5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A flat wire loop of area A lies on a horizontal tabletop in a uniform magnetic field of magnitude B that points horizontally, due east. Which statement about the magnetic flux through the loop is correct?
Answer and reasoning
AIt equals BA, as the whole of the loop lies within the field region. A student who thinks the flux is BA whenever a loop is in a field picks this. ΦB = BA only when the field is perpendicular to the loop. Here the field lies along the loop's surface, so no part of it passes through the loop and the flux is zero.
BIt equals B, as flux is just another name for the field's strength. A student who treats flux and field as the same quantity picks this. The field is measured in teslas at a point; the flux is measured in T·m² for a whole surface and depends on the area and orientation of the surface as well as on B.
CIt has no value until the magnetic field begins to change. A student who thinks flux exists only while the field is changing picks this. A steady field gives a steady flux, which can have any value; here it is zero because of the loop's orientation. What needs a change in flux is an induced emf.
DIt is zero, since B⃗ has no component perpendicular to the loop.Correct Magnetic flux describes the component of the field perpendicular to the surface, multiplied by the area. A horizontal field lies along a horizontal loop, so its perpendicular component is zero and so is the flux, however strong the field.
A closed cardboard box sits on a table in a uniform magnetic field that points vertically downward. What is the direction of the area vector of the box's top face?
Answer and reasoning
AVertically downward, along the magnetic field A student who thinks the area vector points along the field picks this. The area vector is set by the surface alone: perpendicular to it and outward for a closed surface. Here it points up, opposite to the field, so the flux through the top face is negative.
BVertically upward, away from the inside of the boxCorrect The area vector is perpendicular to the plane of the surface, and for a closed surface it points outward. The top face is horizontal, so its area vector is vertical, and outward from the box means upward. The field's direction plays no part in defining it.
CHorizontal, lying in the plane of the top face A student who describes a surface's orientation by a direction lying along it picks this. The area vector is perpendicular to the plane of the surface, so for a horizontal face it is vertical.
DIt has no direction, because area is a scalar quantity A student who has only met area as a plain number picks this. For flux, area is treated as a vector: its magnitude is the area and its direction is perpendicular to the surface, outward for a closed surface.
A square wire loop with sides of 0.15 m lies in a horizontal plane in a uniform vertical magnetic field. The field's magnitude increases steadily from 0.20 T to 0.60 T in 0.12 s. What is the magnitude of the emf induced in the loop?
Answer and reasoning
A0.50 V A student who uses the side length in place of the area picks this: (0.15 m)(0.40 T)/0.12 s = 0.50 V. The area of the loop is (0.15 m)² = 0.0225 m².
B3.3 V A student who treats the flux as the field itself divides the change in field by the time: 0.40 T ÷ 0.12 s = 3.3 T/s, reported as volts. The flux is the field times the area, so the area must be included.
C0.075 VCorrect The flux changes by ΔΦB = AΔB = (0.15 m)²(0.60 T − 0.20 T) = (0.0225 m²)(0.40 T) = 0.0090 T·m². By Faraday's law |ε| = |ΔΦB/Δt| = 0.0090 T·m² ÷ 0.12 s = 0.075 V.
D0.0090 V A student who takes the emf to be the change in flux picks this: ΔΦB = 0.0090 T·m². The emf is the rate of change of flux, so this must be divided by the 0.12 s over which the change happens.
Working A = (0.15 m)² = 0.0225 m². ΔΦB = AΔB = (0.0225 m²)(0.60 T − 0.20 T) = 0.0090 T·m². |ε| = |ΔΦB/Δt| = 0.0090 T·m² ÷ 0.12 s = 0.075 V.
The diagram shows a stationary square loop in the plane of the page, inside a region of uniform magnetic field directed into the page. The arrowheads show the direction of the current induced in the loop. Which change is taking place?
Answer and reasoning
AThe field is growing weaker A student who reverses the right-hand rule reads the counterclockwise current as making a field into the page, which would oppose a decrease. Curling the right hand's fingers counterclockwise points the thumb out of the page, so the induced field opposes an increase.
BThe field is steady in strength A student who thinks a large, steady flux drives a current picks this. A steady field gives a constant flux, and a constant flux induces no emf, so there would be no current.
