AP Physics C Electricity and Magnetism Quiz: Electric Current

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Electric CurrentQuestion 1 of 20

A cylindrical conductor has a non-uniform current density given by $J(r) = J_0 (1 - r/R)$ , where $J_0$ is a constant, $R$ is the radius of the conductor, and $r$ is the radial distance from the center. What is the total current $I$ flowing through the conductor?

\pi R^2 J_0

\frac{1}{3} \pi R^2 J_0

\frac{2}{3} \pi R^2 J_0

\frac{1}{2} \pi R^2 J_0

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Current

Practice Electric Current 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 Electric Current, 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

$\pi R^2 J_0$
$\frac{1}{3} \pi R^2 J_0$ (correct answer)
$\frac{2}{3} \pi R^2 J_0$
$\frac{1}{2} \pi R^2 J_0$

Explanation: The total current

I

is found by integrating the current density

J

over the cross-sectional area

A

. We use an infinitesimal ring of area

dA = 2\pi r dr

. The integral is

I = \int_A J dA = \int_0^R J_0 (1 - r/R) (2\pi r dr) = 2\pi J_0 \int_0^R (r - r^2/R) dr = 2\pi J_0 [\frac{r^2}{2} - \frac{r^3}{3R}]_0^R = 2\pi J_0 (\frac{R^2}{2} - \frac{R^3}{3R}) = 2\pi J_0 (\frac{R^2}{2} - \frac{R^2}{3}) = 2\pi J_0 (\frac{R^2}{6}) = \frac{1}{3} \pi R^2 J_0

Question 2

A particle with charge $q$ and mass $m$ is accelerated from rest through a potential difference $V_0$ . It then enters a uniform wire of length $L$ and cross-sectional area $A$ , becoming one of the charge carriers. This single particle's contribution to current depends primarily on which property derived from its initial acceleration?

Its final kinetic energy $qV_0$ after acceleration.
Its average velocity within the wire, the drift velocity. (correct answer)
The time it took to be accelerated across the potential difference.
Its final momentum $\sqrt{2mqV_0}$ after acceleration.

Explanation: Once the particle enters the wire, it becomes a charge carrier. Its motion inside the wire is characterized by frequent collisions with the lattice, resulting in a very small average drift velocity, which determines the current (

I=nqAv_d

). The high speed gained during initial acceleration is quickly lost to collisions, and the particle settles into the slow drift that constitutes the current.

Question 3

A hollow cylindrical conductor has an inner radius of $R_1$ and an outer radius of $R_2$ . A steady current $I$ flows along the conductor, distributed uniformly over its cross-section. What is the magnitude of the current density $J$ within the conductor?

$I / (\pi R_2^2)$
$I / (\pi (R_2 - R_1)^2)$
$I / (\pi (R_2^2 - R_1^2))$ (correct answer)
$I / (2\pi (R_2 - R_1))$

Explanation: The current density

J

is the total current

I

divided by the cross-sectional area

A

through which the current flows. For a hollow cylinder, the cross-sectional area is the area of the outer circle minus the area of the inner circle:

A = \pi R_2^2 - \pi R_1^2 = \pi (R_2^2 - R_1^2)

. Therefore, the current density is

J = I / A = I / (\pi (R_2^2 - R_1^2))

Question 4

A Nichrome wire and an aluminum wire have the same dimensions and are subjected to the same potential difference across their ends. The number density of charge carriers in Nichrome is roughly half that in aluminum, and the resistivity of Nichrome is about 60 times that of aluminum. How does the electron drift speed $v_{d,N}$ in Nichrome compare to the drift speed $v_{d,A}$ in aluminum?

$v_{d,N} \approx 30 v_{d,A}$
$v_{d,N} \approx \frac{1}{30} v_{d,A}$ (correct answer)
$v_{d,N} \approx \frac{1}{60} v_{d,A}$
$v_{d,N} \approx \frac{1}{120} v_{d,A}$

Explanation: Current is

I = V/R

, and resistance is

R = \rho L/A

. So,

I = VA/(\rho L)

. Drift speed is related to current by

I = nqAv_d

, so

v_d = I/(nqA)

. Substituting the expression for

I

, we get

v_d = (VA/(\rho L))/(nqA) = V/(nq\rho L)

. Since

V

and

L

are the same for both wires,

v_d \propto 1/(n\rho)

. Thus,

\frac{v_{d,N}}{v_{d,A}} = \frac{n_A \rho_A}{n_N \rho_N} = \frac{n_A \rho_A}{(0.5 n_A)(60 \rho_A)} = \frac{1}{30}

. So,

v_{d,N} \approx \frac{1}{30} v_{d,A}

Question 5

Wire 1 and Wire 2 are of the same length and cross-sectional area. Wire 1 is made of aluminum (resistivity $\rho_{Al}$ ) and Wire 2 is made of copper (resistivity $\rho_{Cu}$ ), with $\rho_{Al} > \rho_{Cu}$ . If both wires carry the same current, how does the electric field $E_{Al}$ in the aluminum wire compare to the electric field $E_{Cu}$ in the copper wire?

