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Unformatted text preview: _ Department of Physics, Lehigh University
Physics 21 Introductory Physics 11 Spring 2007 Final Exam May 8, 2007
Closed Note 8:00 AM—ll :00 AM
Student's Name Recitation Section Number Recitation Leader's Name The test is a multiplechoice examination. First, check to be sure that you have twelve
page examination, including the cover sheet and the equation sheet. Please answer each
question by circling one answer from the panel of choices available within each question space. Points summary Part 1 Part II Part In Total
1. ___(20) 4. _(l6) 8. _(2‘4)
2. __(20) , 5. ”(20) 9. “(16)
3. _(28) 6. _m(16) 10. __(20)
7. ___(20)
Sub—Total (68) MG?) ____(60) (200) Part I (68 points) 1. (20 points) There is a hollow conducting sphere with
inner radius a and outer radius 1) and has a charge of +3q.
In addition, a charge of 2q is placed at the center of the hollow space inside A. What is the total charge on the inner surface of the
shell? i) 3q; ii) 2q; iii) 4;; iv) q; or v) 2q B. What is the E—ﬁeld at a point where a < 1' < b? . » 1 —2q A .. — 1 q A
E: ; . E: — ;
1) [4??80 I r2 )r 11) £472.90 13 )r
iii)E=[ 1 ][_2q]17;or iv.)E:0
47:50 r C. What is the total charge on the outer surface of the shell? i) Sq; ii) 3q; iii) 0; iv) q; or V.) 3q D. What is the electric ﬁeld at a point where r > b? . »_ 1 7i ~. .. = l l A_ 1) E —[4ﬁ80 I19 )r , 11) E [47rgojirzjr ’ iii E=[ 1 ][_2q]77; or iv) Ex[ 1 Ii];
4%90 r 4.7580 1' B. What force does a particle withcharge —Q at a distance L from the outer surface of
the sphere feel? i)1?=[ 1 I qFQZJF; ii.)ﬁ= ] [—942]?;
47590 (L—b) 47:80 (EHL) iii.)ﬁ=[ 1 I q"92];;or' iv.)F‘=[ 1 1 $2912}?
47:50 (5+L) 47530 b +L 2. (20 points) A charged conducting sphere of radius a and total charge Q sits at the
origin. 7 A. Determine the potential inside the sphere. Use inﬁnity as a reference point:
V (00) = 0 . i) V: Q2; if) V: Q2; iii) V: Q;
47:307' 47:30a 47:80:" iv) V: Q ;or v)V=0.
4moa B. For two points inside the sphere, the point in the center of the sphere will have a
potential that is: i) Larger than the point farther ﬁom the center;
ii) Smaller than the point farther from the center;
iii) Nonzero constant; or iv) Equal to zero C. Determine the electrostatic energy stored in the electric ﬁeld around the sphere. . _ Q _ .. _ Q2
I) IW‘_271’80a2 ’ u) Jud—275.9%:a 2 iii) W=8j§ ;or iv) lWl=SQ
on 758061 D. Now, the sphere is encased in a material with permittivity 5
out to a radius 19 as shown in the ﬁgure. Determine the
potential for the point at the center of the sphere. Use infinity as a reference point: V (00) = 0 .
. 2 2
. 1 1
i) V22 L+L_L i.) Via [_] "*H _[
47: 80b 5a 5b 47: eob 5a mg 1 iv 2g 1 2
m) V_4Ir[£(b+a)] ) V 47r[s(b+a)] 3. (28 points) Consider the circuit diagam shown in
the ﬁgure. A. With the switch S Open, what is the resistance
between point B and C, if all resistors are the
same? V , bait i) 3R; ii) R; iii) 3R ; or iv) 2R. F E D
With S closed, the currents in and out of junction B obey the following relation: i)15=11+16; ii)Iz+I3+I4=Io; 111)Io+11=1*5;or iv)I(,=Iz+Ig+I4 C. All resistors are 209 and the battery emf is 120 V. With S closed, what is the current through the battery? i) 3.43 A; ii) 1371.4 A; 111) 10.5 A; or iv) 5.94 A Consider the circuit shown on the right. D. R The capacitor C is empty of charge at start, and the 1
switch S is closed. What is the current through the
resistor R1 at the instant of switch closing? (8' R2 C
1) g ; mi; 111) 35—; T— R1 + R2 R1 R2
. 6
1V R + R ; or v 0 ) RIRZ ( 1 2) )
What is the current through resistor R2 after a long time?
