AP Physics C Electricity and Magnetism Quiz: Induced Currents And Magnetic Forces

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Induced Currents And Magnetic ForcesQuestion 1 of 20

A rectangular wire loop is pulled at a constant speed $v$ out of a region of uniform magnetic field $B$ . This results in a magnetic braking force of magnitude $F_0$ . If the experiment is repeated with a magnetic field of strength $2B$ , but all other parameters (loop dimensions, resistance, speed) remain the same, what will be the new magnetic braking force?

F_0 / 4

F_0 / 2

2F_0

4F_0

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Induced Currents And Magnetic Forces

Practice Induced Currents And Magnetic Forces 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 Induced Currents And Magnetic Forces, 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

$F_0 / 4$
$F_0 / 2$
$2F_0$
$4F_0$ (correct answer)

Explanation: The magnetic braking force is given by the expression

F = \frac{B^2 L^2 v}{R}

. This equation shows that the force is directly proportional to the square of the magnetic field strength (

F \propto B^2

). Therefore, if the magnetic field strength

B

is doubled, the force will increase by a factor of

2^2 = 4

. The new force will be

4F_0

Question 2

A square conducting loop is moved at a constant velocity $\vec{v}$ from a region of no magnetic field into a region with a uniform magnetic field $\vec{B}$ directed into the page. The velocity vector is perpendicular to the boundary of the field and to one side of the loop. What is the direction of the magnetic force on the loop as it enters the field?

In the direction of $\vec{v}$ , acting to accelerate the loop.
In the direction opposite to $\vec{v}$ , acting to decelerate the loop. (correct answer)
Into the page, parallel to the direction of $\vec{B}$ .
Out of the page, opposite to the direction of $\vec{B}$ .

Explanation: As the loop enters the field, the magnetic flux into the page increases. By Lenz's law, the induced current creates a magnetic field out of the page to oppose this change. The right-hand rule for current loops indicates this requires a counter-clockwise current. The leading edge of the loop (the one inside the field) has an upward current. Using the Lorentz force right-hand rule (

\vec{F} = I\vec{L} \times \vec{B}

), the force on this leading edge is to the left, opposite to the direction of velocity

\vec{v}

. The forces on the top and bottom segments cancel each other out.

Question 3

A square loop of wire is pulled at a constant speed $v$ first into a uniform magnetic field region, and then later pulled out of the same region at the same constant speed $v$ . Let $F_{in}$ be the magnitude of the magnetic force on the loop as it enters, and $F_{out}$ be the magnitude as it exits. How do $F_{in}$ and $F_{out}$ compare?

$F_{in} > F_{out}$ because the direction of induced current weakens the interaction when exiting.
$F_{in} < F_{out}$ because exiting the field requires more work against the induced field.
$F_{in} = F_{out}$ because the magnitude of the rate of change of flux is the same in both cases. (correct answer)
$F_{out} = 0$ while $F_{in} \neq 0$ because induction only occurs when flux increases.

Explanation: The magnitude of the induced emf is given by Faraday's Law,

|\mathcal{E}| = |-\frac{d\Phi_B}{dt}|

. As the loop enters and exits at the same speed

v

, the rate at which the loop's area within the field changes is the same (

dA/dt = Lv

). Therefore, the magnitude of the rate of change of flux,

|d\Phi_B/dt| = B(Lv)

, is identical for both entering and exiting. This means the magnitude of the induced emf and current

I = |\mathcal{E}|/R

are also the same. Since the magnetic force magnitude is

F = ILB

, the forces

F_{in}

and

F_{out}

must be equal in magnitude.

Question 4

A conducting ring is dropped from rest so that it falls through a finite region of uniform magnetic field directed perpendicular to its plane. The force of gravity acts on the ring. Which statement best describes the ring's downward acceleration?

The acceleration is constant and equal to $g$ throughout its fall.
The acceleration is less than $g$ as it enters and exits the field, and equal to $g$ while fully inside. (correct answer)
The acceleration is less than $g$ as it enters, equal to $g$ while inside, and greater than $g$ as it exits.
The acceleration is always less than $g$ as long as any part of the ring is in the field region.

