Electricity and Magnetism Full-Length Practice Paper 1 Answers | AP Physics C Electricity and Magnetism - Grade 9 PDF Download

 Page 1


SECTION I, ELECTRICITY AND MAGNETISM
36. C
The proximity of the charged sphere will induce negative charge to move to the side of the
uncharged sphere closer to the charged sphere. Since the induced negative charge is
closer than the induced positive charge to the charged sphere, there will be a net
electrostatic attraction between the spheres.
37. E
Typically, the formula for power dissipated by a circuit is expressed as P = IV. However, it
can also be written as P = I
2
R or P = V
2
/ R, which would appear to give contradictory
answers to this question. The correct answer is that it cannot be determined because
there’s no way to double the resistance of a circuit without affecting anything else. Ohm’s
Law tells us that V = IR, so we would have to know whether the voltage or current is
changing as a result of the doubled resistance before we can address the change to the
power dissipated.
38. E
When the particle enters the magnetic field, the magnetic force provides the centripetal
force to cause the particle to execute uniform circular motion:
Since v and B are the same for all the particles, the largest r is found by maximizing the
ratio . Furthermore, since the ratio of u/ e (atomic mass unit/magnitude of
elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to
“factor of charge.” The ratio 20/1 is largest, so it will have the largest r.
39. A
The only place with 0 net electric field would be in the center of the rectangle. Moving up
along Line A would give a net electric field pointing to the right. This is because the top
two charges, which would both force a positive particle to the right (in addition to vertical
effects), are closer and therefore more impactful than the bottom two charges, which
would oppose them by trying to make a positive particle move to the left. Similarly,
moving down along Line A would give a net electric field to the left. Moving to the right
along Line B would give a net electric field upward. Moving to the left along Line B would
give a net electric field downward.
Page 2


SECTION I, ELECTRICITY AND MAGNETISM
36. C
The proximity of the charged sphere will induce negative charge to move to the side of the
uncharged sphere closer to the charged sphere. Since the induced negative charge is
closer than the induced positive charge to the charged sphere, there will be a net
electrostatic attraction between the spheres.
37. E
Typically, the formula for power dissipated by a circuit is expressed as P = IV. However, it
can also be written as P = I
2
R or P = V
2
/ R, which would appear to give contradictory
answers to this question. The correct answer is that it cannot be determined because
there’s no way to double the resistance of a circuit without affecting anything else. Ohm’s
Law tells us that V = IR, so we would have to know whether the voltage or current is
changing as a result of the doubled resistance before we can address the change to the
power dissipated.
38. E
When the particle enters the magnetic field, the magnetic force provides the centripetal
force to cause the particle to execute uniform circular motion:
Since v and B are the same for all the particles, the largest r is found by maximizing the
ratio . Furthermore, since the ratio of u/ e (atomic mass unit/magnitude of
elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to
“factor of charge.” The ratio 20/1 is largest, so it will have the largest r.
39. A
The only place with 0 net electric field would be in the center of the rectangle. Moving up
along Line A would give a net electric field pointing to the right. This is because the top
two charges, which would both force a positive particle to the right (in addition to vertical
effects), are closer and therefore more impactful than the bottom two charges, which
would oppose them by trying to make a positive particle move to the left. Similarly,
moving down along Line A would give a net electric field to the left. Moving to the right
along Line B would give a net electric field upward. Moving to the left along Line B would
give a net electric field downward.
40. B
The resistance of a wire made of a material with resistivity ? and with length L and cross-
sectional area A is given by the equation R = ? L/ A. Since Wire B has the greatest length
and smallest cross-sectional area, it has the greatest resistance.
41. E
From the equation F = qvB, we see that 1 tesla is equal to 1 N·s/(C·m). Since
(A) and (C) are eliminated.
Furthermore,
eliminating (B). Finally, since
(D) is also eliminated.
42. C
Gauss’s law states that the total electric flux through a closed surface is equal to (1/e
0
)
times the net charge enclosed by the surface. Both the cube and the sphere contain the
same net charge, so F
C
 must be equal to F
S
.
