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Concept Tests -- Final Review
CTF-1. Calculator question:
105 = ? A) 1E5
B) 10E5
C) something else.
Answer: 1E5 means 1105 = 105
CTF-2. A cube is made larger and larger...
As the cube grows, the ratio
A) increases
Answer:
area
..
volume
B) decreases
C) remains constant.
area
6 L2
6
decreases with increasing L
=
=
3
volume
L
L
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CTF-3. A box with mass M and charge +Q is pushed along a rough surface at constant
velocity by a uniform horizontal electric field of magnitude E. The coefficient of kinetic
friction between the box and the floor is . What is the net force on the box?
A) QE
B) g
D) QE – Mg + FN –  Mg .
C) QE – Mg
E) None of these
M, Q
E
v = constant

Answer: the magnitude of the net force must be zero. This is actually answer D! FN =
Mg and QE =  FN =  Mg, so the sum is zero. But that was a dirty trick so I'll accept
answer E as correct, if you thought that the answer is zero.
CTF-4. A row of positive charges is stationary on the ground. A person with a gaussmeter (which measures the magnetic field) is running to the right along the row of
charges, at the same height as the charges and in front of them (in the diagram below).
Does the person measure a non-zero B-field?
A) Yes
B) No
v
What is the direction of the B-field which the moving observer measures?
A) Up
B) Down
C) Forward direction 
D) backward direction 
E) no direction, B-field zero
Answers: In the frame of reference of the person, the charges are moving to the left,
which constitutes a current to the left. There is a B-field in the person's frame of
reference and its direction is up (at the location of the person).
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CTF-5. A charge q is located at the center of an gaussian cubical box as shown. A
student is asked: What is the electric flux E though the top face of the cube?
Is the answer hard to calculate or easy to calculate?
A: Easy
B: Hard.
q
Answer: It's easy if you remember Gauss’s Law  E  dA 
Qinside
o
. The total flux through
the entire box is q/o. By symmetry, the flux through one face is 1/6 th the flux through
the whole box. Flux = (1/6) q/o.
CTF-6. A capacitor with capacitance C is attached to a battery with voltage V. What is
the flux
z
 
E  da
through the cubical volume shown? (The end faces of the cube are
within the metal plates of the capacitor.
C
A)
CV
o
B)
2CV
o
C) zero
D) None of these.
V
Answer: Zero. In electrostatic situations, the E-field inside a conductor is zero. Since
E=0 on the right and left ends of the
(imaginary)
 cubical
  box, the flux through those ends

is zero. On the sides of the box, E  da  0 , since Eda . So the total flux is zero.
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Since
z
 
E  da  0 , then by Gauss's Law, the total charge must be zero within the box
must be zero, so we have proven that the two plates must have equal and opposite charge
densities.
CTF-7. Points A and B are distances r and 3r respectively from a point charge q. What is
the voltage difference between points A and B?
r
2r
A
q
B
2kq
A)
3r
D)
3kq
4r
B)
kq
3r
C)
8 kq
9 r
E) None
Answer: Voltage due to a point charge is kq/r. The voltage difference between points A
and B is (kq/r – kq/(3r)) = 2kq/3r.
CTF-8. Consider a point in empty space near several charges, which might be positive,
negative, or both. Consider the following statements.
I. The E-field can be zero while the voltage is non-zero.
II. The voltage can be zero while the E-field is non-zero.
Which of these statements can be true?
A) both can be true
B) neither can be true
C) only I can be true
D) only II can by true
Answer: Both can be true.
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CTF-9. A capacitor has a voltage V across its plates. An electron, initially at rest, is
released from at a point very close to the negative plate of a capacitor and it accelerates
toward the positive plate. The electron has charge –e and mass m. There is no gravity in
this problem. What is the final kinetic energy of the electron just before it collides with
the positive plate?
A) m  e  V
B) 2eV
C) (1/2)mV2
D) eV
E) None of these
V
Answer: |KE| = |PE| = |qV| = eV
CTF-10. A resistor with resistance R is plugged into a 120VAC wall socket. The graph
below is either resistance R, voltage V across, current I thru, or power P dissipated in the
resistor vs. time.
?
time
What could the graph be?
A) V only
B) I only
C) V or I only
D) V, I, or P E) Some other combination
Answer: The graph could be V or I only. Power is always positive.
?
P
time
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I or V
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CTF-11. Consider the circuit shown, with the switch initially open. When the switch is
closed, the current I1 through resistor R1
I bat
A) increases, B) decreases,
C) stays the same.
V
R1
R2
I1
When the switch is closed, the current from
the battery Ibat
A) increases, B) decreases,
C) stays the same.
Answers: I1 stays the same. I1 = V/R1 regardless of whether
or not the switch is closed. The battery current increases
when the switch is closed.
CTF-12. When a resistor is plugged into a standard AC
outlet, the fuse blows and all the lights go out. If we want
to repeat this dangerous experiment, and not have the fuse
blow, we need a resistor that is
A) larger
B) smaller
C) Same R, but larger power rating
Answer: larger. At constant voltage (here V = 120 VAC), the power is P = V2/R. Larger
resistance draws less power.
CTF-13. A charge q is released from rest at point in empty space were there
may be a static E- and/or a static B-fields. There are no forces on the charge
except for the forces due to the E and/or B-fields (no gravity, etc.). The
charge is observed for a short while and is seen to move along a curved path.
Consider the following possibilities
I. There is only an E-field present and no B-field.
II. There is only a B-field present and no E-field.
III. There is both an E-field and a B-field present.
q
Which possibilities could account for the observed motion?
A) all three B) I and III only
C) II and III
D) III only
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The correct answer is: There might be only a (non-zero) E-field present and no B-field.
If the particle is in a curved E-field, it will accelerate along a curved path.
It is true that this path could be caused by a B- and E-field acting together, but it is not
true that the path must be caused by a B- and E-field together.
CTF-14. At time to, a particle with charge q has instantaneous velocity
is moving in a uniform magnetic field
y
q
B.
A) True
V
B
r
v o as shown and
or B) False ?
As time increases, the x-component of the velocity vx
remains constant.
x
A) True or B) False: The KE of the particle remains constant as time increases.
Answers: True. The force exerted by the magnetic field is into the page (right-hand rule).
This force has no x-component. Since there is no x-component of the force, the xcomponent of the acceleration must be zero, and the x-component of the velocity must be
zero.
The magnetic field can do no work on the particle since F  v always. The KE can
change only if work is done (by the work-energy theorem: KE = Wnet.) So the KE is
constant.
CTF-15. 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
one edge entirely within the solenoid as
shown.
 
