Physics 142!
Summer 2013
Final Exam
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
The first four questions concern the fundamental equations of electromagnetism,
numbered 1-5 here:
1.
∫ E ⋅ dA = 4π kQenc !
2.
∫ E ⋅ dr = − ∫ (∂ B/ ∂ t) ⋅ dA
3.
∫ B ⋅ dA = 0 !
4.
∫ B ⋅ dr = µ0 ⎡⎣ Ilinked + ε0 ∫ (∂ E/ ∂ t) ⋅ dA ⎤⎦
5. F = q(E + v × B)
1.!
2.!
If none of the quantities in the equations depend on time, which equation forbids
lines of the E-field to form closed curves?
!
1 only.
√ !
2 only.
!
4 only.
!
5 only.
Which equation requires lines of the B-field to form closed curves?
!
2 only.
√ !
3 only.
!
2 and 4.
!
5 only.
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Physics 142!
3.!
Summer 2013
Which equations imply the existence of induced fields?
√ !
4.!
!
2, 4 and 5.
!
All of 1-4.
!
All of 1-4.
Which equations give the basic principles for electric generators and motors?
!
1 and 3.
!
4 and 5.
√ !
!
5.!
2 and 4 only.
2 and 5. [2 for generators, 5 for motors.]
1 and 5.
Shown are two AC circuits to control the brightness of a light bulb.
In the top circuit the resistance R can be varied. In the bottom
circuit, the self inductance L can be varied. When the bulb is at
the same brightness in both cases:
!
E
R
The generator supplies less power in the top circuit.
√ !
The generator supplies less power in the bottom circuit.
!
The generator supplies the same power in both circuits.
!
In which case the generator supplies more power depends
on its frequency.!
E
[The inductor consumes no power.]!
L
2
Physics 142!
6.!
Summer 2013
Which is NOT a property of an e-m wave emitted by a small distant source?
√ !
The magnitude of each of the fields in the wave falls of as the inverse
square of the distance. [Inverse first power.]
!
The magnitudes obey E = cB .
!
If the wave moves in the +x-direction, and E is in the z-direction, then B is
in the –y-direction.
!
The energy density in the E-field is equal to that in the B-field.
!
7.!
The yellow light from sodium lamps comes from from two spectral lines at about
600 nm, differing in wavelength by 0.6 nm. Consider a grating with rulings 1500
nm apart.
!
The two lines cannot be seen in 2nd order.
!
To resolve these lines in 1st order, the grating must have at least 2000
rulings.
!
The angle at which these lines will be seen in 1st order is given by
sin θ ≈ 0.6 .
√ !
8.!
None of the above is true.
Which of the following is NOT a consequence of the uncertainty principle?
!
√ !
The state of an atomic electron cannot be specified in terms of its position
and velocity.
No state of an atomic electron can have a definite energy.
!
The program of Newtonian mechanics can only be carried out as an
approximation for large systems.
!
A particle in a bound state cannot have zero average kinetic energy.
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Physics 142!
Summer 2013
Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each
question carries a value of 3 points.
1.!
The decay of a neutron into a proton and a gamma ray does not happen because
a fundamental law would be violated.
√
2.!
T
When a charged balloon is inflated to increase its size, the energy stored in the Efield increases.
T
3.!
T
F
T
F
[Induced fields.]
Since parallel rays entering the eye of a nearsighted person focus in front of the
retina, there is no distance at which one can place an object so that this person’s
eye will make an image on the retina (without corrective lenses).
T
6.!
F [Decreases.]
The electromagnetic fields created directly by charges and currents are not those
that constitute an electromagnetic wave.
√
5.!
√
Motion of a conductor through a non-uniform static B-field is opposed by the
interaction forces between the field and the induced currents in the conductor.
√
4.!
F [Conservation of charge.]
√
F
[Any point between near and far points.]
One feature of the photoelectric effect that agrees with the expectations of
classical physics is that the number of electrons released per second is
proportional to the intensity of the incident radiation.
√
T
F
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Physics 142!
Summer 2013
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method
clearly. Problems carry the point values shown.
1.!
Shown is a capacitor with the space between the plates
half filled by a slab of uniform dielectric constant κ = 2 .