. a .. a a . a
1 ; 11—; 111 —; 1v R+R ;orv 0
)R1+R2 )R1 )R2 )RIR2(1 2) ) Determine the time constant for charging the capacitor in the circuit. i) C/Rl; 11) (1/112; iii)R1R2C/(R1 +112); iv) C/(R1+R2); or v)C/(R1R2)v, . What is the maximum charge on the capacitor? ECR1 _ __) «ECR2 ; iii) 3C(R1+R2);Or iv) 5C(R1+R2) i) , 11
R1 + R2 R1 + R2 R1 R2 Part II (72 points) >
4. (16 points) Both the loop and the wire carry a
steady current I. A. Determine the magnitude of the force on the squareloop.
27: 3 5+0
2
:1) F—‘UOI [1— I]
27: s S+a
m) F=ﬂ9£[l+ 1 ],or
2n: S s+a B. The torque on the loop is: i) Counterclockwise; ii) Clockwise; iii) Zero; or iv) Into the page In the ﬁgure at right, a long straight wire perpendicular to the page carries a current I. A I
current loop in the plane of the page also carries a current I . The loop consists of two quarter—circular
segments, of radius a and 2a, whose common center lies on the currentcarrying wire (notice that 2a
the two radial segments of the loop also cross the current carrying Wire. I 7
C. Determine the magnetic force on each of the 0 two quarter circular segments due to their
interaction with the magnetic ﬁeld produced by the current in the long straight Wire. [2 I2 2 2
1) F: H0 _#0 ii) F— _ [101+IUOI .
8a 16a 8a +1651 ’
2 2
iii) F = 0 iv) F : ”OI + #01
27m 2::(2a)
D. Determine the magnetic force on each of the two radial segments.
2
#012 Spa! .. iv Fzﬂol
i) F— _. 2” 111(2) ii) F 27r 22(a ) 121) F m 0 ) 291a 5. (20 points) A torus has 1000 turns of wire that carry 10 amps. The average radius of
the torus is 1 meter and the radius of the wire turns is 1 cm. Here the permeability of free space is given by #0, A. What is the value of the magnetic ﬁeld B in the torus at the location of the average
radius? i) 0; ii) 200 p0; . iii) 20 pom”, iv) 30 yo/n; or v) 5000 pro/Jr 'A small compact coil is wrapped around the torus. It has 10 turns of wire and a radius of 1 cm. What is the magnetic ﬂux in the coil? i) 10'3 7: B; ii) B; iii) 0; iv) 50 #0; or v) none of these The interior the torus is ﬁlled with a material of permeability/.1. What is the new
magnetic ﬁeld in the torus, WhereB denotes the old value of the magnetic ﬁeld? i) 2 #3; ii) 50000 y/ar, iii) (ta/p0) B; iv) #3; or v) 0 . If now the ﬁeld in the torus changes with time, according to B = Bo t, what is the magnitude of the induced emf in the small coil? i) 10'3 7: BO 1‘; ii) Bo t2; iii) 0; iv) 10'3 2:30; or v) 2x103 7: a, If the radius of the small coil is doubled, what is the value now for the induced emf in
the small coil? i) Doubled; ii) Same as before; iii) Larger by 4; iv) 0; or v) None of these _ View from above
6. (16 points) A conductmg rod of length L = 12 Q G G G .. G 3
cm= mass m = 0.5 kg and resistance R = 30 Q i
slides on a frictionless track, with negligible
resistance, that is tilted at an angle 6? = 30° with G
respect to the horizontal. A constant magnetic
ﬁeld B = 6.0 T points upward, perpendicular to Q
the plane of the tracks. A. If the velocity of the rod is 17, then the Q Q G O
magnitude of the induced emf is: _ i) BLv; ii) BLv/R; ' iii) BLv sin 9, or iv) BLv cos a B. The magnitude and direction of the current
in the rod (in terms of v) is: i) 0.024 v, a —> b; ii) 0.0208 v, b —) a;
iii) 3.52x10“ v, a —> b; or iv) 3.52x10“ v, b —> a C. In terms of v, the component of the magnetic force on the rod that is parallel to the
plane of the track and directed uphill is: i) 0864 v; ii) 0.0173 V; iii) 0.086 v; or iv) 0.015 V D. The terminal velocity of the rod is: i) 0.172 m/s; ii) 141.8 m/s; iii) 283.6 m/s; or iv) inﬁnity 7. (20 points) A laser emits a parallel light beam. The laser beam is initially continuous,
and has a uniform intensity in a circular crosssection of 0.1 cm radius. A. If the beam energy is 5 joules in 1 sec, what is the maximum electric ﬁeld value E9?