CThe field region slides sideways A student who thinks relative motion between a field and a loop always induces a current picks this. While the uniform field still covers the whole loop, sliding it sideways leaves the flux unchanged, so there is no current. If the region's edge slid across the loop, the into-page flux would decrease, and the induced current would be clockwise, not counterclockwise as shown.
DThe field is growing strongerCorrect The current is counterclockwise as seen by the reader, so by the right-hand rule its field inside the loop points out of the page. An induced field opposite to the external field (into the page) opposes an increase in the flux, so the field must be growing stronger.
A metal rod slides at a constant 2.0 m/s along two parallel horizontal conducting rails that are 0.25 m apart. The rails are joined at one end by a 0.50 Ω resistor; the rod and rails have negligible resistance. A uniform 0.40 T magnetic field is perpendicular to the plane of the rails. At the instant the rod is 0.50 m from the resistor, what is the current in the resistor?
Answer and reasoning
A0.80 A A student who uses the 0.50 m distance from the resistor as ℓ gets ε = (0.40 T)(0.50 m)(2.0 m/s) = 0.40 V and I = 0.80 A. The flux changes because the rod sweeps out area at the rate ℓv, where ℓ = 0.25 m is the rod's length between the rails.
B0.40 ACorrect The motional emf is ε = Bℓv = (0.40 T)(0.25 m)(2.0 m/s) = 0.20 V, where ℓ is the length of rod between the rails. The current is I = ε/R = 0.20 V ÷ 0.50 Ω = 0.40 A. The rod's distance from the resistor does not enter.
C0.20 A A student who reports the emf as the current picks this: ε = Bℓv = 0.20 V. The current is the emf divided by the resistance, 0.20 V ÷ 0.50 Ω = 0.40 A.
D0.10 A A student who takes the emf to be the flux through the circuit picks this: ΦB = (0.40 T)(0.25 m × 0.50 m) = 0.050 T·m², divided by 0.50 Ω. The emf is the rate of change of the flux, Bℓv, whatever the flux is at that instant.
Working ε = Bℓv = (0.40 T)(0.25 m)(2.0 m/s) = 0.20 V. I = ε/R = 0.20 V ÷ 0.50 Ω = 0.40 A.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
12.4.A.1 Magnetic flux, ΦBFix
Magnetic flux, ΦB
A scalar that describes how much of a magnetic field passes perpendicularly through a surface: the product of the field component perpendicular to the surface and the surface's area. SI unit: tesla square meter (T·m²).
Perpendicular component of the field
The part of the magnetic field that points along the direction perpendicular to a surface, B cos θ. Only this component contributes to the flux; a field lying along the surface contributes nothing. SI unit: tesla (T).
Students often think Magnetic flux is the same thing as the magnetic field strength, so the flux through a loop is set by B alone, whatever the loop's area. In fact No. The field B is a vector at each point, measured in teslas. The flux ΦB is a scalar for a whole surface, measured in T·m²: it depends on the field, on the surface's area and on how the surface is oriented to the field.
Students often think There is magnetic flux through a loop only while the magnetic field is changing; a steady field gives no flux. In fact No. A steady field that has a component perpendicular to a loop gives a steady, nonzero flux through it. What requires a change is the induced emf, which depends on the rate of change of the flux.
12.4.A.2 ΦB = BA cos θ Fix
ΦB = BA cos θ
The magnetic flux through a flat surface of area A in a uniform field of magnitude B, where θ is the angle between the field and the surface's area vector (the perpendicular to the surface). The flux is greatest, BA, when the field is perpendicular to the surface (θ = 0°) and zero when the field lies along the surface (θ = 90°).
Area vector, A⃗
A vector whose magnitude is the area of a flat surface and whose direction is perpendicular to the plane of the surface. For each face of a closed surface it points outward, away from the enclosed volume. SI unit of its magnitude: square meter (m²).
Sign of the magnetic flux
Positive when the field has a component parallel to the area vector (0° ≤ θ < 90°), negative when the field has a component antiparallel to it (90° < θ ≤ 180°), and zero when the field lies along the surface.
Students often think The magnetic flux through a loop is BA as long as the loop is in the field, whatever its orientation. In fact No. Only the component of the field perpendicular to the loop counts: ΦB = BA cos θ, where θ is the angle between the field and the loop's area vector. The flux is BA only when the field is perpendicular to the loop, and zero when the field lies along the loop.
