Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
Signature:
Section 01I, MTWRF 10:05 – 11:30 p. m.
Name:
ID number:
ANSWER 20 OUT THE 25 OF MULTIPLE CHOICE QUESTIONS
THERE IS NO SCANTRON SHEET
ENTER YOUR ANSWERS IN THE SPACES BELOW LIKE 1. A or 2. B
CLEARLY CROSS OUT THE FIVE YOU LEAVE BLANK LIKE 1. XXX
1. B
6. A
11. A
16. B
21. E
2. C
7. A
12. D
17. E
22. B
3. D
8. B
4. E
9. C
5. A
10. E
13. C
14. A
15. B
18. C
23. D
19. D
24. A
20. A
25. C
Your signature signifies that you will obey the HONOR CODE
You may be asked to show your photo ID during the exam.
NO CELL PHONES, NO CALCULATORS, NO PENS
ONLY USE A PENCIL. NO BACKPACKS
THERE IS A SEPARATE FORMULA SHEET
Physics 207 FINAL FORM 4 ANSWER KEY
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
[I] This test has 25 multiple choice questions. Correct choices are marked with
[X]
[1] In a series R-L-C circuit with maximum current Imax , frequency
ω, inductance
L, capacitance
C, and impedance Z, the average power PR in the resistor,
and PL in the inductor are
I2
2
2
[A] PR = Imax
Z, PL = Imax
ωL [X] PR = max
R,
PL = 0
2
2
2
2
2
I
Imax
Imax
Imax
Z,
ωL
[D]
R,
[C] PR = max
P
=
P
=
P
=
L
R
L
2
2
2
2 ωL
2
Imax
[E] PR = 2 Z, PL = 0
~ at the center
[2] Use the Biot-Savart law to find the magnitude of the field B
of the circular loop of current shown in the figure to the rleft below.
[A] B =
µ0 I
2πr
[B] B =
µ0 I
4πr 2
[X] B =
µ0 I
2r
[D] B =
µ0 I
4πr 3
[E] B = 0
dl
r
I
B
[3] An Erlenmeyer flask shown in the figure to the right above is immersed in
a uniform vertical magnetic field of magnitude B0 in the upward direction.
The circular horizontal top has area A and the circular horizontal base has
area 3A. It . The outward magnetic flux through the side surface is
[A] zero [B] −2B0 A [C] 3B0 A [X] 2B0 A [E] −3B0 A
[4] An electromagnetic wave traveling in the positive x direction has a magnetic
~ = Bm cos(kx − ωt + δ)k̂. The electric field of the wave is given by
field B
~ = cBm cos(kx − ωt + δ)k̂ [B] E
~ = cBm sin(kx − ωt + δ)k̂
[A] E
~ = cBm sin(kx − ωt + δ)̂ [D] E
~ = cBm cos(kx − ωt + δ)ı̂
[C] E
~ = cBm cos(kx − ωt + δ)̂
[X] E
[5] Three equal charges q are fixed at the corners of an equilateral triangle.
The side of the triangle has a length a . Find the work done by an external
agent to bring the three charges (with zero kinetic energy) from infinity to
their positions on the triangle.
[X]
3q 2
4π0 a
[B]
−q 2
4π0 a
[C]
2q 2
4π0 a
[D]
−3q 2
4π0 a
Physics 207 FINAL FORM 4 ANSWER KEY
[E]
q2
4π0 a
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
[6] In the figure to the left below the coil forms a closed circuit with no battery.
The magnet is moving toward the coil. The current I induced in the coil is
[X] opposite to the direction shown and the magnet is repelled
[B] opposite to the direction shown and the magnet is attracted
[C] in the direction shown and the magnet is attracted
[D] in the direction shown and the magnet is repelled
[E] no current is induced in the coil
B
y
C
R
R
I
i(t)
L
x
v
S
N
z
I
I
E (t)
[7] The figure at the center above shows a counterclockwise current I in a
circular loop of radius R that lies flat on the paper (on the x − y plane).
~ = B x̂. Find the total
The loop is immersed in a uniform magnetic field B
magnetic force on the loop using x̂, ŷ, ẑ notation.
~ = ~0 [B] F
~ = −2πIRB ẑ [C] F
~ = 2πIRB ẑ [D] F
~ = −2IRB ẑ
[X] F
~ = 2IRB ẑ
[E] F
[8] In the AC circuit above, R = 100 Ω, L = 0.4 H, and C = 2.5 × 10−8 F.
