AP Physics C Electricity and Magnetism Quiz: Electromagnetic Induction

12 questions · exam conditions

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Electromagnetic InductionQuestion 1 of 12

An ideal transformer has $N_p=200$ and $N_s=800$ with $V_p=9.0\,\mathrm{V}$ ; calculate induced secondary voltage magnitude.

36\,\mathrm{V}

, using

V_s=V_p(N_s/N_p)

for a step-up transformer.

2.3\,\mathrm{V}

, from using

V_s=V_p(N_p/N_s)

with ratio reversed.

18\,\mathrm{V}

, from using

N_s/N_p=2

due to misreading turns.

36\,\mathrm{W}

, correct number but watts are incorrectly reported as voltage.

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electromagnetic Induction

Practice Electromagnetic Induction in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electromagnetic Induction, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An ideal transformer has $N_p=200$ and $N_s=800$ with $V_p=9.0\,\mathrm{V}$ ; calculate induced secondary voltage magnitude.

$36\,\mathrm{V}$ , using $V_s=V_p(N_s/N_p)$ for a step-up transformer. (correct answer)
$2.3\,\mathrm{V}$ , from using $V_s=V_p(N_p/N_s)$ with ratio reversed.
$18\,\mathrm{V}$ , from using $N_s/N_p=2$ due to misreading turns.
$36\,\mathrm{W}$ , correct number but watts are incorrectly reported as voltage.

Explanation: This question tests understanding of step-up transformer operation using the fundamental voltage-turns relationship. For an ideal transformer, V_s = V_p × (N_s/N_p), where the secondary has more turns than the primary (N_s > N_p) in a step-up configuration. Calculating: V_s = 9.0 V × (800/200) = 9.0 V × 4 = 36 V. Choice A is correct because it properly identifies this as a step-up transformer with a 4:1 turns ratio, increasing voltage proportionally. Choice B represents the common error of inverting the turns ratio, which would make this a step-down transformer. Students often struggle with ratio direction, so emphasize that the voltage ratio follows the turns ratio directly: if secondary turns are 4× primary turns, then secondary voltage is 4× primary voltage. Use physical demonstrations showing how transformers trade voltage for current while ideally conserving power.

Question 2

A conducting rod moves right at $4.0\ \text{m/s}$ through $B=0.60\ \text{T}$ out of page; induced current direction?

Clockwise, because induced $\vec{B}$ must oppose increasing out-of-page magnetic flux through the loop. (correct answer)
Counterclockwise, because induced $\vec{B}$ must oppose increasing out-of-page magnetic flux through the loop.
Clockwise, because induced $\vec{B}$ must reinforce increasing out-of-page magnetic flux through the loop.
No current, because induction requires a changing magnetic field, not motion through a field.

Explanation: This question tests understanding of electromagnetic induction and Lenz's Law when a conductor moves through a magnetic field, specifically determining induced current direction. When a conducting rod moves through a magnetic field, the motional EMF ε = BLv induces current whose magnetic field opposes the change in flux, as dictated by Lenz's Law. In this scenario, a rod moves right at 4.0 m/s through a 0.60 T field directed out of the page, increasing the out-of-page flux through any closed loop formed by the rod. Choice A is correct because as the rod moves right, the out-of-page flux through the loop increases, so by Lenz's Law, the induced current must flow clockwise to create an into-page magnetic field that opposes this flux increase. Choice B is incorrect because counterclockwise current would create an out-of-page field, reinforcing rather than opposing the flux change, violating Lenz's Law. Encourage students to follow a systematic approach: determine flux change direction, identify what induced field would oppose it, then use the right-hand rule to find current direction. Practice with fields both into and out of the page to build confidence with three-dimensional thinking.

Question 3

A transformer has $N_p=200$ , $N_s=800$ , and $V_p=24\ \text{V}_{\text{rms}}$ ; calculate induced secondary voltage magnitude.

