AP Physics C Electricity and Magnetism Quiz: Magnetic Flux

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Magnetic FluxQuestion 1 of 19

The magnetic field perpendicular to the plane of a $4.0$ cm by $4.0$ cm square loop of wire is given by $B(t) = 3.0t - 1.5t^2$ , where $B$ is in tesla and $t$ is in seconds. What is the magnitude of the magnetic flux through the loop at $t=2.0$ s?

9.6 \times 10^{-3}

4.8 \times 10^{-3}

Zero

2.4 \times 10^{-3}

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Magnetic Flux

Practice Magnetic Flux 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 Magnetic Flux, 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

$9.6 \times 10^{-3}$ Wb
$4.8 \times 10^{-3}$ Wb
Zero (correct answer)
$2.4 \times 10^{-3}$ Wb

Explanation: First, calculate the area of the loop:

A = (0.040 \text{ m})^2 = 0.0016 \text{ m}^2

. Next, evaluate the magnetic field at

t=2.0

B(2.0) = 3.0(2.0) - 1.5(2.0)^2 = 6.0 - 1.5(4.0) = 6.0 - 6.0 = 0

T. Since the magnetic field is zero at this instant, the magnetic flux

\Phi_B = BA

is also zero.

Question 2

A square coil with $N$ turns and side length $s$ is in a uniform magnetic field $B$ . The axis of the coil (normal to its plane) is initially parallel to the field. The coil is rotated by $90^\circ$ so its axis is perpendicular to the field. What is the magnitude of the magnetic flux through the entire coil in this final position?

$N B s^2$
$\frac{N B s^2}{2}$
Zero (correct answer)
$\frac{\pi}{2} N B s^2$

Explanation: The magnetic flux through the entire coil is

N

times the flux through a single turn,

\Phi_{coil} = N(BA \cos\theta)

. Here,

A = s^2

. In the final position, the axis of the coil is perpendicular to the field, meaning the angle

\theta

between the normal vector and the field is

90^\circ

. Since

\cos(90^\circ) = 0

, the magnetic flux through the coil is zero.

Question 3

A uniform magnetic field $\vec{B} = B_0 \hat{k}$ exists in space. What is the magnitude of the magnetic flux through the curved surface of a hemisphere of radius $R$ whose circular base lies in the xy-plane and is centered at the origin?

Zero
$B_0 \pi R^2$ (correct answer)
$2 B_0 \pi R^2$
$4 B_0 \pi R^2$

Explanation: According to Gauss's law for magnetism, the total magnetic flux through any closed surface is zero. Consider the closed surface formed by the hemisphere and its flat circular base.

\Phi_{closed} = \Phi_{curved} + \Phi_{base} = 0

. Therefore,

\Phi_{curved} = -\Phi_{base}

. The flux through the flat base of area

A=\pi R^2

in the xy-plane is

\Phi_{base} = \vec{B} \cdot \vec{A}_{base}

. With the outward normal for the base pointing in the

-z

direction,

\vec{A}_{base} = -\pi R^2 \hat{k}

, so

\Phi_{base} = (B_0 \hat{k}) \cdot (-\pi R^2 \hat{k}) = -B_0 \pi R^2

. The flux through the curved surface is thus

\Phi_{curved} = -(-B_0 \pi R^2) = B_0 \pi R^2

Question 4

A flat, rectangular loop of wire is placed in a region of uniform magnetic field. For which orientation of the loop is the magnitude of the magnetic flux through it a maximum?

The plane of the loop is parallel to the magnetic field.
The plane of the loop is perpendicular to the magnetic field. (correct answer)
The plane of the loop makes an angle of $45^\circ$ with the magnetic field.
The normal to the plane of the loop makes an angle of $45^\circ$ with the magnetic field.

Explanation: The magnetic flux is

\Phi_B = BA \cos\theta

, where

\theta

is the angle between the field and the normal to the loop's plane. The magnitude is maximum when

|\cos\theta|

is maximum, which occurs at

\theta = 0^\circ

180^\circ

. In this case, the normal vector is parallel to the field, meaning the plane of the loop is perpendicular to the field.

Question 5

A circular loop of wire with area $A$ is placed in a uniform magnetic field of strength $B$ . What is the magnetic flux through the loop when the angle between the magnetic field vector and the normal to the plane of the loop is $60^\circ$ ?

