Electricity and Magnetism Full-Length Practice Paper 2 Answers | AP Physics C Electricity and Magnetism - Grade 9 PDF Download

 Page 1


SECTION I, ELECTRICITY AND MAGNETISM
36. E
The equivalent capacitance of capacitors connected in parallel is the sum of each
capacitor. Therefore: C
eq
 = 3 + 3 + 3 = 9 µF.
37. B
Using the equation P = IV, we see current and voltage are inversely proportional if power
is held constant. Since the voltage in the question doubled, the current must be halved.
38. E
The flux is proportional to the number of field lines that pass through the surface of the
sphere. Since the background electric field E is uniform, every background field line that
enters the sphere also exits the sphere. Therefore, it has no contribution to the flux. The
field lines from the +2 Q charge will radiate outward and every field line that exists due to
the +2 Q will pass through the sphere, regardless of R. Therefore, the flux only depends on
the value of Q. This is in accordance with Gauss’s law.
39. D
The potential difference can be calculated using the equation that relates the potential
difference to the electric field, . The limits of the integration will be from 0
to 5, because the radius is 5 for the point (3, 4).
40. C
Determine the direction of the force on the charge using the right-hand rule. Consider the
leftmost point of the circle. The force must be toward the center of the circle, to the right,
and the magnetic field is out of the page, so the velocity must be up. This indicates the
particle is moving clockwise. However, the right-hand rule gives the direction for a
positive charge. Since the actual charge moving is negative, the direction should be
reversed. Thus, the negative charge is moving counterclockwise. The force on the charge
is F = qvB, and use Newton’s Second Law to solve for the acceleration as shown below.
Choice (C) has these two pieces of information correct.
41. D
Page 2


SECTION I, ELECTRICITY AND MAGNETISM
36. E
The equivalent capacitance of capacitors connected in parallel is the sum of each
capacitor. Therefore: C
eq
 = 3 + 3 + 3 = 9 µF.
37. B
Using the equation P = IV, we see current and voltage are inversely proportional if power
is held constant. Since the voltage in the question doubled, the current must be halved.
38. E
The flux is proportional to the number of field lines that pass through the surface of the
sphere. Since the background electric field E is uniform, every background field line that
enters the sphere also exits the sphere. Therefore, it has no contribution to the flux. The
field lines from the +2 Q charge will radiate outward and every field line that exists due to
the +2 Q will pass through the sphere, regardless of R. Therefore, the flux only depends on
the value of Q. This is in accordance with Gauss’s law.
39. D
The potential difference can be calculated using the equation that relates the potential
difference to the electric field, . The limits of the integration will be from 0
to 5, because the radius is 5 for the point (3, 4).
40. C
Determine the direction of the force on the charge using the right-hand rule. Consider the
leftmost point of the circle. The force must be toward the center of the circle, to the right,
and the magnetic field is out of the page, so the velocity must be up. This indicates the
particle is moving clockwise. However, the right-hand rule gives the direction for a
positive charge. Since the actual charge moving is negative, the direction should be
reversed. Thus, the negative charge is moving counterclockwise. The force on the charge
is F = qvB, and use Newton’s Second Law to solve for the acceleration as shown below.
Choice (C) has these two pieces of information correct.
41. D
For a particle traveling in uniform circular motion, 2p r = vT because the particle travels
one revolution in one period. The period is the inverse of the frequency. Applying this
information and solving for f is shown below.
42. C
The energy stored in a capacitor is given by the equations . Select
the last part of the equation because the charge, in millicoulombs, and the capacitance, in
microfarads, are given. The solution is shown below
43. D
The electric field is uniform in between the plates, so the force and the acceleration are
constant. Relating the potential difference to the electric field gives us ? V = - Ex,
where x is the separation of the plates. The definition of the electric field is the force
divided by the charge. Use these two pieces of information and Newton’s Second Law to
solve for the acceleration in terms of V.
This shows the acceleration is directly proportional to the potential difference V.