$96\ \text{V}_{\text{rms}}$ using $V_s=V_p(N_s/N_p)=24(800/200)$ for an ideal transformer. (correct answer)
$6.0\ \text{V}_{\text{rms}}$ using $V_s=V_p(N_p/N_s)$ , inverting the turns ratio by mistake.
$9.6\ \text{V}_{\text{rms}}$ from $24(80/200)$ , dropping a zero in $N_s$ during substitution.
$96\ \text{A}$ found from $V_p(N_s/N_p)$ , giving current units instead of induced voltage units.

Explanation: This question tests understanding of electromagnetic induction in transformers, specifically calculating secondary voltage in a step-up transformer configuration. The transformer equation Vs = Vp(Ns/Np) shows that voltage is proportional to the turns ratio, with Ns > Np creating a step-up transformer that increases voltage. In this scenario, a transformer with 200 primary turns and 800 secondary turns has 24 V applied to the primary, creating a 4:1 step-up ratio. Choice A is correct because it properly applies the transformer equation: Vs = 24(800/200) = 24(4) = 96 V, recognizing this steps up the voltage by a factor of 4. Choice B is incorrect because it inverts the turns ratio, using Vp(Np/Ns) = 24(200/800) = 6.0 V, which would represent a step-down transformer instead of the actual step-up configuration. Encourage students to first identify whether Ns > Np (step-up) or Ns < Np (step-down) before calculating. Always use Ns/Np for the voltage ratio, never the inverse, to avoid sign errors.

Question 4

A nearby solenoid current increases at $dI/dt=3.0\ \text{A/s}$ ; a loop has mutual inductance $M=0.040\ \text{H}$ ; induced EMF magnitude?

$0.12\ \text{V}$ using $|\varepsilon|=M\,|dI/dt|=0.040\times3.0$ for mutual induction. (correct answer)
$0.75\ \text{V}$ using $\varepsilon=\mu_0 M\,dI/dt$ , incorrectly inserting an extra $\mu_0$ factor.
$0.012\ \text{V}$ from $0.040\times0.30$ , dropping a factor of $10$ in $dI/dt$ .
$0.12\ \text{A}$ found from $M\,dI/dt$ , giving current units rather than induced EMF units.

Explanation: This question tests understanding of electromagnetic induction through mutual inductance, where changing current in one circuit induces EMF in a nearby circuit. Mutual inductance M quantifies the coupling between two circuits, with the induced EMF given by ε = -M(dI/dt), where dI/dt is the rate of current change in the primary circuit. In this scenario, a solenoid's current increases at 3.0 A/s, and a nearby loop has mutual inductance M = 0.040 H with the solenoid, creating magnetic coupling between them. Choice A is correct because it properly applies the mutual inductance formula: |ε| = M|dI/dt| = 0.040 × 3.0 = 0.12 V, recognizing that henries times amperes per second yields volts. Choice B is incorrect because it incorrectly inserts an extra μ₀ factor, confusing the mutual inductance formula with field calculations, yielding an erroneous result. Encourage students to remember that mutual inductance M already incorporates all geometric and material factors. Practice dimensional analysis to verify that H·(A/s) = V, reinforcing the physical meaning of inductance.

Question 5

A transformer has $N_p=500$ , $N_s=50$ , and $V_p=120\ \text{V}_{\text{rms}}$ ; calculate induced secondary voltage magnitude.

$12\ \text{V}_{\text{rms}}$ using $V_s/V_p=N_s/N_p=50/500$ for an ideal transformer. (correct answer)
$1.2\ \text{V}_{\text{rms}}$ from $120\times(50/5000)$ , adding an extra zero to $N_p$ .
$1200\ \text{V}_{\text{rms}}$ using the inverted ratio $V_s/V_p=N_p/N_s$ by mistake.
$12\ \text{T}$ found by multiplying turns ratio by $B$ , giving magnetic-field units, not voltage.

