AP Physics 2: Algebra-Based Practice Quiz: Magnetism and Moving Charges

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 15 questions to check your progress.

Question 1 of 15

According to the provided principles, which of the following is a source of a magnetic field?

A)A stationary protonB)A moving neutronC)A moving electronD)A stationary electric dipole

All Questions (15)

According to the provided principles, which of the following is a source of a magnetic field?

A) A stationary proton

B) A moving neutron

C) A moving electron

D) A stationary electric dipole

Correct Answer: C

Content point 3 states, "A single moving charged object produces a magnetic field." An electron is a charged object, and when it is moving, it produces a magnetic field. A stationary proton is charged but not moving, and a neutron has no charge.

A uniform magnetic field exists in a region of space. Under which condition will a charged particle entering this field experience a magnetic force?

A) The particle is at rest within the field.

B) The particle moves with a velocity component perpendicular to the magnetic field.

C) The particle moves with a velocity parallel to the magnetic field.

D) The particle has zero net charge.

Correct Answer: B

Content point 5 gives the force magnitude as $F_B = qvB\sin\theta$. For the force to be non-zero, $q$, $v$, and $B$ must be non-zero, and the angle $\theta$ between velocity and the magnetic field must not be 0° or 180°. A velocity component perpendicular to the field ensures $\sin\theta$ is not zero.

A proton enters a uniform magnetic field with a velocity $v$ and experiences a magnetic force $F_B$. If the proton had entered the same field with a velocity of $2v$ at the same angle, what would be the magnitude of the new magnetic force?

A) $F_B/2$

B) $F_B$

C) $2F_B$

D) $4F_B$

Correct Answer: C

According to the equation $F_B = qvB\sin\theta$ (Content point 5), the magnitude of the magnetic force is directly proportional to the magnitude of the charged object's velocity, $v$. If the velocity is doubled while all other factors remain constant, the force is also doubled.

A charged particle moves through a uniform magnetic field at an angle of 90° to the field lines and experiences a force of magnitude $F_{max}$. If the same particle entered the field at the same speed but at an angle of 30° to the field lines, what would be the magnitude of the force exerted on it?

A) $F_{max}$

B) $F_{max}/2$

C) $F_{max}\sqrt{3}/2$

D) Zero

Correct Answer: B

The force is given by $F_B = qvB\sin\theta$ (Content point 5). At 90°, $F_{max} = qvB\sin(90^\circ) = qvB$. At 30°, the new force is $F' = qvB\sin(30^\circ) = qvB(1/2) = F_{max}/2$.

A particle with a positive charge is moving eastward in a region where the magnetic field is directed northward. What is the direction of the magnetic force exerted on the particle?

A) Downward

B) Upward

C) Westward

D) Southward

Correct Answer: B

According to the right-hand rule (Content point 6), the direction of the force is perpendicular to both the velocity and the magnetic field. Pointing your fingers in the direction of velocity (east) and curling them towards the magnetic field (north) results in your thumb pointing upward.

A charged particle is moving with a constant velocity through a region with a uniform magnetic field. Under which of the following conditions is the magnetic force on the particle equal to zero?

A) The particle's velocity is perpendicular to the magnetic field.

B) The particle's velocity is parallel to the magnetic field.

C) The particle is negatively charged.

D) The magnetic field is very strong.

Correct Answer: B

The magnitude of the magnetic force is $F_B = qvB\sin\theta$ (Content point 5). The force is zero when $\sin\theta = 0$. This occurs when the angle $\theta$ between the velocity vector and the magnetic field vector is 0° or 180°, meaning the velocity is parallel or anti-parallel to the magnetic field.

A proton is shot into a uniform magnetic field. For which orientation of the proton's velocity vector relative to the magnetic field vector will the magnetic force on the proton be at its maximum?

A) The velocity vector is at a 45° angle to the magnetic field vector.

B) The velocity vector is parallel to the magnetic field vector.

C) The velocity vector is anti-parallel to the magnetic field vector.

D) The velocity vector is perpendicular to the magnetic field vector.

Correct Answer: D

The magnetic force is given by $F_B = qvB\sin\theta$ (Content point 5). The sine function has a maximum value of 1 when the angle $\theta$ is 90°. Therefore, the force is maximum when the velocity is perpendicular to the magnetic field.

An alpha particle (charge +2e) and a proton (charge +e) enter the same magnetic field with the same velocity, perpendicular to the field lines. How does the magnitude of the magnetic force on the alpha particle, $F_\alpha$, compare to the force on the proton, $F_p$?

A) $F_\alpha = F_p$

B) $F_\alpha = 2F_p$

C) $F_\alpha = F_p/2$

D) $F_\alpha = 4F_p$

Correct Answer: B

The magnitude of the force is proportional to the magnitude of the charge, $q$, as shown in the equation $F_B = qvB\sin\theta$ (Content point 5). Since the alpha particle has twice the charge of the proton (+2e vs +e) and all other variables (v, B, $\theta$) are the same, the force on the alpha particle will be twice the force on the proton.