$BA$
$\frac{\sqrt{3}}{2} BA$
$\frac{1}{2} BA$ (correct answer)
Zero

Explanation: Magnetic flux is given by

\Phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta

, where

\theta

is the angle between the magnetic field vector and the area vector (which is normal to the loop's plane). With

\theta = 60^\circ

\cos(60^\circ) = 1/2

, so the flux is

\frac{1}{2} BA

Question 6

A uniform magnetic field is given by the vector $\vec{B} = (4.0 \hat{i} - 3.0 \hat{j})$ T. A flat, square surface with side length $2.0$ m lies in the yz-plane. What is the magnetic flux through this surface?

$12$ Wb
$-12$ Wb
$16$ Wb (correct answer)
$20$ Wb

Explanation: The area of the square is

A = (2.0 \text{ m})^2 = 4.0 \text{ m}^2

. Since the surface lies in the yz-plane, its area vector is perpendicular to that plane, meaning it points in the x-direction. We can write

\vec{A} = 4.0 \hat{i} \text{ m}^2

. The magnetic flux is the dot product

\Phi_B = \vec{B} \cdot \vec{A} = ((4.0 \hat{i} - 3.0 \hat{j}) \text{ T}) \cdot (4.0 \hat{i} \text{ m}^2) = (4.0)(4.0) = 16

Wb.

Question 7

A long, straight wire carries a current $I$ . A rectangular loop of wire with width $w$ and length $L$ is positioned such that its sides of length $L$ are parallel to the wire. The side closer to the wire is at a distance $a$ from it. The magnetic field due to the wire is $B = \frac{\mu_0 I}{2\pi r}$ , where $r$ is the perpendicular distance from the wire. Which integral correctly gives the magnitude of the magnetic flux through the loop?

$\frac{\mu_0 I}{2\pi a} (Lw)$
$\int_0^w \frac{\mu_0 I L}{2\pi r} dr$
$\int_a^{a+w} \frac{\mu_0 I L}{2\pi r} dr$ (correct answer)
$\int_a^{a+w} \frac{\mu_0 I}{2\pi r^2} L dr$

Explanation: The magnetic field is not uniform over the area of the loop; it depends on the distance

r

from the wire. To find the total flux, we must integrate. Consider a thin strip of the loop of length

L

and width

dr

at a distance

r

from the wire. The area of this strip is

dA = L dr

. The flux through this strip is

d\Phi_B = B dA = (\frac{\mu_0 I}{2\pi r}) (L dr)

. To find the total flux, we integrate this expression over the width of the loop, from

r=a

r=a+w

Question 8

A square loop of wire with side length $s$ lies in the xy-plane with one corner at the origin and its sides aligned with the positive x and y axes. The loop is in a non-uniform magnetic field given by $\vec{B} = C y \hat{k}$ , where $C$ is a positive constant. What is the total magnetic flux through the loop?

Zero
$C s^3$
$\frac{1}{2} C s^3$ (correct answer)
$\frac{1}{2} C s^2$

Explanation: The magnetic field varies with

y

. We must integrate to find the flux. The area element can be taken as a thin horizontal strip of width

s

(in the x-direction) and height

dy

. So,

d\vec{A} = s \, dy \, \hat{k}

. The flux through this strip is

d\Phi_B = \vec{B} \cdot d\vec{A} = (C y \hat{k}) \cdot (s \, dy \, \hat{k}) = C s y \, dy

. Integrating from

y=0

y=s

gives

\Phi_B = \int_0^s C s y \, dy = C s [\frac{y^2}{2}]_0^s = \frac{1}{2} C s^3

Question 9

Loop 1 is a square with side length $2R$ . Loop 2 is a circle with radius $R$ . Both loops are in the same uniform magnetic field, and their planes are oriented perpendicular to the field lines. What is the ratio of the magnetic flux through Loop 1 to the magnetic flux through Loop 2, $\frac{\Phi_1}{\Phi_2}$ ?