Explanation: This question tests understanding of electromagnetic induction in transformers, specifically the relationship between primary and secondary voltages based on the turns ratio. A transformer operates on the principle that the voltage ratio equals the turns ratio: Vs/Vp = Ns/Np, allowing voltage to be stepped up or down while conserving power in an ideal transformer. In this scenario, a transformer with 500 primary turns and 50 secondary turns has 120 V applied to the primary, creating a step-down transformer. Choice A is correct because it properly applies the transformer equation: Vs = Vp(Ns/Np) = 120(50/500) = 12 V, recognizing this is a 10:1 step-down transformer. Choice C is incorrect because it inverts the turns ratio, using Np/Ns instead of Ns/Np, yielding 1200 V and creating a step-up transformer instead of step-down. Encourage students to check whether the transformer steps up or down by comparing Ns to Np before calculating. Reinforce that secondary voltage is always proportional to the secondary-to-primary turns ratio, not the inverse.

Question 6

A circular loop ( $A=0.010\ \text{m}^2$ ) in $B=0.40\ \text{T}$ has its normal rotate from $0^\circ$ to $60^\circ$ ; determine flux change.

$-2.0\times10^{-3}\ \text{Wb}$ because $\Delta\Phi=BA(\cos60^\circ-\cos0^\circ)$ for the loop. (correct answer)
$+2.0\times10^{-3}\ \text{Wb}$ because $\Delta\Phi=BA(\cos0^\circ-\cos60^\circ)$ is used incorrectly.
$-4.0\times10^{-3}\ \text{Wb}$ because $\cos60^\circ$ is mistakenly taken as $0$ instead of $0.5$ .
$-2.0\times10^{-3}\ \text{V}$ because flux change is reported in volts instead of webers.

Explanation: This question tests understanding of magnetic flux and how it changes when a loop's orientation relative to a magnetic field changes. Magnetic flux is given by Φ = BA cos θ, where θ is the angle between the magnetic field and the loop's normal vector, making orientation crucial for flux calculations. In this scenario, a circular loop with area 0.010 m² in a 0.40 T field rotates so its normal changes from 0° to 60° relative to the field, changing the flux through the loop. Choice A is correct because the flux change is ΔΦ = BA(cos 60° - cos 0°) = (0.40)(0.010)(0.5 - 1) = -2.0×10⁻³ Wb, properly using cos 60° = 0.5 and recognizing flux decreases. Choice C is incorrect because it treats cos 60° as 0 instead of 0.5, calculating ΔΦ = BA(0 - 1) = -4.0×10⁻³ Wb, doubling the actual flux change. Encourage students to memorize key angle values: cos 0° = 1, cos 60° = 0.5, cos 90° = 0. Practice drawing diagrams showing the loop normal and field direction to visualize the angle θ clearly.

Question 7

A $200$ -turn square loop ( $A=0.020\ \text{m}^2$ ) in $B=0.50\ \text{T}$ rotates at $\omega=120\ \text{rad/s}$ ; what induced EMF magnitude?

$0.24\ \text{V}$ using $\varepsilon=NBA\omega$ but with $A$ misread as $0.0020\ \text{m}^2$ .
$240\ \text{V}$ using $\varepsilon=NBA\omega$ with $N=200$ , $B=0.50\ \text{T}$ , $A=0.020\ \text{m}^2$ . (correct answer)
$120\ \text{V}$ using $\varepsilon=BA\omega$ and incorrectly omitting the $N=200$ turns factor.
$240\ \text{A}$ computed as $I=NBA\omega$ , giving current units instead of induced EMF units.

Explanation: This question tests understanding of electromagnetic induction as described by Faraday's Law, specifically calculating the peak induced EMF in a rotating coil. Electromagnetic induction involves generating an electromotive force when a coil rotates in a magnetic field, with the peak EMF given by ε = NBAω, where N is the number of turns, B is the magnetic field strength, A is the area, and ω is the angular velocity. In this scenario, a 200-turn square loop with area 0.020 m² rotates at 120 rad/s in a 0.50 T magnetic field, which continuously changes the magnetic flux through the loop. Choice B is correct because it accurately calculates the peak EMF as ε = (200)(0.50)(0.020)(120) = 240 V, properly including all factors. Choice C is incorrect because it omits the number of turns N = 200, yielding only 120 V, a common mistake when students forget that each turn contributes to the total EMF. Encourage students to write out the formula explicitly and check units: V = (turns)(T)(m²)(rad/s) = V. Practice identifying all relevant parameters in rotating coil problems to avoid missing factors.