44. E
Ampere’s law is . For an ideal solenoid the magnetic field is uniform and
directed along the central axis within the solenoid and zero outside. In this case, only
segment bc contributes to the integral because ad is outside the solenoid where the field
is zero and ab and cd are perpendicular to the field, so the dot product of those segments
with B is zero. The number of times the current will pass through our path is equal to the
number of coils in the solenoid. This gives us the equation .
45. E
The electric field will radiate outward from both +3 Q and + Q, and the field due to the
+3 Q charge will be greater because point P is the same distance for both charges. This
Page 3


SECTION I, ELECTRICITY AND MAGNETISM
36. E
The equivalent capacitance of capacitors connected in parallel is the sum of each
capacitor. Therefore: C
eq
 = 3 + 3 + 3 = 9 µF.
37. B
Using the equation P = IV, we see current and voltage are inversely proportional if power
is held constant. Since the voltage in the question doubled, the current must be halved.
38. E
The flux is proportional to the number of field lines that pass through the surface of the
sphere. Since the background electric field E is uniform, every background field line that
enters the sphere also exits the sphere. Therefore, it has no contribution to the flux. The
field lines from the +2 Q charge will radiate outward and every field line that exists due to
the +2 Q will pass through the sphere, regardless of R. Therefore, the flux only depends on
the value of Q. This is in accordance with Gauss’s law.
39. D
The potential difference can be calculated using the equation that relates the potential
difference to the electric field, . The limits of the integration will be from 0
to 5, because the radius is 5 for the point (3, 4).
40. C
Determine the direction of the force on the charge using the right-hand rule. Consider the
leftmost point of the circle. The force must be toward the center of the circle, to the right,
and the magnetic field is out of the page, so the velocity must be up. This indicates the
particle is moving clockwise. However, the right-hand rule gives the direction for a
positive charge. Since the actual charge moving is negative, the direction should be
reversed. Thus, the negative charge is moving counterclockwise. The force on the charge
is F = qvB, and use Newton’s Second Law to solve for the acceleration as shown below.
Choice (C) has these two pieces of information correct.
41. D
For a particle traveling in uniform circular motion, 2p r = vT because the particle travels
one revolution in one period. The period is the inverse of the frequency. Applying this
information and solving for f is shown below.
42. C
The energy stored in a capacitor is given by the equations . Select
the last part of the equation because the charge, in millicoulombs, and the capacitance, in
microfarads, are given. The solution is shown below
43. D
The electric field is uniform in between the plates, so the force and the acceleration are
constant. Relating the potential difference to the electric field gives us ? V = - Ex,
where x is the separation of the plates. The definition of the electric field is the force
divided by the charge. Use these two pieces of information and Newton’s Second Law to
solve for the acceleration in terms of V.
This shows the acceleration is directly proportional to the potential difference V.
44. E
Ampere’s law is . For an ideal solenoid the magnetic field is uniform and
directed along the central axis within the solenoid and zero outside. In this case, only
segment bc contributes to the integral because ad is outside the solenoid where the field
is zero and ab and cd are perpendicular to the field, so the dot product of those segments
with B is zero. The number of times the current will pass through our path is equal to the
number of coils in the solenoid. This gives us the equation .
45. E
The electric field will radiate outward from both +3 Q and + Q, and the field due to the
+3 Q charge will be greater because point P is the same distance for both charges. This
vector diagram is shown below. The resultant will be in the + x and + y directions.
46. E
The electric potential can be calculated using the equation . Because there are
only positive charges, the electric potential due to each charge will be positive, so the sum
cannot be zero.
47. B
When the switch S is opened, the equivalent resistance, R
eq
, of the parallel circuit
increases. This results in the increase of the R
eq
 of the entire circuit. When the R
eq
 of the
entire circuit increases, the total current, I, must decrease. This is because the total
potential difference, V, is constant. The decrease in total I results in a decrease in
the Vacross the 54 O resistor. Since the total V has not changed, while the V across the 54
O has decreased, the V across the 10 O resistor must increase. This results in an increase
in the current through the 10 O resistor.
