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Unformatted text preview: Physics 2220
Spring 2008
Examination 3 Name: lie)! I Student ID:
Instructor (circle one): Littler ' Matteson Weathers This test consists of 10 multiplechoice questions and 6 freeresponse problems, for a total
of 110 points (so that 10 points of extra credit are possible). To receive credit for the free
response problems, you must show all of your work on thepages" provided. Don’t hand in
‘ any extra sheetsikor other paper. You may also earn partial‘creditj for the some of the
<:mnltiple choice problems if you show your work. _ . , _ " I ' 7 Suggested procedure for solving the problems: 1. Read each problem carefully and make sure you know what is being asked before
starting the problem.  Draw a ﬁgure for the problem. List the parameters given. Write down the equations to be used. Solve for the answer symbolically. Substitute numbers into you ﬁnal equation and circle your answer. F‘SJ'PP’P WORK THE EASY PROBLEMS FIRSTll NOTE: The value of each question is given next to the question number. Please circle the
correct answer. ’ ' Section 1 : Questions 1. (5) If a planar coil of wire is placed in a constant magnetic ﬁeld, an emf can be induced in the
coil (a) if the coil is rotated about an axis parallel to the magnetic ﬁeld direction and the plane
of the coil.
if the coil is rotated about an axis perpendicular to the magnetic ﬁeld direction and
parallel to the plane of the coil. (0) both (a) and (b) (d) An emf will not be induced; the magnetic ﬁeld is constant. ._ Wm loo
Hitic W? 2. (5) An electromagnetic wave is made up of
(a) oscillating electrons.
(b) oscillating electric ﬁelds.
‘oscillating magnetic ﬁelds. oscillating electric and magnetic ﬁelds. 3. (5) Which of the following electromagnetic waves has the lowest frequency?
(a) Ultraviolet light
(b) Visible light
(0 Infrared radiation
(ﬂ) Microwaves
(e) Gamma rays 4. (5) A transformer has 100 turns in its primary and 200 turns in its secondary. If the
primary is connected to a 100V Egsource, what is the voltage across the secondary? (a) 50 V
b 200V _ . _. ‘
(d) 400V 114m W‘i’o b; (nitrite—V”? “award mm fiimﬂ'h P i (e) 1_00V WW §%}5ﬁ,w ”mommy" indwcuanmf mime
Ming» amt , have n M {Mr 11493513 ﬂax
+6 W aIvotiwaL mm . ~ 5. (5) Two identical magnets are dropped into two long hollow tubes. One of the magnets
moves slowly down its tube, while the other undergoes essentially free fall. Which of the
following could explain this behavior?
(a) One of the magnets is dropped with its north pole entering the tube ﬁrst, while the
other is dropped south pole ﬁrst.
(b) One of the tubes has horizontal slots cut through its wall almost all the way around its
circumference at several places along its length ’ ‘A (c) There IS a large electromagnet at the end of one tube, attracting that tube’ 5 magnet.
.Only one of the tubes 15 made of conducting material.’ WMW rm maimllole ﬁvngjhwgfm W 711mm “HM/um 6. (5) In an LC circuit, when the current is at its maximum,
(a the magnetic field in the inductor is at a minimum.
‘the magneticﬁeld in the inductor is at a maximum.
(c) the angular frequency of the circuit reaches its minimum.
((1) the charge on the capacitor is at a maximum. Twitlu Wmmalmd ,ﬁvw gradual. Bauer/11. Claudia EB 15 Want/Nordic].
mﬂsaisol; guM‘ﬁm We'lsbw: 7. (5) A displacement current can exist only when
(a) a timevarying magnetic ﬁeld is present.
electrical charges are in motion.
& a timevarying electric ﬁeld is present.
((1) the line integral of the magnetic ﬁeld around a closed path vanishes.
