AP Physics Test: Electricity and Magnetism- Solutions
Multiple Choice Questions
1. A proton moving at constant velocity enters the region between two charged plates,
as shown above. Which of the paths shown correctly indicates the proton's
trajectory after leaving the region between the charged plates?
A. A
B. B
C. C
D. D
E. E
D—While in the region between the plates, the positively charged proton is attracted to the negative plate, so
bends downward. But after leaving the plates, there is no more force acting on the proton. Thus, the proton
continues in motion in a straight line by Newton's first law.
2. Which of the following quantities is NOT a vector?
A. magnetic field
B. electric force
C. electric current
D. electric field
E. electric potential
E—Fields are vectors because they cause forces, and force is a vector. Current is a vector because charge can
flow in either direction in a wire. However, electric potential is related (by PE = qV ) to potential energy, and
all forms of energy are scalar quantities.
3. Experimenter A uses a very small test charge q0, and experimenter B uses a test charge 2q0 to measure an
electric field produced by stationary charges. A finds a field that is
A. greater than the field found by B
B. the same as the field found by B
C. less than the field found by B
D. either greater or less than the field found by B, depending on the accelerations of the test charges
E. either greater or less than the field found by B, depending on the masses of the test charges
B—An electric field exists regardless of the amount of charge placed in it, and regardless of whether any
charge at all is placed in it. So both experimenters must measure the same field (though they will measure
different forces on their test charges).
4. Two isolated particles, A and B, are 4 m apart. Particle A has a net charge of 2Q, and B has a net charge
of Q. The ratio of the magnitude of the electrostatic force on A to that on B is
A. 4:1
B. 2:1
C. 1:1
D. 1:2
E. 1:4
C—Ah, the Newton's third law question. Whether we're talking about cars or charges, the force on object A due
to object B is always the same as the force on object B due to object A. Thus the forces described must be equal.
5. A uniform electric field points right to left. A small metal ball charged to +2 mC hangs at a 30° angle
from a string of negligible mass, as shown above. The tension in the string is
measured to be 0.1 N. What is the magnitude of the electric field? (sin 30° = 0.50;
cos 30° = 0.87; tan 30° = 0.58).
A. 25 N/C
B. 50 N/C
C. 2500 N/C
D. 5000 N/C
E. 10,000 N/C
A-The charge is in equilibrium, so the horizontal component of the tension must equal the electric force. This
horizontal tension is 0.1 N times sin 30° (not cos because 30° was measured from the vertical), or 0.05 N. The
electric force is qE, where q is 0.002 C. So the electric field is 0.050/0.002. reduce the expression by moving the
decimal to get 50/2, or 25 N/C.
6. 1.0 nC is deposited on a solid metal sphere of diameter 0.30 m. What is the magnitude of the electric
field at the center of the sphere?
A. zero
B. 25 N/C
C. 100 N/C
D. 200 N/C
E. 400 N/C
A—In a metal sphere, charge is free to move, so the charge ends up on the surface evenly distributed. The
electric fields due to each of these charges cancel at the center because for every charge causing a field
pointing one way, another charge on the other side of the sphere will cause an equal field pointing the other
way. (It's worth knowing that the electric field ANYWHERE inside a conductor is always zero; the justification
for this is done in Physics C with Gauss's law.)
7. The parallel plate capacitor above consists of identical rectangular plates of dimensions a × b, separated
by a distance c. To cut the capacitance of this capacitor in half, which of these
quantities should be doubled?
A. a
B. b
C. c
D. ab
E. abc
C—Capacitance of a parallel plate capacitor is εoA/d, where A is the area of the plates and d is the separation
between plates. To halve the capacitance, we must halve the area or double the plate separation. The plate
separation in the diagram is labeled c, so double distance c.
8. Two identical capacitors are connected in parallel to an external circuit. Which of the following
quantities must be the same for both capacitors?
