Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
Section B1, MTWRF 10:05 – 11:30 p. m.
Name:
Signature:
ID number:
DO FOUR OUT THE FIVE OF PROBLEMS [I], [II], [III], [IV], [V]
PROBLEMS [I]-[V] ARE ALL MULTIPLE CHOICE
IF YOU DO MORE THAN FOUR, THE LAST ONE WILL BE IGNORED
THERE IS NO SCANTRON SHEET
ENTER YOUR ANSWERS IN THE SPACES BELOW LIKE 1. A or 2. B
SEPARATE THE FRONT PAGE TO TURN IT IN. KEEP THE OTHER PAGES
[I]
[II]
[III]
[IV]
[V]
1. A
1. C
1. E
1. D
1. D
2. D
2. A
2. B
2. C
2. C
3. B
3. B
3. A
3. B
3. A
4. E
4. D
4. C
4. E
4. B
5. C
5. E
5. B
5. A
5. E
YOU MAY SEPARATE THE PAGES AT THE END WITH THE FORMULAS
Your signature signifies that you will obey the HONOR CODE
You may be asked to show your photo ID during the exam.
NO CALCULATORS, NO PHONES
YOU MAY LEAVE THE ROOM ONLY AFTER YOU TURN IN YOUR EXAM PAPER
Physics 207 TEST 1
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
[I] This problem has five multiple choice questions. Correct answers are marked with [X]
Enter your answers in the front page.
~
[I.1] An electromagnetic wave E(x,
t) = Em ŷ sin[(10m−1 )x − ωt] has a wavelength λ and
frequency f given by
[B] λ = 0.1 m, f = 10c [C] λ = 0.1 m, f = 20πc
[X] λ = 0.2π m, f = 5c
π
[D] λ = 0.2 πm, f = 10πc [E] λ = 0.1 m, f = 10c
[I.2] In class it was demonstrated that a polarizing filter placed in front of the screen of a
computer laptop will make the display invisible if the absorption axis
[A] is vertical [B] is horizontal [C] is in any (every) direction
[X] goes from the upper left corner to the lower right corner of the screen
[E] no such demonstration was done in class
[I.3] A circular loop of radius R carries a current I in the counterclockwise direction as shown
in the figure to the left below. The magnetic field at the center of the loop is
µ0 I
0I
0I
[A] B = µ2R
into the paper [X] B = µ2R
out of the paper [C] B = 2πR
out of the paper
µ0 I
[D] B = 2πR into the paper
C
R
R
R
A
R1
switch
R
R
R
B
C
V
R2
L
0
A
C
C
B
I
D
C
[I.4] In the RL circuit shown at the center above the switch has been open for a long time. Then
the switch is closed. Immediately after the switch is closed what is the current through
i the battery, and what is the Emf across the inductor? You do not need to solve any
differential equations to get the answer.
[A] i =
V0
R1 ,
[D] i =
V0
R1 +R2
Emf = 0
[B] i =
, Emf = V0
V0
R1 ,
Emf = V0
[X] i =
V0
R1 +R2
[C] i =
, Emf =
V0
R1 +R2
, Emf = 0
V 0 R2
R1 +R2
[I.5] The figure to the right above shows 5 resistors and 4 capacitors. Find the equivalent
resistance and the equivalen capacitance between points A and B.
[A] Req = 5R and Ceq = 4 C [B] Req =
[D] Req =
R
2
and Ceq =
Physics 207 TEST 1
5C
3
R
5
and Ceq =
C
4
[[X] Req =
R
2
and Ceq =
3C
5
[E] None of the above
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
[II] This problem has five multiple choice questions. Correct answers are marked with [X]
Enter your answers in the front page.
[II.1] Find the work W needed to be done by an external agent to assemble a configuration of 4
charges q at the corners of a square of side a as shown to the left in the figure below.