A beam of charged particles travels in a straight line through a region of space containing a uniform magnetic field. Which of the following statements could explain this observation?

A) The particles must be neutral.

B) The particles must be stationary.

C) The velocity of the particles is parallel to the magnetic field.

D) The magnetic field must be zero.

Correct Answer: C

For a charged particle to travel in a straight line (i.e., be undeflected), the magnetic force must be zero. According to $F_B = qvB\sin\theta$, this occurs if q=0, v=0, B=0, or $\sin\theta=0$. The problem states the particles are charged and moving through a field, so q, v, and B are non-zero. Therefore, the only possibility is that $\sin\theta=0$, which means the velocity is parallel to the magnetic field.

An electron is moving to the right in a uniform magnetic field that is directed out of the page. What is the direction of the magnetic force on the electron?

A) Upward

B) Downward

C) Into the page

D) To the left

Correct Answer: B

Using the right-hand rule for a positive charge (Content point 6), pointing fingers to the right (velocity) and curling them toward the direction out of the page (magnetic field) results in the thumb pointing upward. However, since the electron has a negative charge, the direction of the force is opposite to the direction given by the right-hand rule. Therefore, the force is directed downward.

A particle with charge $q$ and velocity $v$ enters a magnetic field $B$ at an angle $\theta$ and experiences a force $F$. A second particle with charge $2q$ and velocity $v/2$ enters the same field at the same angle $\theta$. What is the magnitude of the force on the second particle?

A) $F/2$

B) $F$

C) $2F$

D) $4F$

Correct Answer: B

The force is given by $F_B = qvB\sin\theta$ (Content point 5). For the first particle, $F = qvB\sin\theta$. For the second particle, the force is $F' = (2q)(v/2)B\sin\theta = qvB\sin\theta$. Therefore, the new force $F'$ is equal to the original force $F$.

The magnitude of the magnetic force on a moving charged particle does NOT depend on which of the following?

A) The mass of the particle

B) The speed of the particle

C) The magnitude of the particle's charge

D) The angle between the particle's velocity and the magnetic field

Correct Answer: A

The equation for the magnitude of the magnetic force is $F_B = qvB\sin\theta$ (Content point 5). This equation shows the force depends on the charge (q), velocity (v), magnetic field (B), and the angle ($\theta$). The mass of the particle does not appear in this equation and therefore does not affect the magnitude of the magnetic force.

A charged particle moving with speed $v$ at an angle of 30° to a magnetic field $B$ experiences a force $F$. If the particle's speed is doubled to $2v$ and the magnetic field strength is halved to $B/2$, while the angle remains 30°, what is the new force on the particle?

A) $F/2$

B) $F$

C) $2F$

D) $4F$

Correct Answer: B

The initial force is $F = qvB\sin(30^\circ)$. The new force is $F' = q(2v)(B/2)\sin(30^\circ)$. Simplifying, $F' = qvB\sin(30^\circ)$. Thus, the new force is equal to the original force $F$.

A charged particle moves in a uniform magnetic field. Which statement correctly describes the relationship between the magnetic force vector $\vec{F}_B$, the velocity vector $\vec{v}$, and the magnetic field vector $\vec{B}$?

A) $\vec{F}_B$ is parallel to $\vec{v}$ and perpendicular to $\vec{B}$.

B) $\vec{F}_B$ is parallel to the plane containing $\vec{v}$ and $\vec{B}$.

C) $\vec{F}_B$ is perpendicular to the plane containing $\vec{v}$ and $\vec{B}$.

D) $\vec{F}_B$ is parallel to $\vec{B}$ and perpendicular to $\vec{v}$.

Correct Answer: C

Content point 6 states that the direction of the force is perpendicular to *both* the direction of the magnetic field and the velocity of the charge. A vector that is perpendicular to two other vectors is, by definition, perpendicular to the plane that contains those two vectors.

A particle of charge $q$ moves with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. The magnitude of the magnetic force on the particle is given by $F_B = qvB\sin\theta$. Which expression correctly represents the component of the particle's velocity that is perpendicular to the magnetic field?

A) $v\sin\theta$

B) $v\cos\theta$

C) $F_B / (qB)$

D) $F_B / (qv)$

Correct Answer: A

The equation for the magnetic force can be written as $F_B = q(v\sin\theta)B$. The term $v\sin\theta$ represents the component of the velocity vector $\vec{v}$ that is perpendicular to the magnetic field vector $\vec{B}$. The force only depends on this perpendicular component of the velocity. Option C is also equal to $v\sin\theta$ by rearranging the formula, but option A is the direct geometric definition.