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21111
Instructor: Profs. Andrew Rinzler, Paul Avery, Selman Hershfield
PHYSICS DEPARTMENT
PHY 2049
Exam 3
April 7, 2010
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
H
~A=
Gauss’ law: Φ = n̂ · E
q
r̂ (point charge)
4π²o r2
~ =
E
R
dq
r̂ (general)
4π²o r2
E=
σ
(plane)
2²o
~ 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
~ · d~s
E
C=
U=
Resistors: i =
a
V =
q2
2C
V =
∂V
,
∂x
∂V
,
∂y
Ex = −
Ey = −
K²o A
(parallel-plate)
d
u=
dq
= jA
dt
q
(point charge)
4π²o r
R=
q = CV (1 − e−t/RC ) (charging)
1
²o E 2
2
V
i
R
dq
(general)
4π²o r
Ez = −
∂V
∂z
C = C1 + C2 (parallel)
1
1
1
=
+
(series)
C
C1
C2
R=
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
L = N ΦB /i (definition)
UB =
1 2
Li
2
uB =
B2
2µo
H
H
~ · d~s = − dΦB
E
dt
di
L = µo n2 Al (solenoid)
E = −L
dt
~A=
Induction: ΦB = n̂ · B
~ dA
n̂ · B
i=
E=
E
(1 − e−t/τL )
R
i = io e−t/τL
E1 = −M
τL =
L
R
di2
dt
L = L1 + L2 (series)
Vs
Ns
=
Vp
Np
1
1
1
=
+
(parallel)
L
L1
L2
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AC Circuits: ω = √
tan φ =
1
(LC circuit)
LC
XL − XC
R
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
H
p
XL = ωL,
XC =
¡
¢1/2
ω 0 = ω 2 − (R/(2L))2
~ · n̂dA = 0
B
H
H
θc = sin−1
~ =B
~ m sin(~k · ~r − ωt)
B
pr =
Pavg =
1
IEm cos φ
2
1
di
, vL = L ,
ωC
dt
vC =
cos φ =
R
Z
q
C
~ · d~s = −N dΦB
E
dt
~ · d~s = µo ²o dΦE + µo ienc
B
dt
~= 1E
~ ×B
~
S
µo
n1 sin θ1 = n2 sin θ2
I
(total absorption)
c
i = I sin(ωt − φ) (driven RLC)
R2 + (XL − XC )2
~ · n̂dA = qenc
E
²0
E
1
=√
B
µo ²o
~ =E
~ m sin(~k · ~r − ωt)
E
pr =
H
Z=
E = Em sin(ωt)
1 2
E
cµo rms
n2
θB = tan−1
n1
I = Savg =
n2
n1
~m ⊥ B
~ m ⊥ ~k
E
2I
(total reflection)
c
p=
id = ²o
dΦE
dt
Em
Erms = √
2
I=
Ps
4πr2
c = ω/k = f λ
U
(mom. carried by EM radiation of energy U )
c
1. The figure shows a simple optical fiber with a plastic core (n1 = 1.59) surrounded
by a plastic sheath (n2 = 1.54). A light ray in air is incident from the left with an
angle θ relative to the fiber axis. What is the maximum value of θ that will ensure
total internal reflection?
(1) 51.0◦
(2) 39.0◦
(3) 49.5◦
(4) 75.6◦
(5) 23.3◦
2. The magnetic flux entering the circular bottom face of a right circular cylinder is 23.4 T · m2 and the magnetic flux
exiting the sidewall of the cylinder is 63.9 T · m2 . The magnetic flux on the top face must be
(1)
(2)
(3)
(4)
(5)
40.5
2.73
40.5
87.4
87.4
T · m2 ,
T · m2 ,
T · m2 ,
T · m2 ,
T · m2 ,
exiting
entering
entering
entering
exiting
3. In the figure R = 20 Ω, L = 8H, and the ideal battery has a voltage of V = 18
volts. Right after the switch is closed, what is the rate of change of the current?
(1)
(2)
(3)
(4)
(5)
2.25 A/s
2.5 A/s
0.90 A/s
1.25 A/s
0.44 A/s
4. Two LC oscillator circuits (1 & 2) have the same capacitance (C1 = C2 ) but
different inductances (L1 6= L2 ). The plot shows the voltage across the capacitor as a function of time for the respective circuits (dashed curve is for circuit
2) . This implies that
(1) L1 > L2
(2) C1 < L2
(3) C2 < L1
(4) L1 = L2
(5) L2 > L1
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5. A blue-giant star has a luminosity (total power emission) of 2.5 × 1029 W. What is Brms at a distance equal to the earth’s
distance from the sun (1.5 × 108 km)?
(1) 2.4 µT
(2) 216 µT
(3) 18.5 µT
(4) 61 µT
(5) 6.5 µT
6. Refer to the previous problem. A black, spherical dust particle of radius 150 µm experiences a gravitational force that
is 5 times larger than the radiation force. What luminosity is required for the forces to be exactly balanced?