1) 5 «7:30); ii) 316 / mac)”; iii) 5 / 2; iv) 0; or v) 5 B. The laser is now pulsed to give the same energy in one second, but there are ten
uniform pulses per second of duration 1 millisecond. What is the maximum electric
ﬁeld value now in terms of the initial electric ﬁeld value E0? i) 10 E0; ii) 100 E0; iii) E0/ 100; iv) E0 / 2; or v) 25 E0 C. If the maximum electric ﬁeld value is E0, the beam is ﬁJlly reﬂected at a mirror, what
is the radiation pressure imparted on the surface? i) 0; ii) a. 1302; iii) (20/ u0)mE02/c; iv) £2.12; or v)€0p0 E, D. The beam passes through an optical glass with medium permittivity e = 3 so and
permeability p. = no, what is the new value for the velocity of light in the beam? i) 10 0; ii) 0.577 c; iii) 1.732 0; iv) 3 c; or v) 5 c E. If the wavelength of the laser light is 515 nm in a vacuum, what is the wavelength of
light in this glass medium? i) 298 nm; ii) 515 nm; iii) 891 nm; iv) 172; or v) 1288 nm Part III (60 points) ' 8. (24 points) Consider a long glass ﬁber that is 150 pm in diameter. Its index of
refraction [1g is 1.55 at the wavelength of 450 nm and 1.53 at 550 min. A. A pulse of light is spread out along the axis of propagation as a Gaussian function of distance in the form of Eo exp( 362/09), where or is a constant and the center of the
pulse is positioned at one end of the ﬁber (x = 0) at time t = 0 5.. What is the
expression for the light pulse that is traveling in the positive x direction in the ﬁber as
a plane wave? i) E. eXp[(X—ct)2/0t2]; ii) E0 exp[—(xct/ng)2/0t2]; iii) E0 exp[—X2/(0Lct)2]; iv) E.) €Xp(X2/CL2) sin(kx—ct); or v) If0 exp[(xct)/a2] B. Assume that the above traveling pulse describes a propagating photon. If its temporal
width is 2.0 ps, what will be the smallest uncertainty in determining its energy? i)3.3><10'4eV; ii)3.3x10‘2ev; ai)1.1x10'1ev; iv)2.5x10“6ev; or v)0eV C. A narrow pulse of white light is launched at one end of the ﬁber at t = 0 3. How far
apart will the two wavelength components of the pulse, one at 450 nm and another at 550 nm, be separated alter traveling in the ﬁber for 5x10‘12 s?
i) 1.27x10'5 m; ii) 1.27x10“2 m; iii) 1.5x10‘3 1n; iv) 2.54x10"2 m; or iv) 3.0 m D. A ray of light enters the light ﬁber such that it makes at an angle of 15° with respect
to the long axis of the ﬁber within the ﬁber (see the ﬁgure). Calculate the distance the ray travels between two successive 15 0., 7
reﬂections off the sides of the ﬁber. 4—; :' .1—
Assume that the wavelength is 550 nm. ' 90.15111; ii)3.93x10‘5m; iii) 1.50X10'4m; iv)5.6x104m;or v)5.0x10"°m E. What is the critical angle of incidence within the ﬁber medium above which the
incident beam of light is totally reﬂected at the ﬁberair interface when the wavelength of the light is 450 nm?