Students often think The angle θ in ΦB = BA cos θ is measured between the field and the surface of the loop. In fact No. θ is the angle between the field and the area vector, which is perpendicular to the surface. If the field makes an angle α with the plane of the loop, then θ = 90° − α and ΦB = BA sin α.
12.4.A.3 Induced emf, ε Fix
Induced emf, ε
The potential difference produced in a conducting loop or rod by a changing magnetic flux; in a closed circuit it drives an induced current. SI unit: volt (V).
Faraday's law
The magnitude of the emf induced in a loop equals the magnitude of the rate of change of the magnetic flux through it: |ε| = |ΔΦB/Δt|. The emf depends on how fast the flux changes, not on how large the flux is.
Ways of changing the flux
Since ΦB = BA cos θ, the flux through a loop changes when the field's magnitude B changes, when the area A inside the field changes (as when a rod slides along rails), or when the angle θ changes (as when a loop rotates).
Students often think The induced emf equals the total change in flux, whatever the time interval, so how quickly the change happens does not matter. In fact No. The emf equals the rate of change of flux, |ΔΦB/Δt|. The same change in flux produces a larger emf when it happens faster and a smaller emf when it happens more slowly.
Students often think The induced emf is set by the flux through the loop itself, not by its rate of change, so a large or steady flux induces a large emf. In fact No. The emf depends on how fast the flux is changing, not on its value. A large but steady flux induces no emf at all, and a small flux that is changing quickly induces a large one.
12.4.A.4 Lenz's law Fix
Lenz's law
The direction of an induced emf is such that the current it drives produces a magnetic field that opposes the change in flux that caused it. This is the meaning of the minus sign in ε = −ΔΦB/Δt.
Induced magnetic field
The magnetic field produced by an induced current. Inside the loop it points opposite to the external field when the flux is increasing, and in the same direction as the external field when the flux is decreasing.
Right-hand rule for a current loop
When the fingers of the right hand curl in the direction of the current around a loop, the thumb points in the direction of the magnetic field that the current produces inside the loop. A counterclockwise current in the plane of the page produces a field out of the page (⊙) inside the loop.
Students often think The magnetic field of an induced current always points opposite to the external magnetic field. In fact No. It opposes the change in flux, not the field. When the flux is increasing it points opposite to the external field; when the flux is decreasing it points in the same direction as the external field, tending to keep the flux from falling.
Students often think An emf is induced only when a conductor moves relative to a magnetic field, so a loop at rest cannot have an induced current, and moving a field past a loop always induces one. In fact No. An emf is induced whenever the magnetic flux through a loop changes, whatever the cause. A loop at rest in a field whose strength changes has an induced emf; a loop moved around inside a uniform field, with no change in flux, has none.
12.4.A.5 Motional emf, ε = Bℓv Fix
Motional emf, ε = Bℓv
The emf across a conducting rod of length ℓ moving with speed v perpendicular both to its length and to a uniform magnetic field B. On conducting rails, the area of the circuit grows at the rate ℓv, so the flux changes at the rate Bℓv. SI unit: volt (V).
Students often think The length ℓ in ε = Bℓv is the length of the circuit along the rails, from the rod to the resistor. In fact No. ℓ is the length of the rod between the rails (the rails' separation). The area of the circuit grows at the rate ℓv however far the rod is from the end, so the emf does not depend on that distance.
Students often think The induced emf and the induced current are the same quantity, so the emf value can be reported as the current. In fact No. The emf is a potential difference; the current it drives depends also on the resistance of the circuit: I = ε/R. An emf of 1.5 V across 3.0 Ω drives 0.50 A.
9 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 9
A square wire loop with sides of 0.20 m is in a uniform magnetic field of magnitude 0.40 T. The plane of the loop makes an angle of 30° with the direction of the field. What is the magnitude of the magnetic flux through the loop?
Answer and reasoning
A1.4 × 10⁻² T·m² A student who puts the angle between the field and the plane into cos θ picks this: (0.40 T)(0.040 m²)(cos 30°) = 1.4 × 10⁻² T·m². θ is measured from the perpendicular to the plane, so here θ = 60°.
B1.6 × 10⁻² T·m² A student who takes the flux as BA whatever the orientation picks this: (0.40 T)(0.040 m²) = 1.6 × 10⁻² T·m². The loop is tilted, so only the field component perpendicular to it counts, which halves the flux here.
