---
title: "AP Physics 2 Unit 12 Practice Test"
description: "A timed AP Physics 2 Unit 12 (Magnetism and Electromagnetism) practice test in the AP format, drawn from 860 multiple-choice questions and 45 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-physics-2-revised/unit-test/12"
type: "unit-test"
subject: "AP Physics 2"
unit: "Unit 12"
---

# AP Physics 2 Unit 12 Practice Test

## Overview

A timed AP Physics 2 Unit 12 (Magnetism and Electromagnetism) practice test in the AP format, drawn from 860 multiple-choice questions and 45 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

## Unit FRQs by Type

- [FRQ 1 – Mathematical Routines](/ap-physics-2-revised/frqs/mathematical-routines?unit=12): 13 questions
- [FRQ 2 – Translation Between Representations](/ap-physics-2-revised/frqs/translation-between-representations?unit=12): 6 questions
- [FRQ 3 – Experimental Design](/ap-physics-2-revised/frqs/experimental-design?unit=12): 12 questions
- [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-2-revised/frqs/qualitative-quantitative-translation?unit=12): 14 questions

## Practice Questions

### Question 1

A straight wire segment of length 0.50 m carries a current of 4.0 A. It is placed in a uniform magnetic field of magnitude 0.30 T. If the angle between the direction of the current and the magnetic field is 30°, what is the magnitude of the magnetic force exerted on the wire?

- (A) 0.30 N
- (B) 0.60 N
- (C) 0.52 N
- (D) 1.20 N

**Answer: (A)** Calculating the force requires applying the equation F = ILB sin(θ). Substituting the given values yields F = (4.0 A)(0.50 m)(0.30 T) sin(30°) = 0.60(0.5) = 0.30 N.

### Question 2

A circuit consists of a 10 V battery connected to a 5.0 Ω resistor. A 0.50 m straight section of the connecting wire is positioned perpendicular to a uniform 0.40 T magnetic field. What is the magnitude of the magnetic force on this wire segment?

- (A) 0.40 N
- (B) 2.00 N
- (C) 0.80 N
- (D) 0.10 N

**Answer: (A)** Solving this requires a two-step pathway: first using Ohm's law (I = V/R) to find current, then calculating force (F = ILB). With I = 10 V / 5.0 Ω = 2.0 A, the force is F = (2.0 A)(0.50 m)(0.40 T) = 0.40 N.

### Question 3

A metal rod with a mass of 0.060 kg and length 0.30 m carries a current of 4.0 A. It floats horizontally in a uniform magnetic field perpendicular to the rod. Using g = 10 m/s², what is the magnitude of the magnetic field required to balance the gravitational force?

- (A) 0.50 T
- (B) 0.05 T
- (C) 2.00 T
- (D) 0.15 T

**Answer: (A)** Balancing forces requires setting the magnetic force equal to the gravitational force (ILB = mg) and solving for B. Rearranging to B = mg/(IL) gives B = (0.060 kg × 10 m/s²) / (4.0 A × 0.30 m) = 0.60 / 1.2 = 0.50 T.

### Question 4

Two long, straight, parallel wires are separated by 0.10 m. Wire 1 carries 4.0 A and Wire 2 carries 5.0 A in opposite directions. What is the magnitude of the magnetic force exerted on a 3.0 m length of Wire 2? (μ₀ = 4π × 10⁻⁷ T·m/A)

- (A) 1.2 × 10⁻⁴ N
- (B) 4.0 × 10⁻⁵ N
- (C) 6.0 × 10⁻⁵ N
- (D) 1.2 × 10⁻⁶ N

**Answer: (A)** Parallel wires exert a force on each other given by $F = \frac{\mu_0 I_1 I_2 L}{2\pi r}$. Substituting gives $F = \frac{(4\pi\times10^{-7})(4.0)(5.0)(3.0)}{2\pi(0.10)} = \frac{(2\times10^{-7})(60)}{0.10} = 1.2\times10^{-4}$ N.

### Question 5

A circular loop of wire with radius $$R$$ carries a constant current $$I$$, creating a magnetic dipole moment of magnitude $$\mu_0$$. If the radius of the loop is increased to $$2R$$ while the current remains unchanged, what is the new magnitude of the magnetic dipole moment?

- (A) $$4\mu_0$$
- (B) $$2\mu_0$$
- (C) $$\mu_0$$
- (D) $$\frac{1}{4}\mu_0$$

**Answer: (A)** Connecting the dipole moment formula $$\mu = IA$$ to the area of a circle $$A = \pi R^2$$ reveals a quadratic dependence on radius. Since $$\mu \propto R^2$$, doubling the radius increases the moment by a factor of $$2^2 = 4$$.

