13.2 Electromagnetic Induction

1. ANonzero while at rest; opposite in direction while it is pulled out
   A student who thinks an emf is induced whenever a field passes through the coil picks this. With the magnet at rest the flux is steady, so dΦ_B/dt = 0 and there is no emf, however strong the field.
2. BZero while at rest; the same direction as before while it is pulled out
   A student who thinks the induced field always opposes the magnet's field predicts the same direction both times. The induced field opposes the change: it opposes the magnet's field while the flux increases, but points along it while the flux decreases, so the emf reverses.
3. CLargest while at rest; the same direction as before while it is pulled out
   A student who takes the emf to follow the flux itself picks this: the flux is largest with the magnet inside and keeps the same direction throughout. The emf depends on the rate of change of flux, which is zero while the magnet is at rest and changes sign when the flux starts to decrease.
4. DZero while at rest; opposite in direction while it is pulled out Correct
   An emf is induced only while the flux through the coil changes. At rest, the flux is large but steady, so the emf is zero. Pushing the magnet in increases the flux; pulling it out decreases it, so by Lenz's law the induced emf, and the current, reverse.

CED 13.2.A.1 · Read this in Fix

1. AThe field's magnitude increases while the loop is held at rest. Correct
   Curling the right hand's fingers clockwise points the thumb into the page, so the induced current makes a field into the page inside the loop. That opposes an increase in the out-of-page flux, so the out-of-page field must be increasing.
2. BThe field's magnitude falls while the loop is held at rest.
   A student who reverses the right-hand rule thinks a clockwise current makes a field out of the page, which would oppose a decrease. A clockwise current makes a field into the page inside the loop, which opposes an increase in the out-of-page flux.
3. CThe field stays constant while the loop is held at rest.
   A student who thinks a field through a loop is enough to induce a current picks this. With a constant field and a loop at rest, the flux does not change and no current is induced.
4. DThe field stays constant while the loop slides within it.
   A student who thinks motion through a field always induces an emf picks this. Sliding within a uniform field leaves the flux through the loop unchanged, so there is no induced current.

Working Clockwise (viewed) current → induced B into page inside the loop → opposes increasing out-of-page flux.

CED 13.2.A.2.i · Read this in Fix

1. AAn electric force directed along a radius, toward the solenoid's axis
   A student who pictures the induced field like the field of a charged wire picks this. The induced field has no charges to start or end on; around a solenoid its field lines are circles, so the force is tangential.
2. BAn electric force tangent to a circle around the solenoid's axis Correct
   The increasing flux inside the solenoid induces an electric field outside it, whose field lines are circles centered on the axis. The electron at rest feels the electric force −eE⃗ of this field, tangent to such a circle, even though B is negligible where it is.
3. CA magnetic force, as the flux through its circle around the axis changes
   A student who calls the force magnetic because the cause is a changing magnetic flux picks this. The magnetic force q(v⃗ × B⃗) on a charge at rest is zero; the force is exerted by the induced electric field.
4. DNo force, as the magnetic field is negligible where the electron is
   A student who thinks induced effects occur only where B is nonzero picks this. The electron lies on a circle enclosing changing flux, so the induced electric field at its position is not zero.

CED 13.2.A.3 · Read this in Fix

1. AIn a vacuum those of higher frequency travel faster than the others.
   A student who thinks more energetic waves move faster picks this. The speed in free space is fixed by ε₀ and μ₀ alone; a higher frequency means a shorter wavelength, not a higher speed.
2. BThey cannot cross a vacuum, as any wave needs a medium to carry it.
   A student who generalizes from sound picks this. An electromagnetic wave consists of oscillating electric and magnetic fields, which exist in empty space; light crosses the vacuum between the Sun and Earth.
3. CIn a vacuum their speed depends on the speed of the source emitting them.
   A student who adds the source's speed to the wave's, as for objects thrown from a moving vehicle, picks this. The speed in free space is the constant c, whatever the motion of the source.
4. DIn a vacuum all of them travel at the same speed, whatever their frequency. Correct
   Maxwell's equations show that the electric and magnetic fields obey wave equations with one wave speed in free space, c = 1/√(ε₀ μ₀) = 3.00 × 10⁸ m/s, which depends on no property of the wave. Different frequencies have different wavelengths, not different speeds.