I.
the charge stored on the capacitor
II.
the voltage across the capacitor
III.
the capacitance of the capacitor
A. I only
B. II only
C. II and III only
D. I and III only
E. I, II, and III
E—Connected in parallel means that the voltage for each capacitor must be the same. The capacitors are
identical, so the capacitance of each is the same. Thus, by Q = CV, the charge on each must be the same, too.
9.Three resistors are connected to a 1.0 V battery, as shown in the diagram above. What is the current through
the 2.0 Ω resistor?
A. 0.25 A
B. 0.50 A
C. 1.0 A
D. 2.0 A
E. 4.0 A
B—Voltage is always the same across parallel resistors. Because the 2.0Ω resistor is hooked directly to the
battery, the voltage across it is 1.0 V. By Ohm's law, I = V/R = 0.50 amps.
10.What is the voltage drop across R3 in the circuit diagrammed
here?
A. 10 V
B. 10 V
C. 20 V
D. 30 V
E. 50 V
F. 100 V
C—The 5 A of current splits into two paths. The path through R2 takes 2 A, so the path through R3 takes 3 A.
Now use Ohm's law: V = (3 A) (10 Ω) = 30 V.
11.Three resistors are connected to an ideal battery as shown in
the diagram above. The switch is initially open. When the switch
is closed, what happens to the total voltage, current, and
resistance in the circuit?
D—The voltage must stay the same because the battery by definition provides a constant voltage. Closing the
switch adds an extra parallel branch to the network of resistors. Adding a parallel resistor REDUCES the total
resistance. By Ohm's law, if voltage stays the same and resistance decreases, total current must increase.
12.On which of the following physics principles does Kirchoff 's loop rule rest?
D. conservation of charge
E. conservation of mass
F.
conservation of energy
G. conservation of momentum
H. conservation of angular momentum
C—Kirchoff 's loop rule says that the sum of voltage changes around a closed circuit loop is zero. When a
charge gains or loses voltage, it is also gaining or losing energy. The fact that the charge must end up with the
same energy it started with is a statement of energy conservation.
Questions 19 and 20
A uniform magnetic field B is directed into the page. An electron enters this
field with initial velocity v to the right.
19. Which of the following best describes the path of the electron while it is
still within the magnetic field?
A. It moves in a straight line.
B. It bends upward in a parabolic path.
C. It bends downward in a parabolic path.
D. It bends upward in a circular path.
E. It bends downward in a circular path.
E-A charge in a uniform magnetic field takes a circular path because the magnetic force changes direction, so
it's always perpendicular to the charge's velocity. Here the magnetic force by the right-hand rule is initially
down. (Did you think the force on the charge was up? An electron has a negative charge, so the right hand rule
gives a direction opposite to the direction of the force.)
20. The electron travels a distance d, measured along its path, before exiting the magnetic field. How much
work is done on the electron by the magnetic field?
A. evBd
B. zero
C. –evBd
D. evBd sin (d/2π)
E. –evBd sin (d/2π)
B—Work is force times parallel displacement. Because the magnetic force is always perpendicular to the
particle's direction of motion, no work can be done by the magnetic field.
21.The circular wire shown above carries a current I in the counterclockwise direction. What will be the
direction of the magnetic field at the center of the wire?
A. into the page
B. out of the page
C. down
D. up
E. counterclockwise
B—The magnetic field caused by a straight wire wraps around the wire by the right-hand
rule. Consider just little parts of the curvy wire as if the parts were momentarily straight. The part of the wire at
the top of the page produces a B field out of the page in the region inside the loop; the parts of the wire at the
bottom of the page or on the side of the page also produce B fields out of the page inside the loop. These fields
reinforce each other at the center, so the net B field is out of the page.
22.Two parallel wires carry currents in opposite directions, as shown above. In what direction will the righthand wire (the wire carrying current I2) experience a force?
