Magnetic Fields and Forces — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Magnetic force on moving charges and current-carrying wires, magnetic field generation by long straight wires and solenoids, Faraday's and Lenz's laws of electromagnetic induction, and induced EMF in moving conducting rods.

You should already know: AP Physics 1 or equivalent.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Magnetic Fields and Forces?

Magnetic fields are vector fields produced by moving electric charges (including aligned electron spin in permanent magnets) that exert non-contact forces on other moving charges or current-carrying conductors. As one of the four fundamental forces, the electromagnetic force (of which magnetic force is a component) makes up 15-20% of the AP Physics 2 exam content, per the official Course and Exam Description, and is tested in both multiple-choice and free-response sections. Common related terms include electromagnetism and magnetic induction.

2. Magnetic force on charges and currents

Magnetic force only acts on charges moving relative to the magnetic field: stationary charges experience zero magnetic force, a key detail examiners frequently test. For a single point charge moving through a uniform magnetic field, the magnitude of the force is given by: $$F_B=qvB\sin \theta$$ Where:

• $q$ = magnitude of the moving charge (C)
• $v$ = speed of the charge (m/s)
• $B$ = magnitude of the magnetic field (tesla, T)
• $\theta$ = angle between the velocity vector and magnetic field vector

The direction of the force is determined by the right-hand rule (RHR) for positive charges: point the fingers of your right hand in the direction of velocity, curl them toward the direction of the magnetic field, and your thumb points in the direction of the force. For negative charges (e.g., electrons), reverse the direction of the force given by the RHR, or use your left hand instead.

For a current-carrying wire (a collection of moving charges), the total magnetic force is: $$F_B=ILB\sin \theta$$ Where $I$ = current (A), $L$ = length of the wire segment inside the magnetic field (m), and $\theta$ = angle between the current direction and magnetic field vector.

Worked Example: A helium nucleus (charge $q=3.2\times 10^{-19}$ C) moves at $1.5\times 10^6$ m/s at a 30° angle to a 0.6 T uniform magnetic field. Calculate the magnitude of the magnetic force acting on the nucleus. Solution: Substitute values directly into the point charge force formula: $$F_B=(3.2\times 10^{-19})(1.5\times 10^6)(0.6)(\sin 30^{\circ })$$ $\sin 30^{\circ }=0.5$, so $F_B=1.44\times 10^{-13}$ N.

3. Magnetic field of a long wire and solenoid

Just as moving charges experience force in magnetic fields, moving charges (i.e., electric currents) produce magnetic fields around them. For an infinitely long straight current-carrying wire, the magnitude of the magnetic field at a radial distance $r$ from the wire is: $$B=\frac{{\mu }_{0}I}{2\pi r}$$ Where ${\mu }_0=4\pi \times 10^{-7}$ T·m/A is the permeability of free space (provided on the AP Physics 2 equation sheet). The field forms concentric circles around the wire, with direction given by the right-hand grip rule: grip the wire with your right hand, point your thumb in the direction of the current, and your curled fingers show the direction of the magnetic field lines.

A solenoid is a tightly wound coil of wire that produces a nearly uniform magnetic field inside the coil, with negligible field outside. The magnitude of the uniform internal magnetic field is: $$B={\mu }_{0}nI$$ Where $n=\frac{N}{L}$ is the number of turns of wire per unit length (N = total turns, L = total length of the solenoid). The direction of the internal field is found by the right-hand grip rule for coils: grip the solenoid with your right hand, curl your fingers in the direction of the current around the coil, and your thumb points to the north pole of the solenoid and the direction of the internal magnetic field.

Worked Example: Calculate the magnetic field inside a 0.4 m long solenoid with 1200 total turns carrying a 2.5 A current. Solution: First calculate turns per unit length: $n=1200/0.4=3000$ turns/m. Substitute into the solenoid field formula: $$B=(4\pi \times 10^{-7})(3000)(2.5)\approx 9.42\times 10^{-3}\text{ T}=9.42\text{ mT}$$

4. Faraday's law and Lenz's law

Electromagnetic induction is the process where a changing magnetic flux through a closed conducting loop induces an electromotive force (EMF), which drives an induced current if the loop is part of a complete circuit. First, magnetic flux ${\Phi }_B$ is the measure of magnetic field passing through a given area, defined as: $${\Phi }_B=BA\cos \theta$$ Where $A$ = area of the loop, $\theta$ = angle between the magnetic field vector and the normal vector (line perpendicular to the plane of the loop). The unit of flux is the weber (1 Wb = 1 T·m²).

Faraday's Law of Induction states the magnitude of the induced EMF is proportional to the rate of change of magnetic flux through the loop, and the number of turns in the coil: $$\epsilon =-N\frac{\Delta {\Phi }_B}{\Delta t}$$ Where $N$ = number of turns in the coil, $\frac{\Delta {\Phi }_B}{\Delta t}$ = rate of change of magnetic flux over time.

The negative sign in Faraday's law is explained by Lenz's Law, which gives the direction of the induced current: the induced current will produce its own magnetic field that opposes the change in magnetic flux that caused the induced current. A common exam trap is assuming the induced field opposes the original magnetic field, rather than the change in flux: if flux is increasing, the induced field points opposite the original field; if flux is decreasing, the induced field points in the same direction as the original field.

