4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The figure shows a conducting rod sliding at constant velocity on conducting rails that are connected by a resistor, in a uniform magnetic field, with the data labeled. The rod and rails have negligible resistance. What is the magnitude of the magnetic force exerted on the rod?
Answer and reasoning
A1.6 × 10⁻² N A student who uses the flux through the circuit, Bℓx = 0.16 T·m², in place of its rate of change picks this: I = 0.080 A. The emf is the rate of change of flux, Bℓv, which depends on the rod’s speed, not on its distance from the resistor.
B5.2 × 10⁻¹ N A student who uses the whole length of the circuit, 2.6 m, in F = IℓB picks this. The force on the rod acts on the rod’s own length, 0.50 m; forces on the rails act on the rails.
C1.0 × 10⁻¹ NCorrect The rod’s motion induces ε = Bℓv = (0.40 T)(0.50 m)(5.0 m/s) = 1.0 V, which drives I = ε/R = 0.50 A around the circuit. The field then exerts FB = IℓB = (0.50 A)(0.50 m)(0.40 T) = 1.0 × 10⁻¹ N on the rod, opposing its motion.
D5.0 × 10⁻¹ N A student who takes the force to be the power dissipated, ε²/R = 0.50 W, picks this. Power is the rate at which the force does work, FB v; FB = (0.50 W)/(5.0 m/s) = 0.10 N.
Working ε = Bℓv = (0.40 T)(0.50 m)(5.0 m/s) = 1.0 V. I = ε/R = 1.0 V/2.0 Ω = 0.50 A. FB = IℓB = (0.50 A)(0.50 m)(0.40 T) = 0.10 N = 1.0 × 10⁻¹ N. Distractors: flux Bℓx = 0.16 T·m² used as the emf, I = 0.080 A, F = 1.6 × 10⁻² N; ℓ taken as the whole circuit length, 2(0.50 m) + 2(0.80 m) = 2.6 m, F = (0.50 A)(2.6 m)(0.40 T) = 5.2 × 10⁻¹ N; force taken as the power dissipated, ε²/R = 0.50 W, reported as 5.0 × 10⁻¹.
A flat, closed, rectangular coil spins about an axis that lies in the plane of the coil and is perpendicular to a uniform magnetic field. There is no friction. Which statement describes the effect of the magnetic forces on the current induced in the coil?
Answer and reasoning
AThey exert a torque that opposes the rotation, so the coil slows down.Correct The rotation changes the flux, so a current is induced. The forces on opposite sides of the coil are equal and opposite, but they do not act along the same line, so they form a couple whose torque opposes the rotation (Lenz’s law); the kinetic energy lost becomes thermal energy in the coil.
BThey exert a torque that aids the rotation, so the coil speeds up. A student who thinks the force on an induced current helps the motion that produces it picks this. The torque opposes the rotation; a coil that sped up while also heating its own wire would be creating energy.
CThey have no effect, because the forces on opposite sides cancel. A student who applies ‘equal and opposite forces cancel’ picks this. The forces cancel as a net force, but they act along different lines, so together they exert a torque that slows the rotation.
DThey act only until the coil’s own field cancels the external field. A student who thinks the induced current’s field grows until it cancels the external field picks this. It never does: as long as the coil turns, the flux through it keeps changing, a current is induced, and the torque keeps opposing the rotation.
Three rectangular loops of identical size and shape move at the same constant velocity into a region of uniform magnetic field. Loops 1 and 2 are closed, and loop 2 has four times the resistance of loop 1. Loop 3 is identical to loop 1 except for a small gap cut in it. F₁, F₂ and F₃ are the magnitudes of the magnetic forces on the loops while each is entering the field. Which ranking is correct?
Answer and reasoning
AF₂ > F₁ > F₃ = 0 A student who thinks a larger resistance brakes the motion more strongly picks this. A larger resistance means a smaller induced current and so a smaller force: F₂ is one quarter of F₁.