CED 13.2.A.4 · Read this in Fix

1. Aε(t) = BA cos(ωt)
   A student who takes the emf to be the flux itself picks this; its unit is T·m², not V. The emf is the negative of the rate of change of this flux, which brings in the factor ω and turns cos(ωt) into sin(ωt).
2. Bε(t) = −BAω cos(ωt)
   A student who measures the angle from the plane of the loop uses 90° − ωt in cos θ and gets Φ_B = BA sin(ωt), so ε = −BAω cos(ωt). The angle in Φ_B = BA cos θ is measured from the area vector, so Φ_B = BA cos(ωt).
3. Cε(t) = BAω sin(ωt) Correct
   The angle between B⃗ and the area vector is θ = ωt, so Φ_B = BA cos(ωt). Then ε = −dΦ_B/dt = −BA(−ω sin(ωt)) = BAω sin(ωt). The emf is zero when the flux is largest (t = 0) and largest when the plane of the loop is parallel to the field.
4. Dε(t) = BA(1−cos ωt)/t
   A student who divides the flux change since t = 0 by the elapsed time picks this: −[BA cos(ωt) − BA]/t. That is the average emf since t = 0; because the flux does not change at a constant rate, the emf at time t is −dΦ_B/dt = BAω sin(ωt).

Working θ(t) = ωt between B⃗ and the area vector, so Φ_B = BA cos(ωt). ε = −dΦ_B/dt = BAω sin(ωt). Checks: ε(0) = 0 when the flux is largest; units T·m²·s⁻¹ = V. Distractors: Φ_B itself; angle taken from the plane, Φ_B = BA sin(ωt), ε = −BAω cos(ωt); average since t = 0, −ΔΦ_B/t = BA(1 − cos ωt)/t.

CED 13.2.A.1 · Read this in Fix

1. Aε₃ > ε₂ > ε₁
   A student who multiplies the solenoid's field by each loop's own area picks this, ranking by loop size. Outside the solenoid there is no field, so the extra area of loop 3 adds no flux: loops 2 and 3 have the same flux and the same emf.
2. Bε₂ = ε₃ > ε₁ Correct
   The emf in each loop is the rate of change of the flux through it. Loop 1, inside the solenoid, encloses only part of the solenoid's cross section. Loops 2 and 3 both surround the whole cross section; outside it the field is negligible, so they enclose the same flux, larger than loop 1's, and have equal emfs.
3. Cε₁ > ε₂ = ε₃
   A student who thinks an emf is induced only in a loop that lies where the field is picks this, giving loops 2 and 3 no emf. Loops 2 and 3 surround the changing flux inside the solenoid, so they have an emf even though the field at their wires is zero.
4. Dε₁ = ε₂ = ε₃
   A student who thinks the emf depends only on how fast the solenoid's field changes gives all three loops the same emf. The emf depends on the rate of change of flux, and loop 1 encloses less of the solenoid's cross section than loops 2 and 3.

Working Φ₁ = B(πr₁²) with r₁ < R; Φ₂ = Φ₃ = B(πR²) since B = 0 outside. ε ∝ dΦ/dt, so ε₂ = ε₃ > ε₁.