Explanation: As the ring enters the field, the changing magnetic flux induces a current, which results in an upward magnetic braking force opposing gravity. The net downward force is less than

mg

, so the downward acceleration is less than

g

. While the ring is fully inside the uniform field, the magnetic flux is constant, so no current is induced and no magnetic force acts. The only force is gravity, so the acceleration is

g

. As the ring exits, the flux changes again, inducing another upward braking force, and the acceleration is again less than

g

Question 5

A rectangular loop of wire enters a region of uniform magnetic field at a constant speed $v$ , resulting in a magnetic braking force $F$ . If the experiment is repeated with a new loop made of a wire with twice the resistivity but otherwise identical dimensions, moving at the same speed $v$ , what will be the new braking force?

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

Explanation: The magnetic braking force is given by

F = \frac{B^2 L^2 v}{R}

. The resistance

R

of a wire is directly proportional to its resistivity

\rho

(

R = \rho \frac{\text{length}}{\text{area}}

). If the resistivity doubles, the total resistance

R

of the loop also doubles. Since the force

F

is inversely proportional to the resistance

R

, doubling the resistance will cause the braking force to be halved. The new force will be

F/2

Question 6

A square loop of wire is moving at a constant speed $v$ through a non-uniform magnetic field $B(x)$ that is directed perpendicular to the loop and varies only with position $x$ . At a particular instant, the leading edge is at $x_2$ and the trailing edge is at $x_1$ . The existence of a net magnetic force on the loop is a direct consequence of which of the following conditions?

The average magnetic field over the loop's area is non-zero.
The magnetic field is different at the locations of the leading and trailing edges. (correct answer)
The speed of the loop is constant throughout its motion.
The resistance of the loop is finite and non-zero.

Explanation: A net magnetic force arises from two effects: an induced current and the interaction of that current with the field. In this scenario, a net motional EMF is induced only if the EMF on the leading edge (

\mathcal{E}_2 = B(x_2)Lv

) is different from the EMF on the trailing edge (

\mathcal{E}_1 = B(x_1)Lv

). This difference, which drives the current, exists only if

B(x_2) \neq B(x_1)

. Furthermore, the forces on these edges (

F_2=ILB(x_2)

and

F_1=ILB(x_1)

) will only be unbalanced if the field strengths are different. Therefore, the fundamental condition for a net force is that the magnetic field is different at the two edges.

Question 7

A conducting rod of mass $m$ and resistance $R$ slides without friction down two parallel conducting rails, which are a distance $L$ apart and tilted at an angle $\theta$ to the horizontal. The rails are connected at the bottom by a wire of negligible resistance. A uniform magnetic field $\vec{B}$ is directed vertically upward.

After a long time, the rod slides down the rails at a constant terminal velocity $v_T$ . What is the expression for this terminal velocity?

$v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta}$ (correct answer)
$v_T = \frac{mgR \tan\theta}{B^2 L^2}$
$v_T = \frac{mgR}{B^2 L^2 \sin\theta}$
$v_T = \frac{mgR \sin\theta}{B^2 L^2}$

Explanation: At terminal velocity, the component of the gravitational force along the rails (

mg \sin\theta

) is balanced by the component of the magnetic braking force opposing it. The magnetic flux perpendicular to the loop is

\Phi_B = (B \cos\theta)(Lx)

, where

x

is the distance along the rails. The induced emf is

\mathcal{E} = |d\Phi_B/dt| = BLv\cos\theta

. The current is

I = \mathcal{E}/R

. The magnetic force on the rod is

F_B = ILB

, directed horizontally. The component of this force up the incline is

F_{B, \parallel} = F_B \cos\theta = (ILB)\cos\theta = \frac{(BLv\cos\theta)LB\cos\theta}{R} = \frac{B^2 L^2 v \cos^2\theta}{R}

. Setting

mg \sin\theta = F_{B, \parallel}

and solving for

v=v_T

gives

v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta}

Question 8

A rectangular conducting loop is pushed at a constant velocity $\vec{v}$ completely through and out of a finite region of uniform magnetic field $\vec{B}$ directed out of the page. What is the direction of the net magnetic force on the loop as its leading edge exits the field, while its trailing edge is still inside?