4q 2
4π0 a
[A] W =
[B] W =
6q 2
4π0 a
[X] W =
4q 2
4π0 a
+
2
2q√
4π0 2a
[D] W =
4q 2
4π0 a2
[E] W = 0
y
q1
R
q4
i
i(t)
L
12 V
a
2Ω
+
2Ω
−
C
q3
q2
a
E (t)
x
[II.2] In the AC R-L circuit shown at the center in the figure above the applied emf is
E(t) = Em cos ωt. Keeping Em constant, as the frequency ω increases, the current through
the resistor
[X] decreases [B] does not change [C] increases
[D] may increase or decrease depending on the value of the original frequency.
[II.3] In the R-C circuit shown to the right in the figure above the capacitor is charged with a
voltage of 4 Volts before the switch is closed. The switch is closed at t = 0. Immediately
after the switch is closed, what is the current i through the battery? You do not have to
solve any differential equations to get the answer.
[A] i = 0 [X] i = 2 A [C] i = 3 A [D] i = 4 A [E] i = 5 A
[II.4] The figure to the left below shows a magnetic field increasing in time as Bz (t) = B0 +B1 Tt ,
where B0 , B1 , and T are all positive. The magnetic field is uniform in space with cylindrical
symmetry. Find the force on the charge q located at rest at the point shown.
[A] F~ = qπr2 B1 ̂ [B]F~ = −qπr2 B1 ̂ [C] F~ = qrB1 ̂ [X] F~ = −qrB1 ̂ [E] F~ = qrB1 ı̂
T
T
2T
2T
Bz
Ez
y
z axis out of
the paper
r
q
z
EC
C
y
z axis out of
the paper
r
x
loop
2T
x
BC
C
loop
[II.5] The figure to the right above shows an electric field decreasing in time Ez (t) = E0 − E1 Tt ,
where E0 , E1 , and T are all positive. The electric field is uniform in space with cylindrical
symmetry . Use Ampere-Maxwell’s law to find the induced magnetic field BC in the
direction shown at a distance r from the center of symmetry.
[A] BC =
[D] BC =
µ0 0 rE1
2T
µ0 0 E1
− T
[B] BC = −µ0 0 πr2 ET1
Physics 207 TEST 1
[X] BC =
[C] BC = µ0 0 πr2 ET1
−µ0 0 rE1
2T
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
[III] This problem has five multiple choice questions. Correct answers are marked with [X]
Enter your answers in the front page.
[III.1] A uniform electric field is directed along the +x-axis with Ex = 1000 V/m . If the electric
potential at x = 0 m is +1000 V, what is the electric potential at x = 2 m?
[A] 2000 V [B] 1000 V [C] 500 V [D] 0 V [X] -1000 V
[III.2] A proton is accelerated through a potential difference of 1000 V. The proton gains a kinetic
energy of
[A] 1000 eV if it passes from the negative to the positive terminal of the accelerator.
[X] 1000 eV if it passes from the positive to the negative terminal of the accelerator.
[C] 1000 J if it passes from the negative to the positive terminal of the accelerator.
[D] 1000 J if it passes from the positive to the negative terminal of the accelerator.
[E] 1600 eV if it passes from the positive to the negative terminal of the accelerator.
[III.3] A metal sheet with several empty vertical slits as shown in the figure to the left below was
used in class
[X] to show that an electromagnetic wave with horizontal electric field will pass through
it
[B] to show that an electromagnetic wave with vertical electric field will pass through it
[C] to show the wavelength of an electromagnetic wave
[D] used in class to show radiation pressure
[E] no such object was used in class
[III.4] Two large sheets of charge have surface charges densities σ1 and σ2 as shown at the center
in the figure below. Far from the edges, neglecting fringing fields, the uniform electric field
between them has Ex given by
2
2
2
2
2
[B] Ex = A σ1+σ
[X] Ex = σ12−σ
[D] Ex = A σ12−σ
[E] Ex = σ12+σ
[A] Ex = σ1+σ
0
0
0
0
0
y
a
σ2
σ1
⨂
empty slits
⨂
A
A
L
⨂
⨂
⨂
⨂
v
⨂
m
e
t
a
l
B
m
e
t
a
l
⨂
⨂
⨂
b
x
L
x
z
[III.5] The figure to the right above shows a metallic rod moving at constant speed in the positive
x direction (vx > 0) through a uniform magnetic field that points into the paper. An emf
E is produced and charges accumulate on the ends a and b.
