Induced Currents and Magnetic Forces - AP Physics C Electricity and Magnetism

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 ( $P_{dissipated}$ )?
Answer: $P_{ext} = P_{dissipated}$ , because the work done by the external force is converted entirely into thermal energy. 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, $P_{ext} = P_{dissipated}$ .
What is the direction of the net magnetic force exerted by Loop 1 on Loop 2?
Answer: Radially outward, away from the common center. The clockwise current $I_1$ in Loop 1 creates a magnetic field into the page through Loop 2. Since $I_1$ is decreasing, the magnetic flux into the page is decreasing. By Lenz's law, the induced current $I_2$ in Loop 2 will flow clockwise to oppose this change by creating a magnetic field into the page. The magnetic field from Loop 1 varies with radius, being stronger closer to the wire of Loop 1. Since both loops carry clockwise current, each current element of Loop 2 experiences a force due to the non-uniform field of Loop 1. The net effect is a radially outward force on Loop 2, as the magnetic field from Loop 1 tends to push the induced current away from its source.
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?
Answer: $W = 2 \frac{B^2 L^2 v W}{R}$ 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.
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?
Answer: $F_{net} = mg - \frac{B^2 L^2 v}{R}$ 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, $F_{net} = F_g - F_B = mg - \frac{B^2 L^2 v}{R}$ .
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?
Answer: The velocity decreases as it enters the field, becomes constant while fully inside, and decreases again as it exits. 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.
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?
Answer: It increases in magnitude when the flux changes more rapidly in time. 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.
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?
Answer: $F/2$ 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 $F/2$ .
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?
Answer: $F = \frac{B^2 L^2 v}{R}$ 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.
A square conducting loop is released from rest and falls into a region of uniform magnetic field perpendicular to its plane. The loop reaches a terminal velocity $v_T$ . How would the terminal velocity change if the square loop were replaced by a rectangular loop with the same mass, same total resistance, and the same horizontal width, but twice the vertical height?
Answer: The terminal velocity would be the same. Terminal velocity is reached when the upward magnetic braking force equals the downward gravitational force: $F_B = mg$ . The magnetic force is $F_B = \frac{B^2 L^2 v_T}{R}$ , where $L$ is the horizontal width of the loop. Thus, the terminal velocity is $v_T = \frac{mgR}{B^2 L^2}$ . The problem states that mass $m$ , resistance $R$ , and horizontal width $L$ are all the same for the new loop. The vertical height does not appear in the expression for the terminal velocity, so it remains unchanged.
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?
Answer: A tangential force creating a torque that opposes the rotation. 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 $I d\vec{r}$ is $d\vec{F} = I d\vec{r} \times \vec{B}$ . 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.
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?
Answer: The induced force on the rod opposes the upward motion 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.
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?
Answer: $F_{in} = F_{out}$ because the magnitude of the rate of change of flux is the same in both cases. 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.
A solid circular conducting disk is moving at a constant velocity into a uniform magnetic field directed into the page. Eddy currents are induced in the disk. Which statement correctly describes the net magnetic force on the disk due to these eddy currents?
Answer: The net force is opposite to the direction of motion, producing a braking effect on the disk. As the portion of the disk inside the field increases, the magnetic flux into the page increases. Lenz's law dictates that the induced eddy currents will flow in a direction to create a magnetic field that opposes this increase. These circulating currents interact with the external magnetic field via the Lorentz force ( $\vec{F} = I\vec{L} \times \vec{B}$ ). A detailed analysis of the forces on the current paths shows that the net result is a magnetic force directed opposite to the velocity of the disk, providing a braking effect.
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?
Answer: The acceleration is less than $g$ as it enters and exits the field, and equal to $g$ while fully inside. 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 $g$ .
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?
Answer: Clockwise around the loop, so the rod's current is downward. 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.
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?
Answer: Opposite to the direction of $\vec{v}$ . 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.
Which statement best describes the net magnetic force on the loop as it passes completely through the field region?
Answer: A braking force acts on the loop only as it enters the field region and as it exits the field region. 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.
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?
Answer: $v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta}$ 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 $v_T = \frac{mgR \sin\theta}{B^2 L^2 \cos^2\theta}$ .
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?
Answer: It is proportional to $d\Phi/dt$ , so faster flux change gives larger $|\mathcal{E}|$ . 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.
What is the magnitude and direction of the net magnetic force on the small loop?
Answer: There is no net magnetic force or net torque on the small loop. The magnetic field inside an ideal long solenoid is uniform and directed along its axis: $B(t) = \mu_0 n I(t) = \mu_0 n k t$ . A current is induced in the small loop due to the changing flux. However, the net force on a closed current loop in a uniform magnetic field is always zero. The integral of the force element $I d\vec{l} \times \vec{B}$ over a closed loop is $I (\oint d\vec{l}) \times \vec{B}$ , and $\oint d\vec{l} = 0$ . Furthermore, because the loop is coaxial with the field, its magnetic moment is parallel to the field, so the torque ( $\vec{\tau} = \vec{\mu} \times \vec{B}$ ) is also zero.

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