CED 13.2.A.1 · Read this in Fix

1. A1.2 × 10⁻² V Correct
   The loop's area is (0.20 m)² = 0.040 m². At t = 5.5 s the field is falling from 0.30 T at 5 s to 0 at 6 s, a slope of −0.30 T/s. |ε| = A|dB/dt| = (0.040 m²)(0.30 T/s) = 1.2 × 10⁻² V.
2. B6.0 × 10⁻³ V
   A student who takes the emf to be the flux at that instant picks this: B = 0.15 T at 5.5 s, and (0.040 m²)(0.15 T) = 6.0 × 10⁻³ T·m². The emf depends on the slope of the B–t graph at 5.5 s, not on the value of B.
3. C1.1 × 10⁻³ V
   A student who divides the field at 5.5 s by the elapsed time picks this: (0.040 m²)(0.15 T ÷ 5.5 s) = 1.1 × 10⁻³ V. That is the average rate of change since t = 0, which includes a rise and a steady interval; the emf at 5.5 s uses the slope there, 0.30 T/s.
4. D6.0 × 10⁻² V
   A student who uses the side length in place of the area picks this: (0.20 m)(0.30 T/s) = 6.0 × 10⁻² — which does not have the units of emf. The area of the loop is (0.20 m)² = 0.040 m².

Working A = (0.20 m)² = 0.040 m². From the graph, between t = 5 s and 6 s, B falls from 0.30 T to 0: dB/dt = −0.30 T/s. |ε| = A|dB/dt| = 0.040 × 0.30 = 0.012 V = 1.2 × 10⁻² V.

CED 13.2.A.1.i · Read this in Fix

1. A6.0 × 10⁻³ V
   A student who takes the emf to scale with the radius rather than the area gets 2 × ½ = 1, the same emf. The area of a circle is πr², so doubling the radius quadruples the area.
2. B3.0 × 10⁻³ V
   A student who thinks the emf depends only on how fast the field changes halves the emf. The emf is the rate of change of flux, so the four times larger area of loop 2 must be included.
3. C1.2 × 10⁻² V Correct
   With a fixed area, |ε| = A|dB/dt|. Doubling the radius makes the area π(2r)² four times as large; halving the rate of change of the field halves the emf. ε₂ = 4 × ½ × 6.0 × 10⁻³ V = 1.2 × 10⁻² V. The fields' magnitudes at that instant do not matter.
4. D2.4 × 10⁻² V
   A student who takes the emf to follow the flux uses the equal fields and the four times larger area: 4 × 6.0 × 10⁻³ V. The emf depends on the rate of change of the field, which is halved, not on its value at that instant.

Working |ε| = A|dB/dt|. ε₂/ε₁ = (A₂/A₁)(rate₂/rate₁) = 4 × 0.5 = 2. ε₂ = 2 × 6.0 × 10⁻³ V = 1.2 × 10⁻² V.

CED 13.2.A.1.i · Read this in Fix

1. A3.0 × 10⁻¹ V
   A student who uses the side along the direction of motion picks this: (0.50 T)(0.20 m)(3.0 m/s) = 0.30 V. The area inside the field grows by a strip whose length is the leading edge, 0.10 m, perpendicular to the velocity.
2. B1.5 × 10⁻¹ V Correct
   The field is constant, so the emf comes from the changing area inside the field. The leading edge, 0.10 m long, sweeps new area at (0.10 m)(3.0 m/s) = 0.30 m²/s. |ε| = B|dA/dt| = (0.50 T)(0.30 m²/s) = 1.5 × 10⁻¹ V.
3. C5.0 × 10⁻³ V
   A student who takes the emf to be the flux at that instant picks this: (0.50 T)(0.10 m × 0.10 m) = 5.0 × 10⁻³ T·m². The emf is the rate at which that flux is changing, B times dA/dt.
4. D1.0 × 10⁻² V
   A student who takes the emf to be the whole change in flux as the loop enters picks this: (0.50 T)(0.020 m²) = 1.0 × 10⁻² T·m². The emf is the rate of change of flux, which needs the speed.

Working A_in increases at dA/dt = (leading edge)(v) = (0.10 m)(3.0 m/s) = 0.30 m²/s. |ε| = B dA/dt = (0.50 T)(0.30 m²/s) = 0.15 V.

