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Instructor: Profs. Paul Avery, Selman Hershfield
PHYSICS DEPARTMENT
PHY 2049
Final Exam
April 23, 2011
Name (print, last first):
Signature:
On my honor, I have neither given nor received unauthorized aid on this examination.
YOUR TEST NUMBER IS THE 5-DIGIT NUMBER AT THE TOP OF EACH PAGE.
DIRECTIONS
(1) Code your test number on your answer sheet (use 76–80 for the 5-digit number). Code your name on your
answer sheet. DARKEN CIRCLES COMPLETELY. Code your student number on your answer sheet.
(2) Print your name on this sheet and sign it also.
(3) Do all scratch work anywhere on this exam that you like. At the end of the test, this exam printout is to be turned in.
No credit will be given without both answer sheet and printout with scratch work most questions demand.
(4) Blacken the circle of your intended answer completely, using a #2 pencil or blue or black ink. Do not
make any stray marks or the answer sheet may not read properly.
(5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing.
Constants: e = 1.6 × 10−19
>>>>>>>>WHEN YOU FINISH <<<<<<<<
Hand in the answer sheet separately.
C mp = 1.67 × 10−27 kg me = 9.1 × 10−31 kg c = 3 × 108 m/s
²o = 8.85 × 10−12 C 2 /N · m2
k = 1/(4π²o ) = 9 × 109 N · m2 /C 2
µo = 4π × 10−7 T · m/A
micro = 10−6
nano = 10−9
|q1 ||q2 |
Coulomb’s Law: |F~ | =
(point charge)
4π²o r2
~
~ =F
Electric field: E
q
~ =
E
~A=
Gauss’ law: Φ = n̂ · E
H
q
r̂ (point charge)
4π²o r2
~ =
E
R
dq
r̂ (general)
4π²o r2
E=
~ dA = qenc
n̂ · E
²o
R
1
1
F~ · d~s = mvf2 − mvi2 = Kf − Ki
2
2
R
For conservative forces Uf − Ui = − F~ · d~s → Ki + Ui = Kf + Uf
Energy: W =
Electric potential: V =
Vb − Va = −
Rb
a
Ex dx = −
U
q
Rb
Capacitors: q = CV
U=
Resistors: i =
a
V =
~ · d~s
E
C=
q2
2C
dq
= jA
dt
V =
∂V
,
∂x
∂V
,
∂y
Ex = −
Ey = −
K²o A
(parallel-plate)
d
u=
R=
q = CV (1 − e−t/RC ) (charging)
q
(point charge)
4π²o r
V
i
1
²o E 2
2
R=
R
dq
(general)
4π²o r
Ez = −
∂V
∂z
C = C1 + C2 (parallel)
1
1
1
=
+
(series)
C
C1
C2
ρL
(wire)
A
P = iV
q = qo e−t/RC (discharging)
R = R1 + R2 (series)
1
1
1
=
+
(parallel)
R
R1
R2
~
~ ×B
~
~
~
Magnetism: F~ = q~v × B
F~ = iL
µ = N iA
~τ = µ
~ ×B
U = −~
µ·B
H
µo i
µo i
µo iN
~ · d~s = µo ienc
~ = µo id~s × r̂
B
B=
, (wire)
(loop center),
(solenoid)
dB
2
4π r
2πR
2R
L
σ
(plane)
2²o
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L = N ΦB /i (definition)
UB =
1 2
Li
2
uB =
B2
2µo
AC Circuits: ω = √
tan φ =
H
i=
I=
Em
Z
q = Qo e−Rt/(2L) cos(ω 0 t + φ)
First 2 Maxwell’s Eqs.:
Last 2 Maxwell’s Eqs.:
EM Waves: c =
I = Io cos2 θ
E=
E
(1 − e−t/τL )
R
H
Z=
1 1
1
2
+ = =
p
i
f
r
i = io e−t/τL
E = Em sin(ωt)
p
E1 = −M
τL =
L
R
di2
dt
i = I sin(ωt − φ) (driven RLC)
R2 + (XL − XC )2
H
~ · n̂dA = qenc
E
²0
~ · n̂dA = 0
B
H
XL = ωL,
XC =
θc = sin−1
~ =B
~ m sin(~k · ~r − ωt)
B
pr =
1 2
E
cµo rms
n2
θB = tan−1
n1
Interference: ∆L = mλn (constructive)
Pavg =
1
IEm cos φ
2
1
di
, vL = L ,
ωC
dt
n2
n1
~m ⊥ B
~ m ⊥ ~k
E
p=
id = ²o
vC =
cos φ =
R
Z
q
C
dΦE
dt
Em
Erms = √
2
I = Savg =
2I
(total reflection)
c
n1
n2
n2 − n1
+
=
p
i
r
1
1
1
=
+
(parallel)
L
L1
L2
~ · d~s = −N dΦB
E
dt
~ · d~s = µo ²o dΦE + µo ienc
B
dt
~= 1E
~ ×B
~
S
µo
L = L1 + L2 (series)
Vs
Ns
=
Vp
Np
¡
¢1/2
ω 0 = ω 2 − (R/(2L))2
n1 sin θ1 = n2 sin θ2
I
(total absorption)
c
Images:
H
E
1
=√
B
µo ²o
~ =E
~ m sin(~k · ~r − ωt)
E
pr =
~ dA