C8.0 × 10⁻³ T·m²Correct The area is A = (0.20 m)² = 0.040 m². The angle in ΦB = BA cos θ is measured from the area vector, which is perpendicular to the plane, so θ = 90° − 30° = 60°. ΦB = (0.40 T)(0.040 m²)(cos 60°) = 8.0 × 10⁻³ T·m².
D4.0 × 10⁻² T·m² A student who uses the side length in place of the area picks this: (0.40 T)(0.20 m)(cos 60°) = 4.0 × 10⁻² — which does not even have the units of flux. The area of the square is (0.20 m)² = 0.040 m².
Working A = s² = (0.20 m)² = 0.040 m². The field makes 30° with the plane, so it makes θ = 90° − 30° = 60° with the area vector. ΦB = BA cos θ = (0.40 T)(0.040 m²)(0.50) = 8.0 × 10⁻³ T·m².
The diagram shows three flat square loops, X, Y and Z, seen edge-on in the same uniform magnetic field B⃗. The area of each loop is labeled. Which ranks the magnitudes of the magnetic flux through the loops?
Answer and reasoning
AX = Y > ZCorrect Only the field component perpendicular to each loop counts. X is perpendicular to the field: Φ = BA. Y's plane makes 30° with the field, so the field makes 60° with Y's area vector: Φ = B(2A)cos 60° = BA. Z's plane is parallel to the field: Φ = 0.
BY = Z > X A student who takes the flux as BA whatever the orientation ranks by area alone: Y and Z (2A) above X (A). Tilting Y halves its flux, and Z, lying along the field, has no flux at all.
CZ > Y > X A student who measures θ from the plane of each loop gets this: X (90°, cos = 0), Y (30°, 2BA cos 30°) and Z (0°, 2BA). The angle is measured from the perpendicular to the plane, which reverses the order: Z, lying along the field, has zero flux.
DX = Y = Z A student who treats flux as the field itself says the flux is the same because the field is the same. The flux also depends on each loop's area and on how it is oriented to the field.
Working X: plane perpendicular to B⃗, θ = 0°, Φ = BA. Y: plane at 30° to B⃗, so θ = 60° between B⃗ and the area vector, Φ = B(2A)cos 60° = BA. Z: plane parallel to B⃗, θ = 90°, Φ = 0. Ranking: X = Y > Z.
The diagram shows a side view of a closed box-shaped surface in a uniform magnetic field B⃗. Which statement about the magnetic flux through face L is correct?
Answer and reasoning
AIt is negative, as B⃗ points opposite to L's outward area vector.Correct The area vector of face L points outward from the box, to the left. The field points to the right, antiparallel to it (θ = 180°), so ΦB = BA cos 180° = −BA: the flux is negative. Through face R the field is parallel to the outward area vector, so that flux is positive.
BIt is positive, as L's area vector points in the same direction as B⃗. A student who takes the area vector to point along the field picks this. The area vector of a face of a closed surface points outward, and for face L that is to the left, opposite to B⃗.
CIt is zero, as L's area vector lies in the face, perpendicular to B⃗. A student who thinks the direction describing a surface lies along it picks this. The area vector is perpendicular to the face; for L it is horizontal and points to the left, so the field passes straight through L and the flux is not zero.
DIt has no sign, since flux just counts the field lines that cross L. A student who pictures flux only as a count of field lines picks this. The flux has a sign: positive when the field is parallel to the area vector and negative when antiparallel. The field enters the box through L, antiparallel to L's outward area vector.
The graph shows the magnetic flux Φ through a stationary wire loop as a function of time t. Which ranks the magnitude of the emf induced in the loop during intervals I, II and III?
Answer and reasoning
AII > I = III A student who links the emf to the size of the flux ranks by the height of the graph: the flux is largest throughout II, and averages 0.02 T·m² in both I and III. During II the flux is constant, so no emf is induced at all.
BII > I > III A student who reads the emf from the area under the graph gets this: II (0.12), I (0.04), III (0.02), in T·m²·s. The emf is the slope of the flux–time graph, not the area under it.
CI > II > III A student who ranks the slopes with their signs gets this: +0.02, 0 and −0.04 T·m²/s. The question asks for magnitudes: the steep fall in III gives the largest emf, and the flat interval II gives none.