CED 13.2.A.1.ii · Read this in Fix

1. Aε₂ > ε₃ > ε₁
   A student who ranks by the flux through each loop picks this: loop 2 is entirely in the field, loop 3 two-thirds and loop 1 one-third. The emf depends on the rate of change of flux: loop 2's flux is not changing at all.
2. Bε₁ = ε₂ = ε₃
   A student who thinks every loop moving through a field has an emf Bℓv picks this. Loop 2 moves through the field, but the flux through it is constant, so its emf is zero.
3. Cε₂ > ε₁ = ε₃
   A student who adds the emfs of both sides of loop 2 gives it twice the emf of the others. The two sides' emfs act in opposite senses around the loop and cancel, as Faraday's law for the whole loop confirms.
4. Dε₁ = ε₃ > ε₂ Correct
   The field is uniform, so the emf depends on how fast the area inside the field changes. Loop 1 is entering and loop 3 is leaving: in each, one side is sweeping area into or out of the field at the same rate, so |ε| = Bsv for both, where s is the side length. Loop 2 is entirely inside, its flux is constant, and its emf is zero.

Working Loops 1 and 3 (side s): one side sweeps area into or out of the field at rate sv, |ε| = Bsv. Loop 2: constant flux, ε = 0.

CED 13.2.A.1.ii · Read this in Fix

1. A6.0 × 10⁻¹ V Correct
   For one turn, dΦ_B/dt = 2βt = 2(2.5 × 10⁻³ Wb/s²)(3.0 s) = 1.5 × 10⁻² V. The 40 turns are in series, so |ε| = N|dΦ_B/dt| = 40 × 1.5 × 10⁻² V = 0.60 V.
2. B1.5 × 10⁻² V
   A student who gives the coil the emf of a single turn picks this. Each of the 40 turns has this emf, and they add because the turns are in series.
3. C9.0 × 10⁻¹ V
   A student who uses the flux instead of its rate of change picks this: 40 × (2.5 × 10⁻³)(3.0)² = 0.90, in Wb, not V. The emf needs dΦ_B/dt = 2βt.
4. D3.0 × 10⁻¹ V
   A student who divides the flux by the time picks this: 40 × βt²/t = 40βt = 0.30 V. That is the average rate of change since t = 0; because the flux grows as t², the rate at t = 3.0 s is twice that average.

Working dΦ_B/dt = 2βt = 2(2.5 × 10⁻³)(3.0) = 0.015 Wb/s per turn. |ε| = N dΦ_B/dt = 40 × 0.015 = 0.60 V.

CED 13.2.A.1.iii · Read this in Fix

1. A16Nμ₀nπa²α
   A student who multiplies the solenoid's field by the coil's own area, π(4a)², picks this. Between radius 2a and 4a there is no field, so only the solenoid's cross section, π(2a)², carries flux.
2. B8Nμ₀nπaα
   A student who uses the coil's circumference, 2π(4a), in place of an area picks this; it does not even have the units of emf. The flux through each turn is the field times the area through which it passes, π(2a)².
3. C4Nμ₀nπa²α Correct
   Inside the solenoid B = μ₀nI, and outside it is negligible, so the flux through each turn of the coil is Φ_B = μ₀nI·π(2a)² = 4μ₀nIπa², using the solenoid's cross section, not the coil's. dΦ_B/dt = 4μ₀nπa²α per turn, and the N turns in series give |ε| = 4Nμ₀nπa²α.
4. D0
   A student who thinks there can be no emf where the field is zero picks this: the coil's wire lies outside the solenoid, where B is negligible. The coil encloses the changing flux inside the solenoid, so there is an emf around it.

Working B_in = μ₀nI; Φ per turn = μ₀nI·π(2a)² = 4μ₀nIπa² (B = 0 for 2a < r < 4a). |ε| = N dΦ/dt = 4Nμ₀nπa²α. Units: (T·m/A)(1/m)(m²)(A/s) = V. Distractors (sympy): coil area π(4a)² → 16Nμ₀nπa²α; circumference 2π(4a) → 8Nμ₀nπaα (units V/m, not V).