The current is not doubled, though, because that would require the V across the 10 O
resistor to have doubled. This would only have occurred if the V across the 54 O resistor
had halved. That would have required the circuit’s current to have halved as well. This
would have occurred only if the 10 O had a much greater resistance.
48. A
The equation for the resistance of a wire is , and A = p r
2
. If the length is twice as
long, the wire will have twice the resistance and if the radius is half, the wire will have 4
times the resistance per unit length because the radius is squared in the equation.
Conceptually this makes sense also because the charge is trying to flow through a cross-
sectional area one-fourth as large. These two factors mean the total resistance is 8 times
the resistance of wire Y.
49. C
For the region 0 < r < R the electric potential is constant. For the potential to be constant,
the electric field must be zero because .
50. E
Page 4


SECTION I, ELECTRICITY AND MAGNETISM
36. E
The equivalent capacitance of capacitors connected in parallel is the sum of each
capacitor. Therefore: C
eq
 = 3 + 3 + 3 = 9 µF.
37. B
Using the equation P = IV, we see current and voltage are inversely proportional if power
is held constant. Since the voltage in the question doubled, the current must be halved.
38. E
The flux is proportional to the number of field lines that pass through the surface of the
sphere. Since the background electric field E is uniform, every background field line that
enters the sphere also exits the sphere. Therefore, it has no contribution to the flux. The
field lines from the +2 Q charge will radiate outward and every field line that exists due to
the +2 Q will pass through the sphere, regardless of R. Therefore, the flux only depends on
the value of Q. This is in accordance with Gauss’s law.
39. D
The potential difference can be calculated using the equation that relates the potential
difference to the electric field, . The limits of the integration will be from 0
to 5, because the radius is 5 for the point (3, 4).
40. C
Determine the direction of the force on the charge using the right-hand rule. Consider the
leftmost point of the circle. The force must be toward the center of the circle, to the right,
and the magnetic field is out of the page, so the velocity must be up. This indicates the
particle is moving clockwise. However, the right-hand rule gives the direction for a
positive charge. Since the actual charge moving is negative, the direction should be
reversed. Thus, the negative charge is moving counterclockwise. The force on the charge
is F = qvB, and use Newton’s Second Law to solve for the acceleration as shown below.
Choice (C) has these two pieces of information correct.
41. D
For a particle traveling in uniform circular motion, 2p r = vT because the particle travels
one revolution in one period. The period is the inverse of the frequency. Applying this
information and solving for f is shown below.
42. C
The energy stored in a capacitor is given by the equations . Select
the last part of the equation because the charge, in millicoulombs, and the capacitance, in
microfarads, are given. The solution is shown below
43. D
The electric field is uniform in between the plates, so the force and the acceleration are
constant. Relating the potential difference to the electric field gives us ? V = - Ex,
where x is the separation of the plates. The definition of the electric field is the force
divided by the charge. Use these two pieces of information and Newton’s Second Law to
solve for the acceleration in terms of V.
This shows the acceleration is directly proportional to the potential difference V.
44. E
Ampere’s law is . For an ideal solenoid the magnetic field is uniform and
directed along the central axis within the solenoid and zero outside. In this case, only
segment bc contributes to the integral because ad is outside the solenoid where the field
is zero and ab and cd are perpendicular to the field, so the dot product of those segments
with B is zero. The number of times the current will pass through our path is equal to the
number of coils in the solenoid. This gives us the equation .
45. E
The electric field will radiate outward from both +3 Q and + Q, and the field due to the
+3 Q charge will be greater because point P is the same distance for both charges. This
vector diagram is shown below. The resultant will be in the + x and + y directions.
46. E
The electric potential can be calculated using the equation . Because there are
only positive charges, the electric potential due to each charge will be positive, so the sum
cannot be zero.
