Magnetic Fields — AP Physics C: E&M Unit Overview

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

Covers: The full scope of the AP Physics C: E&M Magnetic Fields unit, including magnetic forces on charges and currents, Biot-Savart Law for arbitrary current distributions, and Ampère’s Law for symmetric magnetic field calculations.

You should already know: Vector cross product operations and right-hand rules, properties of steady electric current, electric field symmetry and Gauss’s Law for electrostatics.

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. Why This Unit Matters

This unit makes up 20–25% of the total AP Physics C: E&M exam score, appearing regularly in both multiple-choice and free-response sections. It is the critical bridge between your prior knowledge of electrostatics and steady currents, and the upcoming topic of electromagnetic induction, which often accounts for the longest and highest-weight free-response question on the exam. Magnetic interactions are the foundation of all modern electric technologies, from electric motors to MRI scanners to power transmission, so the concepts here connect directly to real-world engineering applications. Unlike electric fields, which exert force on any charge regardless of motion, magnetic fields only interact with moving charges, creating a new set of rules that build on your vector math and symmetry skills from electrostatics. Mastery of this unit is required to solve any induction problem, as you will need to calculate magnetic fields and their forces to find induced emf and current.

2. Unit Concept Map

The three subtopics of this unit build sequentially from fundamentals of interaction to general calculation to symmetric shortcuts, mirroring the structure of electrostatics:

1. Magnetic Fields and Magnetic Forces: This is the foundational subtopic. It introduces the definition of the magnetic field vector $\vec{B}$, the Lorentz force law for moving charges, and the force law for current-carrying wires. It answers the first core question of the unit: If I already know the magnetic field in a region, how does it affect moving charges and currents? You will also learn right-hand rules for cross product direction, which are used across all subsequent subtopics.
2. Biot-Savart Law: This is the general law for calculating the magnetic field produced by any steady current distribution, analogous to Coulomb’s Law for electric fields from charge distributions. It works for any shape or size of current, but always requires integrating over individual current elements to get the total B-field.
3. Ampère’s Law: This is the symmetry-based shortcut for calculating B-fields, analogous to Gauss’s Law for electric fields. It only works for current distributions with high symmetry (cylindrical, planar, translational), but when applicable, it eliminates the need for a full multi-dimensional Biot-Savart integral.

Nearly all AP exam problems on this unit combine at least two of these subtopics, so the sequential build is intentional: you cannot calculate a force if you cannot first find the B-field, and you will not earn full credit if you use the wrong method to find the B-field.

3. Guided Tour of a Unit-Style Exam Problem

Below we walk through a typical multi-part AP exam problem that combines multiple subtopics, showing how each subtopic contributes to the full solution.

Problem Statement

A long straight conducting wire of radius $R$ carries a uniform steady current $I$ along its central axis. A second long straight wire, carrying current $2I$ in the opposite direction, runs parallel to the first wire with its center a distance $d>R$ from the first wire's center. Find (a) the magnitude of the magnetic field inside the first wire at radius $r<R$, and (b) the magnitude of the force per unit length between the two wires.