CED 13.2.A.1.iii · Read this in Fix

1. A2.0 × 10⁻² V
   A student who thinks the number of turns does not affect the emf applies only the area change: ¼ × 8.0 × 10⁻² V. The emfs of the turns add, so tripling N triples the emf.
2. B1.2 × 10⁻¹ V
   A student who takes the emf to scale with the radius rather than the area gets 3 × ½ = 1.5 times the emf. The area is πr², so halving the radius quarters it.
3. C2.4 × 10⁻¹ V
   A student who thinks the emf of each turn depends only on how fast the field changes applies only the change in N: 3 × 8.0 × 10⁻² V. The flux through each turn depends on its area, which is one-quarter as large.
4. D6.0 × 10⁻² V Correct
   |ε| = N A|dB/dt|. Tripling N triples the emf, and halving the radius makes the area π(r/2)² one-quarter as large: ε = 3 × ¼ × 8.0 × 10⁻² V = 6.0 × 10⁻² V.

Working ε ∝ N r². ε_new = 8.0 × 10⁻² V × 3 × (1/2)² = 6.0 × 10⁻² V.

CED 13.2.A.1.iii · Read this in Fix

1. AClockwise, with its field pointing out of the page inside the loop
   A student who reverses the right-hand rule takes the wire's field at the loop to be out of the page, has the induced field point out of the page to oppose its decrease, and then (reversing the rule again) links that field to a clockwise current. The wire's field at the loop is into the page, and a clockwise current's field is into the page too.
2. BClockwise, with its field pointing into the page inside the loop Correct
   By the right-hand rule, an upward current makes a field into the page on its right, where the loop is. As I decreases, this into-page flux decreases, so the induced field points into the page to oppose the decrease. Curling the right hand's fingers so that the thumb points into the page gives a clockwise current.
3. CCounterclockwise, with its field pointing out of the page inside the loop
   A student who thinks the induced field always opposes the external field picks this. The flux is decreasing, so the induced field points the same way as the wire's field, into the page, and the current is clockwise.
4. DNo current, since neither the wire nor the loop is moving
   A student who thinks induction requires motion picks this. The decreasing current in the wire changes the flux through the loop, so a current is induced even though nothing moves.

Working Upward I → B into page on the right. I decreasing → into-page flux decreasing → induced B into page → clockwise.

CED 13.2.A.2.ii · Read this in Fix

1. AThe induced field has to cancel the external field, so that the net field stays zero.
   A student who thinks the induced field opposes the field itself picks this. The induced field opposes only the change in flux; when the flux is decreasing it points along the external field, and the net field is not kept at zero.
2. BThe induced current makes the loop repel a magnet, whichever way the magnet moves.
   A student who thinks a loop always repels a magnet picks this. A receding magnet is attracted by the induced current; the force opposes the relative motion, which is a consequence of Lenz's law, not a justification of it.
3. CThe minus sign in Faraday's law is a convention, chosen to fix this direction.
   A student who thinks the minus sign is only bookkeeping picks this. The sign states a physical fact, required by conservation of energy; with the opposite sign an induced current would amplify itself.
4. DIf the induced field aided the change, the current would grow unaided and create energy. Correct
   An induced field that reinforced the change in flux would increase the change, which would increase the induced current further, and so on without any external work: energy would be created from nothing. Conservation of energy requires the induced current to oppose the change, so external work must be done to change the flux.

CED 13.2.A.2.i · Read this in Fix

1. AR²β/(2r)
   A student who uses the flux through the whole solenoid, πR²βt, for a circle inside it picks this. A circle of radius r < R encloses only the flux through πr²; R²β/(2r) is the field outside the solenoid.
2. Bβ/(2πr)
   A student who thinks the emf depends only on how fast the field changes sets E(2πr) equal to β, leaving out the area πr² through which the field passes; the result does not even have the units of electric field. The circulation equals the rate of change of flux, πr²β.
3. Crβ/2 Correct
   By symmetry the induced field is tangent to a circle of radius r centered on the axis and has the same magnitude all around it. ∮E⃗ · dℓ⃗ = E(2πr), and the flux through that circle is βt(πr²), so E(2πr) = πr²β and E = rβ/2.
4. D0
   A student who thinks every electric field has zero circulation picks this: by symmetry a circular field with zero circulation must vanish. The induced field has ∮E⃗ · dℓ⃗ = −dΦ_B/dt, which is not zero here.