Question 8

A $150$ -turn loop ( $A=0.030\ \text{m}^2$ ) in $B=0.20\ \text{T}$ rotates at $f=10\ \text{Hz}$ ; what induced EMF magnitude?

$56.5\ \text{V}$ using $\varepsilon_0=NBA\omega$ with $\omega=2\pi f=62.8\ \text{rad/s}$ . (correct answer)
$9.0\ \text{V}$ using $\varepsilon_0=NBAf$ , incorrectly using $f$ instead of $\omega$ in Faraday's law.
$5.65\ \text{V}$ from $56.5/10$ , dividing by frequency again without physical justification.
$56.5\ \text{A}$ computed as $NBA\omega$ , giving current units rather than induced EMF units.

Explanation: This question tests understanding of electromagnetic induction in rotating coils, specifically calculating peak EMF when given rotation frequency rather than angular velocity. For a coil rotating at frequency f in a magnetic field, the peak EMF is ε₀ = NBAω where ω = 2πf, making the complete formula ε₀ = NBA(2πf). In this scenario, a 150-turn loop with area 0.030 m² rotates at 10 Hz in a 0.20 T field, requiring conversion from frequency to angular velocity: ω = 2π(10) = 62.8 rad/s. Choice A is correct because it properly calculates ε₀ = (150)(0.20)(0.030)(62.8) = 56.5 V, correctly converting frequency to angular velocity before applying Faraday's law. Choice B is incorrect because it uses frequency directly without converting to angular velocity, calculating (150)(0.20)(0.030)(10) = 9.0 V, missing the 2π factor entirely. Encourage students to always convert frequency to angular velocity using ω = 2πf before applying rotational EMF formulas. Practice problems with both f and ω given to reinforce when conversion is needed.

Question 9

A $0.30\ \text{m}$ rod moves right at $6.0\ \text{m/s}$ through $B=0.80\ \text{T}$ into page; induced current direction?

Counterclockwise, because induced $\vec{B}$ must oppose increasing into-page flux through the loop. (correct answer)
Clockwise, because induced $\vec{B}$ must oppose increasing into-page flux through the loop.
Counterclockwise, because induced $\vec{B}$ must reinforce increasing into-page flux through the loop.
No current, because uniform $B$ produces zero induction regardless of rod motion in the field.

Explanation: This question tests understanding of electromagnetic induction and Lenz's Law, specifically determining the direction of induced current when a conductor moves through a magnetic field. When a conducting rod moves through a magnetic field, it experiences a motional EMF given by ε = BLv, and Lenz's Law dictates that the induced current creates a magnetic field opposing the change in flux. In this scenario, a 0.30 m rod moves right at 6.0 m/s through a 0.80 T field directed into the page, increasing the flux through any closed loop formed by the rod. Choice A is correct because as the rod moves right, the flux into the page through the loop increases, so by Lenz's Law, the induced current must flow counterclockwise to create an out-of-page magnetic field that opposes this increase. Choice B is incorrect because clockwise current would create an into-page field, reinforcing rather than opposing the flux change, violating Lenz's Law. Encourage students to use the right-hand rule systematically: first determine the direction of flux change, then find what induced field would oppose it, and finally determine the current direction. Practice with various orientations to build intuition for Lenz's Law applications.

Question 10

A Faraday disc of radius $0.10\ \text{m}$ spins at $600\ \text{rpm}$ in uniform $B=0.50\ \text{T}$ ; induced center-edge voltage magnitude?

$0.31\ \text{V}$ using $\Delta V=\tfrac12 B\omega R^2$ with $\omega=600(2\pi/60)$ . (correct answer)
$0.63\ \text{V}$ using $\Delta V=B\omega R^2$ , incorrectly missing the $1/2$ factor.
$0.031\ \text{V}$ using $\omega=600/60$ , forgetting the $2\pi$ conversion from rpm to rad/s.
$0.31\ \text{T}$ computed from $\tfrac12 B\omega R^2$ , giving magnetic-field units, not volts.

