Force on Moving Charges in Magnetic Fields — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: The magnetic Lorentz force law, right-hand rule for force direction, uniform circular motion of charges in magnetic fields, velocity selection, and core problem-solving techniques for AP Physics 2 exam questions.

You should already know: Vector direction rules for cross products, uniform circular motion kinematics, electric force on point charges.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Force on Moving Charges in Magnetic Fields?

This topic is part of Unit 5: Magnetism and Electromagnetic Induction, which accounts for 17-23% of the total AP Physics 2 exam score, with this subtopic making up roughly 3-5% of the total exam, appearing regularly in both multiple choice (MCQ) and free response (FRQ) sections. A moving electric charge generates its own magnetic field, so when it moves through an external magnetic field, the external field exerts a magnetic force on the charge—this is the fundamental interaction we study here. This force is distinct from electric force, which acts on charges regardless of motion, and gravitational force, which is negligible for subatomic charges. Unlike electric force, magnetic force is always perpendicular to both the velocity of the charge and the external magnetic field vector, which gives it a unique property: it never does work on a moving charge, because work requires force parallel to displacement. Common synonyms for this interaction are the magnetic Lorentz force (when combined with electric force, it forms the full Lorentz force law). AP exams expect you to calculate magnitude and direction, connect the force to circular motion, and apply the concept to real devices like velocity selectors and mass spectrometers.

2. Lorentz Force Law (Magnitude and Direction)

The magnetic component of the Lorentz force describes the force on a moving charge in an external magnetic field, and its magnitude is given by: $$F_B=|q|vB\sin \theta$$ where $F_B$ is the magnetic force magnitude, $q$ is the charge of the particle, $v$ is the speed of the particle, $B$ is the magnitude of the external magnetic field, and $\theta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$. Intuitively, this formula matches key observations: if the charge is not moving ($v=0$), there is no force, which is the key difference from electric force that acts on stationary charges. If the charge moves parallel or antiparallel to the magnetic field ($\theta =0^{\circ }$ or $180^{\circ }$, $\sin \theta =0$), there is also no force. The maximum force occurs when $\theta =90^{\circ }$, so $F_B=|q|vB$, which is the most common case on the AP exam. For direction, use the right-hand rule: point the fingers of your right hand along $\vec{v}$, curl your fingers towards $\vec{B}$, and your thumb points in the direction of $\overrightarrow{F_B}$ for a positive charge. For a negative charge, the force points in the exact opposite direction.

Worked Example

An alpha particle (charge $+3.2\times 10^{-19}\text{C}$) moves at $3.5\times 10^4\text{m/s}$ through a $0.75\text{T}$ magnetic field. The velocity of the alpha particle makes a 45° angle with the magnetic field vector. What is the magnitude of the magnetic force on the alpha particle?

1. Identify all given values: $q=3.2\times 10^{-19}\text{C}$, $v=3.5\times 10^4\text{m/s}$, $B=0.75\text{T}$, $\theta =45^{\circ }$.
2. Recall the magnetic force magnitude formula: $F_B=|q|vB\sin \theta$.
3. Substitute values: $\sin 45^{\circ }\approx 0.707$, so $F_B=(3.2\times 10^{-19})(3.5\times 10^4)(0.75)(0.707)$.
4. Calculate the product: $F_B\approx 5.9\times 10^{-15}\text{N}$.

Exam tip: Always confirm the sign of the charge before reporting direction. If the problem gives an electron or other negative particle, explicitly reverse the direction you get from the right-hand rule—this is the most frequent direction error on AP MCQs.

3. Uniform Circular Motion of Charges in Uniform Magnetic Fields

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force is always perpendicular to the particle's velocity, so it acts as a centripetal force that causes the particle to move in a uniform circular path. This combination of magnetic force and circular motion is one of the most commonly tested problem types on the AP Physics 2 exam. We can derive the radius of the circular path by equating magnetic force to centripetal force: $$|q|vB=\frac{m{v}^2}{r}$$ Cancel $v$ from both sides and rearrange to solve for $r$: $$r=\frac{mv}{|q|B}$$ We can also derive the period of revolution (time to complete one full circle): substitute $r$ into the definition of period $T=\frac{2\pi r}{v}$, and $v$ cancels out to give: $$T=\frac{2\pi m}{|q|B}$$ A key takeaway here is that the period does not depend on the speed or radius of the particle, which is the core operating principle of cyclotrons used to accelerate particles for medical and research applications.

