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Final Exam solutions - Physics 1120 - Fall, 2007
1. A balloon of mass M has been positively charged ( its
charge is + Q ) and is held up to the ceiling.
The ceiling polarizes, so the balloon sticks to it.
The friction coefficient between the balloon and the ceiling is
μ , and we will assume that the electric field, E, created by the
ceiling is uniform and directed upwards.
What is the (minimum) magnitude E of this electric field?
A) E = μQ/Mg
B) E = Mg/μQ
C) E = μMg/Q
D) E = Q/Mg
E) E = Mg/Q
The upward force would be F=QE, the downward force is just Mg. Those are the only
forces acting in the vertical direction. (Friction is horizontal, and thus totally irrelevant
for this question!) QE = Mg means E = Mg/Q.
The next two questions refer to this situation:
Two positively charged particles, labeled 1 and 2, are
placed a distance R apart in empty space and are released
from rest.
Particle 1 has mass m and charge + Q ;
particle 2 has mass 2 m and charge + Q/.
Each particle feels only the static Coulomb force due to the other particle. (There is no
gravity, no friction, nor any other forces in this problem.)
2. How do the magnitudes of the initial accelerations of the two particles compare?
[a 1
= magnitude of initial acceleration of particle 1,
a
2
= magnitude of initial acceleration of particle 2]
A) a 1
= 2 a 2
B) a 1
= (1/2)a 2
C) a 1
= 4 a 2
D) a 1
= a 2
E) None of these.
The forces on each must be equal (and opposite) by Newton III. But F=ma, so the smaller
mass has the larger acceleration, by the ratio of masses.
3. As the particles continue to move apart after their release, the speed of each particle...
A) decreases as time goes by
B) increases as time goes by
C) stays the same.
The force is repulsive, and although it gets smaller and smaller, it is still always
repulsive. A positive force means a positive acceleration, forever. Acceleration means
increasing speed in this case. SPEED just keeps on getting bigger and bigger!
The following two questions refer to this situation:
Two negative charges are each located a distance r from the
origin, as shown.
Note that the upper charge is twice as strong as the one to
the right. (-2Q compared to - Q)
4. At the origin, the direction of electric field is ...
A) Up and right at exactly a 45 degree angle to the +x axis.
B) Down and left at exactly a 45 degree angle to the - x axis.
C) Straight up the page
D) The field has no direction, because the field is zero
E) None of these: the field points in some other direction than the
choices given above.
There are two Field vectors at the origin , one pointing straight right, one pointing
straight up with twice the magnitude. The sum of those two arrows is indeed "up and
right", but it is NOT at a 45 degree angle! (The two forces would have to be equal in
magnitude to get that angle)
5. ( Assume, as usual, that Voltage is zero at infinity.) At the origin, given the charge
configuration above, the VOLTAGE is ....
A) +kQ/r B) - kQ/r C) - 2kQ/r D) - 3kQ/r E) None of these choices is correct
Voltage is just the sum of kQ/r, here that would give - kQ/r-2kQ/r.
6. A small plastic bead has a +q charge on its
bottom side and a – q charge on its top side (it’s a
permanent electric dipole).
The bead is placed in an electric field represented by
the field line diagram shown.
At the instant shown in the diagram, what is the
direction of the net force on the bead due to this
Electric field?
A) the net force is zero
B) right → C) left ←
D) down ↓ E) up ↑
We've seen this problem on an earlier exam! The force on the + charge is up (in the
direction of the field lines) and STRONG because the field lines are more concentrated
there. The force on the - side is down but weaker, so the upward force wins. The net force
is up. This is an unstable situation (the bead, if "jiggled" by any OTHER forces, would
sooner or later flip around, and only THEN would the force be downward. But at the
instant shown, it's unambiguously up)
The next TWO problems refer to the circuit at right.
There are two ideal batteries (voltage V each), and
three identical ideal bulbs. Initially the capacitor is
uncharged, and the switch is open.
10. Immediately after the switch is closed, the
absolute value of the voltage difference across bulb
#1 is equal to:
A) 2V B) 4V/
C) V D) 2V/3 E) zero
Just after the switch closes, the capacitor acts like a
"short circuit", so the total circuit has two batteries in series (V+V) and two bulbs in
series. The voltage difference across bulbs 1 and 2 together is thus 2V, but across just
ONE of the two bulbs is half that, V.
11. A very long time after the switch is closed, the absolute value of the potential
difference across the capacitor is equal to
A) 2V B) 4V/3 C) V D) 2V/3 E) zero
Now no current can flow through the capacitor (after a long time, it's charged up!) so
you have 3 identical bulbs. Total voltage is V+V = 2V, so it must be 2V/3 across each
individual bulb. This is the voltage across each of them, in particular, it's the voltage
across the 3rd one (which is the same as the voltage across the capacitor)
12. An ideal battery is attached to three identical
ideal bulbs, as shown in the circuit to the right.