Worked Example: A 20-turn circular coil of area 0.02 m² is placed with its face perpendicular to a uniform magnetic field that decreases from 0.9 T to 0.1 T in 0.4 s. Calculate the magnitude of the induced EMF. Solution: First calculate the change in flux: $\Delta {\Phi }_B=\Delta B\cdot A=(0.1-0.9)(0.02)=-0.016$ Wb. The magnitude of the rate of change is $|0.016/0.4|=0.04$ Wb/s. Substitute into Faraday's law: $$|\epsilon |=20\times 0.04=0.8\text{ V}$$

5. Induced EMF in moving rods

Motional EMF is a specific case of Faraday's law where a conducting rod moves through a uniform magnetic field, creating a potential difference across its ends even if it is not part of a closed circuit. As the rod moves, free charges inside experience a magnetic force that pushes positive charges to one end of the rod and negative charges to the other, creating an internal electric field. At equilibrium, the electric force on charges equals the magnetic force: $qE=qvB$, so $E=vB$. The potential difference (EMF) across the rod length $L$ is $\epsilon =EL=vBL$. The full formula for motional EMF when the velocity is at an angle $\theta$ to the magnetic field is: $$\epsilon =vBL\sin \theta$$ Maximum EMF is produced when the velocity is perpendicular to both the magnetic field and the length of the rod ($\theta =90^{\circ }$, $\sin \theta =1$). If the rod is part of a closed circuit, this EMF drives an induced current.

Worked Example: A 0.6 m long aluminum rod moves at 3 m/s at a 90° angle to a 0.4 T magnetic field, with the velocity perpendicular to the rod length. Calculate the induced EMF, and the induced current if the rod is part of a circuit with total resistance 3 Ω. Solution: Substitute into the motional EMF formula: $\epsilon =3\times 0.4\times 0.6=0.72$ V. The induced current is $I=\epsilon /R=0.72/3=0.24$ A.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Using the right-hand rule for negative charges without reversing the force direction. Why: Students memorize the RHR for protons and forget electrons have negative charge. Correct move: Either use your left hand for negative charges, or apply the standard RHR then flip the force direction by 180°.
• Wrong move: Using the angle between the magnetic field and the loop face in the magnetic flux formula, instead of the angle between the field and the loop normal. Why: Misinterpretation of the $\theta$ definition. Correct move: Always measure $\theta$ relative to the line pointing straight out of the loop plane, not the loop surface itself.
• Wrong move: Using total turns $N$ instead of turns per unit length $n$ in the solenoid magnetic field formula. Why: Mixing up the long wire and solenoid formulas. Correct move: Check units: $n$ has units of turns per meter, so ${\mu }_{0}nI$ gives units of tesla, while ${\mu }_{0}NI$ does not.
• Wrong move: Applying Lenz's law to oppose the original magnetic field instead of the change in flux. Why: Oversimplification of the "oppose" rule. Correct move: First identify if flux is increasing or decreasing: induced B field fights the change, not the original field.
• Wrong move: Calculating magnetic force or motional EMF when velocity is parallel to the magnetic field. Why: Forgetting that $\sin 0^{\circ }=0$. Correct move: If velocity and B field are parallel, magnetic force is zero, no EMF is induced.

7. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A 3 A current flows through a 0.2 m long straight wire placed in a uniform 0.5 T magnetic field, oriented parallel to the field lines. What is the magnitude of the magnetic force on the wire? A) 0 N B) 0.15 N C) 0.3 N D) 0.6 N Solution: The angle between the current direction and magnetic field is 0°, so $\sin 0^{\circ }=0$. Substitute into the wire force formula: $F_B=3\times 0.2\times 0.5\times 0=0$ N. Correct answer: A. The 0.3 N option is a distractor for students who forget the angle dependence.

Question 2 (Free Response Part A)

Calculate the magnitude of the magnetic field 0.05 m from a long straight wire carrying a 12 A current. Show all steps. Solution: Use the long wire magnetic field formula: $$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{(4\pi \times 10^{-7})(12)}{2\pi (0.05)}$$ Cancel $\pi$ from numerator and denominator: $B=\frac{(2\times 10^{-7})(12)}{0.05}=4.8\times 10^{-5}$ T, or 48 μT. Partial credit is awarded for writing the correct formula first, so always show your starting equation.

Question 3 (Multi-Part Free Response)

A square conducting loop of side length 0.15 m, total resistance 4 Ω, is pulled out of a uniform 0.3 T magnetic field at a constant speed of 2 m/s. The magnetic field is perpendicular to the plane of the loop. a) Calculate the magnitude of the induced EMF in the loop as it exits the field. b) Calculate the magnitude of the induced current. c) State the direction of the induced current, justifying your answer with Lenz's law. Solutions: a) Use the motional EMF formula, as the side of the loop cutting the field lines acts as a moving rod: $\epsilon =vBL=2\times 0.3\times 0.15=0.09$ V. b) Induced current: $I=\epsilon /R=0.09/4=0.0225$ A, or 22.5 mA. c) Direction: Assume the original magnetic field points into the page. As the loop exits the field, the flux into the page decreases. Lenz's law states the induced current will produce a magnetic field that opposes this decrease, so the induced B field points into the page. Using the right-hand grip rule for loops, a clockwise current produces an inward B field, so the current flows clockwise.

8. Quick Reference Cheatsheet

9. What's Next

Magnetic fields and forces are a foundational topic for the rest of the AP Physics 2 curriculum: you will use these concepts to understand AC circuit components like inductors, the operation of transformers and generators, the propagation of electromagnetic waves, and even modern physics applications like mass spectrometry and particle accelerator design. Lenz's law and Faraday's law in particular are frequently combined with circuit analysis in 10-point free response questions, so mastering these rules is critical for scoring a 5 on the exam.

If you are stuck on any of the concepts, worked examples, or practice questions in this guide, you can ask Ollie, our AI tutor, for personalized explanations, targeted practice problems, or step-by-step walkthroughs tailored to your specific learning gaps. You can also find more AP Physics 2 topic guides, full-length practice tests, and flashcards on the homepage to reinforce your mastery before exam day.