n̂ · B
1
(LC circuit)
LC
XL − XC
R
H
~ · d~s = − dΦB
E
dt
di
L = µo n2 Al (solenoid)
E = −L
dt
~A=
Induction: ΦB = n̂ · B
I=
Ps
4πr2
c = ω/k = f λ
U
(mom. carried by EM radiation of energy U )
c
1 1
1
1
1
+ = = (n − 1)( − )
p
i
f
r1
r2
∆L = (m + 21 )λn (destructive)
m=−
λn = λ/n
i
p
n = 1 in air
∆L = d sin(θ) (2-slit), 2L1 − 2L2 (interferometer), 2t (thin film) with extra λ/2 when reflecting off higher index
Diffraction: d sin(θ) = mλ (grating)
2d sin(θ) = mλ (X-ray)
sin θ = 1.22λ/d (aperture)
1. Four charges are placed on the corners of a square as shown in the figure. All
of the charges have the same magnitude, but as indicated in the figure some
of the charges are positive and some are negative. In which of the three cases
is the electric field at the center of the square zero?
(1) A,B
(2) A only
(3) A,C
(4) C only
(5) B only
2. One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other
slit. As a result:
(1)
(2)
(3)
(4)
(5)
the
the
the
the
the
dark fringes get brighter and the bright ones get darker
fringes just get dimmer
fringe system disappears
bright fringes get brighter and the dark ones get darker
dark fringes just get brighter
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3. As shown in the figure, light is incident at an angle θ1 from a material with
n1 = 1.1 to a material with n2 = 1.3. What is the smallest value of θ1 so that
there is total internal reflection at the interface between n2 and n3 = 1?
(1) 65.4◦
(2) 57.8◦
(3) 42.1◦
(4) 72.3◦
(5) 50.3◦
4. A current moves from left to right in a horizontal copper wire. Which statement is true?
(1)
(2)
(3)
(4)
(5)
Power dissipation in the wire is directly proportional to both current and resistance
The drift velocity of the electrons is slightly greater than the random motion speed of the electrons
The electric field points from right to left
The resistivity of the copper wire does not depend on its length
The electrons in the wire move from left to right
5. In a television tube electrons are accelerated through a voltage of 5000 volts. What is the speed of the electrons assuming
that they were initially at rest? (Since this is Physics 2, do not use special relativity to calculate the answer even though
these speeds are a fraction of the speed of light.)
(1) 2 × 107 m/s
(2) 3 × 107 m/s
(3) 1 × 107 m/s
(4) 6 × 107 m/s
(5) 4 × 107 m/s
6. An object is placed 30 cm to the left of a diverging lens of focal length −15 cm. A converging lens of focal length 30 cm
is placed 30 cm to the right of the first lens. Where is the final image located relative to lens 2, and what is the overall
magnification m?
(1)
(2)
(3)
(4)
(5)
20 cm to the right; m = −0.5
30 cm to the right; m = −2.67
40 cm to the right; m = +1.5
120 cm to the right; m = −1.0
80 cm to the right; m = +2.5
7. An electron is launched with velocity v in a uniform magnetic field. The angle θ between the velocity vector ~v and the
magnetic field direction is between 0 and 90◦ . As a result, the electron follows a helix, its velocity vector returning to
its initial value in a time interval of:
(1) 2πm/(eB sin θ)
(2) 2πm/(eB)
(3) 2πmv/(eB)
(4) 2πmv sin θ/(eB)
(5) 2πmv cos θ/(eB)
8. An RLC circuit is driven by a voltage of amplitude 12 V and has R = 4Ω, L = 2 H, and C = 100µF. At resonance what
is the amplitude of the voltage oscillations across the inductor?