Opposite to the direction of $\vec{v}$ . (correct answer)
In the direction of $\vec{v}$ .
Out of the page, parallel to $\vec{B}$ .
There is no net magnetic force on the loop.

Explanation: As the loop exits the field, the magnetic flux out of the page is decreasing. According to Lenz's law, the induced current will flow in a direction that creates a magnetic field out of the page to oppose this change. A right-hand rule analysis shows this requires a counter-clockwise current. The trailing edge of the loop is still inside the field and has an upward-directed current. Using the Lorentz force rule (

\vec{F} = I\vec{L} \times \vec{B}

), an upward current in an out-of-page magnetic field results in a force directed opposite to the velocity. This is a braking force.

Question 9

A conducting loop is given an initial velocity $\vec{v}_0$ to the right and enters a region of uniform magnetic field directed into the page. There are no other forces acting on the loop, such as gravity or friction. What happens to the loop's velocity as it moves through the field?

The velocity remains constant at $\vec{v}_0$ because the net magnetic force is always zero.
The velocity decreases as it enters the field, becomes constant while fully inside, and decreases again as it exits. (correct answer)
The velocity decreases continuously as long as any part of the loop is in the magnetic field.
The velocity decreases as it enters and then increases back to $v_0$ as it exits due to energy conservation.

Explanation: As the loop enters the field, a change in flux induces a current and a magnetic braking force, which opposes the motion and causes the loop to slow down. While fully inside the uniform field, the flux is constant, so there is no induced current and no magnetic force; the loop moves at a constant, reduced velocity. As the loop exits, the flux changes again, inducing another braking force that causes the loop to slow down further. The kinetic energy lost is dissipated as thermal energy in the loop.

Question 10

A conducting rod of length $L$ is rotated with constant angular velocity $\omega$ about a pivot at one of its ends, in a plane perpendicular to a uniform magnetic field $\vec{B}$ . This induces a current $I$ which flows radially along the rod (assuming a complete circuit). What is the resulting magnetic force on the rod?

A radial force directed away from the pivot, stretching the rod.
A radial force directed toward the pivot, compressing the rod.
A tangential force creating a torque that assists the rotation.
A tangential force creating a torque that opposes the rotation. (correct answer)

Explanation: Due to the rotation, a motional emf is induced, driving a current

I

along the rod's length. The magnetic force on a current element

I d\vec{r}

d\vec{F} = I d\vec{r} \times \vec{B}

. With the current

I

flowing radially and the field

\vec{B}

perpendicular to the plane of rotation, the force

d\vec{F}

is tangential. By Lenz's law, this force must oppose the motion that creates the current. Therefore, the force is directed tangentially opposite to the direction of motion, creating a magnetic torque that opposes the angular velocity.

Question 11

A rectangular conducting loop of resistance $R$ is pulled with a constant velocity $v$ through a region of uniform magnetic field $B$ directed into the page. The field exists in a region of width $2L$ , where $L$ is the length of the loop side parallel to the velocity.

Which statement best describes the net magnetic force on the loop as it passes completely through the field region?

A constant braking force acts on the loop during the entire time it is moving within the field region.
A braking force acts on the loop only as it enters the field region and as it exits the field region. (correct answer)
A braking force acts on the loop as it enters, and a propelling force of equal magnitude acts on it as it exits.
A braking force acts as the loop enters, and no magnetic force acts while it is fully inside or as it exits.

Explanation: A net magnetic force on the loop results from an induced current, which is caused by a change in magnetic flux through the loop. The flux changes only when the area of the loop inside the field is changing, which occurs as the loop enters and as it exits the region. While the loop is fully inside the uniform field, the flux is constant, so there is no induced current and no net magnetic force. In both entering and exiting, Lenz's law dictates that the force opposes the motion, thus it is a braking force in both instances.

