# Force on Moving Charges in Magnetic Fields

> AP Physics 2 · Unit 5: Magnetism and Electromagnetic Induction
> Source: https://www.owlsprep.com/study/ap-physics-2-u5-force-on-moving-charges-in/

This module covers the magnetic Lorentz force law, right-hand rule for direction, circular motion of charges in uniform fields, and velocity selection—core content regularly tested on AP Physics 2 MCQ and FRQ.

**Prerequisites:** Vector cross product direction rules; Uniform circular motion kinematics; Electric force on point charges

## Learning objectives

- Calculate magnitude and direction of magnetic force on moving charges
- Analyze uniform circular motion of charges in magnetic fields
- Apply the velocity selector principle to solve problems
- Avoid common exam pitfalls for this topic

## The Magnetic Lorentz Force Law: Magnitude and Direction

When a moving electric charge travels through an external magnetic field, the field exerts a magnetic force on the charge. Unlike electric force (which acts on charges regardless of motion) and gravitational force (negligible for subatomic particles), magnetic force is always perpendicular to both the velocity of the charge and the magnetic field vector, so it never does work on a moving charge.

**Magnetic Lorentz Force** — Magnitude of the magnetic force on a moving charge, where $q$ is particle charge, $v$ is speed, $B$ is magnetic field magnitude, and $\theta$ is the angle between $\vec{v}$ and $\vec{B}$

*Notation:* $F_B = |q| v B \sin\theta$

*Example:* Force is zero if $v=0$ or $\vec{v}$ is parallel/antiparallel to $\vec{B}$, maximum when $\theta=90^\circ$

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 $\vec{F}_B$ for a positive charge. For negative charges, the force points in the 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:
2. $$q = 3.2 \times 10^{-19}\ \text{C}, v=3.5 \times 10^4\ \text{m/s}, B=0.75\ \text{T}, \theta=45^\circ$$
3. Recall the magnetic force magnitude formula:
4. $$F_B = |q|vB \sin\theta$$
5. Substitute values, using $\sin 45^\circ \approx 0.707$:
6. $$F_B = (3.2 \times 10^{-19})(3.5 \times 10^4)(0.75)(0.707)$$
7. Calculate the final force magnitude:
8. $$F_B \approx 5.9 \times 10^{-15}\ \text{N}$$

> **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.

## 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 is one of the most commonly tested problem types on the AP Physics 2 exam.

**Derivation:** Derive the radius of the circular path and period of revolution

*Starting from:* Equate maximum magnetic force ($\theta=90^\circ$) to centripetal force for circular motion

1. Set forces equal:
2. $$|q|vB = \frac{mv^2}{r}$$
3. Cancel $v$ from both sides and rearrange for $r$:
4. $$r = \frac{mv}{|q|B}$$
5. Substitute $r$ into the definition of period $T = \frac{2\pi r}{v}$:
6. $$T = \frac{2\pi m}{|q|B}$$

*Conclusion:* The period $T$ is independent of the particle's speed or path radius, which is the core operating principle of cyclotrons.

**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:
3. $$r = \frac{mv}{|q|B}$$
4. Substitute given values:
5. $$r = \frac{(9.11 \times 10^{-31})(2.1 \times 10^8)}{(1.6 \times 10^{-19})(1.0 \times 10^8)}$$
6. Calculate the final radius:
7. $$r \approx 1.2 \times 10^{-11}\ \text{m}$$

> **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.

## 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. This is the core principle behind mass spectrometry, used to identify the mass of unknown particles.

**Derivation:** Derive the speed of undeflected particles

*Starting from:* Net force is zero for undeflected particles, so electric and magnetic force magnitudes are equal

1. Write magnitudes of electric and magnetic force:
2. $$F_E = qE, \quad F_B = qvB$$
3. Set equal for zero net force:
4. $$qE = qvB$$
5. Cancel charge $q$ from both sides:
6. $$v = \frac{E}{B}$$

*Conclusion:* Only particles moving at speed $v = E/B$ pass through undeflected, and this result holds for both positive and negative charges.

**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 entering the selector.
2. Set forces equal, cancel $q$ to get the undeflected speed formula:
3. $$v = \frac{E}{B}$$
4. Substitute values to calculate speed:
5. $$v = \frac{5.0 \times 10^3}{0.10} = 5.0 \times 10^4\ \text{m/s}$$
6. Confirm that this result is independent 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.

## AP-Style Practice Problems

**Worked example:** 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)

1. Apply the right-hand rule for a positive charge: point fingers along velocity (to the right), curl fingers toward the magnetic field (into the page). Thumb points upward, which is the force direction for a positive charge.
2. Reverse the direction for the negative charge, giving a final direction of vertically downward.
3. The correct answer is option B.

**Worked example:** 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. (a) Calculate the speed of undeflected particles exiting the selector. (b) Calculate the difference in radius of the circular paths of the two isotopes. (c) Explain why the magnetic field does not change the speed of the particles.

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

## Common pitfalls

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

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Magnetic Force Magnitude | $F_B = \|q\| v B \sin\theta$ | $\theta$ = angle between $\vec{v}$ and $\vec{B}$; $F_B=0$ if $v=0$ or $\vec{v} \parallel \vec{B}$ |
| Vector Lorentz Force | $\vec{F}_B = q \vec{v} \times \vec{B}$ | Direction from right-hand rule; reverse direction for negative $q$ |
| Radius of Circular Path (perpendicular B) | $r = \frac{mv}{\|q\|B}$ | Derived from equating $F_B$ to centripetal force |
| Period of Circular Motion | $T = \frac{2\pi m}{\|q\|B}$ | Independent of speed $v$ and radius $r$; core of cyclotron operation |
| Undeflected Speed (Velocity Selector) | $v = \frac{E}{B}$ | Crossed E and B fields; independent of $q$ and $m$ |
| Work Done by Magnetic Force | $W_B = 0$ | Always true, because $\vec{F}_B \perp$ displacement |

## What's next

This topic is the fundamental building block for all magnetic force interactions in AP Physics 2 Unit 5, and is a prerequisite for every subsequent magnetism topic on the exam. 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, both regularly tested on the AP exam, and this topic also sets up core vector relationships for electromagnetic induction later in Unit 5.

- [Electromagnetic Induction](https://www.owlsprep.com/study/ap-physics-2-u5-electromagnetic-induction/)
- [Force on Current-Carrying Wire in Magnetic Field](https://www.owlsprep.com/study/ap-physics-2-u5-force-on-current-carrying-wire/)
- [Magnetic Fields Due to Currents](https://www.owlsprep.com/study/ap-physics-2-u5-magnetic-fields-due-to-currents/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-physics-2-u5-force-on-moving-charges-in/