Working Circle radius r < R: ∮E·dℓ = E·2πr = dΦ/dt = d(βt·πr²)/dt = πr²β → E = rβ/2. Distractors: flux through the whole solenoid, E(2πr) = πR²β → R²β/(2r); area left out, E(2πr) = β → β/(2πr) (units T/(s·m), not V/m); zero circulation → 0.

CED 13.2.A.3 · Read this in Fix

1. AR²β/(2r) Correct
   Take a circle of radius r through P, centered on the axis. By symmetry E is tangent to it and uniform in magnitude, so ∮E⃗ · dℓ⃗ = E(2πr). The flux through the circle is only the solenoid's, βt(πR²), since the field outside is negligible. E(2πr) = πR²β, so E = R²β/(2r).
2. Brβ/2
   A student who multiplies the field by the area of the circle through P, πr², picks this. Outside radius R there is no field, so the flux enclosed is βt(πR²).
3. C0
   A student who thinks an induced electric field exists only where the magnetic field is picks this. The circle through P encloses changing flux, so the circulation of E⃗ around it is not zero, and by symmetry E is not zero at P.
4. Dβ/(2πr)
   A student who thinks the emf depends only on how fast the field changes sets E(2πr) equal to β, leaving out the solenoid's cross-sectional area πR²; the result does not even have the units of electric field. The circulation equals the rate of change of the enclosed flux, πR²β.

Working Circle radius r > R: E·2πr = d(βt·πR²)/dt = πR²β → E = R²β/(2r). Distractors: area of the circle through P, πr² → rβ/2; area left out, E(2πr) = β → β/(2πr) (units T/(s·m), not V/m); no field at P → 0.

CED 13.2.A.3 · Read this in Fix

1. Aβℓ(d + 2vt) Correct
   The flux is Φ_B = (βt)ℓ(d + vt) = βℓ(dt + vt²), and both the field and the area grow. Differentiating gives |ε| = βℓ(d + 2vt): ℓ(d + vt)β from the growing field plus βt(ℓv) from the growing area.
2. Bβℓ(d + vt)
   A student who divides the flux gained since t = 0, βtℓ(d + vt), by the time taken picks this. The flux does not grow at a constant rate, so that average is not the emf at time t; it leaves out the growth in area, B(dA/dt) = βtℓv.
3. Cβℓvt
   A student who takes the emf of a sliding rod to be its motional emf Bℓv alone picks this: (βt)ℓv. The field is also growing, which adds ℓ(d + vt)β to the rate of change of flux.
4. Dβℓ(dt + vt²)
   A student who takes the emf to be the flux through the circuit picks this. The emf is the rate of change of the flux, dΦ_B/dt; this expression is in Wb, not V.

Working Φ_B = BA = (βt)ℓ(d + vt) = βℓ(dt + vt²). |ε| = dΦ_B/dt = βℓ(d + 2vt); equivalently A(dB/dt) + B(dA/dt) = ℓ(d + vt)β + βt(ℓv). Units: (T/s)(m)(m) = Wb/s = V. Distractors (checked with sympy; with β = ℓ = d = v = 1 and t = 2 the options give 5 (key), 3, 2, 6): flux gained since t = 0 divided by the time taken, Φ_B(t)/t → βℓ(d + vt) (the same as A dB/dt alone); motional emf of the rod alone, Bℓv → βℓvt; the flux itself → βℓ(dt + vt²).

CED 13.2.A.1 · Read this in Fix

1. A√3Bv²t Correct
   The rod is vt from O, and its length between the rails is vt tan 60° = √3vt, so the triangle’s area is (√3/2)v²t². The field is constant, so |ε| = B dA/dt = √3Bv²t, which is also Bℓv for the rod’s length at time t.
2. B√3Bv²t/2
   A student who divides the flux at time t, (√3/2)Bv²t², by the time since the start picks this. The area grows as t², not at a constant rate, so that average is half of the emf at time t.
3. CBv²t
   A student who finds the rate of change of area from the side along the direction of motion, the first rail’s length vt, picks this: (vt)v. The new area is a strip along the rod, whose length is √3vt, swept at speed v.
4. D√3Bv²t²/2
   A student who takes the emf to be the flux through the circuit picks this. The emf is the rate at which that flux changes; this expression is in Wb, not V.