Question 12

A rectangular loop of width $L$ , length $W$ , and resistance $R$ is pulled at a constant speed $v$ from a region with no magnetic field, completely through a region of width $2W$ with a uniform magnetic field $B$ , and into a region with no field on the other side. How much work is done by the external force pulling the loop?

$W = 0$
$W = \frac{B^2 L^2 v W}{R}$
$W = 2 \frac{B^2 L^2 v W}{R}$ (correct answer)
$W = \frac{B^2 L^2 v^2 W}{R}$

Explanation: Work is done by the external force only when it must oppose a magnetic braking force. This occurs as the loop enters the field (over a distance

W

) and as it exits the field (over a distance

W

). The magnitude of the braking force is constant during these phases and is given by

F = \frac{B^2 L^2 v}{R}

. The work done by the external force is

W = F \times d

. The total work is the sum of the work done entering and exiting:

W_{total} = F \cdot W + F \cdot W = 2FW = 2 \frac{B^2 L^2 v W}{R}

. No work is done against a magnetic force while the loop is fully inside the field.

Question 13

According to the scenario, a solenoid's current ramps up, increasing magnetic flux through a nearby coil; Faraday's Law $\mathcal{E}=-d\Phi/dt$ induces a current whose magnetic field opposes the increasing flux (Lenz's Law), as in transformer action. According to Faraday's Law, what happens to the induced EMF when the magnetic field changes?

It increases in magnitude when the flux changes more rapidly in time. (correct answer)
It becomes smaller when the flux changes faster because opposition cancels EMF.
It depends only on the coil's area, not on how fast flux changes.
It is nonzero only if the magnetic field direction reverses each cycle.

Explanation: This question tests AP Physics C concepts of electromagnetic induction, specifically understanding how EMF magnitude relates to flux change rate in transformer coupling. Faraday's Law establishes that induced EMF magnitude is directly proportional to the rate of magnetic flux change: |ε|=|dΦ/dt|, fundamental to transformer operation. In the provided scenario, a solenoid's increasing current creates rising flux through a nearby coil, demonstrating this principle through mutual inductance. Choice A is correct because it accurately states that EMF magnitude increases when flux changes more rapidly, reflecting the derivative relationship in Faraday's Law. Choice B is incorrect because it suggests faster changes produce smaller EMF, contradicting the mathematical relationship. To help students: Use graphical representations showing steeper flux-time slopes producing larger EMF values, and connect to transformer efficiency at different frequencies. Watch for: Students confusing Lenz's Law (direction) with Faraday's Law (magnitude), or thinking opposition reduces EMF size.

Question 14

Based on the description, a metal rod of length $L$ moves upward at speed $v$ through a uniform magnetic field $\vec{B}$ to the right, with the rod perpendicular to both $\vec{v}$ and $\vec{B}$ . Charges in the rod experience the magnetic force $q\,\vec{v}\times\vec{B}$ , separating charges and creating a motional EMF. For this geometry, $\mathcal{E}=BLv$ , and if the rod is part of a closed circuit, a current flows. Lenz's Law predicts a magnetic force on the rod opposing its motion. How does Lenz's Law apply to the situation described?

The induced force on the rod opposes the upward motion (correct answer)
The induced force on the rod points upward, aiding motion
No force occurs because $\vec{B}$ is uniform
Current flows without motion because charges separate instantly

Explanation: This question tests AP Physics C concepts of electromagnetic induction, specifically understanding motional EMF and the resulting magnetic forces on moving conductors. Electromagnetic induction occurs when a conductor moves through a magnetic field, with free charges experiencing a magnetic force that creates charge separation and motional EMF. In the provided scenario, a metal rod moving upward through a rightward magnetic field induces an EMF of magnitude BLv, and if part of a circuit, current flows through the rod. Choice A is correct because Lenz's Law requires that the magnetic force on this current-carrying rod (F = IL × B) opposes the upward motion that causes the induction. Choice B is incorrect because it violates Lenz's Law by suggesting the induced effects would aid rather than oppose the motion. To help students: Use the right-hand rule twice - first for charge separation (v × B), then for force on current (I × B). Emphasize energy conservation: mechanical work against the magnetic force equals electrical power generated. Watch for: Students correctly finding current direction but incorrectly determining the force direction on that current.