BF₁ = F₂ > F₃ = 0 A student who thinks the induced current is fixed by the motion, whatever the resistance, picks this. The motion fixes the emf; the current is the emf divided by the resistance, so loop 2 carries less current and feels less force.
CF₁ = F₂ = F₃ > 0 A student who thinks the induced emf itself drags the loops, whether or not a current flows, picks this. The force is exerted on a current; with a gap, loop 3 carries no current and feels no magnetic force, and loop 2 carries less current than loop 1.
DF₁ > F₂ > F₃ = 0Correct All three loops have the same emf, Bℓv. The current is the emf divided by the resistance, so loop 2 carries one quarter of loop 1’s current, and loop 3, with a gap, carries none. The force F = IℓB follows the current: F₁ = 4F₂, and F₃ = 0.
Working The three loops have the same rate of change of flux and so the same emf, ε = Bℓv. I₁ = ε/R, I₂ = ε/(4R), I₃ = 0 (no closed path). F = IℓB, so F₁ = 4F₂ and F₃ = 0: F₁ > F₂ > F₃ = 0.
A conducting rod of mass m and length ℓ slides on frictionless horizontal rails connected by a resistor of resistance R, in a uniform vertical magnetic field of magnitude B; all other resistances are negligible. At time t = 0 the rod has speed v₀, and no horizontal force other than the magnetic force acts on it. Which expression gives the rod’s speed v at time t?
Answer and reasoning
Av₀eB²ℓ²t/(mR) A student who takes the magnetic force to act along the motion picks this: m dv/dt = +(B²ℓ²/R)v. The force on the induced current opposes the motion, so the rod slows down; a speed growing without limit would create energy.
Bv₀e−B²ℓ²t/(mR)Correct The magnetic force is FB = B²ℓ²v/R, opposite to the velocity, so m dv/dt = −(B²ℓ²/R)v. The rate of decrease is proportional to v itself, which gives exponential decay from v₀ with time constant mR/(B²ℓ²).
Cv₀(1 − B²ℓ²t/(mR)) A student who keeps the force at its initial value, B²ℓ²v₀/R, picks this: a constant deceleration. The force is proportional to the speed, so it weakens as the rod slows, and the speed decays exponentially instead of reaching zero at a definite time.
Dv₀mR/(mR + B²ℓ²v₀t) A student who takes the retarding force to be the power dissipated, ε²/R = B²ℓ²v²/R, picks this: m dv/dt = −B²ℓ²v²/R. That expression is a power, not a force; the force is B²ℓ²v/R.
Working At speed v: ε = Bℓv, I = Bℓv/R, FB = IℓB = B²ℓ²v/R, opposing the motion. Newton’s second law: m dv/dt = −(B²ℓ²/R)v. Separating variables and integrating from v₀ at t = 0: ln(v/v₀) = −B²ℓ²t/(mR), so v = v₀e−B²ℓ²t/(mR). Distractors (checked with sympy): force held at its initial value, v = v₀ − (B²ℓ²v₀/(mR))t = v₀(1 − B²ℓ²t/(mR)); force taken to aid the motion, v₀eB²ℓ²t/(mR); force taken as the power ε²/R, m dv/dt = −B²ℓ²v²/R, v = v₀mR/(mR + B²ℓ²v₀t).
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
13.3.A.1 Magnetic force on an induced current Fix
Magnetic force on an induced current
Once a current is induced in a conductor, the magnetic field already present exerts a force on the moving charge carriers, and so on the conductor: F⃗B = ∫I(dℓ⃗ × B⃗). For a straight segment of length ℓ perpendicular to a uniform field, FB = IℓB. SI unit: newton (N).
Force on a curved segment in a uniform field
In a uniform field, ∫dℓ⃗ along a curved segment equals the straight vector from its start to its end, so the magnetic force on the segment equals the force on a straight wire joining its ends and carrying the same current.
Students often think A conductor moving through a magnetic field feels a magnetic drag force because of the emf induced in it, whether or not a current can flow. In fact No. The magnetic force on the conductor is the force on the current in it, F⃗B = ∫I(dℓ⃗ × B⃗). A loop with a gap, or a loop inside a uniform field whose flux is not changing, has no induced current, so there is no magnetic force on it, although emfs may be induced in its parts.
