Magnetic Fields and Magnetic Forces — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Magnetic field definition and SI units, Lorentz force law, force on current-carrying wires, torque on current loops, right-hand direction rules, and motion of charged particles in uniform magnetic fields, with worked examples for all common AP exam problem types.

You should already know: Cross product vector mathematics, kinematics of circular motion, current as flow of moving charge.

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 C: E&M 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 Magnetic Fields and Magnetic Forces?

This topic is the core foundation of Unit 4 (Magnetic Fields) for AP Physics C: E&M, and accounts for approximately 10–15% of the total exam score per the official College Board CED. It appears in both MCQ and FRQ sections, often as a standalone MCQ or the opening segment of a longer FRQ that connects to Ampere’s law or Faraday’s law later in the exam. A magnetic field ($\vec{B}$) is a vector field that exerts force only on moving electric charge; unlike electric fields, magnetic forces never do work on charge because force is always perpendicular to displacement. The SI unit for magnetic field is the tesla (T), where $1\text{ T}=1\text{ N}/(\text{Acdotpm})$. Magnetic forces are the forces exerted by magnetic fields on moving charge, grouped into three common testable cases: force on a single moving point charge, force on a bulk current-carrying wire, and torque on a current-carrying loop (the operating principle of electric motors and galvanometers). This topic also covers the trajectory of charged particles in uniform B-fields, a common exam context for mass spectrometry problems.

2. The Lorentz Force Law

The Lorentz force law describes the total force exerted on a charged particle moving through combined electric and magnetic fields, and it is the starting point for all magnetic force calculations. The general form is: $$\vec{F}=q\left(\vec{E}+\vec{v}\times \vec{B}\right)$$ where $q$ is the particle’s charge, $\vec{v}$ is its instantaneous velocity, $\vec{E}$ is the electric field, and $\vec{B}$ is the magnetic field. If no electric field is present, the law simplifies to $\vec{F}=q\vec{v}\times \vec{B}$. The magnitude of the magnetic force is $F=qvB\sin \theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Direction follows the right-hand rule for cross products: for positive $q$, point the fingers of your right hand along $\vec{v}$, curl them toward $\vec{B}$, and your thumb points to the force direction; for negative $q$, the force is exactly opposite this direction. A key property: magnetic force is always perpendicular to velocity, so it changes only the direction of motion, not speed, so magnetic forces never do work.

Worked Example

A proton with charge $q=1.6\times 10^{-19}$ C moves at $v=2.0\times 10^6$ m/s along the +x axis, through a uniform magnetic field $\vec{B}=0.50$ T $\hat{k}$ (out of the page). What is the magnitude and direction of the magnetic force on the proton?

1. Identify given vectors: $\vec{v}=2.0\times 10^6\hat{i}$ m/s, $\vec{B}=0.50\hat{k}$ T, $q>0$.
2. Calculate the cross product: $\vec{v}\times \vec{B}=(2.0\times 10^6\hat{i})\times (0.50\hat{k})=1.0\times 10^6(\hat{i}\times \hat{k})=-1.0\times 10^6\hat{j}$ m·T/s.
3. Multiply by charge: $\vec{F}=(1.6\times 10^{-19})(-1.0\times 10^6\hat{j})=-1.6\times 10^{-13}\hat{j}$ N.
4. Final result: magnitude $1.6\times 10^{-13}$ N, direction along the negative y-axis.

Exam tip: Always explicitly check the sign of the charge before writing your final direction. AP MCQs almost always include an option that matches the magnitude but has the wrong direction for negative charges to catch this common mistake.

3. Force on Current-Carrying Wires

A current in a wire is simply a continuous stream of moving charges, so we can derive the force on a wire directly from the Lorentz force law. For an infinitesimal segment of wire of length $d\vec{l}$, where $d\vec{l}$ points in the direction of conventional current, the total magnetic force on the moving charges in the segment is $d\vec{F}=Id\vec{l}\times \vec{B}$, where $I$ is the current in the wire. Integrating over the full length of the wire gives: $$\vec{F}=I\left(\int d\vec{l}\right)\times \vec{B}$$ for a uniform magnetic field. For a straight wire of total length $L$, this simplifies to $\vec{F}=I\vec{L}\times \vec{B}$, with magnitude $F=ILB\sin \theta$, where $\theta$ is the angle between $\vec{L}$ (current direction) and $\vec{B}$. A key result: the net force on any closed current loop in a uniform magnetic field is zero, because the integral of $d\vec{l}$ around a closed loop is zero.

Worked Example

A 30 cm long straight wire carries 2.0 A of current along the +y axis, in a uniform magnetic field $\vec{B}=1.2$ T $\hat{i}+0.8$ T $\hat{k}$. Find the magnitude of the net magnetic force on the wire.