A switch in parallel with bulb 3 is originally
open and is then closed.
When the switch is closed, what happens to the
brightness of bulbs 1 and 2?
A) #1 brightens, #2 brightens
B) #1 brightens, #2 stays same
C) #1 stays same, #2 stays same
D) #1 stays same, #2 brightens
E) None of these is correct
When you close the switch, the resistance of the circuit goes down, so more current
comes out of the battery. But wait, bulb 1 is directly across the battery, the current
through it can NOT CHANGE (because it always has "V" volts across it!) So that extra
current must go through bulb 2. So 1 stays the same, 2 brightens.
13. This circuits shown on the right has five identical ideal light
bulbs (labeled 1-5).
Rank the 5 bulbs from brightest to dimmest
(hint: look carefully at the figure!)
(A) 1=5>2>3=4 (B) 1>2>3=4>5 (C) 1>2>5>3=
(D) 1>2=3=4>5 (E) None of these is correct!
All the current goes through 1, it's definitely the brightest.
3=4 (they're in parallel, same Δ V across each means same brightness! ).
5 gets MORE current than 3 or 4 individually, because it gets the SUM of their currents.
So 1> 5>3=4. We just need to think about #2, where does it fit in?
Let's see: Δ V(2)= the SUM of the Δ V(3) and Δ V(5). But a sum is greater than the parts,
so Δ V(2)> Δ V(5). That means 2 is brighter than 5!
(But it still has to be dimmer than 1, which had ALL the current).
So I've got 1>2>5>3=4.
14. Which schematic diagram best represents the
realistic circuit shown on the right
The two bulbs on the right are in series with each
other. But the two of them are in PARALLEL with the
single one on the left. That's exactly circuit A!
E) None of these four schematic circuit choices is equivalent to the "realistic" one!
A)
B)
C)
D)
18. A long straight wire, carrying a current I to the right, is located below a stationary
rectangular conducting loop. The straight wire and loop are in the same plane, and the
entire straight wire is moving downward, away from the loop, with speed v, as shown.
The induced current in the stationary rectangular loop is
A) zero
B) clockwise
C) counter-clockwise
Lenz' law says "fight the change". The wire makes flux OUT of the page up where the
loop is (by the right hand rule), and this is decreasing. The loop will generate its own flux
to try to "keep" that flux, that's counterclockwise!
19. The same current I is flowing through two long (ideal)
solenoids, labeled A and B, which both have circular cross
section.
Solenoid B has twice the diameter of A, half the length of
A, and half as many turns (coils) as A.
What is the ratio of the total magnetic energy contained in
solenoid B to that in solenoid A, that is, what is U
B
/U
A
? (Hint: for a solenoid B = μ
o
n I )
A) 1 (total stored magnetic energy is same in each)
B) 2 C) 4 D) 8 E) None of these is correct
We learned in class that total stored energy is proportional to B^2*volume. Here, we
have half as many turns but half the length, so "n" is the SAME for both! So the B field is
the same. Thus, the answer is merely the ratio of volumes of the two solenoids.
That would be area*length, which is 4 times bigger area for B, but half the length, that
makes a total of just 2 times the volume.
20. An electron and a proton, both with the same initial velocity
o
v
uur
enter a region with a uniform magnetic field B into the page, as
shown. Each one undergoes semi-circular motion in the field and
exits the field some distance d from the entry point.
(The diagram shows the path for just one of the two particles.)
Consider the following statements, and decide if they are T or F.
i) The particle shown in the picture must be the negative one (the electron)
ii) The distance "d" for the proton will be greater than "d" for the electron.
A) Both i and ii are true B) i is true, but ii is false C) i is false, but ii is true
D) Both i and ii are false
The right hand rule says that a positive particle moving right into a B-field INTO the
page (as shown) would curve the other way. So this must be the negative one.
We learned that R = Mv/qB, and since a proton has much more mass, all other quantities
being the same, the proton would have the larger radius.
I
I
I
A
B
d
21. Three ideal polaroid filters, labeled 1, 2, and 3, are placed between a source of
unpolarized light and an observer. As shown, the pass axes of the three filters are at
angles of 0
o
o
, and 90
o
relative to the vertical.
The middle filter (2) is then removed. What does the observer see?
A) No light when filter 2 is present and some light when filter 2 is removed.
B) Some light when filter 2 is present and no light when filter 2 is removed.