When the ac current i(t) = Im sin ωt operates at the resonant frequency of
the circuit, the maximum current amplitude is Imax = 0.2 A. Find ω, and
the amplitude of the voltage VL across the inductor.
[A] ω = 108 rad/s, VL = 8 × 106 V [X] ω = 104 rad/s, VL = 800 V
[C] ω = 104 rad/s, VL = 20 V [D] ω = 104 rad/s, VL = 50 V
[E] ω = 104 rad/s, VL = 0 V
[9] Which of the following equations is Faraday’s Law?
H
R ρ
H
H
R ∂ B~
~ · n̂ dA =
~ · n̂ dA = 0 [X]
~ ·d~l = −
[A] S E
dV
[B]
B
E
· n̂ dA
V 0
S
C
S ∂t
H
R
R
~
~ · d~l = µ0 J~ · n̂ dA + µ0 0 ∂ E · n̂ dA
[D]
B
C
S
S
∂t
[10] The acceleration of a point charge q at the origin is a(t) = a0 cos ωt. The
magnitude of the radiation electric field produced by the accelerating charge
~ r, t) at time t a large distance r from the charge varies as (v c)
E(~
[A]
[X]
qa0 cos ω(t−r/c)
r2
qa0 cos ω(t−r/c)
r
[B]
qa0 cos ωt
r
[C]
qa0 cos ωt
r2
[D]
Physics 207 FINAL FORM 4 ANSWER KEY
qa0 cos ω(t+r/c)
r
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
[11] In class it was demonstrated that a polarizing filter in front of the screen of
a laptop computer will make the display invisible if the absorption axis
[X] goes from the lower left corner to the upper right corner of the screen
[B] is vertical [C] is horizontal [D] is in any (every) direction
[E] no such demonstration was done in class
[12] The figure to the left below shows a point charge Q located exactly at the
center of a cube. The outward electric flux through one face of the cube is
100 V·m. The total outward electric flux through the other five faces is
[A] -100 V·m [B] 100 V·m [C] -500 V·m [X] 500 V·m [E] 600 V·m
y
Bz
Ez
y
Q
z axis out of
the paper
r
x
q
z
EC
z
C
y
z axis out of
the paper
r
x
loop
x
BC
C
loop
[13] The figure at the center above 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.
qrB1
1
~
[A] F~ = qπr2 BT1 ̂ [B] F~ = −qπr2 BT1 ̂ [X] F~ = −qrB
2T ̂ [D] F = 2T ̂ [E] 0
[14] 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.
0 rE1
[X] BC = − µ0 2T
[B] BC = −µ0 0 πr2 ET1 [C] BC = µ0 0 πr2 ET1
0 rE1
[D] BC = − µ0 T0 E1 [E] BC = µ0 2T
[15] The rest energy of a proton mp c2 = 9.4 × 108 eV where 1 eV=1.6 × 10−19 J.
If relativity were not true (that is if KE = mv 2 /2) , find what accelerating
voltage Vacc would be enough to accelerate the proton starting from rest to
a final speed equal to the speed of light, that is vf inal = c.
[A] Vacc = 4.7 × 108 eV [X] Vacc = 4.7 × 108 V [C] Vacc = 9.4 × 108 V
[D] Vacc = 9.4 × 108 eV [E] Vacc = 2.4 × 108 eV
Physics 207 FINAL FORM 4 ANSWER KEY
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
[16] A metal sheet with several empty vertical slits as shown in the figure to the
left below was used in class
[A] to show that electromagnetic waves cannot pass through it.
[X] to show that an electromagnetic wave with horizontal electric field will
pass through it
[C] to show that an electromagnetic wave with vertical electric field will
pass through it
[D] to show radiation pressure [E] no such object was used in class
σ2
σ1
empty slits
A
A
C
m
e
t
a
l
L
m
e
t
a
l
R
vx
L
x
x
[17] Two large sheets of charge have surface charges densities σ1 and σ2 as shown
at the center in the figure above. Neglecting edge effects, the field between
the plates is
2
2
2
[A] Ex = σ1+σ
[B] Ex = σ12+σ
[C] Ex = σ1−σ
0
0
0
1
2
[D] Ex = σ22−σ
[X] Ex = σ12−σ
0
0
[18] In the figure to the right above a wire of length L immersed in a uniform
~ out of the paper is moving with speed vx > 0 along two
magnetic field B
parallel rails connected to a resistance R. The induced current iC is
[A] iC = − BLv
R
2
B 2 L2 v
[X] iC
R
2 2 2
iC = B LR v
−3
[B] iC =
2
=
BLv
R
[D] iC = − B RL v [E]
[19] In an oscillating LC circuit, L = 10 H and C = 4×10−5 F. The maximum
charge on the capacitor is qmax = 2 × 10−6 C. Find the maximum current.