(e) the line integral of the electric ﬁeld around a closed path does not vanish. 3,491.11... IWWw/téj—i‘ chgejat MWMwMEtW
QWWB‘QIEI‘OVMHMM‘M 8. (5) The net magnetic flux through a closed surface will be nonzero under which of the
following circumstances? (Circle all that apply.) (a) A bar magnet is placed inside the surface. (b) A bar magnet is placed half in and half out of the surface, so that the surface passes
between its poles. (c) A bar magnet is cut in half, and one of the halves is placed inside the surface. (d) A loop of wire is placed inside the surface, and current is passed through the wire. e) A charged, insulating sphere is spun around inside the surface. None of the above. §i§d~31=0 (WEWQIWM) 9. (5) Radio waves are propagating in the ~—j direction. At a particular instant, the electric ﬁeld
is observe to be in the +k direction. What is the direction of the magnetic ﬁeld at this
instant? ' , lO. (5) A series RLC circuit consists of a 40 Q resistor, a 20 HF capacitor, and a 60 mH inductor.
What is the impedance of this circuit “when driven at its resonant frequency? V, . ' (a)i1‘7 r2“ ‘ ,
b 831E; ‘ A‘ _ ,, :5,
409 A‘f MW, wL= 5E,W Z’K (d)25£2 (e) on Section II—Problems Show all work on each problem. Credit will not be given unless work 15 shown. Remember
that ﬁelds and forces are vectors. .direCtion IS part Of the answer. ‘ . 1. (10) A 5000 kg spacecraft (capsule and sail) employs a 1.00 X106n12 perfectly reﬂecting
surface to provide propulsion for travelzling through the Inner solar System. If the aVerage
solar intensity is constant at 1370 W/m2 for a trip starting from Earth and assuming ideal
conditions (no gravitational effects constant acceleration, sail facing the Sun etc. ), ﬁnd: (a) the force exerted on the sail, 2: I370 iii/mt IND/inc
2.9?Bix109w/5 .A A: .. 23 ’ _ ’ ,
?= 231“ =§s F=?'A" —¢m A” ‘ 9J4N WE'VV‘LSM (b) the acceleration of the spacecraft, ' m L
A a: F = 9.14M = 1‘33110'3Mz Maggi/MSW 9:“ 500003 (c) the time required to pass the Moon, which 18 3. 84 x 108 m from the Earth. oi= ﬁat: => 47 F 23,934me 548x105 —?S’oda.3 I 83w .__._...__.. 2. (10) Consider the circuit shown, Where E  15 V, L— — lOmH, and R  15 Q. (a) What 1s the induCtiVe time censtant of the circuit? lxio’zﬂ
' IS'SL 73* e]; = = “72*? (b) Calculate the current in the circuit 200 us after the switch is closed. I= IMO5%) W im‘ {g s 1— 1%0 e We») 0.35711 (c). How long does it take the current to reach 40% of its maximum value? 3. (10) A circular coil of 15 turns and radius of 20 cm is wrapped around a 1 m solenoid of
radius 10 cm and 300 turns/meter The curl ent in the solenoid varies according to the _ expression I(t)= (20A) cos[(100 s"l)t]. (a) Find the induced emf 1n the 15 turn coil as a function of time. liHurn coil Esau/Joni : W I; h E  with“ '  t dt SEW 0M " , ﬁsgmulsda
«N; ADMIAM ”4w I= Imc'amt Nc 1o MI sanwt jam wt”: = #5 100/ 20A 4ND 715% 30%1r Mm )5th
“; 0:35?! s‘m[(lOOs l)t] n " (b) If the 15 turn coil were tilted at an angle of 30° with respect to the vertical, would the
induced emf change? Why or Why not? No, hawthmwc aﬁux’thrauz/it‘th IFMMIWMW. (WWW
(Si “datum (.0st WWO”) Tun' “530° > ('5th 4. (10) A generator produces an rms current of 50 A at 600 V. If the voltage is stepped up to
17,000 V by an ideal transformer and transmitted a long distance through a power line of
total resistance of 12 ohms, determine the percentage of p0wer lost in transmission to a
resistive load. (hint: % power lost is (P105: / P) X 100). IhMlM‘fW/JW, Spewwiu WWW. “#1 Wind“ rw5ﬁue, ﬂu
WQCWi51MMchW14L AV ., _, S‘IAV WIZAV .> LE L‘A— AV2: 'l’muw lost ‘mfmJnJSs‘u/u u ?‘ ILK um. k , Haematww ‘3 (1.333;) Pm ,  AV. 1;. l’,  __ ..