43. E
The particle would experience oscillations between the positions x = 30 cm and x = 70
cm. At the position x = 50 cm, it would have no net force acting on it, but this would be
similar to a spring having no net force as it passes through the equilibrium position. It
will continue moving until an opposing force reduces its speed to 0, which would occur
at x = 30 cm. It would then be pushed to right until it reached x = 70 cm, and the cycle
would repeat infinitely.
44. A
The Table of Information provides the following conversion factor: 1 e = 1.6 × 10
-19
 C.
Multiplying the magnitude of charge, 1 C, by this conversion factor gives the following:
There are 6.25 × 10
18
 electrons in 1 coulomb of negative charge.
45. E
Resistors R
2
, R
3
, and R
4
 are in parallel, so their equivalent resistance, R
2-3-4
, satisfies
Page 3


SECTION I, ELECTRICITY AND MAGNETISM
36. C
The proximity of the charged sphere will induce negative charge to move to the side of the
uncharged sphere closer to the charged sphere. Since the induced negative charge is
closer than the induced positive charge to the charged sphere, there will be a net
electrostatic attraction between the spheres.
37. E
Typically, the formula for power dissipated by a circuit is expressed as P = IV. However, it
can also be written as P = I
2
R or P = V
2
/ R, which would appear to give contradictory
answers to this question. The correct answer is that it cannot be determined because
there’s no way to double the resistance of a circuit without affecting anything else. Ohm’s
Law tells us that V = IR, so we would have to know whether the voltage or current is
changing as a result of the doubled resistance before we can address the change to the
power dissipated.
38. E
When the particle enters the magnetic field, the magnetic force provides the centripetal
force to cause the particle to execute uniform circular motion:
Since v and B are the same for all the particles, the largest r is found by maximizing the
ratio . Furthermore, since the ratio of u/ e (atomic mass unit/magnitude of
elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to
“factor of charge.” The ratio 20/1 is largest, so it will have the largest r.
39. A
The only place with 0 net electric field would be in the center of the rectangle. Moving up
along Line A would give a net electric field pointing to the right. This is because the top
two charges, which would both force a positive particle to the right (in addition to vertical
effects), are closer and therefore more impactful than the bottom two charges, which
would oppose them by trying to make a positive particle move to the left. Similarly,
moving down along Line A would give a net electric field to the left. Moving to the right
along Line B would give a net electric field upward. Moving to the left along Line B would
give a net electric field downward.
40. B
The resistance of a wire made of a material with resistivity ? and with length L and cross-
sectional area A is given by the equation R = ? L/ A. Since Wire B has the greatest length
and smallest cross-sectional area, it has the greatest resistance.
41. E
From the equation F = qvB, we see that 1 tesla is equal to 1 N·s/(C·m). Since
(A) and (C) are eliminated.
Furthermore,
eliminating (B). Finally, since
(D) is also eliminated.
42. C
Gauss’s law states that the total electric flux through a closed surface is equal to (1/e
0
)
times the net charge enclosed by the surface. Both the cube and the sphere contain the
same net charge, so F
C
 must be equal to F
S
.
43. E
The particle would experience oscillations between the positions x = 30 cm and x = 70
cm. At the position x = 50 cm, it would have no net force acting on it, but this would be
similar to a spring having no net force as it passes through the equilibrium position. It
will continue moving until an opposing force reduces its speed to 0, which would occur
at x = 30 cm. It would then be pushed to right until it reached x = 70 cm, and the cycle
would repeat infinitely.
44. A
The Table of Information provides the following conversion factor: 1 e = 1.6 × 10
-19
 C.
Multiplying the magnitude of charge, 1 C, by this conversion factor gives the following:
There are 6.25 × 10
18
 electrons in 1 coulomb of negative charge.
45. E
Resistors R
2
, R
3
, and R
4
 are in parallel, so their equivalent resistance, R
2-3-4
, satisfies
Since this is in series with R
1
, the total resistance in the circuit is
The current provided by the source must therefore be
Drawing a series of equivalent circuits can make it easier to see what the resistance
should be at each step.