[A] imax = 2 × 10−6 A [B] imax = 50 A [C] imax = 5 × 10−3 A
[X] imax = 10−2 A [E] Insufficient information
[20] A source of electromagnetic waves radiates uniformly in all radial directions
with Psource = 240π W. The intensity 2 meters away from the source is
W
W
W
15
[B] I = 15 W [C] I = 60π m
[D] I = 15
[X] I = 15 m
2
2
π m2 [E] I = π W
Physics 207 FINAL FORM 4 ANSWER KEY
June 26 2015
B
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
[21] An electromagnetic wave E(x, t) = Em ŷ sin[kx − ωt] has a frequency f =
109 Hz. The wavelength λ is
1
m [C] λ = 2π
[A] λ = 6πm [B] λ = 6π
3 m [D] λ = 3m [X] λ = 0.3m
[22] The figure to the left below shows two polarizing filters with their planes
parallel to the x − y plane.The first has its polarizing axis (pass axis) along
the y axis. The√second has its pass axis at an angle θ = 60◦ from the y
axis (sin 60◦ = 23 , cos 60◦ = 12 ). Unpolarized light of intensity I1 in the z
direction falls on the first sheet. The intensities I2 and I3 are
[A] I2 = I1 , I3 = 41 I2 [X] I2 = 12 I1 , I3 = 14 I2 [C] I2 = 12 I1 , I3 =
[D] I2 = I1 , I3 = √12 I2 [E] I2 = 12 I1 , I3 = 34 I2
√1 I2
2
y
1
I1 unpolarized
pass axis
x
2
θ
pass axis
I2 polarized
E(t)
3
I3 polarized
z
[23] The figure to the right above shows an electric dipole antenna. The strength
of the radiation in the directions of the arrows 1, 2, 3 is
[A] approximately the same in directions 1, 2, 3
[B] strong 1, medium 2, weak 3 [C] weak 1, strong 2, weak 3
[X] weak 1, medium 2, strong 3 [E] strong 1, weak 2, strong 3
[24] Given the electric potential in volts V (x, y, z) = ax2 y 2 , find the electric
~
field E(x,
y, z)
~ = −2axy 2 ı̂ − 2ax2 y̂ [B] E
~ = −axy 2 ı̂ − ax2 y̂
[X] E
~ = −a x3 y2 ı̂−a x2 y3 ̂ [D] E
~ = 2axy 2 ı̂+2ax2 y̂ [E] E
~ = axy 2 ı̂+ax2 y̂
[C] E
3
2
[25] A spherically symmetric cloud of radius R has a uniform charge density ρ0 .
Find the radial component Er of the electric field inside the cloud of charge
at a distance r from the center of the cloud, where r < R.
ρ0 R3
ρ0 r
ρ0 r
ρ0
ρ0 R3
[A] Er = 3
[B]
E
=
[X]
E
=
[D]
E
=
[E]
E
=
2
r
r
r
r
0
30
4π0 r 2
0 r 2
0r
Physics 207 FINAL FORM 4 ANSWER KEY
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
FORMULAE SHEET FINAL
2
1
= 9 × 109 NC·m
g = 10 m/s2 , e = 1.6 × 10−19 C, k = 4π
2 .
0
8
The speed of light c = 3 × 10 m/s
mass of electron me = 9.1 × 10−31 kg, mass of proton mp = 1.67 × 10−27 kg,
Coulomb’s Law, Electric Fields (N/C)
q1 q2 ~
~Q =
F~ =
F~q = q E,
E
4π0 r 2
kQ
r 2 r̂,
Electric Field Flux Φ, and Gauss’ Law
I
Φclosed
surface
with qinside
~ · n̂ dA = qinside
E
0
ZS
Z
Z
=
ρ dV, or qinside =
σ dA, or qinside =
λ dl
=
V
A
Spherical symmetry : Er =
qinside
,
4πo r2
L
Cylindrical symmetry : Er =
λinside
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
~ are F~total = 0, U = −~
~ ~τ = p~ × E.