c
a
What is B  dl ?
z
B
A) Bc
B) 2B(c + a)
C)
2Bc
D) None of these
The solenoid has n turns per length and carries a current I. What is
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z
 
B  dl ?
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A)  o nI
B)  o ncI
C)  o n( a  c) I
D) None of these.
v v
Answers: ò
the bottom of the loop contributes to the line integral.
ÑB ×da = Bc . Only

On the sides B  dl , so B  d l = 0. On the top of the loop (outside the solenoid), B = 0.
By Ampere's Law,
v
v
òÑB×da =
0 n c I
CTF-16. A horizontal loop of wire is in a vertical magnetic field. (caused by an external
magnet). The magnetic field varies smoothly from pointing upward to pointing
downward, as shown. (The B-field gradually gets smaller, goes to zero, then grows in the
other direction.)
At the moment when the external magnetic field in the loop is zero, is there an induced
current in the loop?
A) Yes
B) No
At the moment when the external B-field in the loop is zero, what is the direction of the
induced current?
A) Clockwise (as seen from above)
B) Counterclockwise
C) No direction because induced current is reversing
Answers: Yes. Even though the B-field is instantaneously zero, it is still in the process
of changing (from up to down). So the time-rate of change of B is non-zero, and by
Faraday's Law, there must be an emf and an induced current. The induced current is
counterclockwise.
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CTF-17.
A single circular loop of wire of radius R surrounds a long solenoid of radius r, turns per
length n, carrying a current I. The diagram shows the end-on
view of the situation. What is the magnetic flux through the
R
single wire loop?
A: onI R2
B: oI r2/(2R)
C: R2(oI/2R)
D: onI r2
I
r
solenoid
E: None of these.
Answer: onI r2, NOT onI R2.
The B-field is non-zero only within the solenoid, so only the area of the of the solenoid
 
contributes to the integral B  da .
z
CTF-18. The following questions refer
A
to the circuit shown which has two
resistors, both with resistance R, an
B
L
V
inductor with inductance L, a battery
with constant voltage V, and a switch
R
R
which can be in position A or B.
Suppose the switch has been in position
A for a long time and is then switched to
position B.
1. Immediately after the switch is thrown to B, what is the current in the inductor?
A) zero
B) V/R C) V/(2R)
D) 2V/R
E) None
2. Immediately after the switch is thrown to B, what is the voltage across the inductor
L? a)
A) zero
B) V
C) 2V D) V/2
E) None
Answers: The current thru L just after the switch goes to B must be the same as when
the switch was at A: I = V/R
The voltage across the inductor is 2V. By Kirchhoff's voltage law, the voltage across
the inductor L must be the same magnitude as the voltage across the two resistors: V = I
(2R) = (V/R)(2R) = 2V.
CTF-19. A hand-cranked electrical generator is attached to a resistor R. The generator is
turned at a constant rate which causes an AC current to flow in the resistor. Suddenly,the
value of the resistance R increases. Does the generator get easier or harder to turn?
A) Easier.
B) Harder.
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C) No change in difficulty.
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Answer: Easier. The power is P = IV (V = emf = constant amplitude AC, since crank is
turned at constant rate). Current I = V/R. So P = I V = V2/R. As R increases, the power
P decreases. Less power means crank is easier to turn.
CTF-20. An radio tower is emitting a single-wavelength (monochromatic) EM wave in
all directions (isotropically). A technician measures the electric field amplitude E at 100
meters from the tower and again at 400m from the tower.
The intensity of an EM wave is proportional to the square of the amplitude of the E-field:
I  E2. What is the ratio E100/E400?
E measured
at 100m
A) 2
B) 4
C) 16
E measured
at 400m
D) 162 = 256
Answer: 4 The intensity is 16 times greater at 100 m than at 400 m, but the E-field is
only 4 time greater.
CTF-21. A head lamp of power P sends rays of light in the forward
direction only. At a distance R in front of the lamp, the intensity of
the light is _________
A) greater than
P
4p R 2
(fill in the blank)
B) less than
C) equal to
Answer: greater than
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CTF-22. A ray of light is sent into an optic fiber at an angle such that the angle with the
normal to the pipe is . The fiber has an inner core with index ni and an outer cladding
with index no. If no/ni is not large enough there will be no internal reflection. In order for
internal reflection to possibly occur, the ratio no/ni must be
no

A) positive
D) less than one
ni
B) negative C) greater than one
E) equal to one
Answer: less than one
CTF-23. Is the image formed by the lens in your eye a real imagage or a virtual image?
A) real
B) virtual
There is a real image formed on your retina by your eye lens.
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