[A] E = BLv pushing downward, with positive charges at a and negative charges at b.
[X] E = BLv pushing upward, with positive charges at a and negative charges at b.
[C] E = BLv pushing downward, with negative charges at a and positive charges at b.
[D] E = BLv pushing upward, with negative charges at a and positive charges at b.
[E] No emf is produced
Physics 207 TEST 1
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
[IV] This problem has five multiple choice questions. Correct answers are marked with [X]
Enter your answers in the front page.
[IV.1] In class a cathode ray tube was used to demonstrate
[A] Faraday’s law [B] Ampere’s law [C] the Biot-Savart law
[X] the magnetic force on moving charges [E] Coulomb’s law
[IV.2] The electric potential V (in Volts) at a point in space with coordinates (x, y, z) is given
by V (x, y, z) = 2x2 y 2 + xyz, where x, y and z are in meters. What is the electric field (in
units of Volts/meter) at the point (1 meter, 0 meter, 1 meter)?
~ = 4ı̂ + ̂ + k̂ [B] E
~ = −4ı̂ − ̂ − k̂ [X] E
~ = −̂ [D] E
~ = 0 [E] E
~ = −1
[A] E
[IV.3] If in the figure to the left below the the magnet is moving toward the coil, a current I is
induced in the coil
[A] in the direction shown and the magnet is attracted
[X] in the direction shown and the magnet is repelled
[C] opposite to the direction shown and the magnet is repelled
[D] opposite to the direction shown and the magnet is attracted
[E] no current is induced in the coil
B
y
R
R
v
I
i(t)
C
x
N
S
z
I
E (t)
[IV.4] The figure at the center above shows a circular loop of radius R that lies flat on the paper
(on the x−y plane). It carries a counterclockwise current I as shown. The loop is immersed
in a uniform magnetic field of magnitude B that points toward the right in the paper , so
~ = B x̂. Find the total magnetic force on the loop using x̂, ŷ, ẑ notation.
B
~ = −2πIRB ẑ [B] F
~ = 2πIRB ẑ [C] F
~ = −2IRB ẑ [D] F
~ = 2IRB ẑ [X] F
~ = ~0
[A] F
[IV.5] In the AC circuit above, R = 100 Ω, L = 0.4 H, and C = 2.5 × 10−8 F. When the ac source
E(t) = Em cos ωt operates at the resonant frequency of the circuit, the maximum current
amplitude is Imax = 0.2 A. Find ω, the amplitude Em , and the amplitude of the voltage
VL across the inductor.
[X] ω = 104 rad/s, Em = 20 V, VL = 800 V
[B] ω = 108 rad/s, Em = 20 V, VL = 8 × 106 V
[C] ω = 104 rad/s, Em = 20 V, VL = 20 V
[D] ω = 104 rad/s, Em = 10 V, VL = 800 V
[E] ω = 104 rad/s, Em = 20 V, VL = 0 V
Physics 207 TEST 1
June 21 2013
L
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
[V] This problem has five multiple choice questions. Correct answers are marked with [X]
Enter your answers in the front page.
~ r, t) produced by an accelerating charge at time t a
[V.1] The magnitude of the electric field E(~
large distance r from the charge, varies
[A] as r12 where r is measured from the position of the charge at the same time t.
1
r
[C] as r12
[X] as 1r
[B] as
[E] as
1
r
where r is measured from the position of the charge at the same time t.
where r is measured from the position of the charge at the earlier time tr = t−r/c.
where r is measured from the position of the charge at the earlier time tr = t−r/c.
where r is measured from the position of the charge at the later time tr = t + r/c.
[V.2] A source of electromagnetic waves radiates uniformly in all radial directions with a power
Psource = 240 W. Find the electric field amplitude E0 at a distance r = 2 m from the
source.