46. E
The current through R
4
 is  the current through R
1
, and R
4
 = R
1
, so
47. D
Note that all answers have the same number, so for this problem, don’t even worry about
multiplying the numbers—just focus on the units. Because the time constant for an RC
circuit is equal to the product of resistance and capacitance, t = RC, this product has the
dimensions of time. If you don’t know that off the top of your head, just be sure to
carefully break the farad and ohm into their subunits, simplifying where possible:
1 F × 1 O = 1 C/V × 1 V/A = 1 C/A = 1 s
48. B
Apply Faraday’s Law of Electromagnetic Induction:
Page 4


SECTION I, ELECTRICITY AND MAGNETISM
36. C
The proximity of the charged sphere will induce negative charge to move to the side of the
uncharged sphere closer to the charged sphere. Since the induced negative charge is
closer than the induced positive charge to the charged sphere, there will be a net
electrostatic attraction between the spheres.
37. E
Typically, the formula for power dissipated by a circuit is expressed as P = IV. However, it
can also be written as P = I
2
R or P = V
2
/ R, which would appear to give contradictory
answers to this question. The correct answer is that it cannot be determined because
there’s no way to double the resistance of a circuit without affecting anything else. Ohm’s
Law tells us that V = IR, so we would have to know whether the voltage or current is
changing as a result of the doubled resistance before we can address the change to the
power dissipated.
38. E
When the particle enters the magnetic field, the magnetic force provides the centripetal
force to cause the particle to execute uniform circular motion:
Since v and B are the same for all the particles, the largest r is found by maximizing the
ratio . Furthermore, since the ratio of u/ e (atomic mass unit/magnitude of
elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to
“factor of charge.” The ratio 20/1 is largest, so it will have the largest r.
39. A
The only place with 0 net electric field would be in the center of the rectangle. Moving up
along Line A would give a net electric field pointing to the right. This is because the top
two charges, which would both force a positive particle to the right (in addition to vertical
effects), are closer and therefore more impactful than the bottom two charges, which
would oppose them by trying to make a positive particle move to the left. Similarly,
moving down along Line A would give a net electric field to the left. Moving to the right
along Line B would give a net electric field upward. Moving to the left along Line B would
give a net electric field downward.
40. B
The resistance of a wire made of a material with resistivity ? and with length L and cross-
sectional area A is given by the equation R = ? L/ A. Since Wire B has the greatest length
and smallest cross-sectional area, it has the greatest resistance.
41. E
From the equation F = qvB, we see that 1 tesla is equal to 1 N·s/(C·m). Since
(A) and (C) are eliminated.
Furthermore,
eliminating (B). Finally, since
(D) is also eliminated.
42. C
Gauss’s law states that the total electric flux through a closed surface is equal to (1/e
0
)
times the net charge enclosed by the surface. Both the cube and the sphere contain the
same net charge, so F
C
 must be equal to F
S
.
43. E
The particle would experience oscillations between the positions x = 30 cm and x = 70
cm. At the position x = 50 cm, it would have no net force acting on it, but this would be
similar to a spring having no net force as it passes through the equilibrium position. It
will continue moving until an opposing force reduces its speed to 0, which would occur
at x = 30 cm. It would then be pushed to right until it reached x = 70 cm, and the cycle
would repeat infinitely.
44. A
The Table of Information provides the following conversion factor: 1 e = 1.6 × 10
-19
 C.
Multiplying the magnitude of charge, 1 C, by this conversion factor gives the following:
There are 6.25 × 10
18
 electrons in 1 coulomb of negative charge.
45. E
Resistors R
2
, R
3
, and R
4
 are in parallel, so their equivalent resistance, R
2-3-4
, satisfies
Since this is in series with R
1
, the total resistance in the circuit is
The current provided by the source must therefore be
Drawing a series of equivalent circuits can make it easier to see what the resistance
should be at each step.
46. E
The current through R
4
 is  the current through R
1
, and R
4
 = R
1
, so
47. D
Note that all answers have the same number, so for this problem, don’t even worry about
multiplying the numbers—just focus on the units. Because the time constant for an RC
circuit is equal to the product of resistance and capacitance, t = RC, this product has the
dimensions of time. If you don’t know that off the top of your head, just be sure to
carefully break the farad and ohm into their subunits, simplifying where possible:
1 F × 1 O = 1 C/V × 1 V/A = 1 C/A = 1 s
48. B
Apply Faraday’s Law of Electromagnetic Induction:
Since r = 1 m, the value of E at t = 1 s is E = 1 N/C.