Guided Step-by-Step Tour

1. Map problem parts to unit subtopics: First, part (a) asks for a magnetic field produced by a current distribution. The first wire has cylindrical symmetry: it is infinitely long, the current is uniform around the axis, so the B-field only depends on $r$, not angle or position along the wire. This matches the use case for Ampère's Law (third subtopic), rather than the general but more calculation-heavy Biot-Savart Law (second subtopic). For part (b), once we have the B-field from the first wire, we need to calculate the force on the second current-carrying wire, which uses the magnetic force rules from the first subtopic Magnetic Fields and Magnetic Forces.
2. Apply Ampère's Law for part (a): Ampère's law states $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enclosed}}$. We choose a circular Amperian loop of radius $r<R$ concentric with the wire. $\vec{B}$ is tangent to the loop everywhere, so the dot product simplifies to $B\oint dl=B(2\pi r)$. The enclosed current is proportional to the area: $I_{\text{enclosed}}=I\frac{\pi r^2}{\pi R^2}=I\frac{r^2}{R^2}$. Solving gives $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$, which matches the expected linear dependence inside a uniform current.
3. Find B-field at the second wire's location: Repeating Ampère's Law for $r=d>R$, enclosed current is just $I$, so $B=\frac{{\mu }_{0}I}{2\pi d}$, as expected for an infinite wire.
4. Apply magnetic force rules for part (b): The force on a current-carrying wire of length $L$ is $|\vec{F}|=I_{\text{wire}}LB\sin \theta$, where $\theta$ is the angle between the current direction and $\vec{B}$. For parallel wires, the current of the second wire is perpendicular to the B-field from the first, so $\sin \theta =1$. Substituting values: $|F|=(2I)L\left(\frac{{\mu }_{0}I}{2\pi d}\right)=\frac{{\mu }_0{I}^{2}L}{\pi d}$. Force per unit length is $\frac{|F|}{L}=\frac{{\mu }_0{I}^2}{\pi d}$.

If the wire were an arbitrary shape (e.g., a finite bent wire), we would have used Biot-Savart Law to calculate the B-field instead of Ampère's.

Exam tip for unit problems: Always break the problem into steps: first find the B-field (using the correct method for the symmetry), then calculate the force or other quantity from the B-field. AP graders award points for each step separately.

4. Cross-Cutting Common Pitfalls

These are the most common unit-wide traps that trip up students across all three subtopics:

• Wrong move: Mixing up the order of vectors in the cross product for force, writing $\vec{F}=q\vec{B}\times \vec{v}$ instead of $q\vec{v}\times \vec{B}$, leading to reversed force direction. Why: Students confuse which vector is the source vs the moving charge, and swap right-hand rule finger order across all cross product problems. Correct move: Always write the cross product in standard Lorentz force order first: $\vec{F}=q\vec{v}\times \vec{B}$ for charges, $\vec{F}=I\vec{L}\times \vec{B}$ for wires, with current/velocity first.
• Wrong move: For Biot-Savart, integrating only the magnitude of $d\vec{B}$ instead of integrating individual vector components. Why: Students get used to scalar integrals from symmetric problems and forget B-field is a vector that cancels across components. Correct move: For any non-symmetric Biot-Savart problem, split $d\vec{B}$ into x, y, z components before integrating, and integrate each component separately.
• Wrong move: Applying Ampère's Law to a non-symmetric current distribution to get a quick answer. Why: Students see Ampère's is faster than Biot-Savart and overuse it, just like overusing Gauss's Law for non-symmetric charge distributions. Correct move: Before applying Ampère's Law, confirm that $\vec{B}$ is constant in magnitude and parallel to your Amperian loop everywhere. If not, you must use Biot-Savart.
• Wrong move: Miscalculating enclosed current for Ampère's Law by using total current instead of only current inside the Amperian loop. Why: Students confuse total current in the system with enclosed current, especially for problems with multiple conductors or inside/outside regions. Correct move: Always explicitly write $I_{\text{enclosed}}$ as the net current passing through the area bounded by your Amperian loop before solving for B.
• Wrong move: Forgetting to flip the force direction for negative charges after applying the right-hand rule. Why: Right-hand rules are defined for positive charges, and students forget the sign of $q$ changes the direction. Correct move: After finding the direction of $\vec{v}\times \vec{B}$, multiply by the sign of $q$ to get the final force direction.

5. Quick Check (When To Use Which Subtopic)

Test your understanding of tool selection by matching each scenario to the correct subtopic:

1. Find the magnetic force on an electron moving at constant speed through a uniform known magnetic field.
2. Find the magnetic field produced by a finite straight wire segment at a point off the axis of the wire.
3. Find the magnetic field inside a long uniformly wound solenoid carrying steady current.
4. Find the magnetic force per unit length between two parallel current-carrying wires, given their currents and separation.

