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AP Physics C: Electricity and Magnetism Practice Quiz: Inductance

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 11 questions to check your progress.

Question 1 of 11

According to the provided text, what is the primary characteristic of inductance?

All Questions (11)

According to the provided text, what is the primary characteristic of inductance?

A) The ability to store energy in an electric field.

B) The opposition to the flow of a steady electrical current.

C) The tendency of a conductor to oppose a change in electrical current.

D) The generation of a magnetic field from a stationary charge.

Correct Answer: C

The content explicitly states, 'Inductance is the tendency of a conductor to oppose a change in electrical current.' Option B describes resistance, and option A describes capacitance.

An inductor is a component in an electrical circuit. In what form does it store energy?

A) As potential energy in an electric field.

B) As kinetic energy of the moving charges.

C) As chemical energy within its core material.

D) As energy within the magnetic field generated by the current.

Correct Answer: D

The provided text states, 'Inductors store energy in the magnetic field that is generated by current in the inductor.'

A scientist is designing a solenoid inductor and wants to increase its inductance without changing the number of turns or the core material. Based on the equation $L=\frac{\mu_{core}N^{2}A}{\ell}$, which of the following changes would achieve this goal?

A) Increasing the length (ℓ) of the solenoid.

B) Decreasing the cross-sectional area (A) of the solenoid.

C) Increasing the cross-sectional area (A) of the solenoid.

D) Using a thinner wire for the turns.

Correct Answer: C

The formula for the inductance of a solenoid is $L=\frac{\mu_{core}N^{2}A}{\ell}$. Inductance (L) is directly proportional to the cross-sectional area (A). Therefore, increasing the area will increase the inductance. Increasing the length (ℓ) would decrease the inductance.

The number of turns in a solenoid-shaped inductor is doubled, while its length, cross-sectional area, and core material remain constant. How does the new inductance compare to the original inductance?

A) It is halved.

B) It is doubled.

C) It is quadrupled.

D) It remains the same.

Correct Answer: C

Inductance (L) is proportional to the square of the number of turns (N²), as shown in the equation $L=\frac{\mu_{core}N^{2}A}{\ell}$. If N is doubled, the new inductance will be proportional to (2N)², which is 4N². Therefore, the inductance is quadrupled.

An inductor is connected in a circuit with a time-varying current. According to the equation $\mathcal{E}_{induced}=-L\frac{dI}{dt}$, the magnitude of the induced EMF across the inductor is greatest at the moment when:

A) the current is at its maximum steady value.

B) the current is zero and not changing.

C) the rate of change of the current is at its maximum.

D) the energy stored in the inductor is zero.

Correct Answer: C

The equation $\mathcal{E}_{induced}=-L\frac{dI}{dt}$ shows that the induced EMF is directly proportional to the rate of change of current (dI/dt). Therefore, the EMF will have the greatest magnitude when the current is changing most rapidly.

An inductor with an inductance of 4.0 H has a steady current of 3.0 A flowing through it. What is the total energy stored in the inductor's magnetic field?

A) 6.0 J

B) 12 J

C) 18 J

D) 36 J

Correct Answer: C

The energy stored in an inductor is given by the equation $U_{L}=\frac{1}{2}LI^{2}$. Plugging in the values: $U_{L}=\frac{1}{2}(4.0 \text{ H})(3.0 \text{ A})^{2} = \frac{1}{2}(4.0)(9.0) = 18 \text{ J}$.

The current flowing through a 10 H inductor is increased uniformly from 0 A to 5 A in 2 seconds. What is the magnitude of the induced EMF in the inductor?

A) 2.5 V

B) 5 V

C) 25 V

D) 50 V

Correct Answer: C

The induced EMF is given by $\mathcal{E}_{induced}=-L\frac{dI}{dt}$. The rate of change of current, $\frac{dI}{dt}$, is $\frac{5 \text{ A} - 0 \text{ A}}{2 \text{ s}} = 2.5 \text{ A/s}$. The magnitude of the EMF is $|-L\frac{dI}{dt}| = |-(10 \text{ H})(2.5 \text{ A/s})| = 25 \text{ V}$.

An ideal solenoid has an inductance L. A new solenoid is constructed with the same number of turns and core material, but its length is halved and its cross-sectional area is doubled. What is the inductance of the new solenoid?

A) L/4

B) L

C) 2L

D) 4L

Correct Answer: D

The original inductance is $L=\frac{\mu_{core}N^{2}A}{\ell}$. The new solenoid has length $\ell' = \ell/2$ and area $A' = 2A$. The new inductance is $L'=\frac{\mu_{core}N^{2}A'}{\ell'} = \frac{\mu_{core}N^{2}(2A)}{(\ell/2)} = 4\frac{\mu_{core}N^{2}A}{\ell} = 4L$.

An air-core solenoid inductor is modified by inserting a ferromagnetic core, which has a magnetic permeability ($\\mu_{core}$) much greater than that of air. How does this modification affect the inductor's properties?

A) The inductance decreases significantly.

B) The inductance increases significantly.

C) The inductance remains unchanged, but it can store more energy.

D) The inductance remains unchanged, but the induced EMF is greater for the same change in current.

Correct Answer: B

According to the formula $L=\frac{\mu_{core}N^{2}A}{\ell}$, the inductance L is directly proportional to the magnetic permeability of the core, $\\mu_{core}$. Since the ferromagnetic core has a much higher permeability than air, the inductance will increase significantly.

An inductor stores an amount of energy U when the current flowing through it is I. If the current is decreased to I/2, what is the new amount of energy stored in the inductor?

A) 4U

B) 2U

C) U/2

D) U/4

Correct Answer: D

The energy stored in an inductor is given by $U_{L}=\frac{1}{2}LI^{2}$. Energy is proportional to the square of the current (I²). If the current is changed to I/2, the new energy $U'_{L}$ will be proportional to $(I/2)² = I²/4$. Therefore, the new energy is U/4.

Inductor 1 is a solenoid with N turns and length ℓ. Inductor 2 is a solenoid made with the same wire, core material, length, and cross-sectional area, but with 3N turns. If the same steady current I flows through both inductors, what is the ratio of the energy stored in Inductor 2 to the energy stored in Inductor 1 (U₂/U₁)?

A) 1/3

B) 3

C) 9

D) 1/9

Correct Answer: C

First, find the relationship between the inductances. $L_1 \propto N^2$ and $L_2 \propto (3N)^2 = 9N^2$. Therefore, $L_2 = 9L_1$. Next, use the energy storage equation, $U_{L}=\frac{1}{2}LI^{2}$. The ratio of the energies is $\frac{U_2}{U_1} = \frac{\frac{1}{2}L_2 I^2}{\frac{1}{2}L_1 I^2} = \frac{L_2}{L_1} = \frac{9L_1}{L_1} = 9$.