$E_{Al} < E_{Cu}$
$E_{Al} > E_{Cu}$ (correct answer)
$E_{Al} = E_{Cu}$
The comparison depends on the magnitude of the current.

Explanation: The relationship between electric field, resistivity, and current density is

E = \rho J

. The current density is

J = I/A

. Since both wires have the same current

I

and area

A

, their current densities are equal. Therefore, the electric field is directly proportional to the resistivity (

E \propto \rho

). Since aluminum has a higher resistivity than copper (

\rho_{Al} > \rho_{Cu}

), the electric field in the aluminum wire is greater than in the copper wire (

E_{Al} > E_{Cu}

Question 6

Based on the text, what is the current if $V=9\,\text{V}$ and $R=3\,\Omega$ ?

$0.33\,\text{A}$
$12\,\text{A}$
$27\,\text{A}$
$3.0\,\text{A}$ (correct answer)
$6.0\,\text{A}$

Explanation: This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current calculation using Ohm's Law. Electric current is the flow of electric charge in a circuit, calculated as the ratio of voltage to resistance. In this problem, we have a voltage of 9 V across a resistance of 3 Ω, requiring straightforward application of I = V/R. Choice D is correct because I = V/R = 9 V / 3 Ω = 3.0 A, properly applying Ohm's Law to find the current. Choice C is incorrect because it multiplies voltage and resistance (9 × 3 = 27), which is a common algebraic error when students forget the correct relationship or confuse multiplication with division. To help students: Create a triangle diagram with V at the top and I and R at the bottom to visualize the relationships. Watch for: Students who multiply quantities instead of dividing, or who forget to check their units (amperes) for reasonableness.

Question 7

Based on the text, how does adding a resistor in parallel affect the total current in the circuit?

Total current decreases because resistance increases
Total current increases because total resistance decreases (correct answer)
Total current stays the same in all parallel circuits
Total current becomes zero because current is used up
Total current increases because voltage increases automatically

Explanation: This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current behavior in parallel circuits. Electric current in parallel circuits divides among branches, with the total current being the sum of individual branch currents. When adding a resistor in parallel, you create an additional path for current flow, which affects the circuit's total resistance and current. Choice B is correct because adding a parallel resistor decreases the total circuit resistance (1/R_total = 1/R₁ + 1/R₂), and with constant voltage, decreased resistance leads to increased total current according to Ohm's Law. Choice A is incorrect because it applies series circuit logic to a parallel situation, assuming that adding components always increases total resistance, which is a fundamental misconception about parallel circuits. To help students: Use the analogy of multiple lanes on a highway - more lanes (parallel paths) allow more traffic (current) flow. Practice calculating equivalent resistance for parallel combinations and emphasize that parallel resistance is always less than the smallest individual resistance.

Question 8

Based on the text, what is the current if $V=12\,\text{V}$ and $R=6\,\Omega$ ?

$0.5\,\text{A}$
$2.0\,\text{A}$ (correct answer)
$18\,\text{A}$
$72\,\text{A}$
$6.0\,\text{A}$

Explanation: This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current and circuit analysis using Ohm's Law. Electric current is the flow of electric charge in a circuit, typically measured in amperes, and is calculated using the relationship I = V/R. In this problem, we're given a voltage of 12 V and a resistance of 6 Ω, requiring direct application of Ohm's Law. Choice B is correct because I = V/R = 12 V / 6 Ω = 2.0 A, which properly applies the fundamental relationship between voltage, current, and resistance. Choice D is incorrect because it multiplies voltage and resistance (12 × 6 = 72), which is a common error when students confuse the formula. To help students: Emphasize memorizing Ohm's Law in all three forms (I = V/R, V = IR, R = V/I) and practice unit analysis. Watch for: Students multiplying instead of dividing, or confusing which quantity goes in the numerator versus denominator.

Question 9

Based on the text, what is the current if $V=18\,\text{V}$ and $R=9\,\Omega$ ?