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

Explanation: The magnetic flux is

\Phi = BA

, since the field is uniform and perpendicular to the planes. For Loop 1 (square), the area is

A_1 = (2R)^2 = 4R^2

. For Loop 2 (circle), the area is

A_2 = \pi R^2

. The ratio of the fluxes is

\frac{\Phi_1}{\Phi_2} = \frac{BA_1}{BA_2} = \frac{4R^2}{\pi R^2} = \frac{4}{\pi}

Question 10

The magnetic flux through a surface is calculated using the surface integral $\Phi_B = \int \vec{B} \cdot d\vec{A}$ . This integral form, rather than the simpler $\Phi_B = \vec{B} \cdot \vec{A}$ , is generally required under which of the following conditions?

Only when the magnetic field is changing with time.
Only when the surface is a closed surface.
When either the magnetic field is non-uniform over the surface or the surface is not flat. (correct answer)
Only when the magnetic field vector is not perpendicular to the surface.

Explanation: The integral form is a general definition. It simplifies to the algebraic form

\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta

only if the magnetic field

\vec{B}

is uniform over the entire surface and the surface is flat (planar). If either the field varies from point to point on the surface (is non-uniform) or the surface is curved, the integral is necessary to sum up the contributions of

d\Phi_B = \vec{B} \cdot d\vec{A}

over all the infinitesimal area elements

d\vec{A}

that make up the surface.

Question 11

A rectangular loop of wire has initial width $w_0$ and length $L$ . The loop is in a uniform magnetic field $B$ perpendicular to its plane. The loop is then stretched at a constant rate such that its width becomes $2w_0$ while its length remains $L$ . What is the magnetic flux through the loop as a function of its instantaneous width $w$ ?

$B L w$ (correct answer)
$B L w_0$
$B(L+w)$
$\frac{B L}{w}$

Explanation: The magnetic flux is given by

\Phi_B = BA

, where A is the instantaneous area of the loop, because the field is uniform and perpendicular to the loop's plane. The area of the rectangular loop at any instant is its length times its instantaneous width,

A = Lw

. Therefore, the magnetic flux as a function of the width

w

\Phi_B = B(Lw) = BLw

Question 12

A long solenoid of radius $R$ has $n$ turns per unit length and carries a current $I$ . The magnetic field inside is uniform and given by $B=\mu_0 n I$ . A smaller, coaxial circular loop of radius $r < R$ is placed inside the solenoid. What is the magnetic flux through the smaller loop?

$\mu_0 n I \pi R^2$
$\mu_0 n I \pi r^2$ (correct answer)
Zero, because the loop is inside the solenoid.
$\mu_0 n I (2\pi r)$

Explanation: The magnetic field inside the solenoid is uniform with magnitude

B=\mu_0 n I

and is directed along the axis. The flux is calculated through the area of the smaller loop,

A = \pi r^2

. Since the field is uniform and perpendicular to the plane of the inner loop, the flux is

\Phi_B = BA = (\mu_0 n I)(\pi r^2)

Question 13

A long straight wire carries a current $I$ . A square loop of side length $a$ is placed a distance $a$ from the wire. The magnetic flux through the loop is calculated to be $\Phi_0$ . If the current in the wire is doubled to $2I$ and the side length of the loop is halved to $a/2$ (while keeping the closest side at distance $a$ ), what is the new magnetic flux through the loop?

Less than $\Phi_0$ (correct answer)
Equal to $\Phi_0$
Greater than $\Phi_0$
The relationship cannot be determined without knowing the value of $a$ .

Explanation: The flux is

\Phi = \int B dA

. Initially,

\Phi_0 = \int_a^{2a} (\frac{\mu_0 I}{2\pi r}) (a \, dr) = \frac{\mu_0 I a}{2\pi} \ln(2)

. The new flux is

\Phi_{new} = \int_a^{a+a/2} (\frac{\mu_0 (2I)}{2\pi r}) (\frac{a}{2} \, dr) = \frac{\mu_0 I a}{2\pi} \int_a^{1.5a} \frac{dr}{r} = \frac{\mu_0 I a}{2\pi} \ln(1.5)

. Since

\ln(1.5) < \ln(2)

, the new flux is less than the original flux

\Phi_0

Question 14

A circular loop is held in a uniform magnetic field. The flux through the loop is non-zero. The loop is then reshaped into a square of equal perimeter while remaining in the same field and orientation. How does the magnitude of the magnetic flux through the square compare to the original flux through the circle?

The flux is greater through the square.
The flux is the same for both shapes.
The flux is less through the square. (correct answer)
The relationship depends on the strength of the magnetic field.