Students often think The magnetic force on an induced current is found from the rate of change of the field, as F = Iℓ(dB/dt), because the changing field is what produces the effect. In fact No. The rate of change of the field sets the emf and so the current; the force on a segment is then FB = IℓB, using the value of the field at that instant.
13.3.A.2 Segments inside the field Fix
Segments inside the field
Magnetic forces act only on the segments of a loop that are inside the field. When a loop is entirely inside a uniform field, the forces on opposite sides cancel and the net force is zero; when only part of the loop is inside, the forces need not cancel.
Torque from forces on a loop
Equal and opposite forces on opposite sides of a loop that do not act along the same line exert a torque, which can give the loop an angular acceleration even though the net force on it is zero.
Students often think The magnetic force on an induced current acts along the whole length of the circuit, so ℓ in FB = IℓB is the total length of wire that carries the current. In fact No. The force on the rod is the force on the rod’s own length inside the field. Forces on other parts of the circuit, such as the rails, act on those parts, not on the rod, and parts outside the field experience no magnetic force at all.
Students often think When a loop leaves a field region, the induced current reverses, so the magnetic force on the loop reverses as well and acts in the direction of its motion. In fact No. As the loop leaves, the induced current reverses, but the segment still in the field is now the trailing side, so F⃗ = Iℓ⃗ × B⃗ still points opposite to the velocity. The force opposes the motion both while the loop enters and while it leaves.
13.3.A.3 Magnetic braking Fix
Magnetic braking
The magnetic force on an induced current opposes the motion (or the change) that produces the current, in keeping with Lenz’s law and energy conservation: the kinetic energy the conductor loses becomes thermal energy in the circuit’s resistance.
Retarding force on a rod on rails, FB = B²ℓ²v/R
For a rod of length ℓ moving at speed v perpendicular to a uniform field B, in a circuit of resistance R: ε = Bℓv, I = Bℓv/R and FB = IℓB = B²ℓ²v/R, opposite to the velocity. The force is proportional to the induced current, which depends on the rate of change of flux, the resistance and the speed.
Students often think The induced current, and so the magnetic force on a loop, depends on how much magnetic flux passes through the loop rather than on how fast the flux is changing. In fact On how fast it changes. The induced emf has magnitude |dΦB/dt|, so the induced current is |dΦB/dt|/R: a large but steady flux induces no current, and a small flux that changes quickly can induce a large one. The magnetic force follows the current.
Students often think The retarding force on a moving rod equals the power dissipated in the circuit, ε²/R, since both measure how strongly the circuit resists the motion. In fact No. Force and power are different quantities, in N and W. The power dissipated, ε²/R, equals the rate at which the retarding force does work on the rod, FB v, so FB = ε²/(Rv) = B²ℓ²v/R.
13.3.A.4 Newton’s second law with an induced force Fix
Newton’s second law with an induced force
For a rod of mass m on horizontal frictionless rails with no other horizontal force, m dv/dt = −B²ℓ²v/R, so the speed decays exponentially: v = v₀e−t/τ, with τ = mR/(B²ℓ²).
Terminal speed
A rod falling between vertical rails speeds up until the upward magnetic force equals its weight: B²ℓ²vt/R = mg, so vt = mgR/(B²ℓ²). At that speed the net force, and so the acceleration, is zero.
Students often think A rod set moving on rails slows down at a constant rate, set by the magnetic force at the start, so its speed decreases linearly and reaches zero at a definite time. In fact No. The retarding force B²ℓ²v/R is proportional to the speed, so the deceleration decreases as the rod slows. Newton’s second law, m dv/dt = −(B²ℓ²/R)v, gives an exponential decay, not a linear decrease.