$162\,\text{A}$
$1.0\,\text{A}$
$0.5\,\text{A}$
$27\,\text{A}$
$2.0\,\text{A}$ (correct answer)

Explanation: This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current calculation through direct application of Ohm's Law. Electric current is determined by the ratio of applied voltage to circuit resistance, representing the rate of charge flow. Given a voltage of 18 V and a resistance of 9 Ω, we need to calculate the resulting current using I = V/R. Choice E is correct because I = V/R = 18 V / 9 Ω = 2.0 A, which accurately applies the fundamental relationship between these electrical quantities. Choice A is incorrect because it multiplies voltage and resistance (18 × 9 = 162), a common error that suggests the student doesn't understand that resistance opposes current flow. To help students: Create practice problems with simple integer ratios to build confidence before moving to decimals. Watch for: Students who default to multiplication when unsure, or who don't verify their answers make physical sense.

Question 10

Based on the text, how does current behave in a series vs. parallel circuit?

Series: current splits; Parallel: current is identical everywhere
Series: current is the same through each element; Parallel: current divides among branches (correct answer)
Series: current increases after each resistor; Parallel: current decreases
Series and parallel circuits always carry the same total current
Series: current is zero; Parallel: current is nonzero

Explanation: This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current behavior in series versus parallel circuit configurations. Electric current follows different rules in series and parallel arrangements, which is fundamental to circuit analysis. In series circuits, components share the same current path, while in parallel circuits, current has multiple paths to follow. Choice B is correct because it accurately states that in series circuits, the same current flows through each element (conservation of charge), while in parallel circuits, the total current divides among the available branches according to their resistances. Choice A is incorrect because it reverses the behaviors, claiming current splits in series circuits, which violates the principle of charge conservation in a single path. To help students: Use water pipe analogies - series is like a single pipe where all water flows through each section, while parallel is like pipe branches where flow divides. Watch for: Students who memorize rules backwards or confuse current behavior with voltage behavior in these configurations.

Question 11

The net charge passing through a cross-section of a conductor is described by the function $q(t) = 5t^3 - 2t + 1$ , where $q$ is in coulombs and $t$ is in seconds. What is the instantaneous electric current in the conductor at $t = 2$ s?

$13$ A
$37$ A
$58$ A (correct answer)
$60$ A

Explanation: Electric current

I

is defined as the time rate of change of charge,

I(t) = \frac{dq}{dt}

. Taking the derivative of the given charge function:

I(t) = \frac{d}{dt}(5t^3 - 2t + 1) = 15t^2 - 2

. Evaluating this at

t = 2

s gives

I(2) = 15(2)^2 - 2 = 15(4) - 2 = 60 - 2 = 58

Question 12

A steady current of $2.0$ A flows through a copper wire. Copper has approximately $8.5 \times 10^{28}$ free electrons per cubic meter. If the wire has a cross-sectional area of $2.0 \times 10^{-6} \text{ m}^2$ , what is the approximate drift velocity of the electrons?

$7.4 \times 10^{-5} \text{ m/s}$ (correct answer)
$1.5 \times 10^{-4} \text{ m/s}$
$5.9 \times 10^{3} \text{ m/s}$
$1.2 \times 10^{6} \text{ m/s}$

Explanation: The relationship between current

I

, charge carrier density

n

, elementary charge

q

, cross-sectional area

A

, and drift velocity

v_d

I = nqAv_d

. Solving for

v_d

gives

v_d = \frac{I}{nqA}

. Plugging in the values:

v_d = \frac{2.0 \text{ A}}{(8.5 \times 10^{28} \text{ m}^{-3})(1.6 \times 10^{-19} \text{ C})(2.0 \times 10^{-6} \text{ m}^2)} \approx 7.4 \times 10^{-5} \text{ m/s}

Question 13

Two cylindrical wires, A and B, are made of the same material and have the same length. Wire A has radius $r$ , and wire B has radius $2r$ . They carry the same uniform current $I$ . How does the magnitude of the current density $J_A$ in wire A compare to the magnitude of the current density $J_B$ in wire B?

$J_A = 4J_B$ (correct answer)
$J_A = 2J_B$
$J_A = J_B$
$J_A = \frac{1}{2}J_B$

Explanation: Current density is defined as

J = I/A

, where

A

is the cross-sectional area. The area is given by

A = \pi r^2

. For wire A,

A_A = \pi r^2

. For wire B,

A_B = \pi (2r)^2 = 4\pi r^2 = 4A_A

. Since both wires carry the same current

I

, we have

J_A = I/A_A

and

J_B = I/A_B = I/(4A_A)

. Therefore,

J_A = 4J_B

Question 14

The current in a wire varies with time according to the equation $I(t) = 4 - 2t$ , where $I$ is in amperes and $t$ is in seconds. What is the total charge that passes a point in the wire from $t=0$ s to $t=2$ s?