The left plate is grounded (V = 0) and the right plate is
held at potential V0 by the battery.
a.!
Find the fields E0 in the empty gap and E1 in the
V0
x
slab (magnitude and direction) in terms of V0 and
d. [What is the ratio of their magnitudes?]
b.!
Let the left plate be at x = 0 and the right plate at
x = d as shown. Make a careful plot of V vs. x.
[What is the potential at the left side of the slab?]
c.!
Call the plate area A. Find the stored energy, in terms of V0 , d and A. [Use
d
energy density, or find the charge on the plates.]
d.!
Find the capacitance.!
[20 points]
a.!
The ratio is E1 /E0 = 1/κ = 1/ 2 , and V0 = E0 ⋅ (d / 2) + E1 ⋅ (d / 2) . Solving these two
equations we find E0 =
4V0
2V
and E1 = 0 . Both fields are directed to the left.
3d
3d
b.!
Graph shown at right below. The E-fields are the negative of the slopes.
c.!
We have U = 12 ε 0E02 ⋅ 12 Ad + 12 κε 0E12 ⋅ 12 Ad = 14 ε 0 Ad ⋅ 32 E02 . From (a) we find
U=
2ε 0 A 2
⋅V0 . Or, σ = ε 0E0 = ε 0 ⋅ (4V0 /3d) , so
3d
Q = σ A = (4ε 0 AV0 )3d . Then U = 12 QV0 =
2ε 0 A 2
V
3d 0
V
V0
2V0 / 3
as before.
d.!
Comparing to U = 12 CV02 we see that C =
5
4ε 0 A
.
3d
d/2
d
x
Physics 142!
2.!
Summer 2013
A proton from the sun enters the magnetic field around the earth, with field lines
running from the south magnetic pole to the north magnetic pole. The drawing
shows a region high above the earth, with the B-field lines into the page.
v
⊗
B
a.!
If the proton enters with velocity as indicated, show on the drawing the
circle it will describe, indicating which way the proton runs.
b.!
If the proton also has a velocity component toward the north (into the
page) it will move northward in a helical motion. As it approaches the
north magnetic pole the B-field grows stronger. This causes the northward
motion of the proton to slow, stop and reverse. Explain why. [Consider the
magnetic moment of the proton’s circular loop.]
c.!
What will happen as the proton, now moving in a southward helix,
approaches the south magnetic pole, where the field again gets stronger?
Explain.
[15 points]
a.!
Drawing shown. The initial force is to the right, so the proton goes into a counterclockwise circle.
b.!
The magnetic moment is out of the page (to the south) opposite to the field. When
the field gets stronger the proton loop will be repelled, so its northward motion
will slow, stop and reverse. (The circular motion continues.)
c.!
As it approaches the south pole, its magnetic moment is still opposite to the field,
so the southward motion will slow, stop and reverse. It is trapped in the field.
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Physics 142!
3.!
Summer 2013
The jumping ring demonstration involves a solenoid with an iron
core which creates a strong magnetic field in the z-direction given by
Bz (z,t) = B0 (z) ⋅ cos ω t , where B0 decreases with z above the core so
z
dB0 /dz < 0 . You will investigate the force on the ring.
a.!
Assume the B-field is uniform across the cross-section area A
of the ring. Find the induced emf E(t) in the ring.
b.!
If the ring has resistance R, find the current I(t) and the
magnitude µ(t) of the induced magnetic moment.
c.!
What is the direction of µ if the solenoid field is upward and
increasing? What is its direction if that field is upward but decreasing?
d.!
The upward force on the ring is Fz (t) =
d
[µ(t) ⋅ Bz (t)] . Write this out and
dz
show that its average over a cycle is zero.
!
[Use for averages over a cycle: (sin θ ) av = (cosθ ) av = (sin θ ⋅ cosθ ) av = 0 ;
(sin 2 θ ) av = (cos2 θ ) av = 12 .]
[20 points]
a.!
The flux is Φ = Bz A = B0 Acos ω t , so the emf is E = −dΦ/dt = ω B0 Asin ω t .
b.!
We have I(t) = E /R = (ω B0 A/R)sin ω t . The magnetic moment magnitude is
µ = IA = (ω B0 A 2 /R)sin ω t .
c.!