C) No light when filter 2 is present and still no light when filter 2 is removed.
D) Some light when filter 2 is present and more light when filter 2 is removed.
E) Some light when 2 is present and the same amount of light when filter 2 is removed.
Without #2, we have crossed polarizers, and see nothing. WITH #2, we do see some light.
22. A very large (effectively infinite) plastic slab of thickness T has a uniform positive
charge density ρ (charge per volume, C/m
3
). A student wishes to compute the magnitude
E of the electric field at a distance d (d > T/2) from the center of the slab.
The student writes down Gauss’s Law
E • dA
= Q ( enc ) / #
0
, and sketches the centered,
cylindrical Gaussian surface shown (dashed). The cylinder has length 2d and end caps
each of area A=πr
What is the correct expression for the charge Q
enclosed
Q
enclosed
=... A) A T ρ B) 2A ρ C) 2A d ρ D) 2A T ρ E) A ρ.
Charge enclosed is just ρ times volume. (Charge/volume times volume!). But we have to
be careful, the correct volume is just the volume where there IS charge, which is only
inside the slab. From the picture, that's a volume of area A and width T, thus AT is the
volume. So the answer is A T ρ
observer
light source
(unpolarized)
polarizer 1 2 3
side view
r
2d
ρ
T
perspective
view
area A
26. A 1200W hairdryer (this refers to the average power of the hairdryer) is designed to
be plugged into a standard 120V electrical outlet. What is the maximum number of such
hair dryers that can be plugged in and turned on without tripping the breaker in a circuit
that has a 15 A breaker switch?
Hint: A 15A breaker switch will interrupt the current if the total rms current exceeds 15A.
A) 80 B) 10 C) 4 D) 2 E) 1
P(ave) = V(rms)*I(rms). Here, I(rms) = P(ave)/V(rms) = 1200/120 = 10 A. That means
only ONE hairdryer can be on, a second one makes 20 A which trips the circuit.
The next two problems refer to this situation: a transformer at a power station is designed
to step up the voltage from 120 V (RMS, AC) to 1200 V (RMS, AC).
The voltage oscillates with a frequency of 60 Hz.
27. If the primary (input) side is a coil with 100 turns, how many turns should the
secondary (output) side have?
A) 10 turns B) 20 turns C) 1000 turns D) 1200 turns
E) None of these is correct!
It's step up, so we need MORE turns, in the ratio V(out)/V(in) = 1200/120 = 10:
So we need 10*100 = 1000 turns.
28. If the station delivers an average power of 120 MW (that's MegaWatts), what is the
maximum instantaneous current flowing out of the secondary side?
(Answer to two place precision)
A) 120 kA B) 71 kA C) 100 kA D) 140 kA
E) None of these is correct!
Power is V*I. On the secondary side, V(rms) = 1200 V, so we have
I(rms, secondary) = P(ave)/V(rms) = 120E6/1200 = 1E5 Amps. But that's the rms
current, not the maximum. Maximum is Sqrt[2]*rms, or 1.4E5 = 140 kA.
29. If you halve the period, T of a traveling electromagnetic wave in vacuum, what
happens to the wavelength of that wave?
A) wavelength is the same, it is independent of period.
B) wavelength increases by a factor of 2.
C) wavelength increases by a factor of 4
D) wavelength decreases by a factor of 2
E) wavelength decreases by a factor of 4.
We know lambda * frequency = c = constant, and frequency = 1/T.
So lambda/T = constant. If T is halved, then lambda is halved too, to keep their ratio the
same.
30. A coil of wire with N=100 turns and area A = 0.10 m
2
is oriented so
that its plane is perpendicular to a uniform magnetic field B which is
increasing at a rate of 0.010 T/s.
The coil is connected to a resistor R = 10Ω.
What is the power dissipated in the resistor at the moment when B = 0.1 T?
A) 0.1 W B) 10
W C)
- 3
W D)
W
E) None of these.
Faraday says EMF = - N d(B*A)/dt. Here, N=100, A = .1, and dB/dt = .01.
So EMF = 100.1.01 = 0.1 V.
But power is V^2/R = (.1V)^2/(10 Ohm) = 1E-3 W.
31. A "coaxial cable" consists of a long inner solid wire
with radius R , and an outer (hollow) cylindrical wire
with inner radius 2 R and outer radius 3 R.
The inner wire carries total current I into the page and
the outer (hollow cylindrical) wire carries the same
magnitude current I out of the page.
Both wires have uniform current density.
What is the magnitude of the magnetic field B at the
surface of the outer wire, that is, at r = 3 R?
A) μ 0
I/2πR B) μ
0
I/ 6πR C) μ
0
I/πR D) zero
E) None of these.