Students often think The magnetic field of the induced current grows until it cancels the external field, after which no further emf is induced. In fact No. The induced current’s field opposes the change in flux, but it does not cancel the external field; the emf continues as long as the flux keeps changing. For a falling rod at terminal speed, the flux through the circuit keeps changing at a constant rate, so the emf and current are constant, not zero.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
The figure shows a square conducting loop moving to the right at constant velocity through a region of uniform magnetic field directed into the page. The loop is shown at three positions, 1, 2 and 3, drawn at different heights for clarity. Which choice gives the direction of the net magnetic force on the loop at each position? (← means to the left, → to the right, and 0 means no net force.)
Answer and reasoning
A(1) ←, (2) 0, (3) ←Correct At 1 the increasing flux drives a counterclockwise current, and the force on the right side, the only side in the field, points left. At 2 the flux is constant: no current, no force. At 3 the decreasing flux drives a clockwise current, and the force on the left side, now the only side in the field, again points left. The force opposes the motion whenever there is one.
B(1) ←, (2) ←, (3) ← A student who thinks any conductor moving through a field is dragged back, current or not, picks this. At 2 the flux through the loop is not changing, so no current is induced and the field exerts no force on the loop.
C(1) ←, (2) 0, (3) → A student who reasons that the reversed current at 3 reverses the force picks this. The current does reverse, but the segment in the field is now the left side instead of the right, so the force still points left, opposing the motion.
D(1) →, (2) 0, (3) → A student who thinks the force on an induced current helps the motion that produces it picks this. By Lenz’s law and energy conservation, the force opposes the motion both as the loop enters and as it leaves.
Working 1: the flux into the page is increasing, so the induced current is counterclockwise; the right side, inside the field, carries current up the page, and I ℓ⃗ × B⃗ points left. 2: the flux is constant, so there is no current and no force. 3: the flux is decreasing, so the current is clockwise; the left side, now the only side inside the field, carries current up the page, and the force again points left. Answer (1) ←, (2) 0, (3) ←.
A conducting rod slides at constant speed along conducting rails in a uniform magnetic field perpendicular to the plane of the rails. The rails are connected by a resistor, and all other resistances are negligible. The magnetic force on the rod is 0.60 N. The experiment is repeated with the magnetic field doubled, the rod’s speed tripled and a resistor of twice the resistance. What is the magnitude of the magnetic force on the rod now?
Answer and reasoning
A1.8 N A student who scales the force with the current alone picks this: the current changes by 2 × 3/2 = 3. The field also appears in F = IℓB, so doubling it doubles the force once more.
B3.6 NCorrect The emf is Bℓv, the current Bℓv/R and the force IℓB = B²ℓ²v/R. Doubling B multiplies the force by 4, tripling v by 3, and doubling R divides it by 2: (0.60 N)(4)(3)/2 = 3.6 N.
C7.2 N A student who thinks the resistance does not affect the induced current picks this: (0.60 N)(4)(3). The current is the emf divided by the resistance, so doubling R halves the current and the force.
D1.2 N A student who takes the current to depend on the flux rather than on its rate of change leaves out the speed: (0.60 N)(4)/2. The emf is the rate of change of flux, Bℓv, so tripling the speed triples the current and the force.
Working FB = IℓB with I = Bℓv/R, so FB = B²ℓ²v/R. New force: (0.60 N)(2²)(3)/2 = 3.6 N. Distractors: force taken as proportional to the current alone (current × 2 × 3/2 = 3), 1.8 N; resistance taken to have no effect on the current, (0.60 N)(4)(3) = 7.2 N; speed ignored (current taken to depend on the flux rather than its rate of change), (0.60 N)(4)/2 = 1.2 N.
The figure shows a square conducting loop at rest, with half of it inside a region of uniform magnetic field; the field is increasing, and the data are labeled. What is the magnitude of the net magnetic force on the loop at the instant shown?
Answer and reasoning
A3.2 × 10⁻² N A student who uses the whole area of the loop, 0.040 m², for the flux picks this. Only the half inside the field region has flux, so the emf and current are half as large.
B2.6 × 10⁻² N A student who uses the flux, (0.80 T)(0.020 m²), in place of its rate of change picks this: I = 0.16 A. The emf is the rate of change of flux, (0.50 T/s)(0.020 m²) = 0.010 V.