Question 15

Based on the description, a metal rod moves upward at speed $v$ through a uniform magnetic field to the right, generating motional EMF $\mathcal{E}=B\ell v$ ; the induced current produces a magnetic force opposing the rod's motion (Lenz's Law), relevant to velocity sensors. What is the direction of the induced current in the scenario described?

Into the page through the rod, using the right-hand rule for $\vec v\times\vec B$ .
Out of the page through the rod, using the right-hand rule for $\vec v\times\vec B$ . (correct answer)
No current flows because the rod is not part of a closed circuit.
To the right through the rod, parallel to $\vec B$ in all cases.

Explanation: This question tests AP Physics C concepts of electromagnetic induction, specifically determining induced current direction using the Lorentz force on moving charges. When a conductor moves through a magnetic field, the magnetic force F=qv×B on charges creates charge separation and current flow according to the right-hand rule. In the provided scenario, a metal rod moves upward through a rightward magnetic field, demonstrating this principle through motional EMF generation. Choice B is correct because using the right-hand rule with velocity upward and field rightward, the cross product v×B points out of the page, indicating the force direction on positive charges and thus conventional current direction. Choice A is incorrect because it reverses the cross product direction, misapplying the right-hand rule. To help students: Practice the right-hand rule systematically - index finger for velocity, middle finger for field, thumb shows force on positive charges. Watch for: Students using left-hand rule for positive charges or forgetting that conventional current follows positive charge motion.

Question 16

Based on the description, a loop rotates in a uniform field in a simple generator so $\Phi=BA\cos(\omega t)$ and $\mathcal{E}=-d\Phi/dt$ ; reversing flux change reverses current direction (Lenz's Law), converting mechanical to electrical energy. According to Faraday's Law, what happens to the induced EMF when the magnetic field changes?

It is proportional to $d\Phi/dt$ , so faster flux change gives larger $|\mathcal{E}|$ . (correct answer)
It decreases when $d\Phi/dt$ increases because induction resists change.
It is nonzero only if the coil has a ferromagnetic core inserted.
It depends only on the loop's resistance, not on flux change rate.

Explanation: This question tests AP Physics C concepts of electromagnetic induction, specifically understanding Faraday's Law and its relationship to flux change rate. Faraday's Law states that induced EMF equals the negative rate of change of magnetic flux: ε=-dΦ/dt, making EMF directly proportional to how quickly flux changes. In the provided scenario, a rotating loop in a generator has flux Φ=BAcos(ωt), demonstrating this principle through sinusoidal flux variation. Choice A is correct because it accurately states that EMF is proportional to dΦ/dt, meaning faster flux changes produce larger induced EMF magnitudes. Choice B is incorrect because it suggests an inverse relationship, contradicting Faraday's Law's mathematical form. To help students: Emphasize the derivative relationship - doubling the rate of flux change doubles the EMF, crucial for understanding transformer and generator operation. Watch for: Students confusing the negative sign (which indicates direction via Lenz's Law) with the magnitude relationship.

Question 17

According to the scenario, a metal rod of length $\ell$ slides right at speed $v$ on rails in a uniform $\vec B$ into the page, producing motional EMF $\mathcal{E}=B\ell v$ and a current that creates a magnetic force opposing motion (Lenz's Law), like in rail generators. What is the direction of the induced current in the scenario described?