i) 90.0"; ii) 65.3”; iii) 40.2“; iv) 33.8”; or v) 15.0° F. A collimated beam of light, consisting of two spectral components at 450 nm and 550
nm, respectively, is normally incident on a transmission grating. If the grating has 240 grooves per mm, what will be the angular separation of the two Spectral lines
alter the grating due to diffraction at the grating? i) 0.69“; ii) 1.11“; iii)1.38"; iv) 1.98“; or v) 2.14° 9. (16 points) Two plane plates of glass (rag: 
1.51) of length L = 15 cm are touching at One
end and separated at the other by inserting a
strand of hair of diameter d = 0.25 mm as drawn
at right. Light of wavelength 2. = 550 nm then
reﬂects off the top and bottom faces of the
wedge of air between the two plates. A. If a designates the reﬂection at the top face
and b the reﬂection at the bottom face, then there is a phase change of a“. i) For a only; ii) For I) only; iii) For both a and b; or iv) For neither (I nor .6 B. The distance from the left end, where the two plates touch, to the ﬁrst bright fringe is: i) M; inﬂ; Halli .0: 1.0131
n d d 2 d anL g C. The region between the two plates is next replaced with an optical substance with
index of refraction n5 2 1.53. The distance to the ﬁrst bright fringe is now: i) i; i051“; iii)."i_;or iv) 13’:
11361 d 2 ”5d 2 nsd
D. Three polarizing sheets are stacked with their
polarizing axes pointing vertical, and at angles
6'; : 727’4 and 62 = m’3 relative to the vertical as
drawn. If unpoiarized light of intensity I.3 strikes
the ﬁrst plate, the value of the intensity of the light leaving the 31rd plate is:
i) 0.0625 10; ii) 0.4667 10; T iii) 0.2333 10; or iv) 0.1250 1’0 10 10. (20 points) The steps of a ﬂight of stairs are 20.0 cm high vertically. If a 60kg
person stands with both feet on the same step, what is the gravitational energy of this
person, relative to standing on the ground:
A. When the person is standing on the second step? i) 0.30 J; ii) 235 J; iii) 4700 J; iv) 10500 J; or v) 13.6 eV
B. When the person is standing on the 11‘“ step? i) 118 n J; ii) 118 n2 J; iii) 0.30 n J; iv) 235 nJ; or v) 13.6 11 eV
C. What is the change in energy as the person descends from step 10 to step 2? i) 108.8 J; ii) 944 J; iii) 4.4 J; iv) 1880 eV; or v) 84000 J D. What is the de Broglie wavelength of this person when the person is moving at steady
speed of 0.1 m/s? i) 5.0x10'1l n1; ii) 1.1x10'34 111; iii) 2.0 In; iv) 150 In; or v) 1.0)(10'10 In
E, What is the de Broglie wavelength of a neutron traveling at the speed of 0.1 m/s? i)1.6><10'1°m; ii)1.6x10'6m; iii)3.97><10'6m; iv)5x10'“m;or v)I.6x10'2m 11 Lehigh University
Physics 21 Spring 200'?
Equation Sheet awed of light. in. manna r:
Gravitational Nutshut G
Arngmlm's Number NA
Boilzmalul‘a mutant k;
charge cm ekclmu a
free space permittivity to
{we space permeability Ira gravitational amleralkm 9' F l Q1Q2trsrel Q=f‘[;\1;;!
9411 43% I1—r2ls ('=:0Ki=r_i
l d
F2913 o:
. ;  .._ ‘ 9 __
dEljalrlzkl a‘Qirlfl {up=3CI.—\lcl *5:
E T‘i '6” '“rl “Malt“:
" . .. . 1‘=m ur l'—m=o
B_ iﬂ+ji+§il r 1
a.» 3g .9: “=9: 6;
‘l;.I.a=Jl;.JE .m F=I1" Pxf’R
_; Q _ r Ml‘nlM plnle
1=—L9»w=J—*Q E=rr—+Emdw
:rter Irer—r'l ""I j“ 5'“ _'
.4 l l .l k
Hak:]?r.‘3h .umal:mﬂ'l A r13: .l, .1, .4:
\\'¢rk= frag'5 5': 3! 3: . "'3 = u d ,
dl——EfBNA l'
u'l : Jrcl + Hoéoﬁf E  HA. lit: —=n («+\{n2+H3l
I \‘u' Jr a: min .