Quick Check Answers

1. Magnetic Fields and Magnetic Forces: You already have the B-field, you only need to apply the Lorentz force law.
2. Biot-Savart Law: The finite wire has no translational symmetry along its length, so Ampère's Law cannot simplify the integral.
3. Ampère's Law: The infinite solenoid has perfect translational symmetry, so Ampère's Law gives a simple solution.
4. Both: First use Ampère's Law (faster for symmetric infinite wires) to find the B-field from one wire at the location of the other, then use magnetic force rules to find the net force.

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

Question 1 (Multiple Choice)

Which of the following current distributions cannot have its magnetic field calculated directly via Ampère's Law? A) An infinitely long straight wire of uniform current density B) A solid toroidal ring of uniform current density C) A finite straight wire segment of uniform current density D) An infinitely long hollow cylindrical conducting tube carrying uniform current on its outer surface

Worked Solution: Ampère's Law only gives a simple solution when $\vec{B}$ is constant in magnitude and tangent to the Amperian loop everywhere. A finite straight wire has end effects that make B-field magnitude vary along any concentric loop, so no valid Amperian loop exists for this case. All other options have sufficient symmetry to use Ampère's Law directly. The correct answer is C.

Question 2 (Free Response)

A point particle with charge $q$ and mass $m$ enters a region of uniform magnetic field $B$ with velocity $v$ perpendicular to the direction of $\vec{B}$. (a) Derive the radius of the circular path the particle follows in the magnetic field. (b) Derive the period of the particle's circular motion, and show that it does not depend on the particle's speed $v$ (for non-relativistic speeds). (c) If the magnetic field is produced by a large solenoid with $n$ turns per unit length carrying current $I$, write an expression for $B$ inside the solenoid in terms of given constants.

Worked Solution: (a) The magnetic force provides the centripetal force for circular motion. Magnitude of the magnetic force is $F=qvB$ (since $v\perp B$). Centripetal force is $\frac{m{v}^2}{r}$. Equate the two: $qvB=\frac{m{v}^2}{r}$. Solving for $r$ gives $r=\frac{mv}{qB}$. (b) Period $T$ is time to complete one orbit: $T=\frac{2\pi r}{v}$. Substitute $r$ from part (a): $T=\frac{2\pi (mv/qB)}{v}=\frac{2\pi m}{qB}$. The speed $v$ cancels out, so $T$ is independent of $v$, as required. (c) For an ideal infinite solenoid, Ampère's Law gives the uniform internal magnetic field: $B={\mu }_{0}nI$.

Question 3 (Application / Real-World Style)

In a cathode ray tube (old portable television), electrons are accelerated from rest through a potential difference of 12 kV, then enter a uniform magnetic field of $8.0\times 10^{-4}$ T perpendicular to their velocity. Find the radius of curvature of the electrons' path in the magnetic field. Use $m_e=9.11\times 10^{-31}$ kg, $e=1.60\times 10^{-19}$ C.

Worked Solution: First, use conservation of energy to find the electron speed after acceleration: $eV=\frac{1}{2}{m}_e{v}^2$, so $v=\sqrt{\frac{2eV}{m_e}}=\sqrt{\frac{2(1.60\times 10^{-19})(12000)}{9.11\times 10^{-31}}}\approx 6.49\times 10^7$ m/s. Then use the cyclotron radius formula: $r=\frac{m_{e}v}{eB}=\frac{(9.11\times 10^{-31})(6.49\times 10^7)}{(1.60\times 10^{-19})(8.0\times 10^{-4})}\approx 0.46$ m. A radius of 46 cm matches the physical size of the screen in small portable cathode ray tube televisions, as expected.

7. Quick Reference Cheatsheet

8. See Also (Sub-Topics In This Unit)

• Magnetic Fields and Magnetic Forces
• Biot-Savart Law
• Ampère's Law