A. left
B. right
C. into the page
D. out of the page
E. The right-hand wire will experience no force.
B—The magnetic field produced by I1 is out of the page at the location of wire 2. The current I2 is thus moving
in a magnetic field; the force on that current-carrying wire is given by ILB and the right-hand rule, which
shows a rightward force.
23.A proton moves in a straight line. Which of the following combinations of electric or magnetic fields could
NOT allow this motion?
A. only an electric field pointing in the direction of the proton's motion
B. only a magnetic field pointing opposite the direction of the proton's motion
C. an electric field and a magnetic field, each pointing perpendicular to the proton's motion
D. only a magnetic field pointing perpendicular to the proton's motion
E. only a magnetic field pointing in the direction of the proton's motion
D—An electric field exerts a force on all charged particles parallel to the field; thus situation (A) would slow
the proton down, but not change its direction. Magnetic fields only apply a force if they are perpendicular to a
charge's velocity, so (B) and (E) cannot change the proton's direction. A magnetic field perpendicular to motion
DOES change a proton's direction, but if a suitable electric field is also included, then the electric force can
cancel the magnetic force. So choice D is the only one that must change the direction of the proton.
24. An electron moves to the right in a uniform magnetic field that points into the page. What is the direction of
the electric field that could be used to cause the electron to travel in a straight line?
A. down toward the bottom of the page
B. up toward the top of the page
C. into the page
D. out of the page
E. to the left
A—Use the right-hand rule for the force on a charge. Point in the direction of velocity,
curl the fingers into the page, the thumb points up the page … but this is a negative charge, so the force on the
charge is down the page. Now, the electric force must cancel the magnetic force for the charge to move in a
straight line, so the electric force should be up the page. (E and B fields cannot cancel, but forces sure can.)
The direction of an electric force on a negative charge is opposite the field; so the field should point down,
toward the bottom of the page
25.
26. At what angle from the parallel to the direction of the magnetic field does a wire carrying 8A have to be
oriented so that its 5 m length subject to a 2 T field experiences a force of 2 N?
A. 0°
B. 14.5°
C. 45°
D. 63.5°
E. 90°
B – Use F  IlBsin  : rearranging gives sinθ=0.25. You do not have a calculator, so estimate. There is only
one answer that makes sense.
27.Two parallel wires carry currents I1 and I2 in the same direction and separated by a distance d. The
magnitude of the magnetic force between the wires is F0. What is the force between the wires if each
current is doubled and the separation is quadrupled?
(A) 2F0
(B) 4F0
(C) F0
(D) F0/2
(E) F0/4
C – current-carrying wires
28.An electron with a mass m and charge e enters at a constant speed v a uniform
magnetic field B. What is the radius of the curvature of the electron in the field?
(A)mv/eB
(B) eB/mv
(C) me/vB
(D) mB/ev
(E) zero
A – mass spectrometer
29. A vertical wire carries a current straight up in a region of the magnetic field directed north. What is
the direction of the magnetic force on the current due to the magnetic field?
(A) East
(B) South
(C) North
(D) West
(E) Applied force is zero
D - RHR
30. A horizontal thin wire has a mass m and length L. The wire carries a constant current I. What must be
the direction of the magnetic field in order to cancel the gravitational force?
(A) Left
(B) Right
(C) Down the page (D) Out of the page (E) Into the page
st
D – RHR, Newton’s 1 Law
*Bonus:
1. A proton enters a uniform magnetic field perpendicular to the field
lines. What is the new path of the proton as it passes the field?
(A) A
(B) B
(C) C
(D) D
(E) E
E – mass spectrometer
2. An electron enters a uniform electric field perpendicular to the field lines.
What must be the direction of the magnetic field in order to cancel the electric
force effect?
(A) Left
(B) Right
(C) Top the page
(D) Into the page
(E) Out the page
E - RHR
Solutions http://www.education.com/study-help/article/physics-practiceexammultiple-choice-questions/