### Question 6

A small bar magnet creates a magnetic field of magnitude $$B_0$$ at a distance $$d$$ along its axis. Assuming the magnet behaves as an ideal dipole where the field magnitude is inversely proportional to the cube of the distance ($$B \propto \frac{1}{r^3}$$), what is the field magnitude at a distance $$3d$$?

- (A) $$\frac{B_0}{27}$$
- (B) $$\frac{B_0}{9}$$
- (C) $$\frac{B_0}{3}$$
- (D) $$3B_0$$

**Answer: (A)** Applying the given functional dependence $$B \propto r^{-3}$$ allows for the prediction of the new field value. Tripling the distance results in the field decreasing by a factor of $$3^3 = 27$$.

### Question 7

A solenoid with $$N$$ turns and length $$L$$ carries a current $$I$$, producing a magnetic field of magnitude $$B_0$$ at its center. If the number of turns is doubled to $$2N$$ while the length and current remain constant, what is the new magnetic field magnitude?

- (A) $$2B_0$$
- (B) $$4B_0$$
- (C) $$B_0$$
- (D) $$\frac{B_0}{2}$$

**Answer: (A)** Using the solenoid field equation $$B = \mu_0 \frac{N}{L} I$$ identifies the linear proportionality between field strength and the number of turns. Doubling $$N$$ directly doubles the magnetic field.

### Question 8

A magnetic dipole placed in a uniform magnetic field of strength $$B$$ experiences a maximum torque $$\tau_0$$. If the magnetic field strength is reduced to $$0.5B$$ while the dipole remains unchanged, what is the new maximum torque?

- (A) $$0.5\tau_0$$
- (B) $$\tau_0$$
- (C) $$0.25\tau_0$$
- (D) $$2\tau_0$$

**Answer: (A)** Applying the torque formula $$\tau = \mu B \sin\theta$$ reveals a direct linear relationship between torque and magnetic field strength. Halving the field strength therefore halves the maximum torque.

### Question 9

An electron moves in a circular orbit of radius $$r$$ with speed $$v$$, creating a magnetic dipole moment $$\mu$$. If the speed of the electron is doubled to $$2v$$ while the radius of the orbit remains constant, what is the new magnetic dipole moment?

- (A) $$2\mu$$
- (B) $$4\mu$$
- (C) $$\mu$$
- (D) $$0.5\mu$$

**Answer: (A)** Analyzing the definition $$\mu = IA$$ where $$I = e/T$$ and $$T = 2\pi r/v$$ shows that $$\mu = evr/2$$. Thus, $$\mu$$ is directly proportional to speed $$v$$, so doubling $$v$$ doubles $$\mu$$.

### Question 10

A bar magnet is aligned with an external magnetic field $$B$$, resulting in a magnetic potential energy $$U_0$$ (defined relative to the perpendicular orientation). If the external magnetic field strength is quadrupled to $$4B$$, what is the new potential energy?

- (A) $$4U_0$$
- (B) $$16U_0$$
- (C) $$2U_0$$
- (D) $$U_0$$

**Answer: (A)** Using the potential energy equation $$U = -\vec{\mu} \cdot \vec{B}$$ identifies the linear proportionality between potential energy and field strength. Quadrupling $$B$$ directly quadruples the magnitude of $$U$$.

## Review the Unit

- [12.1 Magnetic Fields](/ap-physics-2-revised/unit-12/1-magnetic-fields/study-guide/8CQ1URzqZQqRb7qQ)
- [12.2 Magnetism and Moving Charges](/ap-physics-2-revised/unit-12/2-magnetism-and-moving-charges/study-guide/EquvYgnfwi2ptpX5)
- [12.3 Magnetism and Current-Carrying Wires](/ap-physics-2-revised/unit-12/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ)
- [12.4 Electromagnetic Induction and Faraday's Law](/ap-physics-2-revised/unit-12/4-electromagnetic-induction-and-faradays-law/study-guide/UqgM4DyzPPfjroi3)

## Other Unit Tests

- [Unit 9: Thermodynamics](/ap-physics-2-revised/unit-test/9)
- [Unit 10: Electric Force, Field, and Potential](/ap-physics-2-revised/unit-test/10)
- [Unit 11: Electric Circuits](/ap-physics-2-revised/unit-test/11)
- [Unit 13: Geometric Optics](/ap-physics-2-revised/unit-test/13)
- [Unit 14: Waves, Sound, and Physical Optics](/ap-physics-2-revised/unit-test/14)
- [Unit 15: Modern Physics](/ap-physics-2-revised/unit-test/15)

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