47. B
When the switch S is opened, the equivalent resistance, R
eq
, of the parallel circuit
increases. This results in the increase of the R
eq
 of the entire circuit. When the R
eq
 of the
entire circuit increases, the total current, I, must decrease. This is because the total
potential difference, V, is constant. The decrease in total I results in a decrease in
the Vacross the 54 O resistor. Since the total V has not changed, while the V across the 54
O has decreased, the V across the 10 O resistor must increase. This results in an increase
in the current through the 10 O resistor.
The current is not doubled, though, because that would require the V across the 10 O
resistor to have doubled. This would only have occurred if the V across the 54 O resistor
had halved. That would have required the circuit’s current to have halved as well. This
would have occurred only if the 10 O had a much greater resistance.
48. A
The equation for the resistance of a wire is , and A = p r
2
. If the length is twice as
long, the wire will have twice the resistance and if the radius is half, the wire will have 4
times the resistance per unit length because the radius is squared in the equation.
Conceptually this makes sense also because the charge is trying to flow through a cross-
sectional area one-fourth as large. These two factors mean the total resistance is 8 times
the resistance of wire Y.
49. C
For the region 0 < r < R the electric potential is constant. For the potential to be constant,
the electric field must be zero because .
50. E
The force between two parallel wires is given by the equation , where x is the
distance between the wires and l is the length of the wires. Therefore, if the currents are
halved the force will be one-fourth the original value.
51. A
The area of the square is given by A = l
2
 = k
2
t
4
. Use Faraday’s law to determine the
induced emf as shown below. Since the question asks for the magnitude of the emf, it is
not necessary to include the negative sign in your answer.
52. D
The potential difference around the loop must be zero according to Kirchhoff’s loop rule.
The current in the circuit is going counterclockwise if the battery is being charged by the
120 V outlet. A counterclockwise loop starting right below the 120 V outlet will produce
the following, 120 - e - (3)(30) = 0. Therefore, e = 30 V.
53. C
Use the right hand rule on each half of the dipole. For each answer choice, the resulting
magnetic forces would be
From these figures, we can see that all three would have 0 net force, but (C) is the only
one that will rotate due to a net torque.
54. C
The magnetic field through the loop decreases when the magnet is pulled away. To
oppose the change in the flux through the loop, the loop will induce its own current that
creates a magnetic field to the right. This will cause the loop to be attracted to the magnet
Page 5


SECTION I, ELECTRICITY AND MAGNETISM
36. E
The equivalent capacitance of capacitors connected in parallel is the sum of each
capacitor. Therefore: C
eq
 = 3 + 3 + 3 = 9 µF.
37. B
Using the equation P = IV, we see current and voltage are inversely proportional if power
is held constant. Since the voltage in the question doubled, the current must be halved.
38. E
The flux is proportional to the number of field lines that pass through the surface of the
sphere. Since the background electric field E is uniform, every background field line that
enters the sphere also exits the sphere. Therefore, it has no contribution to the flux. The
field lines from the +2 Q charge will radiate outward and every field line that exists due to
the +2 Q will pass through the sphere, regardless of R. Therefore, the flux only depends on
the value of Q. This is in accordance with Gauss’s law.
39. D
The potential difference can be calculated using the equation that relates the potential
difference to the electric field, . The limits of the integration will be from 0
to 5, because the radius is 5 for the point (3, 4).
40. C
Determine the direction of the force on the charge using the right-hand rule. Consider the
leftmost point of the circle. The force must be toward the center of the circle, to the right,
and the magnetic field is out of the page, so the velocity must be up. This indicates the
particle is moving clockwise. However, the right-hand rule gives the direction for a
positive charge. Since the actual charge moving is negative, the direction should be
reversed. Thus, the negative charge is moving counterclockwise. The force on the charge
is F = qvB, and use Newton’s Second Law to solve for the acceleration as shown below.
Choice (C) has these two pieces of information correct.
41. D
For a particle traveling in uniform circular motion, 2p r = vT because the particle travels
one revolution in one period. The period is the inverse of the frequency. Applying this
information and solving for f is shown below.