1. Write the length vector: $\vec{L}=0.30\hat{j}$ m, aligned with the current direction.
2. Calculate the cross product: $\vec{L}\times \vec{B}=0.30\hat{j}\times (1.2\hat{i}+0.8\hat{k})=0.30(-1.2\hat{k}+0.8\hat{i})=0.24\hat{i}-0.36\hat{k}$ m·T.
3. Multiply by current: $\vec{F}=2.0(0.24\hat{i}-0.36\hat{k})=0.48\hat{i}-0.72\hat{k}$ N.
4. Find the magnitude: $|\vec{F}|=\sqrt{(0.48{)}^2+(0.72{)}^2}=\sqrt{0.7488}\approx 0.87$ N.

Exam tip: For any curved wire in a uniform magnetic field, the net force equals the force on a straight wire connecting the two endpoints of the curve. This shortcut saves significant time on FRQs.

4. Motion of Charged Particles in Uniform Magnetic Fields

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force that maintains uniform circular motion. Equating the Lorentz force magnitude to centripetal force: $$qvB=\frac{m{v}^2}{r}$$ Rearranging gives the radius of the circular path: $$r=\frac{mv}{qB}=\frac{p}{qB}$$ where $p=mv$ is the particle’s momentum. The cyclotron angular frequency is $\omega =\frac{qB}{m}$, which is independent of the particle’s speed or radius, a property that makes cyclotron particle accelerators work. If the particle has a velocity component parallel to $\vec{B}$, that component experiences no force, so the particle follows a helical path. This behavior is the basis of mass spectrometry, where particles of different mass are separated by their radius of curvature.

Worked Example

A mass spectrometer uses a 0.25 T uniform magnetic field to separate singly ionized ($q=+e$) carbon isotopes. Carbon-12 has mass $1.99\times 10^{-26}$ kg, and enters the B field perpendicular to $\vec{B}$ with speed $1.2\times 10^5$ m/s. What is the diameter of its circular path?

1. Recall the radius formula for perpendicular motion: $r=\frac{mv}{qB}$.
2. Substitute values: $r=\frac{(1.99\times 10^{-26}\text{ kg})(1.2\times 10^5\text{ m/s})}{(1.6\times 10^{-19}\text{ C})(0.25\text{ T})}\approx 0.060$ m = 6.0 cm.
3. The question asks for diameter, not radius: multiply the radius by 2.
4. Final result: diameter $d=2(0.060\text{ m})=0.12$ m = 12 cm.

Exam tip: AP problems intentionally ask for diameter (the quantity measured experimentally in mass spectrometers) much more often than radius. Always confirm what quantity you are asked for before writing your final answer.

5. Torque on Current-Carrying Loops

Even though the net force on a closed current loop in a uniform magnetic field is zero, there is a net torque that rotates the loop to align its dipole moment with the magnetic field. The torque formula is: $$\vec{\tau }=\vec{\mu }\times \vec{B}$$ where $\vec{\mu }$, the magnetic dipole moment of the loop, has magnitude $\mu =IA$, with $I$ the current and $A$ the area of the loop. The direction of $\vec{\mu }$ is along the normal to the plane of the loop, found via right-hand rule: curl your fingers along the direction of current, and your thumb points to $\vec{\mu }$. The magnitude of torque is $\tau =IAB\sin \theta$, where $\theta$ is the angle between $\vec{\mu }$ and $\vec{B}$. The potential energy of the dipole in the B field is $U=-\vec{\mu }\cdot \vec{B}$, so the lowest energy (equilibrium) state is when $\vec{\mu }$ is aligned with $\vec{B}$. This is the operating principle of electric motors and galvanometers.

Worked Example

A rectangular 2.0 cm × 3.0 cm current loop carries 5.0 A of current, and is placed in a 0.4 T uniform magnetic field. The angle between the plane of the loop and the magnetic field is 60 degrees. What is the magnitude of the torque on the loop?

1. Calculate the area of the loop: $A=(0.02\text{ m})(0.03\text{ m})=6.0\times 10^{-4}{\text{ m}}^2$.
2. Find $\theta$, the angle between the normal (dipole moment) and B: $\theta =90^{\circ }-60^{\circ }=30^{\circ }$.
3. Calculate torque: $\tau =IAB\sin \theta =(5.0A)(6.0\times 10^{-4}{\text{ m}}^2)(0.4T)(\sin 30^{\circ })=6.0\times 10^{-4}\text{ Ncdotpm}$.

Exam tip: If the problem gives the angle between the plane of the loop and B, always subtract from 90 degrees to get the correct $\theta$ for the torque formula.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to flip the force direction for negative charges, giving the same direction as a positive charge with identical velocity. Why: Students memorize the right-hand rule for positive charge and do not account for the sign of $q$ in the cross product. Correct move: Always explicitly note the sign of $q$ after finding the right-hand direction, and flip direction if $q$ is negative.
• Wrong move: Stopping after calculating radius when the question asks for the diameter of a particle’s circular path, getting an answer half the correct value. Why: Students default to outputting radius from the common formula, and do not read the question’s request carefully. Correct move: Circle the quantity the question asks for (radius, diameter, momentum, frequency) before starting calculations, and double-check at the end.
• Wrong move: Calculating non-zero net work done by magnetic force, using $W=Fd$. Why: Students confuse magnetic force with other constant forces and forget its key direction property. Correct move: If asked for work done by magnetic force, immediately state it is zero, because force is always perpendicular to displacement.
• Wrong move: Using the angle between the plane of the current loop and B as $\theta$ in the torque formula. Why: Problems often describe orientation relative to the plane, so students mix up plane angle and normal angle. Correct move: Recall $\theta$ is the angle between B and the dipole moment (normal), so $\theta =90^{\circ }-\varphi$, where $\varphi$ is the angle between the plane of the loop and B.
• Wrong move: Claiming net force on a closed current loop is zero in a non-uniform magnetic field. Why: Students memorize the zero net force rule and forget it only applies to uniform B. Correct move: Only use the zero net force rule for uniform B; integrate $d\vec{F}=Id\vec{l}\times \vec{B}$ for non-uniform B to find net force.

7. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

An electron moving with velocity $\vec{v}=2.0\times 10^5\hat{i}$ m/s enters a velocity selector region with uniform crossed fields $\vec{E}=100\hat{j}$ V/m and $\vec{B}=B\hat{k}$ T. The net force on the electron is zero. What is the magnitude of B? A) $2.0\times 10^{-7}$ T B) $5.0\times 10^{-4}$ T C) $2.0\times 10^{-3}$ T D) $5.0\times 10^6$ T

Worked Solution: For zero net force, the magnitude of the electric force must equal the magnitude of the magnetic force: $|qE|=|qvB|$. The charge $q$ cancels out, so $E=vB$, regardless of the sign of $q$ (both forces flip direction for negative charge, so cancellation still holds). Solving for B gives $B=E/v=100/(2.0\times 10^5)=5.0\times 10^{-4}$ T. The correct answer is B.

Question 2 (Free Response)

A straight 25 cm wire of mass 10 grams is hung horizontally from two springs, connected to a battery such that current flows to the west through the wire. A uniform horizontal magnetic field points north, with magnitude 1.0 T. (a) Draw a free-body diagram for the wire, and find the direction of the magnetic force on the wire. (b) The wire is in equilibrium when the current is 0.392 A. What is the net tension force from the two springs on the wire? (c) If the direction of the current is reversed, what will be the new net tension force at equilibrium?

Worked Solution: (a) Free-body diagram includes three forces: gravity downward, tension upward, and magnetic force. Using the right-hand rule: current = west (-x), B = north (+y), so $\vec{F}=I\vec{L}\times \vec{B}=-ILB\hat{z}$ (downward, if +z is up). The magnetic force points downward. (b) Calculate weight: $mg=(0.010\text{ kg})(9.8{\text{ m/s}}^2)=0.098$ N. Magnetic force magnitude: $F_B=ILB=(0.392\text{ A})(0.25\text{ m})(1.0\text{ T})=0.098$ N. Force balance for equilibrium: $T=mg+F_B=0.098+0.098=0.196$ N (or 0.20 N for two significant figures). (c) Reversing current reverses the direction of magnetic force, so $F_B$ points upward. New force balance: $T+F_B=mg$, so $T=mg-F_B=0.098-0.098=0$ N. The magnetic force exactly cancels gravity, so the springs exert zero tension.

Question 3 (Application / Real-World Style)

Modern particle detectors measure the radius of curvature of charged particle tracks to find their momentum. A proton produced in a collision is tracked in a 2.0 T uniform magnetic field, and has a measured track radius of 1.2 meters. Find the momentum of the proton in units of GeV/c (use $c=3.0\times 10^8$ m/s, $e=1.6\times 10^{-19}$ C, $1\text{ GeV}=1.6\times 10^{-10}$ J).

Worked Solution: From the circular motion relation, $p=qBr$. Substitute values: $p=(1.6\times 10^{-19}\text{ C})(2.0\text{ T})(1.2\text{ m})=3.84\times 10^{-19}$ kg·m/s. Multiply by $c$ to get $pc=(3.84\times 10^{-19}\text{ kgcdotpm/s})(3.0\times 10^8\text{ m/s})=1.152\times 10^{-10}$ J. Convert to GeV: $pc=1.152\times 10^{-10}/1.6\times 10^{-10}=0.72$ GeV, so $p=0.72$ GeV/c. This is a typical momentum for a proton produced in a medium-energy particle collision, matching the large 1.2 m radius measured in a detector.

8. Quick Reference Cheatsheet

9. What's Next

This chapter is the foundational prerequisite for all remaining topics in Unit 4 (Magnetic Fields), starting with the calculation of magnetic fields produced by currents via the Biot-Savart law and Ampere's law. To find the force between two parallel current-carrying wires, you need the force on a current-carrying wire from this chapter, combined with the B-field produced by the second wire that you will derive next. Without mastering the direction rules and force formulas from this chapter, you cannot correctly solve problems involving solenoids, motors, or electromagnetic induction in later units. This topic is also the basis for all real-world electromagnetic device problems that commonly appear on AP exam FRQs. Next topics to study:

• Biot-Savart Law
• Ampere's Law
• Magnetic Force Between Parallel Currents
• Faraday's Law of Induction