AP Physics C Electricity and Magnetism Quiz: Inductance

18 questions · exam conditions

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InductanceQuestion 1 of 18

An ideal air-core solenoid has $N$ turns, a length ${\ell}$ , and a cross-sectional area $A$ . If the number of turns is tripled to $3N$ while the length and area remain constant, the new inductance will be...

one-ninth of the original inductance.

one-third of the original inductance.

three times the original inductance.

nine times the original inductance.

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Inductance

Practice Inductance 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 Inductance, 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

one-ninth of the original inductance.
one-third of the original inductance.
three times the original inductance.
nine times the original inductance. (correct answer)

Explanation: The inductance of a long solenoid is given by

L = \frac{\mu_0 N^2 A}{\ell}

. Since the inductance

L

is proportional to the square of the number of turns (

N^2

), tripling

N

3N

will change the inductance by a factor of

(3)^2 = 9

Question 2

A steady current is flowing through a large inductor. The power supply is suddenly disconnected, and the inductor is simultaneously connected across a resistor. Which of the following occurs?

The current immediately drops to zero, and the stored magnetic energy is lost instantly.
The inductor induces a current in the same direction as the original current, which then decays through the resistor. (correct answer)
The inductor induces a current in the opposite direction to the original current, which then charges the resistor.
The magnetic field in the inductor reverses direction, causing an oscillating current to flow through the resistor.

Explanation: An inductor resists changes in current. When the power supply is removed, the current begins to decrease. To oppose this decrease, the inductor induces an EMF that drives a current in the same direction as the original current. This induced current flows through the resistor, and the energy stored in the inductor's magnetic field is dissipated as heat in the resistor, causing the current to decay over time.

Question 3

An air-core solenoid has an inductance $L_0$ . A rod of a paramagnetic material is then fully inserted into the solenoid's core. The inductance of the solenoid will...

increase slightly, because the magnetic permeability of paramagnetic materials is slightly greater than ${\mu_0}$ . (correct answer)
decrease slightly, because the magnetic permeability of paramagnetic materials is slightly less than ${\mu_0}$ .
increase significantly, because paramagnetic materials have very high magnetic permeability.
remain unchanged, because paramagnetic materials do not affect the magnetic field or flux inside a solenoid.

Explanation: The inductance of a solenoid is proportional to the magnetic permeability

{\mu}

of its core material (

L = \frac{\mu N^2 A}{\ell}

). Paramagnetic materials have a magnetic permeability

{\mu}

that is slightly greater than the permeability of free space,

{\mu_0}

. Therefore, inserting a paramagnetic core will cause a slight increase in the inductance.

Question 4

The current $I$ in an inductor with inductance $L = 5.0$ mH is given by the function $I(t) = 3.0t^2 - 4.0t$ , where $I$ is in amperes and $t$ is in seconds. What is the magnitude of the induced EMF in the inductor at time $t=2.0$ s?

10 mV
20 mV
30 mV
40 mV (correct answer)

Explanation: The induced EMF is given by

{\mathcal{E}} = -L \frac{dI}{dt}

. First, find the derivative of the current:

\frac{dI}{dt} = \frac{d}{dt}(3.0t^2 - 4.0t) = 6.0t - 4.0

. At

t=2.0

\frac{dI}{dt} = 6.0(2.0) - 4.0 = 12.0 - 4.0 = 8.0

A/s. The magnitude of the EMF is

|{\mathcal{E}}| = |-(5.0 \times 10^{-3} \text{ H})(8.0 \text{ A/s})| = 40 \times 10^{-3} \text{ V} = 40

mV.

Question 5

The energy stored in the magnetic field of an inductor is $U_L$ . If the steady current flowing through the inductor is doubled, while its inductance remains constant, the new energy stored in the inductor will be...

$\frac{1}{4} U_L$
$\frac{1}{2} U_L$
$2 U_L$
$4 U_L$ (correct answer)

Explanation: The energy stored in an inductor is given by the formula

U_L = \frac{1}{2} L I^2

. Since the energy

U_L

is proportional to the square of the current (

I^2

), doubling the current from

I

2I

will change the stored energy by a factor of

(2)^2 = 4

Question 6

Two long solenoids, A and B, are constructed from the same type of wire and have the same cross-sectional area. Solenoid A has $N$ turns and length ${\ell}$ . Solenoid B has $2N$ turns and length $4{\ell}$ .

What is the ratio of the inductance of solenoid B to the inductance of solenoid A, $L_B / L_A$ ?