DIII > I > IICorrect The magnitude of the emf is the magnitude of the slope of the flux–time graph, |ΔΦ/Δt|. Interval I: 0.04 T·m² in 2 s, 0.02 V. Interval II: no change, 0 V. Interval III: 0.04 T·m² in 1 s, 0.04 V. So III > I > II.
Working |ε| = |slope| of Φ–t. I: (0.04 − 0) T·m²/2 s = 0.02 V. II: 0. III: (0.04 T·m²)/1 s = 0.04 V. Ranking III > I > II.
A circular wire loop of radius r lies perpendicular to a uniform magnetic field. The field's magnitude changes steadily by ΔB over a time interval Δt, inducing an emf of magnitude ε₁ in the loop. A second loop, of radius 2r, lies perpendicular to an identical field whose magnitude changes steadily by the same ΔB over a time interval 2Δt, inducing an emf of magnitude ε₂. What is ε₂/ε₁?
Answer and reasoning
Aε₂/ε₁ = 1 A student who scales the area with the radius rather than its square gets 2/2 = 1. The area of a circle is πr², so doubling r multiplies it by 4.
Bε₂/ε₁ = 2Correct |ε| = ΔΦB/Δt = πr²ΔB/Δt. Doubling the radius multiplies the area, and so the change in flux, by 4; doubling the time interval halves the rate. ε₂/ε₁ = 4/2 = 2.
Cε₂/ε₁ = 4 A student who thinks the emf depends only on how much the flux changes gets 4, from the fourfold area. The change takes twice as long, so the rate of change of flux, and the emf, is only twice as large.
Dε₂/ε₁ = ½ A student who treats flux as the field itself ignores the loop's size: the same ΔB over twice the time gives half the emf. The flux through the larger loop changes four times as much, which more than makes up for the slower change.
A stationary wire loop lies in the plane of the page in a uniform magnetic field directed out of the page (⊙). The graph shows the magnitude B of the field as a function of time t. During which interval or intervals is the current induced in the loop clockwise, as seen by the reader?
Answer and reasoning
ADuring interval I onlyCorrect In I the outward flux increases, so the induced current makes a field into the page inside the loop, opposing the increase: by the right-hand rule that current is clockwise. In II the flux is constant and there is no current. In III the outward flux decreases, so the induced field points out of the page, keeping the flux from falling, and the current is counterclockwise.
BDuring intervals I and III A student who thinks the induced field always points opposite to the external field puts it into the page in both I and III. The induced field opposes the change in flux: in III the outward flux is falling, so the induced field points outward and the current is counterclockwise.
CDuring interval III only A student who reverses the right-hand rule for a loop picks this. Lenz's law gives an induced field into the page in I and out of the page in III; a clockwise current makes a field into the page inside the loop, so the clockwise current is in I, not III.
DDuring none of the intervals A student who thinks an emf needs a conductor moving through a field picks this, since the loop is at rest. The flux through the loop changes in I and III because the field's strength changes, and a changing flux induces an emf whether or not anything moves.
A metal rod rests across two horizontal conducting rails that are joined at one end by a resistor, in a uniform vertical magnetic field. A student gives the rod a brief push along the rails and lets go. Friction and air resistance are negligible. Which claim about the rod's motion after the push, with its justification, is correct?
Answer and reasoning
AIt speeds up, because the force on its induced current points along its motion. A student who expects the induced current to help the motion that caused it picks this. A force along the motion would increase the rod's kinetic energy while the resistor also warms, creating energy. The force on the induced current opposes the motion.
BIt keeps a constant speed, because a steady, uniform field induces no current. A student who looks only at whether B changes picks this. The rod's motion changes the area of the circuit, so the flux ΦB = BA changes even though B is steady, and a current is induced.
CIt slows down, because its induced current feels a magnetic force opposing its motion.Correct As the rod moves, the area of the circuit and so the flux through it change, inducing a current in the rod. By Lenz's law the magnetic force on that current opposes the change, so it points opposite to the rod's velocity and the rod slows; its kinetic energy becomes thermal energy in the resistor.
DIt stops at once, because nothing pushes it forward once the push ends. A student who thinks motion needs a continuing push picks this. With no force opposing it, the rod would keep moving at constant velocity. It does slow, but gradually, because of the magnetic force on the induced current.