√ V
√ V
V
V
V
[A] E0 = 60
[B]
60
2
[X]
E
=
60
[D]
E
=
120
π m [E] E0 = 120 m
0
0
π m
m
m
[V.3] A uniform charge per unit area (call it σ) is located on the outer surface of a hollow
conducting shell with inner radius R and outer radius 2R. The conducting shell surrounds
a point charge Q at its center. Given that the total charge everywhere on the body of the
conducting shell is also Q, what is the correct relation between Q and σ?
Q
Q
[X] σ = 8πR
[B] σ = 4πR
[C] σ = 8πR2 Q [D] σ = 4πR2 Q
2
2
[E] insufficient information to answer
[V.4] A wire of radius R carries a current I uniformly distributed throughout its interior. Which
of the five plots below labeled [A], [B], [C], [D], and [E] best represents the magnitude of
the magnetic field as a function of the distance r measured from the center of the wire?
[A]
[X] [C]
[D]
[E]
B
B
[A]
R
2R
[B]
R
r
[D]
B
R
2R
2R
r
R
B
r
[C]
B
2R
r
[E]
R
2R
r
[V.5] In class a large pendulum with a copper plate at the end was used to demonstrate
[A] the oscillations of a magnetic moment [B] that magnetic monopolies do not exist
[C] conservation of energy [D] the force between a current carrying wire and a magnet
[X] eddy currents
Physics 207 TEST 1
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
FORMULAE SHEET FINAL EXAM
e = 1.6 × 10−19 C, me = 9.1 × 10−31 kg, mp = 1.67 × 10−27 kg, k =
Coulomb’s Law, Electric Fields (N/C)
~
F~ = q1 q2 2 F~q = q E,
4π0 r
~ Q| =
|E
1
4π0
2
= 9 × 109 NC·m
2 .
kQ
r2
~ = n̂ dA
Electric Field Flux Φ, and Gauss’ Law: dA
I
Z
qinside
~
~
Φclosed surface = E · dA =
, qinside =
ρ dV
0
S
V
λinside
qinside
, cylindrical symmetry : Er =
spherical symmetry : Er =
2
4π0 r
2π0 r
Electric Field and sheets of surface charge density σ:
On each side of an infinite sheet En = 2σ0 . However, outside the surface of a conductor,
En = σ0 .
Force, Potential Energy and Torque for an Electric Dipole
The Force, torque, and potential energy of an electric dipole p~ immersed in uniform electric
~ are F~total = 0, U = −~
~ ~τ = p~ × E.
~
field E
p · E,
p
2
Harmonic Oscillator: Fx = −kx, ddt2x + ω 2 x = 0, ω = k/m
agent
f ield
Electric Potential and Energy: Wi→f
= (K + U )f − (K + U )i , Wi→f
= Ui − Uf
Rf
Q
~ · d~s, when E field is uniform V = Ed
,
U
(~
r
)
=
qV
(~
r
),
V
−
V
=
VQ (r) = 4π
E
q
i
f
r
i
0
~ = − ∂V ı̂ −
E
∂x
∂V
∂y
̂ −
∂V
∂z
k̂, Er = − ∂V
∂r
when E field is uniform E =
V
d
Potential energy U and work W to assemble configuration of charges:
X
qi qj
U =W =
4π0 rij
ij pairs
Capacitance Q = CV, U =
series
Q2
2C ,
1
Ceq
=
0 E 2
κo A
2 , C =
d
1
C2 , parallel Ceq
uE =
1
C1
2
+
= C1 + C2
2
2
q
Energy in a capacitor: UC = 2C
= CV2 , energy density in an E field uE = 02E
R
~ σ = 1/ρ, V = iR, R = ρL/A
Circuits i = J~ · n̂ dA, J~ = q+ n+~vD+ + q− n−~vD− = σ E,
Resistors, Power: in series Req = R1 + R2 , in parallel
1
Req
=
1
R1
+
1
R2 ,
P = iV = iE = i2 R
Kirchhoff ’s rules: (1) at a junction, current in equals current out
(2) sum of voltage rises around a loop equals zero.
