This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 9digit 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 1012 C2/Nm2, k = 1/47reo = 8.99 X 109 Nm2/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 200turn 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 105 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 '' 590“? 3 Z'Omx
(it
:(OQI) (DJkn.) (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) 6.4x105T dine
c) 6.4x1010 T 931'”; ‘‘ “Lg—5‘ a}
(.3 8.0x1011T 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 _ 
«awm 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 " sheTr ‘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 ’ rLE Cf " 5
” ' ' ‘
c) 1 x 10 Hz Lo! ____ A .. .. ll ._\
lLLéc é fitaw 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) vms=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 109) 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 (’— saw3 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
Want313 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 92.6 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 baaHeb 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 ...
View
Full
Document
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.
 Fall '11
 RonaldGilman
 Physics

Click to edit the document details