C1.6 × 10⁻² NCorrect Only the half of the loop inside the field has flux: 0.020 m². The emf is (0.020 m²)(0.50 T/s) = 0.010 V and the current 0.10 A. The forces on the top and bottom sides inside the field cancel, leaving the force on the right side: (0.10 A)(0.20 m)(0.80 T) = 1.6 × 10⁻² N.
D1.0 × 10⁻² N A student who puts the rate of change of the field in place of the field in F = IℓB picks this: (0.10 A)(0.20 m)(0.50). The rate of change sets the current; the force uses the field present at that instant, 0.80 T.
Working Only the area inside the field counts: A = (0.20 m)(0.10 m) = 0.020 m². |ε| = A(dB/dt) = (0.020 m²)(0.50 T/s) = 0.010 V; I = ε/R = 0.010 V/0.10 Ω = 0.10 A. The right side (0.20 m, wholly in the field) feels F = IℓB = (0.10 A)(0.20 m)(0.80 T) = 1.6 × 10⁻² N; the forces on the parts of the top and bottom sides inside the field are equal and opposite, and the left side is outside the field. Net force 1.6 × 10⁻² N. Distractors: whole loop area, I = 0.20 A, 3.2 × 10⁻² N; flux BA = 0.016 T·m² used as the emf, I = 0.16 A, 2.6 × 10⁻² N; dB/dt used in place of B in the force, (0.10 A)(0.20 m)(0.50) = 1.0 × 10⁻² N.
A horizontal conducting rod is released from rest between two long, vertical conducting rails that are connected at the top by a resistor, in a uniform horizontal magnetic field perpendicular to the plane of the rails. The rod stays in contact with the rails without friction. A student claims that the rod reaches a constant terminal speed. Which reasoning correctly supports the claim?
Answer and reasoning
AAs the rod falls, the flux through the circuit grows, and the upward force grows with the flux until it equals the weight. A student who links the force to the flux itself gives this reasoning. The flux does grow as the rod falls, but the current and force depend on how fast the flux changes, which is set by the rod’s speed.
BThe induced current’s field grows until it cancels the external field, and after that no further emf is induced. A student who thinks the induced field ends up cancelling the external field gives this reasoning. At terminal speed the flux still changes at a constant rate, so the emf and current are constant, not zero; if they were zero, the rod would speed up again.
CThe resistor opposes the rod’s motion more strongly as the current grows, until this braking equals the weight. A student who pictures the resistor as a brake on the rod gives this reasoning. The upward force is the magnetic force on the current in the rod, and a larger resistance would make it weaker, not stronger.
DAs the rod speeds up, the emf, current and upward magnetic force all grow until that force equals the weight.Correct The emf Bℓv and the current Bℓv/R grow with the speed, and so does the upward magnetic force B²ℓ²v/R. When that force equals mg, the net force and the acceleration are zero, so the speed stays constant.
A flat circular conducting loop of radius a and resistance R lies in the plane of the page. A uniform magnetic field directed into the page fills the region to the right of a straight boundary that passes through the loop’s center, so exactly half of the loop is inside the field. The field’s magnitude is increasing at the constant rate β. At the instant the magnitude is B, what is the magnitude of the net magnetic force on the loop?
Answer and reasoning
Aπ²a³βB/(2R) A student who uses the arc’s length, πa, in F = IℓB picks this. The forces on different parts of the arc point in different directions and partly cancel; their sum equals IB times the straight distance between the arc’s ends, 2a.
B2πa³βB/R A student who uses the loop’s whole area, πa², for the flux picks this. Only the half of the loop inside the field has flux, so the emf and current are half as large.
C0 A student who thinks the forces on a closed loop always cancel picks this. They cancel for a loop wholly inside a uniform field; here only the arc is in the field, and the force on it is not balanced.
Dπa³βB/RCorrect The flux through the loop is B(πa²/2), so the current is (πa²β/2)/R. Only the semicircular arc is in the field. In a uniform field the force on the arc equals the force on a straight wire joining its ends, a diameter 2a: F = I(2a)B = πa³βB/R.