Working At time t the rod is a distance vt from O along the first rail, and the length of rod between the rails is vt tan 60° = √3vt. Area A = (1/2)(vt)(√3vt) = (√3/2)v²t², so Φ_B = (√3/2)Bv²t². |ε| = B dA/dt = √3Bv²t (the same as Bℓv with the rod’s instantaneous length ℓ = √3vt). Units: T(m/s)²(s) = T·m²/s = V. Distractors (checked with sympy; with B = v = 1 and t = 3 the options give 5.20 (key), 2.60, 3, 7.79): flux since t = 0 divided by the time taken, Φ_B/t → √3Bv²t/2; rate of change of area taken from the side along the direction of motion, (vt)v → Bv²t; the flux itself → √3Bv²t²/2.

CED 13.2.A.1.ii · Read this in Fix

1. A1.0 × 10⁻⁵ V Correct
   Integrating over strips parallel to the wire, Φ_B = (μ₀Iℓ/(2π)) ln((x + ℓ)/x), so |ε| = (μ₀Iℓv/(2π))(1/x − 1/(x + ℓ)) = (2.0 × 10⁻⁶ V·m)(1/0.10 − 1/0.20) m⁻¹ = 1.0 × 10⁻⁵ V. Equivalently, the near and far sides have opposing motional emfs, (4.0 − 2.0) × 10⁻⁵ T × 0.10 m × 5.0 m/s.
2. B3.0 × 10⁻⁵ V
   A student who adds the motional emfs of the two sides that move across the field picks this: (4.0 + 2.0) × 10⁻⁵ T × 0.10 m × 5.0 m/s. Both sides move the same way through fields in the same direction, so their emfs act in opposite senses around the loop and partly cancel.
3. C2.0 × 10⁻⁵ V
   A student who counts only the side nearer the wire, where the field is strongest, picks this: B₁ℓv = (4.0 × 10⁻⁵ T)(0.10 m)(5.0 m/s). The far side also moves across the field, and its emf opposes the near side's; the net emf equals the rate of change of flux through the whole loop.
4. D8.9 × 10⁻⁶ V
   A student who takes the flux to be the field at the loop's center, 0.15 m from the wire, times the loop's area picks this, differentiating Φ_B = ℓ²μ₀I/(2πr) with dr/dt = 5.0 m/s. The field varies as 1/r across the loop, so the flux must be integrated over strips parallel to the wire.

Working Let the near side be at distance x, so the far side is at x + ℓ (ℓ = 0.10 m). Flux through strips of width dr parallel to the wire: Φ_B = ∫ₓ^x+ℓ (μ₀I/(2πr)) ℓ dr = (μ₀Iℓ/(2π)) ln((x + ℓ)/x). dΦ_B/dt = (μ₀Iℓ/(2π)) (1/(x + ℓ) − 1/x)(dx/dt), so |ε| = (μ₀Iℓv/(2π))(1/x − 1/(x + ℓ)) = (2 × 10⁻⁷)(20)(0.10)(5.0)(1/0.10 − 1/0.20) = (2.0 × 10⁻⁶)(5.0) = 1.0 × 10⁻⁵ V. Check by motional emf: B₁ = 4.0 × 10⁻⁵ T at the near side, B₂ = 2.0 × 10⁻⁵ T at the far side; their emfs oppose around the loop: (B₁ − B₂)ℓv = 1.0 × 10⁻⁵ V. Distractors: (B₁ + B₂)ℓv = 3.0 × 10⁻⁵ V; near side only, B₁ℓv = 2.0 × 10⁻⁵ V; flux taken as B(center)·ℓ², d/dt gives ℓ²μ₀Iv/(2πr_c²) with r_c = 0.15 m: 8.9 × 10⁻⁶ V.

CED 13.2.A.1 · Read this in Fix