(1) 3 V
(2) 420 V
(3) 6 V
(4) 12 V
(5) 24 V
9. Two concentric shells of negligible thickness have radius r1 = 10 cm and r2 = 20 cm. If the electric field at r = 15 cm
is directed radially outwards with magnitude 1000 V /m and the electric field at r = 25 cm is directed radially inwards
with magnitude 500 V /m, what is the sum of the charges on the two shells?
(1) +2.5 nC
(2) −3.5 nC
(3) −1.0 nC
(4) +3.5 nC
(5) +6.0 nC
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10. A current of 3A flows into circular parallel plate capacitor as shown in the
figure. The plates have radius 1.0 cm. What is the magnitude of the magnetic
field at point P in between the plates at a radial distance of 0.5 cm from the
center of the plates?
(1) 12 × 10−5 T
(2) 3.0 × 10−5 T
(3) 6.0 × 10−5 T
(4) 1.5 × 10−5 T
(5) 0.75 × 10−5 T
11. Sunlight is incident on a diffraction grating with 400 rulings/mm . What is the highest order, m, that a complete visible
spectrum 400 nm < λ < 700 nm can be seen?
(1) 1
(2) 5
(3) 4
(4) 2
(5) 3
12. A concave shaving mirror has a radius of curvature of 30 cm. It is positioned so that the (upright) image of a man’s face
is 3 times the size of the face. How far is the mirror from the face (in cm)?
(1) 40
(2) 20
(3) 10
(4) 3.8
(5) 80
13. A circular loop of wire of radius 2 cm is in the plane of the page. A magnetic field perpendicular to the page is changing
with time according to B(t) = 1 + 3t, where t is measured in seconds, B is measured in Tesla, and positive field is going
out of the page. If the loop has resistance 5Ω, what is the magnitude and direction of the induced current at t = 1
second?
(1)
(2)
(3)
(4)
(5)
0.5 mA clockwise
0.75 mA counterclockwise
1.0 mA clockwise
0.75 mA clockwise
1.0 mA counterclockwise
14. In the figure shown, two long straight wires with separation d carry currents i1
and i2 = 2i1 out of the page, with i1 located at x = 0. At what point on the
x-axis shown is the net magnetic field due to the currents equal to zero?
y
i2
i1
x
d
(1) 2d/3
(2) 2d
(3) d/2
(4) d/3
(5) 3d/2
15. Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a voltage source with 24 V are all in series. The potential difference across
the 2.0 Ω resistor is:
(1) 8 V
(2) 24 V
(3) 12 V
(4) 2 V
(5) 4 V
16. A satellite is 20,000 km above the surface of the Earth. It has a lens of diameter 50 cm. For light of wavelength 500 nm,
what is the smallest resolvable distance based on Rayleigh’s criterion?
(1) 25 m
(2) 2.5 cm
(3) 2.5 m
(4) 25 cm
(5) 250 m
17. A small insect starts walking away from a concave mirror along its central axis. Which statement is true?
(1)
(2)
(3)
(4)
(5)
The
The
The
The
The
image is initially erect, then flips upside down
image grows monotonically larger as the insect moves away
distance from the insect to its image remains constant
image is always upside down
image is always erect
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18. Three charges are placed on the corners of a rectangle as shown in the figure.
If Q1 = −2µC, Q2 = 1µC, and Q3 = 3µC, what is the x component of the
force on Q3 due to Q1 and Q2 ?
(1) +9.8 N
(2) +17 N
(3) +34 N
(4) −4.7 N
(5) −0.4 N
19. A battery is used to charge a series combination of two identical capacitors. If the potential difference across the battery
terminals is V and total charge Q flows through the battery during the charging process then the charge on the positive
plate of each capacitor and the potential difference across each capacitor are, respectively:
(1) Q and V
(2) Q and V /2
(3) Q/2 and V
(4) Q/2 and V /2
(5) 2Q and 2V
20. A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. The intensities
of wavelengths 480 nm and 800 nm and no wavelengths in between are to be maximized in the reflected beam. (Take
the index of refraction of air to be 1.) The thickness of the film is:
(1) 240 nm
(2) 600 nm
(3) 400 nm
(4) 360 nm
(5) 150 nm