~
uniform electric field E
p · E,
p
2
Harmonic Oscillator: Fx = −kx, ddt2x + ω 2 x = 0, ω = k/m
agent
f ield
Work-Energy Theorem: Wi→f
= (K + U )f − (K + U )i , Wi→f
= Ui − Uf
Electric Potential and Energy:
Rf
Q
~ · d~s,
VQ (r) = 4π
,
U
(~
r
)
=
qV
(~
r
),
V
−
V
=
E
q
i
f
i
0r
~ = − ∂V ı̂ −
E
∂x
∂V
∂y
̂ −
∂V
∂z
k̂, Er = − ∂V
∂r ,
when E field is uniform E =
Physics 207 FINAL FORM 4 ANSWER KEY
V
d
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
Capacitance Q = CV , Energy in capacitor U =
CV 2
2
κo A
1
1
1
, series
=
+
, parallel Ceq = C1 + C2
d
Ceq
C1
C2
R
R
Current i = S J~ · n̂ = S Jz (r)2π rdr dA, J~ = q+ n+~vD+ + q− n−~vD− =
C=
Resistors: V = iR, R =
ρL
A ,
in series Req = R1 + R2 , in parallel
1
Req
=
~
E
ρ
1
R1
+
1
R2
Kirchhoff ’s rules: (1) at a junction, current in equals current out
(2) sum of voltage rises around a loop equals zero.
RC circuits
q
Differential equation: dq
solution q(t) = Ae−t/τ + Kτ
dt + τ = K,
−q0 −t/τ CE −t/τ
dq
−t/τ
−t/τ
=
e
+
e
, with τ = RC.
q(t) = q0 e
+CE 1−e
, i(t) =
dt
τ
τ
The voltage across a capacitor cannot change instantaneously.
Magnetic forces and Gauss’ law:
R
~ F~ = i d~` × B, ~τ = µ
~ µ
F~ = q~v × B,
~ × B,
~ = I n̂A
H
~ · dA
~ = 0, dA
~ = n̂ dA
Gauss’, Biot-Savart, and Ampere-Maxwell laws: B
~ = N µ0 i d~` × ~r,
dB
4πr3
I
~ · d~` = µ0 iin = µ0 (iC + iD ), B = µ0 i for ∞ wire
B
2πr
C
where the conduction current iC is given in terms of the conduction current
~ and the displacement current iD is given in terms of the displacement
density J,
current density J~D , as
Z
Z
~
∂E
~
~
~
iC =
J · n̂ dA, and iD =
JD · n̂ dA, with JD = 0
.
∂t
S
S
Faraday’s Law
R
~ · n̂ dA,
ΦB = B
R ∂~B~
R
dΦB
~ · dA~
E
=
−
=
·
n̂
dA
+
B
C
dt
dt
S
S ∂t
S
H
~ + ~v × B)
~ · d~`,
where E = C (E
H
R ∂B
R
~ · n̂ dA and H (~v × B)
~ · d~` = −
~ · d~` = B
~·
with C E
S ∂t
C
S
Physics 207 FINAL FORM 4 ANSWER KEY
~
dA
dt
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
di
Self and Mutual inductance : ΦB = iL, E = −L dt
, Φ2 = M i1 , E2 = −Mdtdi1 .
A coil of length `, and cross sectional area A, with n turns per unit length
has a self inductance L = µ0 n2 `A.
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
Energy in an inductor: UL = 12 Li2 , energy density in a B field is uB =
q
1
L-C circuit: ω = LC
B2
2µ0 .
1
AC circuits: Reactances XL = ωL, XC = ωC
p
In a series R-L-C circuit |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
~
~
Plane Electromagnetic wave: E(x,
t) = Ey (x, t)̂, B(x,
t) = Bz (x, t)k̂, speed c
∂Ey
∂Bz
∂Ey
∂Bz ∂ 2 Ey
∂ 2 Ey
1
=−
, µ 0 0
=−
,
=
µ
,
µ
=
0
0
0
0
∂x
∂t
∂t
∂x
∂x2
∂t2
c2
Sinusoidal Plane wave:
~
~ = 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
~=
~ A,
~ I = P ower with intensity I = Sav = Em Bm = 0 Em c
S
, P = S·d
µ0
Area
2µ0
2
for point source I(r) =
Psource
4πr2
Physics 207 FINAL FORM 4 ANSWER KEY
June 26 2015
Dr. Huerta Phy 207 Summer 1 FINAL FORM 4 ANSWER KEY June 26 2015
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 FINAL FORM 4 ANSWER KEY
June 26 2015