Ampere's law says the integral of B around a circle is equal to the current "poking
through" that circle. Here, there is I(in) and I(out), they go in opposite directions, but
have equal magnitude, so the total current "poking through" the page here is ZERO.
There is thus ZERO integral of B around that outer wire, and by symmetry, B=
everywhere.
32. An electron (charge - e) is released from rest in a region
where there is a uniform E-field and a uniform B-field,
both are pointing along the +x-direction; that is,
E= E
x
i , B = B x
i.
What is the path of the electron after its release?
A) Left and curving upwards
B) Left and curving downwards
C) straight left ← , along the – x direction, forever.
D) straight right →, along the +x direction, forever.
E) Some other motion.
The E field will accelerate the electron (with negative charge) to the LEFT. This is
"antiparallel" to the B field, which thus has no effect. It just goes left, forever!
+Q
– Q
Q
– Q
Befor
e:
After :
36. Two charges, labeled 1 and 2, with charges +Q and – Q,
are a fixed distance apart.
A metal cube, with no net charge on it, is placed between the
charges, as shown in the diagram. What happens to the
magnitude of the net force on charge 1 when the cube is
placed between the charges.
A) net force on 1 is unchanged
B) net force on 1 decreases to zero
C) net force on 1 decreases (but remains non-zero)
D) net force on 1 increases E) we cannot decide if we do not know the numerical
magnitude of the charge Q.
You will polarize that metal cube, making some - charge on the left side, and some + on
the right. So there is an ADDITIONAL force on +Q: a RIGHTWARDS force attracting it
to the nearby negative side, and a much smaller leftwards force repelling it from the back
side. And then there's still the same old rightwards force from - Q that never goes away!
So you have the old right force, plus a new BIG right force (and a small left force), the
result must be an even bigger right forced than you started with.
The next three questions refer to this situation.
A circuit consists of a battery, two resistors, an inductor
and a switch as shown. Initially the switch is open and has
been open for a long time
37. The switch is closed. Immediately after the switch is
closed, what is the magnitude of the current that flows
through the battery?
A) 0 A B) 0.5 A C) 1. A D) 2. A E) None of these is correct!
Right after the switch is closed, the current through the inductor cannot change right
away, so it stays zero. Thus, the circuit is just "battery - R2 - back to battery", and thus
the current through the battery is 10 V / 10 Ohm = 1 Amp.
38. After a very long time, what is the magnitude of the current through the battery?
A) 0 A B) 0.5 A C) 1. A D) 2. A E) None of these is correct!
After a long time, the inductor looks like an ideal wire. So the circuit is just two parallel
resistors, and R(total) = 5 Ohms, so 2 Amps flows out of the battery. (1 Amp = 10V/
Ohms through EACH of the two parallel resistors)
39. After being left closed for a very long time, the switch is opened again.
Immediately after the switch is re-opened, describe the current flowing the rightmost
resistor, labeled R 2
A) 1 A, flowing up ↑ B) 1 A, flowing down ↓ C) 2 A, flowing up ↑
D) 2 A, flowing down ↓ E) 0, no current flows through R 2
at that instant.
The current was 1 Amp flowing DOWN through the inductor in the previous problem - so
that's what it still must be immediately after opening the switch. (You cannot instantly
change I through an inductor).
Where can this current go? It has no choice, it must go up through R2...
40. A long solenoid with many turns per length has a
uniform magnetic field B within its interior.
Consider the imaginary rectangular path of length "c"
and width "a" with the bottom edge entirely within the
solenoid, as shown.
What is the integral of
r
B • d
r
L
around this
rectangular path in the counterclockwise direction?
A) 0 B) 2 B c C) 2 B (c+a) D) B a c E) B c
This fancy integral just means "add up B dot dL around the loop". On the bottom leg, of
length c, you are INSIDE the solenoid, B is parallel to dL, and you get Bc. On the right
and left legs, B is perpendicular to dL, you get nothing due to the dot product. Outside,
B=0, again nothing...
41. A coil of wire carrying current I can rotate freely
about an axis in a uniform magnetic field.
The coil is in the plane of the page, and the magnetic
field is directed from left to right as shown.
If released from rest in the position shown, which
way does it rotate?
(A) the right side will rotate out of the page
(B) the left side will rotate out of the page.
(C) the loop will not rotate at all, but it will feel a
net force to the right
D) the loop will not rotate at all, and it will feel zero net force
E) the loop will not rotate at all, and it will feel a net force up the page.
The right hand rule applied to each leg in turn gives me that the force on the LEFT
current is out of the page, but on the right segment is into the page. That makes a torque,
it rotates.