Worked Example

A neutron star has a surface magnetic field of $1.0\times 10^8\text{T}$. An electron moving perpendicular to the magnetic field has a speed of $2.1\times 10^8\text{m/s}$ (close to the speed of light). What is the radius of the electron's circular path? The electron mass is $9.11\times 10^{-31}\text{kg}$ and charge is $1.6\times 10^{-19}\text{C}$.

1. Confirm velocity is perpendicular to $B$, so $\sin \theta =1$ and all magnetic force acts as centripetal force.
2. Use the derived radius formula: $r=\frac{mv}{|q|B}$.
3. Substitute values: $r=\frac{(9.11\times 10^{-31})(2.1\times 10^8)}{(1.6\times 10^{-19})(1.0\times 10^8)}$.
4. Calculate: $r\approx 1.2\times 10^{-11}\text{m}$, roughly 1/10 the radius of a hydrogen atom.

Exam tip: Do not forget to cancel $v$ when deriving the radius formula. Leaving $v^2$ in your final expression for $r$ is a common algebra error that will cost you points on FRQs.

4. Velocity Selection in Crossed Fields

A velocity selector is a common device that uses crossed electric and magnetic fields (E perpendicular to B) to filter out all particles except those moving at a specific desired speed, regardless of their mass or charge. When a charged particle moves through crossed fields, it experiences an electric force $F_E=qE$ and a magnetic force $F_B=qvB$. For the particle to pass through undeflected, the two forces must be equal in magnitude and opposite in direction, resulting in zero net force. Setting the magnitudes equal gives: $$qE=qvB$$ The charge $q$ cancels out from both sides, leading to the simple result: $$v=\frac{E}{B}$$ Only particles moving at this exact speed will pass through undeflected; faster particles are deflected one way, slower particles the other, and the result holds for both positive and negative charges (both forces reverse direction for negative charges, so they still cancel). This is the core principle behind mass spectrometry, used to identify the mass of unknown particles.

Worked Example

A mass spectrometer uses a velocity selector with $E=5.0\times 10^3\text{N/C}$ and $B=0.10\text{T}$. A beam of unknown charged particles passes through undeflected. What is the speed of the undeflected particles?

1. Confirm the fields are crossed, so electric and magnetic forces are opposite in direction for any particle moving into the selector.
2. Set forces equal: $qE=qvB$, cancel $q$ to get $v=E/B$.
3. Substitute values: $v=\frac{5.0\times 10^3}{0.10}=5.0\times 10^4\text{m/s}$.
4. Confirm: This result is the same regardless of the charge or mass of the particles, so no additional information is needed.

Exam tip: Even if the problem gives you the charge and mass of the particles, do not include them in your calculation of the undeflected speed. They are almost always red herrings designed to test if you know q cancels out.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the right-hand rule for a negative charge and not flipping the final force direction. Why: Students memorize the rule for positive charges and forget that the negative sign of charge reverses the cross product direction. Correct move: Always write the sign of the charge next to your direction work, and explicitly reverse the direction for negative charges before finalizing your answer.
• Wrong move: Leaving $v^2$ in the radius formula for circular motion, writing $r=m{v}^2/(qB)$ instead of simplifying to $r=mv/(qB)$. Why: Students stop after equating force to centripetal force and forget to simplify. Correct move: After setting $qvB=m{v}^2/r$, always cancel one v from both sides before substituting values.
• Wrong move: Calculating non-zero work done by the magnetic force as $W=F_{B}d$. Why: Students memorize work as force times distance and forget the direction property of magnetic force. Correct move: Always recall that magnetic force is always perpendicular to displacement, so work done by magnetic force is always zero.
• Wrong move: Including charge in the undeflected speed calculation for a velocity selector, writing $v=qE/B$. Why: Problems often give charge to test for this mistake, so students assume it must be used. Correct move: Always cancel q when equating electric and magnetic force, regardless of whether q is given.
• Wrong move: Calculating $F_B=qvB$ when velocity is parallel to the magnetic field, forgetting the $\sin \theta$ term. Why: Most AP problems use $\theta =90^{\circ }$, so students get used to dropping $\sin \theta$. Correct move: Always check the angle between v and B before calculating force, and multiply by $\sin \theta$ even if it seems redundant.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A negatively charged particle moving horizontally to the right along the x-axis enters a uniform magnetic field pointing vertically into the plane of the page (negative z-direction). What is the direction of the net magnetic force on the particle? A) Vertically upward (positive y-direction) B) Vertically downward (negative y-direction) C) Horizontally to the left (negative x-direction) D) Horizontally to the right (positive x-direction)

