d) Bisin Izzy—wt) .1.) B {nit fluke-+5" e B0] s1n(kz — wt 4 A ' 50 E vé : K Same. $114 UH.- wt) ’; \o‘w 4"?“ 'A'mm‘ a 3 20. The figure shows the electric field lines due to two charged parallel 22. E clan cans—Fwd 21. metal plates. We conclude that: a. the upper plate is positive and the lower late is ne ative. a positive charge at X would experience the same force if it tY. a positive charge at X experiences a greater force than if it were placed at Z. a positive charge at X experiences less force than if it were placed at Z. a negative charge at X could have its weight balanced by the electrical force. I A magnetic field exerts a force on a charged particle: a) b) 6 d) 8) always. only when the particle moves exactly perpendicular to field 23' lines. wen the article is moving across the field lines. when the particle is moving a ong the field ines. when the particle is moving. §M=sGrB =3 QWLL WORN—5 V #6 => 69 i 0 WW (:6 S'OAGL {g- 9-2.6 A long straight wire is 10 m behind a square loop With 0.1 m sides, as shown in the figure. The straight wire has a current going to the right. The loop has a current going counterclockwise. The wire is fixed in place, while the loop can rotate about an axis through its center. Which of the following is true? a. There is no torque on the loop; it will not rotate. b. If the axis is vertical, the left edge rotates out of the paper towards you. c. If the axis is vertical, the 3 right edge rotates out of aw” «the paper towards you. d. If the axis is horizontal, the top e ge ro ates ou 0 paper towards you. ‘3)c v5 5:}. e. If the axis is horizontal, the bottom edge rotates out of A the paper towards you. {3an e. A capacitor charged to a potential of 12.0 V is put in series with a 7.5 M9 resistor (note 1 M9 = 106 After a time of 4.0 s the potential across the capacitor is 6.0 V. The time constant of the circuit therefore is fiijQ. b) 4.0s “‘2 ‘ c) 8.0s -t/£L WV"? 5.8s .. ’B é:\2e, L, 24. The space between the plates of a parallel plate capacitor is ini- tially filled with air (K. = 1). A battery is connected across the plates of this capacitor, the voltage across the capacitor is V0 and the capacitor plates have a charge i620. With the battery kept connected to the ca acit r ' pletely filled with a dielectric (K. = 2). Which of the following relations concerning the dielectric~filled capacitor voltage V and charge Q is correct? 51) V=Vo/2 b) V=2V =Qo/2 6) Q2520 C?- {2" CD 03 *d 93 O (D C)" § CD (D D C?- [3‘ (D *d y... 93+ CD CD 5- CD 0 O 5 baa-Heb come 9% 3) V a V0 é CQ ‘ Gt. (3. =- 2 Go 25. The figure indicates some electric field lines and three paths by which a charge is moved from point G to point H; all are in the plane of the paper. Path 2 is along the field line from point G to point H. The relative magnitude of the work done by the electric field on the charge for each path is: 26. An infinitely long, straight metal rod has a radius of 0.03 m and a charge per unit length, /\ = 5 x 10‘9 C/m. Find the magnitude of electric field at a distance 0.15 m from the axis of the rod. {‘3 3830»??? 5 Kiwi = “M 29 c) 6N/C inn/E glues Cl) 30 E =- x /J1T£°h mm!» - 5x a“? . .. zwgflmpqz Us .. 600M; Consider two resistors A and B. Resistor A has twice the resistance of resistor B. They are connected in parallel to a battery. The ratio of the power dissipated by A to that by B is: ' C) 1:1 d) 2:1 e) 4:1 RA=QR3$ 9pc:- Pg/z an 7&1P8:Hl Three capacitors are connected by conducting wires as shown. If any other circuitry were connected to the points A and B, the three together would act as an equivalent capacitor of capacitance Ceqv. What is Ceqv? A A a 10m? Equiv. to w 1: Cqu b) 52 [1F 2”F c) 1.6 MP 40w? d) 1.92 [1F e) 2.13 nF B B . C. um Familial GM " Cubv :. 8+}: (OM? 29. Wires A and B are made of the same material, but the radius of 32. A standing electromagnetic wave of wavelength A is set up in a wire A is twice that of B. The current in wire A is four times that cavity of length L with condécting walls aligned along the x— axis in wire B. How are the drift speeds of charge carriers in the two as shown in the figure below. One of the electromagnetic fields in wires related? the cavity is aligned along the z— axis, as shown below, but you a) DA = SUB IA UR PA} are not informed whether it is the electric, or the magnetic field. b) DA = 203 ’ "'- 2 '5" z ' Which of the following is true? 1 g ‘3 $3 a. A = §L and the magnetic UA=1GUB =)u' I z \1 fild' 1‘ d1 th A-=U .3" -.(‘ _. 2;) e 1spoar1ze aong e e) ’UA=4’UB Bra éz‘ngkC zaxis. ' “k; V’s A b. A = %L and the electric 30. What is the field at the origin due to the two infinitely long wires field is polarized along the shown. Each carries a current I in the +2 (out of the paper) or z axis. —2 (into the paper) direction as indicated. c. A = %L and the electric . field is polarized along the 3. L01? A + 1 A y All dimensions 2 axis. 1 a = —a: — b) B = [[358 + 3?] (0’ are m meters (1. A = %L and the magnetic field is polarized along the z .. 27? A axis. C) B = 5—01 _ e. A = L and the electric field polarized along the z axis. d) B = sen—ti — %291 X , e) B = /;—°I[——2$ — 3y] m&vo-\*M5 3) éfawdu 2‘0 .9 ’\ —— 4 vse. $=Mor g=&T_)_a_ )___Mo’,_‘“ 3- - ~ g I aw 1 Ba 3.“— 3 'D (55 had” 31. A voltage 12 V cos(30t) is applied to a 100 Q resistor. What is 2) grahm‘gj q! ,E the rms power dissipated by the resistor? J V k L mj “N a) Need to be given the relative phase of the current to deter— z 1 mine this. :1) A 2 a L- b) 1.44w v =- 61»! cos (301:) x c) 1.02 W __ .. \J 9"”? gm cos3°t a) e) 0.12W - no plans: JIM $0? mush»? .. L _ 1 IL) P— zIa‘Xaf—(fg‘l’ 103;“) [L ...
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This note was uploaded on 01/04/2012 for the course PHYSICS 750:227 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.

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ES3F10 - .019.“ Physics 227 — Final Monday, December...

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