(1) 1.0 × 1030 W
(2) 5.59 × 1029 W
(3) 1.25 × 1030 W
(4) 5.0 × 1028 W
(5) 6.25 × 1031 W
7. In the figure the magnetic flux through the loop increases according to the relation
ΦB = 3 + 6t + 3t2 + t3 , where ΦB is in milliwebers and t is in seconds. What is the
magnitude of the emf (in millivolts) induced in the loop when t = 2 s and the direction
of the induced current? (cw = clockwise, ccw = counterclockwise)
(1)
(2)
(3)
(4)
(5)
35,
35,
30,
34,
30,
ccw
cw
ccw
ccw
cw
8. The primary of an ideal transformer has 100 turns and the secondary has 600 turns. Then:
(1)
(2)
(3)
(4)
(5)
the
the
the
the
the
power in the primary circuit is less than that in the secondary circuit
primary current is six times the secondary current
voltages in the two circuits are the same
frequency in the secondary circuit is six times that in the primary circuit
currents in the two circuits are the same
9. In a sinusoidally driven series RLC circuit the current leads the applied emf. The power dissipated in the resistor can
be increased by:
(1)
(2)
(3)
(4)
(5)
decreasing the amplitude of the driving emf (making no other changes)
decreasing the inductance (making no other changes)
decreasing the capacitance (making no other changes)
decreasing the driving frequency (making no other changes)
increasing the inductance (making no other changes)
10. The rectangular water tank in the figure has width L = 5.00 m. A ray of light
striking the midpoint of the tank at an angle 62.0◦ from the normal refracts and
hits the lower corner of the tank. What is the depth D of the tank? n = 1.333 for
water.
D
L
(1) 1.66 m
(2) 2.21 m
(3) 3.77 m
(4) 3.34 m
(5) 2.83 m
11. In the figure the instantaneous current charging the capacitor is i = 1.2 A.
The circular capacitor plates have a radius of 20 cm. The magnitude of the
magnetic field at the point labeled p that lies between the plates, 15 cm from
the centerline between the plates is
(1) 9.0 × 10−7 T
(2) 1.8 × 10−8 T
(3) 6.6 × 10−7 T
(4) 3.2 × 10−7 T
(5) 4.8 × 10−8 T
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12. An AC power supply drives a 270 mH inductor. The power supply EMF is E = Em sin ωd t, where Em = 12V and
ωd = 120 rad/s. When the instantaneous current in the circuit is a maximum, the instantaneous emf of the power supply
is
(1) +12 V
12
(3) − √ V
2
(2) −12 V
(4) 0 V
12
(5) + √ V
2
13. A solenoid of length 75 cm and radius 10 cm is wound with 300 turns of wire. If the total energy stored in the solenoid
is 400 J, what is the current in the wire?
(1) 410 A
(2) 1200 A
(3) 190 A
(4) 550 A
(5) 84 A
14. For a driven series RLC circuit, as the driving frequency is increased the current amplitude peaks at 3.5A. The inductance
is 200 µH, the capacitance is 12 µF and the emf amplitude is 12V. The resistance must be:
(1) 8.0 Ω
(2) 3.4 Ω
(3) 16 Ω
(4) 32 Ω
(5) 64 Ω
15. A beam of light polarized along the y axis and moving along the +z axis passes through two polarized sheets with axes
of polarization oriented 20◦ and 70◦ relative to the y axis. The final intensity of the beam is measured to be 71 W/m2 .
What is the initial beam intensity?
(1) 26 W/m2
(2) 118 W/m2
(3) 220 W/m2
(4) 690 W/m2
(5) 195 W/m2
16. In the figure a metal rod is forced to move with constant velocity 80 cm/s along
two parallel metal rails connected with a strip of metal at one end. A magnetic
field of magnitude B = 2 T points out of the page. The rails are separated by
L = 20 cm, and the rod has resistance 10 Ω. What is magnitude and direction
of the current in the rod? (cw = clockwise, ccw = counterclockwise)
(1) 3.2 mA, ccw
(2) 32 mA, ccw
(3) 32 mA, cw
(4) 6.4 mA, ccw
(5) 3.2 mA, cw
17. Oxygen:
(A) has an intrinsic magnetic dipole moment,
(B) is paramagnetic,
(C) is attracted to the high field point of a non-uniform magnetic field,
(D) is diamagnetic.
Of these statements only the following are true,
(1) A & C
(2) A, B & C
(3) D
(4) A
(5) B & C
18. The circular region of radius r = 50 cm shown in the figure contains a uniform electric
field pointing into the page that is increasing according to E(t) = 2.4 × 105 t (in units
of V/m), where t is in seconds. For the path indicated by the dotted line what is the
H
~ · d~s, in units of T · m?
magnitude of B
(1) 2.1 × 10−12
(2) 1.4 × 10−12
(3) 7.6 × 10−12
(4) length of dotted line required to solve.
(5) 8.4 × 10−12
19. An LC oscillator oscillates at a frequency of 450 Hz. The capacitor has capacitance of 3.00 µF. The inductor must have
inductance of:
(1) 27.1 mH
(2) 16.5 mH
(3) 80.8 mH
(4) 69.6 mH
(5) 41.7 mH
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20. A battery is hooked up to a series arrangement of a resistor and an inductor. After how many time constants, τL = L/R,
will the energy stored (not the current) in the inductor be 55% of its final value?
(1) 0.60
(2) 0.80
(3) 1.35
(4) 2.21
(5) 0.30