A flat wire loop of area A and resistance R is in a uniform magnetic field of magnitude B, with the plane of the loop perpendicular to the field. During a time interval Δt, the loop is turned through 60° about an axis in its plane, so that its plane ends up at 30° to the field. Which expression gives the magnitude of the total charge that flows around the loop during this interval?
Answer and reasoning
ABAΔt/(2R) A student who takes the emf to be the change in flux itself, BA/2, whatever the time taken, picks this: the current would be BA/(2R), and multiplying by Δt gives BAΔt/(2R). The emf is the rate of change of flux, BA/(2Δt), so the Δt cancels when the charge is found.
BBA/2 A student who reports the induced emf, BA/(2Δt), as the current picks this, multiplying it by Δt. The current is the emf divided by the loop's resistance, so the charge is BA/(2R).
C0 A student who thinks the flux through the loop is BA whatever its orientation picks this: with no change in flux there would be no emf and no charge would flow. Only the field component perpendicular to the loop counts, ΦB = BA cos θ, so turning the loop from θ = 0° to θ = 60° halves the flux.
DBA/(2R)Correct The flux is BA cos θ, with θ measured between the field and the loop's area vector: it falls from BA (θ = 0°) to BA cos 60° = BA/2, a change of BA/2. The average emf is BA/(2Δt), the average current is that emf divided by R, BA/(2RΔt), and the charge is the current multiplied by the time: BA/(2R). Turning the loop faster gives a larger current for a shorter time, and the same charge.
Working ΦB = BA cos θ, with θ the angle between the field and the area vector (perpendicular to the loop). Initially the plane is perpendicular to the field, so θ = 0° and Φi = BA. Turning the loop through 60° makes θ = 60° (the plane is then at 30° to the field), so Φf = BA cos 60° = BA/2 and |ΔΦB| = BA/2. Average emf: |ε| = |ΔΦB/Δt| = BA/(2Δt). Average current: I = |ε|/R = BA/(2RΔt). Charge: q = IΔt = BA/(2R), which does not depend on how quickly the loop is turned. (Errors: flux BA whatever the orientation gives no change and no charge; taking the emf as the change in flux itself gives I = BA/(2R) and q = BAΔt/(2R); reporting the emf as the current gives I = BA/(2Δt) and q = BA/2.)
A metal rod slides without friction at constant speed v along two parallel horizontal conducting rails that are a distance ℓ apart. The rails are joined at one end by a resistor of resistance R; the rod and rails have negligible resistance. A uniform magnetic field of magnitude B is perpendicular to the plane of the rails. What magnitude of applied force, directed along the rails, is required to keep the rod moving at constant speed v?
Answer and reasoning
AB²ℓ²v A student who treats the induced emf Bℓv as the current picks this: (Bℓv)ℓB. The emf drives a current I = ε/R through the resistor, so the force, and the external force that balances it, is B²ℓ²v/R.
BB²ℓ²vR A student who multiplies the emf by the resistance to find the current, I = BℓvR, picks this. ΔV = IR gives I = ε/R: a larger resistance means a smaller current and so a smaller force on the rod.
CB²ℓ²v/RCorrect The moving rod has an induced emf ε = Bℓv, which drives a current I = ε/R = Bℓv/R. The field then exerts a force IℓB = B²ℓ²v/R on the current in the rod, opposing its motion. At constant speed the net force is zero, so the external force must have this same magnitude.
D0 A student who thinks a steady, uniform field cannot induce an emf picks this, since with no current there would be no magnetic force to balance. The circuit's area, and so the flux through it, grows as the rod moves, inducing a current on which the field exerts a force opposing the motion.
Working The rod sweeps out area at the rate ℓv, so the induced emf is ε = Bℓv. The current in the circuit is I = ε/R = Bℓv/R. The field exerts a force on this current in the rod, whose length between the rails ℓ is perpendicular to B: FB = IℓB = B²ℓ²v/R, directed opposite to the rod's velocity (Lenz's law). The speed is constant, so the net force is zero and the external force has magnitude F = B²ℓ²v/R. (Errors: taking the current as ε = Bℓv gives B²ℓ²v; I = εR gives B²ℓ²vR; thinking a steady field induces no emf gives 0.)
Compiled from the AP Physics 2 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