49. A
The magnetic force, F
B
, on the top horizontal wire of the loop must be directed upward
and have magnitude mg. (The magnetic forces on the vertical portions of the wire will
cancel.) By the right-hand rule, the current in the top horizontal wire must be directed to
the right—because B is directed into the plane of the page—in order for F
B
 to be directed
upward; therefore, the current in the loop must be clockwise, eliminating (B) and (D).
Since F
B
 = IxB, we must have
50. C
First, find the difference in potential between the starting location (2 cm away) and the
location in question (1 cm away):
Next, find the change in potential energy.
? PE = q?F = (-5 × 10
-9
)(4500) = -2.25 × 10
-5
 J
Because total energy will be conserved, the change in kinetic energy must be 2.25 x 10
-5
J.
This can be used to find the final speed.
51. A
Call the top wire (the one carrying a current I to the right) Wire 1, and call the bottom
wire (carrying a current 2 I to the left) Wire 2. Then in the region between the wires, the
individual magnetic field vectors due to the wires are both directed into the plane of the
page, so they could not cancel in this region. Therefore, the total magnetic field could not
be zero at either Point 2 or Point 3. This eliminates (B), (C), (D), and (E), so the answer
must be (A). Since the magnetic field created by a current-carrying wire is proportional to
the current and inversely proportional to the distance from the wire, the fact that Point 1
is in a region where the individual magnetic field vectors created by the wires point in
opposite directions and that Point 1 is twice as far from Wire 2 as from Wire 1 implies that
the total magnetic field there will be zero.
52. A
Page 5


SECTION I, ELECTRICITY AND MAGNETISM
36. C
The proximity of the charged sphere will induce negative charge to move to the side of the
uncharged sphere closer to the charged sphere. Since the induced negative charge is
closer than the induced positive charge to the charged sphere, there will be a net
electrostatic attraction between the spheres.
37. E
Typically, the formula for power dissipated by a circuit is expressed as P = IV. However, it
can also be written as P = I
2
R or P = V
2
/ R, which would appear to give contradictory
answers to this question. The correct answer is that it cannot be determined because
there’s no way to double the resistance of a circuit without affecting anything else. Ohm’s
Law tells us that V = IR, so we would have to know whether the voltage or current is
changing as a result of the doubled resistance before we can address the change to the
power dissipated.
38. E
When the particle enters the magnetic field, the magnetic force provides the centripetal
force to cause the particle to execute uniform circular motion:
Since v and B are the same for all the particles, the largest r is found by maximizing the
ratio . Furthermore, since the ratio of u/ e (atomic mass unit/magnitude of
elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to
“factor of charge.” The ratio 20/1 is largest, so it will have the largest r.
39. A
The only place with 0 net electric field would be in the center of the rectangle. Moving up
along Line A would give a net electric field pointing to the right. This is because the top
two charges, which would both force a positive particle to the right (in addition to vertical
effects), are closer and therefore more impactful than the bottom two charges, which
would oppose them by trying to make a positive particle move to the left. Similarly,
moving down along Line A would give a net electric field to the left. Moving to the right
along Line B would give a net electric field upward. Moving to the left along Line B would
give a net electric field downward.
40. B
The resistance of a wire made of a material with resistivity ? and with length L and cross-
sectional area A is given by the equation R = ? L/ A. Since Wire B has the greatest length
and smallest cross-sectional area, it has the greatest resistance.
41. E
From the equation F = qvB, we see that 1 tesla is equal to 1 N·s/(C·m). Since
(A) and (C) are eliminated.
Furthermore,
eliminating (B). Finally, since
(D) is also eliminated.
42. C
Gauss’s law states that the total electric flux through a closed surface is equal to (1/e
0
)
times the net charge enclosed by the surface. Both the cube and the sphere contain the
same net charge, so F
C
 must be equal to F
S
.
43. E
The particle would experience oscillations between the positions x = 30 cm and x = 70
cm. At the position x = 50 cm, it would have no net force acting on it, but this would be
similar to a spring having no net force as it passes through the equilibrium position. It
will continue moving until an opposing force reduces its speed to 0, which would occur
at x = 30 cm. It would then be pushed to right until it reached x = 70 cm, and the cycle
would repeat infinitely.
