AP Physics 2 Unit 12 Practice Test

Pick a unit and take a timed, AP^®-style test built only from that unit's questions — with AI feedback on every free-response answer.

unit 9 unit 10 unit 11 unit 12 unit 13 unit 14 unit 15

Magnetism and Electromagnetism

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Two ways to practice unit 12

standard

The exam experience, sized to one unit — timed like the real thing

20 Multiple Choice Questions

40 min

1 FRQ – FRQ 1 – Mathematical Routines

24 min

1 FRQ – FRQ 2 – Translation Between Representations

24 min

Sections run back-to-back with the same directions, timers, and scoring as the real exam

Total: 1 hr 28 min

customizable

Pick your own mix of question types

multiple choice questions

free response questions (choose as many as you want)

FRQ 1 – Mathematical Routines · 24 minFRQ 2 – Translation Between Representations · 24 minFRQ 3 – Experimental Design · 24 minFRQ 4 – Qualitative/Quantitative Translation · 24 min

Total: 40 min

How scoring works: multiple choice is scored instantly, free-response answers get point-by-point AI feedback, and your result is a percentage for this unit — not a projected AP^® score.

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AP Physics 2 Unit 12 practice questions

Try 10 multiple-choice questions from Unit 12 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

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?
0.30 N
0.60 N
0.52 N
1.20 N
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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?
0.40 N
2.00 N
0.80 N
0.10 N
Show answer
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?
0.50 T
0.05 T
2.00 T
0.15 T
Show answer
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)
1.2 × 10⁻⁴ N
4.0 × 10⁻⁵ N
6.0 × 10⁻⁵ N
1.2 × 10⁻⁶ N
Show answer
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?
$4\mu_0$
$2\mu_0$
$\mu_0$
$\frac{1}{4}\mu_0$
Show answer
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 $3d$ ?
$\frac{B_0}{27}$
$\frac{B_0}{9}$
$\frac{B_0}{3}$
$3B_0$
Show answer
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?
$2B_0$
$4B_0$
$B_0$
$\frac{B_0}{2}$
Show answer
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?
$0.5\tau_0$
$\tau_0$
$0.25\tau_0$
$2\tau_0$
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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?
$2\mu$
$4\mu$
$\mu$
$0.5\mu$
Show answer
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?
$4U_0$
$16U_0$
$2U_0$
$U_0$
Show answer
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 $U$ .

What's on the AP Physics 2 Unit 12 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

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

Questions are drawn from 860 multiple-choice questions and 45 FRQs written for Unit 12, Magnetism and Electromagnetism.

Practice AP Physics 2 Unit 12 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

FRQ 1 – Mathematical Routines (13 questions)
FRQ 2 – Translation Between Representations (6 questions)
FRQ 3 – Experimental Design (12 questions)
FRQ 4 – Qualitative/Quantitative Translation (14 questions)

AP Physics 2 Unit 12 Practice Test

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