Explanation: For a fixed perimeter, a circle encloses the maximum possible area. Let the perimeter be

P

. For the circle,

P=2\pi r

, so

r=P/(2\pi)

, and area

A_{circle}=\pi r^2 = P^2/(4\pi)

. For the square, side length is

s=P/4

, and area

A_{square}=s^2 = P^2/16

. Since

4\pi \approx 12.56 < 16

, we have

A_{circle} > A_{square}

. As flux is proportional to area (

\Phi=BA

), the flux through the square is less than the flux through the circle.

Question 15

A flat surface of area $A$ is in a uniform magnetic field $B$ . The magnetic flux through the surface is $\Phi_B = \frac{1}{2}BA$ . What is the angle between the magnetic field vector and the plane of the surface?

$30^\circ$ (correct answer)
$45^\circ$
$60^\circ$
$90^\circ$

Explanation: The magnetic flux is given by

\Phi_B = BA \cos\theta

, where

\theta

is the angle between the magnetic field and the normal to the surface. We are given

\Phi_B = \frac{1}{2}BA

, so

BA \cos\theta = \frac{1}{2}BA

, which means

\cos\theta = 1/2

. This gives

\theta = 60^\circ

. This is the angle between the field and the normal. The question asks for the angle between the field and the plane of the surface, which is

90^\circ - \theta = 90^\circ - 60^\circ = 30^\circ

Question 16

Which of the following statements provides the best conceptual definition of magnetic flux?

The work done by the magnetic field on a unit charge passing through a surface area.
A measure of the total number of magnetic field lines passing through a given surface area. (correct answer)
The strength of the magnetic field vector at the geometric center of a given surface area.
The total magnetic field generated by the currents enclosed by the boundary of a surface area.

Explanation: Magnetic flux (

\Phi_B

) is conceptually a measure of the amount of magnetic field passing through a surface. It is quantified by counting the net number of magnetic field lines that pierce the surface.

Question 17

What is the net magnetic flux through any closed surface, such as a sphere or a cube, regardless of the magnetic fields or currents present?

Zero. (correct answer)
Proportional to the net current passing through the surface.
Proportional to the net charge enclosed by the surface.
Dependent on the volume enclosed by the surface.

Explanation: This is a statement of Gauss's law for magnetism,

\oint \vec{B} \cdot d\vec{A} = 0

. Since there are no magnetic monopoles, magnetic field lines always form closed loops. Therefore, any field line that enters a closed surface must also exit it, resulting in a net magnetic flux of zero.

Question 18

Based on its fundamental definition involving magnetic field and area, the SI unit of magnetic flux, the weber (Wb), is equivalent to which of the following?

Volt-second ( $\text{V}\cdot\text{s}$ )
Newton per coulomb ( $\text{N}/\text{C}$ )
Tesla-meter squared ( $\text{T}\cdot\text{m}^2$ ) (correct answer)
Ampere per meter ( $\text{A}/\text{m}$ )

Explanation: The fundamental definition of magnetic flux for a uniform field is

\Phi_B = BA \cos\theta

. The unit of magnetic field (

B

) is the tesla (T) and the unit of area (

A

) is the meter squared (

\text{m}^2

). Therefore, the unit of magnetic flux is tesla-meter squared (

\text{T}\cdot\text{m}^2

), which is defined as the weber (Wb). While a volt-second is also equivalent to a weber, it is derived from Faraday's law of induction, not the direct definition of flux.

Question 19

Which of the following correctly contrasts magnetic flux with electric flux?

Magnetic flux lines begin on north poles and end on south poles, while electric flux lines begin on positive charges and end on negative charges.
Magnetic flux through a closed surface must be zero, while electric flux through a closed surface is proportional to the enclosed charge. (correct answer)
Magnetic flux is a vector quantity defined by a cross product, while electric flux is a scalar quantity defined by a dot product.
Magnetic flux only exists when charges are in motion, while electric flux can exist for stationary charges.

Explanation: This statement correctly identifies the key difference embodied by Gauss's laws for electricity and magnetism. Due to the absence of magnetic monopoles, magnetic field lines are always closed loops, so the net magnetic flux through any closed surface is zero. Electric field lines, however, originate and terminate on charges, so the net electric flux through a closed surface is proportional to the net charge enclosed.