[ﬁrm—r: WM" 1 .
= — Lnu I (i)
{7 a lid“ __ 2 3
fm_ 311101 +«] rlu
{:3 + u5 v: Tﬁ: {T=l‘9lwioui % a ﬂag :‘= (3T,.I lixg’slm 3 2 iii: 2 Bl vAJ’  55:; W n = 2:; }. = 2:9 §= leach]?  $5383 T‘ = If! iT=periodi 5‘03, Erma," ?” = gun133a ' 2m = “"3;—
2 _%7::‘flx”l + U(x]‘?{x.l}  iﬁ—gflx'ﬁ 3.00 r 10' unis 6.67 :r 10““ N mafkf
6.02 r 10” nlol" 1.38 r 10‘” UK 1.50 .c 10"” c: 3.55 , 10"2 (1% m“
. \c 10" ‘1‘ mffl
9.50? um!»2 April 24, 2007 Piam's constant I: MEI!!! w ll‘l'” J :5
Planck's mnstantjtﬁwl ll = ME: L055 a 10"" J 5
Electra: that. mass ”lo 9A! x [III3' kg
pmlon rest mm m... [.6726 1 10‘” kg
neutron ml mazea m“ [.5749 l 10"f kg
mmn‘w mass unit :2 1.6605 1 13"?“ kg,
lit13793] l: 5‘99 I 10' N “PFC” sinia i b] = sin1 mi) t on a sin {I sinib‘: é}?  mummg i lwﬂxin a}
= ia'ﬂsﬂ ﬁaC: [wrini 91' RI [parallel]:
l l l. =—+_ tam—um :1 £2 F=r_.t\vB. JF=II£HJB
no MI 1‘ {r  r’] dB=E——T2r—rf ﬁeld of long: win9:13  *2;— 5:=C': tparalblmr R: Eserieal: Semen» = z: + '53 mule? of loan: H = pg! {‘29 . . . l
.\ u = R. 3:. = 41.. .\c: = :F I: 95?. I = mm
RC time mnstant. = RC 1; N, I. _ X,
Li? lime rmlslaul = UR ﬁ _ V: T: _ "C
Qiii fur RLC circuit \m _ 1 _ 1
#0
Go («pl—Hir’zﬁlﬁsd r = p 13 ﬂ = [A Milt? t6} _ oas{K‘CrnlhT silwsin b cainc! + sini: = Scan: ch: [1115 + "333” «cm It“ (in = 16""
n
jihad“ : ulnu — u £2 = L _ ll: Halﬁlloitl B = mt?! FA" JL mlrlloiul I, = I:Q‘\‘c.4 ('1'.
cinumfemw or R‘ll‘t‘lx“
cirmnnfemm‘e ;‘l' cilt'lo
antn .I' circle I.  er" surluce men (I‘ sphem
' = in 3 vnlunle rl whet?
n‘ﬂ +b_I+rv—II —
“a . n+1; “ ‘
'2 )""( 2') .=
_ :1 fa" dz: = _I n'u’l
_ {:2 \_,aﬁ_ + u u + l
r." f
_ i f N = — Into +6421
— —ﬁ _ n + In: E:
\
u‘n‘
— = In a:
u 2): a '3: h
f (135‘ (iffy =/ Sillfo'a I: T
0 0 In: =p3__.'r.2.1n
E = 5‘...) = hf
cm = mm? + :1;in L‘ = 1;"\'*'ouo
E ‘4 l3 \. v {plane wave}
Am; ? n in .. 3.32:} p: M: AM.” (d? Bras“?! nl si.1'.l_l§']=nzs.i1102
l l l S —=(n—l —+—] p:— r R. R. c
l+l_i p q f 12 ...
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