Working Flux: Φ = B(πa²/2), so |ε| = (πa²/2)β and I = πa²β/(2R). The part of the loop inside the field is a semicircular arc. In a uniform field, F⃗ = I∫dℓ⃗ × B⃗ = I(∫dℓ⃗) × B⃗, and ∫dℓ⃗ along the arc is the chord joining its ends, a diameter of length 2a. F = I(2a)B = πa³βB/R. Distractors (checked with sympy): arc length πa used as the length, π²a³βB/(2R); whole loop area in the flux, I = πa²β/R, 2πa³βB/R; forces on a closed loop taken to cancel, 0.
Two long, parallel conducting rails a distance ℓ apart lie in a plane inclined at 30° above the horizontal, and their lower ends are connected by a resistor of resistance R. A uniform magnetic field of magnitude B is directed vertically upward. A conducting rod of mass m is laid across the rails, perpendicular to them, and released. It slides down without friction, staying perpendicular to the rails and in contact with them. The rod and rails have negligible resistance, and g is the acceleration due to gravity. Which expression gives the rod’s terminal speed?
Answer and reasoning
A0.58 mgR/(B²ℓ²) A student who takes the magnetic force to act straight up the incline, against the velocity, balances mg sin 30° with the whole force B²ℓ²v cos 30°/R and picks this. The force IℓB is perpendicular to the vertical field, so it is horizontal, and only its component IℓB cos 30° acts along the rails.
B0.50 mgR/(B²ℓ²) A student who treats the field as if it were perpendicular to the rails takes the emf as Bℓv and the force IℓB as acting along the rails, so balances mg sin 30° = B²ℓ²v/R and picks this. The vertical field is at 30° to the normal of the rails’ plane, which brings a factor cos 30° into the emf and another into the force component.
C1.33 mgR/(B²ℓ²) A student who sets the magnetic retarding force along the rails, B²ℓ²v cos²30°/R, equal to the whole weight mg picks this. Only the component of the weight along the rails, mg sin 30°, drives the rod down the incline, so that is what the magnetic force balances at terminal speed.
D0.67 mgR/(B²ℓ²)Correct Only the field component perpendicular to the rails, B cos 30°, contributes to the flux, so ε = Bℓv cos 30° and I = Bℓv cos 30°/R. The magnetic force IℓB is horizontal, and its component up the incline is IℓB cos 30°. Setting mg sin 30° = B²ℓ²v cos²30°/R gives v = (0.50/0.75) mgR/(B²ℓ²) ≈ 0.67 mgR/(B²ℓ²).
Working Let θ = 30°. The normal to the plane of the rails makes angle θ with the vertical field, so ΦB = BA cos θ. As the rod slides at speed v the enclosed area changes at the rate ℓv, so |ε| = Bℓv cos θ and I = Bℓv cos θ/R. The force on the rod, FB = IℓB = B²ℓ²v cos θ/R, is perpendicular to the rod and to the vertical field, so it is horizontal; its component up the incline is FB cos θ = B²ℓ²v cos²θ/R (its component perpendicular to the incline only changes the normal force). At terminal speed the net force along the incline is zero: mg sin θ = B²ℓ²v cos²θ/R, so v = (sin θ/cos²θ) mgR/(B²ℓ²) = (0.5/0.75) mgR/(B²ℓ²) = 0.667 mgR/(B²ℓ²) ≈ 0.67 mgR/(B²ℓ²). Units: (kg·m/s²)(Ω)/(T²·m²) = m/s. Distractors (checked with sympy): whole force IℓB taken along the rails, mg sin θ = B²ℓ²v cos θ/R, coefficient tan 30° = 0.58; field treated as perpendicular to the rails, mg sin θ = B²ℓ²v/R, coefficient sin 30° = 0.50; retarding force along the rails set equal to the whole weight, mg = B²ℓ²v cos²θ/R, coefficient 1/cos²30° = 1.33.
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