Explanation: This question tests understanding of electromagnetic induction in a Faraday disc (homopolar generator), where a conducting disc rotating in a magnetic field generates voltage between center and edge. For a disc of radius R rotating at angular velocity ω in uniform field B, the induced voltage is ΔV = ½BωR², with the factor of ½ arising from the radial integration of the motional EMF. In this scenario, a disc with radius 0.10 m spins at 600 rpm in a 0.50 T field, requiring conversion of rpm to rad/s: ω = 600(2π/60) = 62.8 rad/s. Choice A is correct because it properly calculates ΔV = ½(0.50)(62.8)(0.10)² = 0.31 V, including both the ½ factor and correct angular velocity conversion. Choice C is incorrect because it uses ω = 600/60 = 10 rad/s, forgetting the 2π factor in converting rpm to rad/s, yielding only 0.031 V. Encourage students to memorize the Faraday disc formula with its distinctive ½ factor. Always convert rpm to rad/s using ω = (rpm)(2π/60), and verify units throughout the calculation.

Question 11

A transformer has $N_p=500$ and $N_s=50$ with $V_p=120\,\mathrm{V_{rms}}$ ; calculate induced secondary voltage magnitude.

$12\,\mathrm{V_{rms}}$ , using $V_s/V_p=N_s/N_p$ for an ideal transformer. (correct answer)
$1200\,\mathrm{V_{rms}}$ , from inverting the turns ratio $N_p/N_s$ incorrectly.
$24\,\mathrm{V_{rms}}$ , from doubling $N_s/N_p$ due to a ratio arithmetic error.
$12\,\mathrm{A}$ , correct magnitude but amperes are mistakenly used instead of volts.

Explanation: This question tests understanding of ideal transformer operation, specifically the voltage transformation ratio relationship. For an ideal transformer, the voltage ratio equals the turns ratio: V_s/V_p = N_s/N_p, which gives V_s = V_p × (N_s/N_p). With N_p = 500, N_s = 50, and V_p = 120 V_rms, we calculate: V_s = 120 × (50/500) = 120 × 0.1 = 12 V_rms. Choice A is correct because it properly applies the transformer equation for a step-down configuration where N_s < N_p. Choice B incorrectly inverts the ratio, yielding 1200 V, which would require a step-up transformer. When teaching transformers, use the mnemonic that voltage and turns are directly proportional on the same side: more turns means higher voltage. Demonstrate with actual transformers how stepping down voltage allows stepping up current while conserving power in ideal cases.

Question 12

A 40-turn coil's flux per turn decreases from $6.0\ \text{mWb}$ to $2.0\ \text{mWb}$ in $0.020\ \text{s}$ ; induced EMF magnitude?

$8.0\ \text{V}$ from $\varepsilon=N\Delta\Phi/\Delta t=40(0.004)/0.020$ using Faraday's law magnitude. (correct answer)
$0.20\ \text{V}$ from $\varepsilon=\Delta\Phi/\Delta t$ , incorrectly omitting the $N=40$ turns factor.
$80\ \text{V}$ from $40(0.040)/0.020$ , misreading $4.0\ \text{mWb}$ as $40\ \text{mWb}$ .
$8.0\ \text{A}$ computed as $I=N\Delta\Phi/\Delta t$ , giving current units instead of volts.

Explanation: This question tests understanding of electromagnetic induction as described by Faraday's Law, specifically calculating induced EMF when magnetic flux through a multi-turn coil changes. Faraday's Law states that the induced EMF equals the negative rate of change of total magnetic flux: ε = -N(dΦ/dt), where N is the number of turns and each turn experiences the same flux change. In this scenario, a 40-turn coil experiences a flux change from 6.0 mWb to 2.0 mWb per turn over 0.020 s, meaning ΔΦ = -4.0 mWb = -0.004 Wb per turn. Choice A is correct because it properly calculates |ε| = N|ΔΦ/Δt| = 40(0.004/0.020) = 8.0 V, including the multiplication by the number of turns. Choice B is incorrect because it omits the factor of N = 40, calculating only 0.004/0.020 = 0.20 V, a common error when students forget that each turn contributes to the total EMF. Encourage students to clearly distinguish between flux per turn and total flux through the coil. Emphasize writing Faraday's Law with N explicitly shown to avoid this common omission.