$1/2$
$1$ (correct answer)
$2$
$4$

Explanation: The inductance of a solenoid is given by

L = \frac{\mu_0 N^2 A}{\ell}

. For solenoid A,

L_A = \frac{\mu_0 N^2 A}{\ell}

. For solenoid B,

L_B = \frac{\mu_0 (2N)^2 A}{4\ell} = \frac{\mu_0 4N^2 A}{4\ell} = \frac{\mu_0 N^2 A}{\ell}

. Therefore,

L_B = L_A

, and the ratio

L_B / L_A = 1

Question 7

A long solenoid with $N$ turns, length ${\ell}$ , and cross-sectional area $A$ has an inductance $L$ . What would be the inductance of a solenoid with $N$ turns, length $2{\ell}$ , and cross-sectional area $A/2$ ?

$L/4$ (correct answer)
$L/2$
$L$
$2L$

Explanation: The inductance of a long solenoid is given by

L = \frac{\mu_0 N^2 A}{\ell}

. The new inductance,

L'

will be

L' = \frac{\mu_0 N^2 (A/2)}{2\ell} = \frac{1}{4} \left( \frac{\mu_0 N^2 A}{\ell} \right) = \frac{L}{4}

Question 8

An ideal solenoid contains a uniform magnetic field $B$ when it carries a current $I$ . The energy density (energy per unit volume) of the magnetic field inside the solenoid is proportional to which of the following quantities?

$B$
$B^2$ (correct answer)
$1/B$
$1/B^2$

Explanation: The energy stored in an inductor is

U_L = \frac{1}{2}LI^2

. For a solenoid,

L=\frac{\mu_0N^2A}{\ell}

and

B=\mu_0nI=\mu_0\frac{N}{\ell}I

, so

I=\frac{B\ell}{\mu_0N}

. Substituting these into the energy equation gives

U_L=\frac{1}{2} \left( \frac{\mu_0N^2A}{\ell} \right) \left( \frac{B\ell}{\mu_0N} \right)^2 = \frac{1}{2\mu_0}B^2(A\ell)

. The volume of the solenoid is

V=A\ell

. The energy density is

u_B = U_L/V = \frac{B^2}{2\mu_0}

. Thus, the energy density is proportional to

B^2

Question 9

Inductance is a property of an electrical conductor that measures its tendency to...

resist the flow of a steady, direct current through it, converting electrical energy into thermal energy.
store electrical charge when a potential difference is applied across it, creating an electric field.
oppose a change in the electric current flowing through it by inducing an electromotive force. (correct answer)
generate a constant magnetic field when a constant current is flowing through it.

Explanation: Inductance (

L

) is defined by the relationship where the induced electromotive force (EMF,

{\mathcal{E}}

) is proportional to the rate of change of current (

dI/dt

), specifically

{\mathcal{E}} = -L(dI/dt)

. This relationship shows that an inductor opposes changes in current.

Question 10

An inductor with an inductance of 2.0 H carries a steady current of 3.0 A. How much energy is stored in the magnetic field of the inductor?

3.0 J
6.0 J
9.0 J (correct answer)
12.0 J

Explanation: The energy stored in an inductor is calculated using the formula

U_L = \frac{1}{2} L I^2

. Substituting the given values:

U_L = \frac{1}{2} (2.0 \text{ H}) (3.0 \text{ A})^2 = \frac{1}{2} (2.0)(9.0) = 9.0

Question 11

The SI unit of inductance is the henry (H). Which of the following combinations of units is equivalent to the henry?

Volt per ampere ( $\text{V/A}$ )
Volt-second per ampere ( $\text{V} \cdot \text{s/A}$ ) (correct answer)
Volt-ampere per second ( $\text{V} \cdot \text{A/s}$ )
Ampere-second per volt ( $\text{A} \cdot \text{s/V}$ )

Explanation: From the formula for induced EMF,

{\mathcal{E}} = -L \frac{dI}{dt}

, we can solve for the units of inductance

L

. The units are

[L] = \frac{[\mathcal{E}]}{[dI/dt]} = \frac{\text{Volts}}{\text{Amperes/Second}} = \frac{\text{V} \cdot \text{s}}{\text{A}}

Question 12

The inductance of a long, air-filled solenoid is $L_0$ . If its length and total number of turns are both doubled, while keeping its cross-sectional area constant, what is its new inductance $L'$ ?

$L_0/2$
$L_0$
$2L_0$ (correct answer)
$4L_0$

Explanation: The inductance of a long solenoid is given by the formula

L = \frac{\mu_0 N^2 A}{\ell}

. The new parameters are

N' = 2N

and

{\ell}' = 2{\ell}

. The new inductance is

L' = \frac{\mu_0 (N')^2 A}{{\ell}'} = \frac{\mu_0 (2N)^2 A}{2{\ell}} = \frac{\mu_0 4N^2 A}{2{\ell}} = 2 \left( \frac{\mu_0 N^2 A}{\ell} \right) = 2L_0

Question 13

An ideal solenoid of length ${\ell}$ and radius $r$ has $N$ turns of wire. Assuming the solenoid is long enough that end effects are negligible, its self-inductance is proportional to...