If B s upward and increasing then µ is downward. If B is upward and decreasing
µ is upward.
d.!
dB02 ω A 2
We have Fz =
⋅
⋅ sin ω t ⋅ cos ω t . Averaged over a cycle this is zero because
dz
R
of the sin ω t ⋅ cos ω t .
7
Physics 142!
4a.!
Summer 2013
We look more carefully at the previous problem, since we know the ring always
jumps upward. Let the ring have self inductance L as well as resistance R.
a.!
Take your answer from (a) of the previous problem for the induced emf.
Write the instantaneous current I(t) in the ring, in terms of B0 , the
impedance Z, and the phase angle φ . Give the formulas for Z and φ .
b.!
Look again at µ(t) and the force. Show that the average of the force Fz
over a cycle is now positive (remember dB0 /dz < 0 ), so the ring jumps.
!
[Use sin(α − β ) = sin α ⋅ cos β − cos α ⋅ sin β .]
[10 points]
a.!
Now the current is I(t) = (Emax /Z) ⋅ sin(ω t − φ ) , where Emax = ω AB0 ,
Z = R 2 + (ω L) 2 , and tan φ = ω L/R . Note that φ > 0 .
b.!
dB02 ω A 2
Now µ(t) = (ω A B0 /Z)sin(ω t − φ ) and Fz =
⋅
⋅ sin(ω t − φ ) ⋅ cos ω t . Since
dz
Z
2
sin(ω t − φ ) ⋅ cos ω t = cos φ ⋅ sin ω t ⋅ cos ω t − sin φ ⋅ cos2 ω t we see that the average over
a cycle is (Fz ) av = −
dB02 ω A 2 1
⋅
⋅ sin φ . Since dB0 /dz < 0 and φ > 0 , we have
dz
Z 2
(Fz ) av > 0 , and the ring always has an upward force.
4b.!
To carry power to a device that operates on 60 Hz AC one uses parallel wires, as
in a lamp cord. To carry power between devices that operate at high frequencies
on uses a coaxial cable. Explain why.
!
[5 points]
The problem is power loss by radiation, which is severe at high frequencies. The coaxial
cable, by confining the fields to the interior of the cable, eliminates radiation.
8
Physics 142!
5a.!
Summer 2013
A resistor carries current as shown. It is a cylinder
of radius a and length  , and has resistance R.
I
•
a.!
At the point shown on the top surface, what
are the fields E and B (magnitude and direction in each case). [What is the
potential difference across the ends of the resistor?]
b.!
What is the Poynting vector S at that point (magnitude and direction)?
c.!
Show that the flux of S inward through the surface equals I 2 R .
[15 points]
a.!
The left end of the resistor is at higher potential, so E is to the right. B is out of the
page. The magnitudes are E = IR / , B = µ0 I / 2π a .
b.!
S is downward, into the resistor, with magnitude S = EB/ µ0 = I 2 R / 2π a .
c.!
The surface area of the resistor is 2π a , so the inward flux is I 2 R as claimed.
5b.!
A beam of unpolarized light in air passes through a
polaroid filter before being reflected from a
horizontal glass plate. When the beam strikes the
θ
plate as a certain angle the reflected beam
disappears. Is the filter’s transmission axis parallel
to the plane made by the rays, or perpendicular to it? How do you know?
[5 points]
If the reflected beam disappears, the angle must be the Brewster angle and the incident
beam must be polarized with E in the plane made by the rays. That must have been the
direction of the filter’s transmission axis.
9
Physics 142!
6.!
Summer 2013
The power used by the cornea-lens system of a relaxed normal eye to focus
parallel rays on the retina is P0 (in D).
a.!
The lens of the normal eye can add sufficient accommodation power Pa to
allow focusing on objects at 0.25 m. How much is Pa (in D)?
b.!
A person’s relaxed eye has focusing power P1 > P0 , so it cannot focus
clearly on objects farther away than distance F. One prescribes corrective
lenses of power ΔP = P0 − P1 . What is ΔP in terms of F?
c.!
A second person’s relaxed eye has focusing power P2 < P0 , so even using
the lens power Pa found in (a) it cannot focus clearly on objects closer
than distance N. One prescribes corrective lenses of power ΔP = P0 − P2 .