EC LTV, LAV ' '(sz) \
"t4 f I 5. (10) A thick glass plate (nglass = 1.62) lies at the bottom of a tank of water (nwate, = 1.33). A
light ray enters the water from air at an angle of 50° with respect to the normal to the surface.
What angle does the ray make with the normal when itis in the glass? Assume that the top
face of the glass plate is horizOntal. L I 9,1:500 _ » ' Inf!“ ,_ ﬂ ‘ _ ._l__, a : o
Vi.sm9.=msm9ﬁ 9:35:19; '3? (93 = answnGﬂ—gsme.) = Mama“ 5,3150) 28.2 .(10) If an R ~— 1. 00— k9. resistor, a C ~— 1. 00,uF capacitor, and an L: 0. 200 H inductor are
connected 1n series with a V= (150 V) sin [(377 s )t] source, ﬁnd (a) the phase angle between the current and the voltage in the source, and __
l (f): Wam[xﬂ Xe]; i '6: [me ELC’]: madam{3175iazmH!;37?s"t¢l0’6f; R R 10009; ,
. ~ (.818"
(b) the maximum current delivered by the source.
V We
11%;: “2% MM 2‘ R2+(ML'Z;E)
W
7. _. 2 H , ,1... f
5 000051,) +(ms  o 3H54'1Vw'él‘ Physics 2220
Spring 2008
Exam 3 Equation Sheet . ~ F ' 
F12 electric 2 kg 9132 7‘ E E electric
7” (10
E — I; q” " E” — k dq “
"‘ e Z 2 z " e "737"
i 'L
, _ 2
"=Fo+ﬁ'ot+:,lz£it2 17:"0+at (182:0:
@electric = /E ‘ (114‘ f2? ' d N = gen:1:sed
B ' .
AV=é—g=—/ Ed§ V=kezﬁ V=ke E
90 A i 7'1' 7"
_~ 4in __. =§K QK*_£9X:
U—‘kE Z Tij — VV: Bani 8y] 62
umque .
pairs
Q 1 1
: __ C . = 0. _ = __
0 AV _ parallel equw. i '1. Cseries equiv. ; Ci
2 1 1 e A
U: :20: 562 AV— _ —0(AV)2 mm = 5601.772 0 = n00 0mm = “Off
Electric dipole: p5 d~q F=ﬁXE =]5'E
Magnetic dipole : [.l, = IA ? = [i x E = —ﬂ B
:62? 1an  nquA J A J 2 0E
V 2
REE3% 12:12:51 P=IAV=IZR=(A;)
R . . _ Z a .___:____ _. :1.
series eqmv. i Rparallel equiv. 1. B11.
ZIinzszt 2 szo
Closed
loop
Y=YOet/T Y=Yo(1—e't/T) 7:30 orT=L/R YE{V,Q,I}
' ' _. —' ' w v  [1,0 Id§X ’F F3 [1.01112
FLorentz:Q(E+'UXB) F=ILIXB (13:471‘ 7'2 l 2 2710,
N > OI
Bsolenoid = MOTI Bstraight wire 2 35:71, (COS 01 “‘ C05 62)
fﬁd§=ml+uoeog—§t— @5=/§~de fﬁd/Eo 5: Ed§=—5‘—l—:—B
__ dI __ NQB __ dI1 _ NQII)” __ __
81,— Ldt L— I 82 —~ N112 dt ZVIlg— [1 ——]\/121 ~JW B2 1 U=~LI2 .u '=— w=——
2 , VB ; 2M0 w/Lc
Y 1 ‘ ” X ~X
Yrms = ~37? XL 5 wL X0 ,2. 35 Z E R2 + (XL — X0)? ¢ = arctan(——LT—q)
1m = A?“ PM = Imam/ms cosqs = IfmsR AV2 = 7%AV1 mm = 13sz
E a 1 a ~ S 23 73
5—0 S=EEXB P—EorT c—Af ITSavg [—47r72,
, c A . .
91 = 61 n E — = —— n151n61 = 712511102
» 11 An
k5 = 1 = 8.988 x 109N  1112/02 e = 1.602 x 10*19 C
471'50 #0 = 47r x 10‘7T  m/A c = 2.998 x 108m/s ...
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 Spring '00
 Littler
 Magnetism

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