Clockwise around the loop, so the rod's current is upward.
Counterclockwise around the loop, so the rod's current is downward.
No current flows because $B$ is uniform and constant in time.
Clockwise around the loop, so the rod's current is downward. (correct answer)

Explanation: This question tests AP Physics C concepts of electromagnetic induction, specifically determining induced current direction in motional EMF scenarios. When a conductor moves through a magnetic field, charges experience a magnetic force that creates charge separation and induces current according to the motional EMF formula ε=Bℓv. In the provided scenario, a metal rod slides right through a magnetic field directed into the page, demonstrating this principle through a classic rail generator setup. Choice D is correct because using the right-hand rule for v×B (velocity right, field into page), the magnetic force on positive charges points downward in the rod, making conventional current flow downward through the rod and clockwise around the circuit. Choice B is incorrect because it reverses the current direction, misapplying the cross product rule. To help students: Practice the right-hand rule systematically - point fingers in velocity direction, curl toward field direction, thumb shows force on positive charges. Watch for: Confusion between force on positive charges versus electron flow direction.

Question 18

A square loop of wire with side length $L$ and resistance $R$ is pulled with constant speed $v$ into a uniform magnetic field $B$ directed perpendicular to the plane of the loop. What is the magnitude of the net magnetic force on the loop as its leading edge enters the field?

$F = \frac{B^2 L^2 v}{R}$ (correct answer)
$F = \frac{BLv}{R}$
$F = B L v$
$F = \frac{B^2 L v^2}{R}$

Explanation: The motional emf induced in the leading edge of the loop is

\mathcal{E} = B L v

. According to Ohm's law, the induced current in the loop is

I = \frac{\mathcal{E}}{R} = \frac{B L v}{R}

. The magnetic force on the leading edge, which carries this current, is given by

F = I L B

. Substituting the expression for

I

gives

F = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}

. The forces on the top and bottom segments cancel.

Question 19

A square conducting loop of mass $m$ , side length $L$ , and resistance $R$ is released from rest. At the instant its top edge enters a uniform magnetic field $B$ directed into the page, the loop has a downward speed $v$ . What is the magnitude of the net force on the loop at this instant?

$F_{net} = mg - \frac{B^2 L^2 v}{R}$ (correct answer)
$F_{net} = mg$
$F_{net} = \frac{B^2 L^2 v}{R}$
$F_{net} = mg + \frac{B^2 L^2 v}{R}$

Explanation: Two forces act on the loop: the downward gravitational force,

F_g = mg

, and the upward magnetic braking force,

F_B

. The magnetic force opposes the motion. Its magnitude is

F_B = I L B = (\frac{BLv}{R})LB = \frac{B^2 L^2 v}{R}

. According to Newton's second law, the net force is the vector sum of these forces. Taking downward as positive,

F_{net} = F_g - F_B = mg - \frac{B^2 L^2 v}{R}

Question 20

A conducting loop is pushed at a constant velocity $v$ into a magnetic field. An external force $F_{ext}$ is required to maintain this constant velocity against the magnetic braking force $F_{mag}$ . What is the relationship between the rate at which the external force does work ( $P_{ext}$ ) and the rate of thermal energy dissipation in the loop ( $P_{dissipated}$ )?

$P_{ext} = P_{dissipated}$ , because the work done by the external force is converted entirely into thermal energy. (correct answer)
$P_{ext} > P_{dissipated}$ , because some energy is stored in the magnetic field of the induced current.
$P_{ext} < P_{dissipated}$ , because the external magnetic field also contributes energy to the system.
The relationship cannot be determined without knowing if the loop's kinetic energy changes.

Explanation: Since the loop moves at a constant velocity, its kinetic energy is constant. By the work-energy theorem for a non-conservative system, the work done by the external force must equal the energy dissipated as heat.

P_{ext} = F_{ext} v

. Since

F_{ext}

balances

F_{mag}

to keep

v

constant,

F_{ext} = F_{mag} = B^2L^2v/R

. Thus,

P_{ext} = (B^2L^2v/R)v = B^2L^2v^2/R

. The dissipated power is

P_{dissipated} = I^2 R = (BLv/R)^2 R = B^2L^2v^2/R

. Therefore,

P_{ext} = P_{dissipated}