Physics 207 TEST 1
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
RC circuits
q(t) = q0 e
−t/τ
−t/τ
+ CE 1 − e
,
i(t) =
dq
−q0 −t/τ CE −t/τ
=
e
+
e
,
dt
τ
τ
with τ = RC.
The voltage across a capacitor cannot change instantaneously.
R
~ F~ = i d~` × B, ~τ = µ
~ µ
Magnetic forces and Gauss’ law: F~ = q~v × B,
~ × B,
~ = I n̂A
H
~ · dA
~ = 0, dA
~ = n̂ dA
Gauss’, Biot-Savart, and Ampere-Maxwell laws:
B
I
N µ0 i ~
~ · d~` = µ0 iin = µ0 (iC + iD ), B = µ0 i for ∞ wire
~
d` × ~r,
B
dB =
3
4πr
2πr
C
~ and
where the conduction current iC is given in terms of the conduction current density J,
the displacement current iD is given in terms of the displacement current density J~D , as
Z
iC =
J~ · n̂ dA, and iD =
S
Faraday’s Law: ΦB =
R
S
~ · n̂ dA,
B
~
∂E
J~D · n̂ dA, with J~D = 0
.
∂t
S
Z
B
E = − dΦ
dt
H
~ + ~v × B)
~ · d~`.
E = (E
di
Self and Mutual inductance : ΦB = iL, E = −L dt
, Φ2 = M i1 , E2 =
of length `, and cross sectional area A, with n turns per unit length
has a self inductance L = µ0 n2 `A.
−M di1
dt .
A coil
The current in an inductor cannot change instantaneously
R-L circuits: DC charging inductor with iL (t = 0) = 0
iL (t) =
Eapplied
(1 − e−t/τL ), τL = L/R.
R
2
B
Energy in an inductor: UL = 12 Li2 , energy density in a B field is uB = 2µ
.
0
q
1
L-C circuit: ω = LC
p
1
AC circuits: In a series R-L-C circuit XL = ωL, XC = ωC
, |Z| = R2 + (XL − XC )2
i(t) = Im cos(ωt), Ea (t) = Em cos(ωt + φ), Im =
Em
|Z| ,
tan φ =
XL −XC
,
R
cos φ =
R
|Z|
vR (t) = VR cos ωt, VR = Im R, vL (t) = VL cos(ωt + 90◦ ), VL = Im XL ,
and vC (t) = VC cos(ωt − 90◦ ), VC = Im XC .
Resonance: XL = XC
Physics 207 TEST 1
June 21 2013
Dr. Huerta
Phy 207 1st Summer Session Final exam Answer key
June 21 2013
Electromagnetic wave: The speed of light c = 3 × 108 m/s
~
~ = Bm sin(kx − ωt − δ)k̂, Bm =
E(x,
t) = Em sin(kx − ωt − δ)ŷ, B
k=
2π
λ ,ω
Em
c
= kc, ω = 2πf .
Energy in EM waves, Ey (x, t), Bz (x, t) = Ey /c :
Z
2
~ ×B
~
E
~
~ · dA,
~ I = P ower with intensity I = Sav = Em Bm = 0 Em c ,
S=
, P = S
µ0
Area
2µ0
2
for point source I(r) =
Psource
4πr2
Polarization: When unpolarized light of intensity I unpolarized passes
through a polarizing filter, the passing, or transmitted intensity is
unpolarized
always Itransmitted = 12 Iin
, no matter what is the direction
of the polarizing axis of the filter.
When light with electric field of amplitude Ein and intensity Iin polarized vertically passes
through a polarizing filter with its polarizing axis (pass axis) at an angle θ from the vertical
direction,
Etransmitted = Ein cos θ, Itransmitted = Iin cos2 θ
The transmitted (or passed) electric field will be in the direction of the polarizing axis.
Physics 207 TEST 1
June 21 2013