What is ΔP in terms of N?
!
[In all cases the distance to the retina is the same. Relate that distance to P0 .]
[15 points]
a.!
The image distance is related to P0 by (in effect)
have
1
= P0 . For an object at 0.25 m we
q
1
1
= 4 + P0 . The additional power is Pa = − P0 = 4 D.
f
f
b.!
We have
c.!
Now
1
1
1
= P1 = + P0 , so ΔP = P0 − P1 = − .
f
F
F
1
1
1
= P2 + 4 = + P0 , or ΔP = P0 − P2 = 4 − .!
f
N
N
10
Physics 142!
7a.!
Summer 2013
White light is reflected at normal incidence from a soap film ( n = 4 /3) in air. It is
observed that no light is reflected for wavelengths 400 nm and 600 nm, but for no
wavelength in between. What is the thickness of the film?!
[10 points]
For 400 nm we have δ 400 = 4π ⋅ (4 /3) ⋅ t / 400 + π , and for 600 nm we have
δ 600 = 4π ⋅ (4 /3) ⋅ t /600 + π . These two must differ by 2π , since there is no wavelength
between them where destructive interference occurs. Thus δ 400 − δ 600 = 2π . This gives
1 ⎞ 3
⎛ 1
t ⋅⎜
−
= , leading to t = 450 nm.
⎝ 400 600 ⎟⎠ 8
7b.!
A pair of objects are 1.22 m apart and are 20 km from a camera with a lens of
diameter 1 cm. If resolution is limited only by diffraction, show that the image
made by the camera resolves these objects with light of λ = 400 nm, but not with
light of λ = 600 nm.
[5 points]
The angular separation of the objects is Δθ = 1.22/ 2 × 10 4 = 6.1 × 10−5 . For 400 nm the
half width of the diffraction peak is δθ = 1.22λ /D = 1.22 ⋅ (400 × 10−9 )/10−2 = 4.88 × 10−5
which is less than Δθ . For 600 nm we get δθ = 7.32 × 10−5 which is greater than Δθ .
11
Physics 142!
8.!
Summer 2013
To estimate the time T since the supernova that created the heavy elements found
in the earth you make the reasonable assumption that the two long-lived isotopes
235
of uranium ( 238
92 U and 92 U ) were created in equal numbers during that
cataclysmic event. The average lifetimes are τ 1 and τ 2 , respectively, where
τ 1 > τ 2 . You are given a sample of uranium and you measure the decay rate
( R = N /τ ) for each isotope.
a.!
Express the current numbers N1 and N 2 in your sample in terms of the
original number N0 of each just after the supernova, T and the lifetimes.
b.!
Express N1 and N 2 in terms of the decay rates and lifetimes.
c.!
Compare the ratio N 2 / N1 from (a) and (b) and solve for T in terms of the
decay rates and lifetimes.
[15 points]
a.!
We have N1 (T) = N0 e −T/τ 1 and N 2 (T) = N0 e −T/τ 2 .
b.!
We have N1 = R1τ 1 and N 2 = R2τ 2 .
c.!
⎛1
⎛R τ ⎞
1⎞
We find N 2 / N1 = R2τ 2 /R1τ 1 = eT/τ 1 −T/τ 2 . Thus T ⎜ − ⎟ = ln ⎜ 2 2 ⎟ , or
⎝ τ1 τ 2 ⎠
⎝ R1τ 1 ⎠
T=
⎛Rτ ⎞
τ 1τ 2
ln ⎜ 1 1 ⎟ . [The experimental numbers give T ≈ 5 × 109 yr.]
τ 1 − τ 2 ⎝ R2τ 2 ⎠
12
Physics 142!
9.!
Summer 2013
You have completed two courses on the description of nature given by physics.
Pick three important things you learned that you didn’t already know, and
explain briefly why you chose these.
[15 points]
Surely you could think of three.
13
Physics 142!
Summer 2013
Final Exam
25
Median = 82
St Dev = 11
Students
20
15
10
5
0
<40
40
50
60
70
80
90
70
80
90
Course Total
25
Median = 81
St Dev = 10
Students
20
15
10
5
0
<40
40
50
60
14