42. C
The energy stored in a capacitor is given by the equations . Select
the last part of the equation because the charge, in millicoulombs, and the capacitance, in
microfarads, are given. The solution is shown below
43. D
The electric field is uniform in between the plates, so the force and the acceleration are
constant. Relating the potential difference to the electric field gives us ? V = - Ex,
where x is the separation of the plates. The definition of the electric field is the force
divided by the charge. Use these two pieces of information and Newton’s Second Law to
solve for the acceleration in terms of V.
This shows the acceleration is directly proportional to the potential difference V.
44. E
Ampere’s law is . For an ideal solenoid the magnetic field is uniform and
directed along the central axis within the solenoid and zero outside. In this case, only
segment bc contributes to the integral because ad is outside the solenoid where the field
is zero and ab and cd are perpendicular to the field, so the dot product of those segments
with B is zero. The number of times the current will pass through our path is equal to the
number of coils in the solenoid. This gives us the equation .
45. E
The electric field will radiate outward from both +3 Q and + Q, and the field due to the
+3 Q charge will be greater because point P is the same distance for both charges. This
vector diagram is shown below. The resultant will be in the + x and + y directions.
46. E
The electric potential can be calculated using the equation . Because there are
only positive charges, the electric potential due to each charge will be positive, so the sum
cannot be zero.
47. B
When the switch S is opened, the equivalent resistance, R
eq
, of the parallel circuit
increases. This results in the increase of the R
eq
 of the entire circuit. When the R
eq
 of the
entire circuit increases, the total current, I, must decrease. This is because the total
potential difference, V, is constant. The decrease in total I results in a decrease in
the Vacross the 54 O resistor. Since the total V has not changed, while the V across the 54
O has decreased, the V across the 10 O resistor must increase. This results in an increase
in the current through the 10 O resistor.
The current is not doubled, though, because that would require the V across the 10 O
resistor to have doubled. This would only have occurred if the V across the 54 O resistor
had halved. That would have required the circuit’s current to have halved as well. This
would have occurred only if the 10 O had a much greater resistance.
48. A
The equation for the resistance of a wire is , and A = p r
2
. If the length is twice as
long, the wire will have twice the resistance and if the radius is half, the wire will have 4
times the resistance per unit length because the radius is squared in the equation.
Conceptually this makes sense also because the charge is trying to flow through a cross-
sectional area one-fourth as large. These two factors mean the total resistance is 8 times
the resistance of wire Y.
49. C
For the region 0 < r < R the electric potential is constant. For the potential to be constant,
the electric field must be zero because .
50. E
The force between two parallel wires is given by the equation , where x is the
distance between the wires and l is the length of the wires. Therefore, if the currents are
halved the force will be one-fourth the original value.
51. A
The area of the square is given by A = l
2
 = k
2
t
4
. Use Faraday’s law to determine the
induced emf as shown below. Since the question asks for the magnitude of the emf, it is
not necessary to include the negative sign in your answer.
52. D
The potential difference around the loop must be zero according to Kirchhoff’s loop rule.
The current in the circuit is going counterclockwise if the battery is being charged by the
120 V outlet. A counterclockwise loop starting right below the 120 V outlet will produce
the following, 120 - e - (3)(30) = 0. Therefore, e = 30 V.
53. C
Use the right hand rule on each half of the dipole. For each answer choice, the resulting
magnetic forces would be
From these figures, we can see that all three would have 0 net force, but (C) is the only
one that will rotate due to a net torque.
54. C
The magnetic field through the loop decreases when the magnet is pulled away. To
oppose the change in the flux through the loop, the loop will induce its own current that
creates a magnetic field to the right. This will cause the loop to be attracted to the magnet
with a force to the left (again to oppose the change in flux). Choice (C) combines these two
pieces of information correctly.
55. A
Use the right-hand rule to determine the force on each part of the loop. A diagram is
shown below with the forces indicated. These forces would produce a torque that causes
the loop to rotate about the indicated axis. This would cause the loop to rotate as shown
in (A).
56. C
First, find the equivalent resistance of the entire circuit.
R
2+3
 = R
2
 + R
3
          = 8 O + 16 O = 24 O
Then use Ohm’s law to solve for current.