Worked Solution: First, apply the right-hand rule for a positive charge: point the fingers of the right hand along the velocity (to the right), then curl fingers toward the magnetic field direction (into the page). The thumb points vertically upward, which is the force direction for a positive charge. Since the particle is negatively charged, the force direction is reversed. This gives a final direction of vertically downward. The correct answer is B.

Question 2 (Free Response)

A beam of charged particles contains singly ionized neon-20 and neon-22 atoms (charge $+e=1.6\times 10^{-19}\text{C}$ for both, masses $m_{20}=20\text{u}$ and $m_{22}=22\text{u}$, where $1\text{u}=1.66\times 10^{-27}\text{kg}$). The beam passes through a velocity selector with crossed fields $E=8.0\times 10^4\text{N/C}$ and $B=0.20\text{T}$, then enters a second uniform $0.20\text{T}$ magnetic field perpendicular to the velocity of the particles. (a) Calculate the speed of the particles that exit the velocity selector undeflected. (b) Calculate the difference in radius of the circular paths of the two isotopes in the second magnetic field. (c) Explain why the magnetic field does not change the speed of the particles as they move along their circular paths.

Worked Solution: (a) For undeflected particles, net force is zero, so $qE=qvB$. Cancel $q$ to get $v=E/B$. Substitute values: $v=(8.0\times 10^4)/0.20=4.0\times 10^5\text{m/s}$. (b) For circular motion, $r=mv/(qB)$. The radius difference is $\Delta r=(v/(qB))(m_{22}-m_{20})$. $m_{22}-m_{20}=2\text{u}=3.32\times 10^{-27}\text{kg}$. Substitute: $\Delta r=\frac{(4.0\times 10^5)(3.32\times 10^{-27})}{(1.6\times 10^{-19})(0.20)}=0.0415\text{m}=4.2\text{cm}$. (c) Magnetic force is always perpendicular to the velocity (and instantaneous displacement) of the particle. Work done by a force is $W=Fd\cos \theta$, and $\theta =90^{\circ }$, so $\cos \theta =0$ and work done by magnetic force is zero. By the work-energy theorem, zero work means no change in kinetic energy, so speed remains constant.

Question 3 (Application / Real-World Style)

A cyclotron in a university lab uses a uniform 1.2 T magnetic field to accelerate deuterium nuclei (mass $3.34\times 10^{-27}\text{kg}$, charge $+1.6\times 10^{-19}\text{C}$) for nuclear physics experiments. Calculate the period of revolution of the deuterium nuclei, and explain the relevance of this value to cyclotron operation.

Worked Solution: Use the period formula for charged particles in magnetic fields: $$T=\frac{2\pi m}{qB}$$ Substitute values: $$T=\frac{2\pi (3.34\times 10^{-27})}{(1.6\times 10^{-19})(1.2)}\approx 1.1\times 10^{-7}\text{s}$$ This period is independent of the deuterium's speed, so the cyclotron can use a constant-frequency electric field to accelerate the particles every half period, even as the particles speed up and spiral outward to larger radii.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the fundamental building block for all magnetic force interactions in Unit 5, and is a prerequisite for every subsequent magnetism topic in AP Physics 2. Next, you will apply the force rule for individual moving charges to derive the force on current-carrying wires in magnetic fields, which is just the net sum of magnetic forces on the many moving charge carriers in the wire. Without mastering the direction and magnitude rules for individual charges here, you will not be able to correctly solve for force on wires or torque on current loops, which are regularly tested on the AP exam. This topic also sets up the core vector relationships needed for electromagnetic induction, the second major part of Unit 5.

• Force on Current-Carrying Wires
• Torque on Current Loops
• Electromagnetic Induction
• Faraday's Law of Induction