44. A
The Table of Information provides the following conversion factor: 1 e = 1.6 × 10
-19
 C.
Multiplying the magnitude of charge, 1 C, by this conversion factor gives the following:
There are 6.25 × 10
18
 electrons in 1 coulomb of negative charge.
45. E
Resistors R
2
, R
3
, and R
4
 are in parallel, so their equivalent resistance, R
2-3-4
, satisfies
Since this is in series with R
1
, the total resistance in the circuit is
The current provided by the source must therefore be
Drawing a series of equivalent circuits can make it easier to see what the resistance
should be at each step.
46. E
The current through R
4
 is  the current through R
1
, and R
4
 = R
1
, so
47. D
Note that all answers have the same number, so for this problem, don’t even worry about
multiplying the numbers—just focus on the units. Because the time constant for an RC
circuit is equal to the product of resistance and capacitance, t = RC, this product has the
dimensions of time. If you don’t know that off the top of your head, just be sure to
carefully break the farad and ohm into their subunits, simplifying where possible:
1 F × 1 O = 1 C/V × 1 V/A = 1 C/A = 1 s
48. B
Apply Faraday’s Law of Electromagnetic Induction:
Since r = 1 m, the value of E at t = 1 s is E = 1 N/C.
49. A
The magnetic force, F
B
, on the top horizontal wire of the loop must be directed upward
and have magnitude mg. (The magnetic forces on the vertical portions of the wire will
cancel.) By the right-hand rule, the current in the top horizontal wire must be directed to
the right—because B is directed into the plane of the page—in order for F
B
 to be directed
upward; therefore, the current in the loop must be clockwise, eliminating (B) and (D).
Since F
B
 = IxB, we must have
50. C
First, find the difference in potential between the starting location (2 cm away) and the
location in question (1 cm away):
Next, find the change in potential energy.
? PE = q?F = (-5 × 10
-9
)(4500) = -2.25 × 10
-5
 J
Because total energy will be conserved, the change in kinetic energy must be 2.25 x 10
-5
J.
This can be used to find the final speed.
51. A
Call the top wire (the one carrying a current I to the right) Wire 1, and call the bottom
wire (carrying a current 2 I to the left) Wire 2. Then in the region between the wires, the
individual magnetic field vectors due to the wires are both directed into the plane of the
page, so they could not cancel in this region. Therefore, the total magnetic field could not
be zero at either Point 2 or Point 3. This eliminates (B), (C), (D), and (E), so the answer
must be (A). Since the magnetic field created by a current-carrying wire is proportional to
the current and inversely proportional to the distance from the wire, the fact that Point 1
is in a region where the individual magnetic field vectors created by the wires point in
opposite directions and that Point 1 is twice as far from Wire 2 as from Wire 1 implies that
the total magnetic field there will be zero.
52. A
If the mass of the + q charge is m, then its acceleration is
Because a should decrease as r increases, you can eliminate (C) and (E). The graph in (A)
best depicts an inverse-square relationship between a and r.
53. C
Once the switch is closed, the capacitors are in parallel, which means the voltage
across C
1
 must equal the voltage across C
2
. Since V = Q/ C,
This makes sense qualitatively: C
2
 has twice the capacitance of C
1
, so it should hold twice
the charge. The total charge, 24 µC + 12 µC = 36 µC, must be redistributed so that 
. Therefore, we see that  = 24 µC (and  = 12 µC).
54. E
Because v is parallel to B, the charges in the bar feel no magnetic force, so there will be no
movement of charge in the bar and no motional emf.
55. E
Along the line joining the two charges of an electric dipole, the individual electric field
vectors point in the same direction, so they add constructively. At the point equidistant
from both charges, the total electric field vector has magnitude
56. E
Start with basic dimensional analysis. Choices (A) and (B) can be eliminated because
those expressions do not yield units of capacitance. Imagine placing equal but opposite
charges on the spheres, say, + Q on the inner sphere and - Q on the outer sphere. Then the
electric field between the spheres is radial and depends only on + Q, and its strength is
Therefore, the potential difference between the spheres is