0 C
2 C
4 C (correct answer)
8 C

Explanation: Current is the rate of flow of charge,

I = dq/dt

. To find the total charge

Q

that passes a point, we must integrate the current with respect to time:

Q = \int_{t_1}^{t_2} I(t) dt

. For the given interval,

Q = \int_{0}^{2} (4 - 2t) dt = [4t - t^2]_0^2 = (4(2) - 2^2) - (0) = 8 - 4 = 4

Question 15

For an ohmic conductor, the relationship between the magnitude of the electric field $E$ inside it and the magnitude of the current density $J$ is given by $E = \rho J$ , where $\rho$ is the resistivity. If a wire of length $L$ and cross-sectional area $A$ is connected to a battery of emf $V$ , which expression represents the current density $J$ ?

$V / (\rho L)$ (correct answer)
$V \rho / L$
$V A / (\rho L)$
$V / (\rho A)$

Explanation: For a uniform wire of length

L

connected to a potential difference

V

, the electric field inside is uniform with magnitude

E = V/L

. Substituting this into the equation

E = \rho J

, we get

V/L = \rho J

. Solving for the current density

J

gives

J = V / (\rho L)

Question 16

A steady current flows through a conical conductor of uniform resistivity. The conductor's circular cross-section has a radius that increases linearly from left to right. How does the drift velocity of the charge carriers vary along the length of the conductor from left to right?

It increases because the electric field becomes stronger.
It remains constant because the current is steady.
It decreases because the cross-sectional area increases. (correct answer)
It increases and then decreases, following the shape of the conductor.

Explanation: For a steady current

I

, the current is the same at all cross-sections. The current is related to drift velocity

v_d

I = nqAv_d

. Since

I

n

, and

q

are constant,

v_d

is inversely proportional to the cross-sectional area

A

. As the radius increases from left to right, the area

A

increases, and therefore the drift velocity

v_d

must decrease.

Question 17

The current density vector $\vec{J}$ in a conductor is related to the number density of charge carriers $n$ , their charge $q$ , and their drift velocity $\vec{v}_d$ . Which of the following equations correctly represents this relationship?

$\vec{J} = \frac{n q}{\vec{v}_d}$
$\vec{J} = n q \vec{v}_d$ (correct answer)
$\vec{J} = \frac{\vec{v}_d}{n q}$
$\vec{J} = n q^2 \vec{v}_d$

Explanation: Current density

J

is the current per unit area. Since

I = nqAv_d

, dividing by area

A

gives

J = nqv_d

. As vectors, the direction of

J

is defined by the direction of the drift velocity of positive charges (or opposite to negative charges). For a charge

q

(which can be positive or negative), the drift velocity vector

\vec{v}_d

determines the direction, so

\vec{J} = n q \vec{v}_d

Question 18

In the absence of an electric field, the free electrons in a metallic conductor move randomly with high speeds due to thermal energy. When an electric field is applied, a net current is produced. Why is the resulting drift speed of the electrons typically very small (e.g., $10^{-4}$ m/s)?

The elementary charge of an electron is extremely small, limiting the effect of the electric force.
The number density of free electrons is very low, so few electrons contribute to the current.
Electrons frequently collide with the ions of the metallic lattice, which repeatedly resets their net forward motion. (correct answer)
The applied electric field is mostly canceled by an opposing field from induced surface charges.

Explanation: While the electric field accelerates the electrons, they constantly collide with the stationary ions in the crystal lattice. Each collision effectively randomizes the electron's velocity, and it must be accelerated again by the field. This start-and-stop motion, averaged over many electrons and collisions, results in a very slow net drift speed in the direction opposite the field.

Question 19

A potential difference is applied across a cylindrical conductor, resulting in an electric field $E$ and a current density $J$ . If the temperature of the conductor increases, its resistivity $\rho$ increases. Assuming the applied potential difference remains constant, what happens to the current density $J$ ?

It increases, because the charge carriers move faster at higher temperatures.
It remains the same, because the electric field is unchanged.
It decreases, because the increased resistivity impedes the flow of charge. (correct answer)
It becomes zero, as the conductor loses its conductive properties.

Explanation: The relationship is

E = \rho J

. The electric field is

E = V/L

. Since the potential difference

V

and length

L

are constant, the electric field

E

remains constant. From

J = E/\rho

, if the resistivity

\rho

increases due to the temperature increase, the current density

J

must decrease.

Question 20

Which of the following physical quantities is most analogous to electromotive force (emf) in an electrical circuit?

The force exerted by a pump that circulates water in a closed pipe system. (correct answer)
The total volume of water contained within a closed pipe system.
The friction or drag that opposes the flow of water in a pipe system.
The kinetic energy of the water flowing through a pipe system.

Explanation: Electromotive force (emf) is the work done per unit charge by a source (like a battery) to move charges and maintain a potential difference, thus sustaining current. In the water-pipe analogy, a pump provides the pressure difference (analogous to potential difference) that does work on the water to make it flow, which is analogous to the role of emf in a circuit.