$\frac{N r^2}{\ell}$
$\frac{N^2 r^2}{\ell}$ (correct answer)
$\frac{N^2 r}{\ell}$
$\frac{N r}{\ell^2}$

Explanation: The inductance of a long solenoid is

L = \frac{\mu_0 N^2 A}{\ell}

. The cross-sectional area is

A = \pi r^2

. Therefore,

L = \frac{\mu_0 N^2 (\pi r^2)}{\ell}

. Ignoring the constants

{\mu_0}

and

{\pi}

, the inductance is proportional to

\frac{N^2 r^2}{\ell}

Question 14

Which of the following statements best distinguishes the behavior of an ideal inductor from that of an ideal resistor in a DC circuit with a steady current?

The inductor stores energy in its electric field, while the resistor stores energy in its magnetic field.
Both dissipate energy as heat, but the inductor does so at a rate proportional to the current squared.
The resistor causes a constant potential drop proportional to the current, while the inductor has zero potential drop.
The inductor dissipates no energy, while the resistor causes a potential drop and dissipates energy continuously. (correct answer)

Explanation: In a DC circuit with a steady current (

dI/dt = 0

), an ideal inductor has no induced EMF (

\mathcal{E} = -L(0) = 0

) and behaves like a wire with zero resistance, thus dissipating no energy. An ideal resistor always has a potential drop

V=IR

and dissipates power as heat (

P=I^2R

) as long as current flows. The correct answer reflects this difference in behavior under steady DC conditions.

Question 15

A circuit contains an ideal inductor and a switch connected to a battery. When the switch is suddenly closed, the inductor generates a 'back EMF'. This induced EMF is a direct consequence of...

the inductor's internal resistance converting electrical energy to heat, which opposes current flow.
the conservation of charge, requiring the current to build up slowly from zero at a uniform rate.
the magnetic flux created by the rising current, which itself induces an EMF that opposes this change in flux. (correct answer)
the capacitance of the windings of the inductor, which must charge before current can flow steadily.

Explanation: According to Faraday's law of induction and Lenz's law, as the current begins to flow and increase, it creates a changing magnetic flux through the inductor's coils. This changing flux induces an EMF that opposes the change, which in this case is the increase in current. This opposition is the source of the back EMF.

Question 16

A constant potential difference of 12 V is maintained across a 4.0 mH inductor. What must be the rate of change of current through the inductor?

3.0 A/s
48 A/s
3000 A/s (correct answer)
The current must be constant, so the rate of change is zero.

Explanation: A potential difference across an ideal inductor is due to a changing current. The relationship is

|\mathcal{E}| = L |\frac{dI}{dt}|

(ignoring internal resistance). To maintain a constant potential difference, the current must be changing at a constant rate. Rearranging for the rate of change:

|\frac{dI}{dt}| = \frac{|\mathcal{E}|}{L} = \frac{12 \text{ V}}{4.0 \times 10^{-3} \text{ H}} = 3000

A/s.

Question 17

Immediately after a switch is closed to connect an unpowered series RL circuit to a DC battery, the inductor behaves most like...

a simple connecting wire with zero resistance, allowing maximum current to flow immediately.
an open circuit with infinite resistance, allowing essentially zero initial current to flow. (correct answer)
a capacitor that is fully charged, preventing any further current from flowing.
a resistor whose resistance is equal to the internal resistance of the battery.

Explanation: At the instant the switch is closed (

t=0^+

, the current begins to change from zero. The inductor opposes this change by inducing a back EMF equal in magnitude to the battery's EMF. This opposition is so strong initially that it prevents any significant current from flowing, making the inductor behave like an open circuit or a break in the wire.

Question 18

A 500-turn inductor has an inductance of 25 mH. If a steady current of 4.0 A flows through it, what is the magnetic flux through a single turn of the inductor?

$5.0 \times 10^{-5}$ Wb
$2.0 \times 10^{-4}$ Wb (correct answer)
$0.10$ Wb
$50$ Wb

Explanation: Inductance is related to the total flux linkage (

N\Phi_B

) by

L = \frac{N\Phi_B}{I}

, where

\Phi_B

is the flux through a single turn. Solving for

\Phi_B

gives

\Phi_B = \frac{LI}{N}

. Plugging in the values:

\Phi_B = \frac{(25 \times 10^{-3} \text{ H})(4.0 \text{ A})}{500} = \frac{0.10 \text{ H} \cdot \text{A}}{500} = 2.0 \times 10^{-4}

Wb.