Induction is an HL-only topic. An emf is induced whenever the magnetic flux linkage through a circuit changes, and Faraday's law gives its size as the rate of change of flux linkage. Lenz's law, the minus sign, follows from conservation of energy.
The questions cover a rod moving through a field, ε = BvL, magnets falling through or oscillating near coils, and coils rotating in a uniform field, where the emf is sinusoidal and its peak depends on the number of turns, the area, the field and the angular speed.
48 questions
274 marks
Paper 1A: 26
Paper 1B: 8
Paper 2: 14
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22 practice questions on D.4 Induction
1D-1A-09
Faraday's law·D.4 Induction (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two adjacent regions of uniform magnetic field of the same flux density B are separated by a straight boundary. The field is directed into the page on the left of the boundary and out of the page on the right. A flat square coil of N turns and side L moves to the right at constant speed v, in the plane of the page, with two of its sides parallel to the boundary. Each region is wider than the coil.
What is the magnitude of the emf induced in the coil while it is crossing the boundary?
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Notes
Step 1While the coil crosses the boundary, its leading side moves through the outward field and its trailing side through the inward field; each side of each turn has an emf BvL.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The fields are opposite, so the two emfs act in the same sense round the coil and add. (Equivalently, the flux linkage changes from +NBL² to −NBL², a change of 2NBL², in a time L/v.)
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Step 3ε = 2NBLv.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes that the emfs in the two sides cancel, as they do for a coil moving inside a single uniform field. Here the two fields are opposite, so the emfs add.
BThis is the emf for a coil entering a field from a field-free region, where only one side cuts field lines. Here the flux linkage changes by twice as much in the same time.
CCorrect: both sides cut field lines and, because the fields are opposite, their emfs add: 2NBLv.
DThis counts the effect of the opposite field twice: it uses the change in flux linkage 2NBL² and then also adds the emfs of the two sides.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; Faraday's law ε = −NΔΦ/Δt (rectangular coils moving in and out of fields) Command term: Determine
2D-1A-10
Motional emf·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A conducting rod is pulled at constant speed v along two parallel horizontal rails that are joined by a resistor of resistance R. A uniform vertical magnetic field acts on the rod. The resistances of the rod and the rails are negligible.
The speed of the rod is doubled and the resistor is replaced by one of resistance 2R. What is (new electrical power dissipated)/(original electrical power dissipated)?
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Notes
Step 1The induced emf is ε = BvL, so doubling the speed doubles the emf.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The power dissipated is P = ε²/R: the emf squared gives a factor 4 and the doubled resistance gives a factor ½.
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Step 3New power/original power = 4 × ½ = 2.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the factor for the external force needed, F = BIL = B²L²v/R (×2 for the speed, ×½ for the resistance), not for the power.
BCorrect: P ∝ v²/R, so the factor is 2²/2 = 2.
CThis ignores the change of resistor: ε² ∝ v² alone gives 4.
DThis works out I²R using the current for the original resistor (I = BLv/R, so I² × 4) and then multiplies by the new resistance 2R. With 2R in the circuit the current does not change at all.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; Lenz's law as a consequence of energy conservation Command term: Determine
3D-1A-14
Lenz's law·D.4 Induction (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A bar magnet is dropped along the vertical axis of a horizontal, complete copper ring and falls through it.
I. The magnet takes longer to fall through the complete ring than through an identical ring that has a narrow cut in it.
II. While the magnet approaches the ring, the magnetic force on the magnet is directed upwards.
III. While the magnet moves away below the ring, the magnetic force on the magnet is directed upwards.
Which statements are correct?
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Notes
Step 1The changing flux induces an emf in both rings, but only the complete ring carries a current and so exerts a force on the magnet: I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2By Lenz's law the induced current opposes the change. While the magnet approaches, the flux through the ring increases and the ring repels the magnet: the force on the magnet is upwards, so II is correct.
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Step 3While the magnet moves away, the flux decreases and the current reverses; the ring now attracts the magnet back towards it, which is again upwards, so III is correct. Throughout, kinetic energy of the magnet is transferred to thermal energy in the ring.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis assumes the force on the magnet reverses once it has passed the ring. The current does reverse, but so does the position of the magnet relative to the ring, and the force still opposes the relative motion.
BThis rejects II, as if the ring attracted the approaching magnet. A force that sped the magnet up would create energy from nothing, contrary to Lenz's law.
CThis rejects I. The cut ring has an induced emf but no complete circuit, so there is no current and no opposing force.
DCorrect: the induced current always opposes the relative motion, so the force on the magnet is upwards both before and after it passes the ring.
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation; Faraday's law of induction (relative motion between a conductor and a magnetic field) Command term: Deduce
4D-1A-16
Peak emf of a rotating coil·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A rectangular coil rotates at constant angular speed in a uniform magnetic field. The graph shows how the magnetic flux linkage NΦ of the coil varies with time t.
What is the peak emf induced in the coil?
Flux linkage of the rotating coil against time.Show mark scheme
Marking point
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Notes
Step 1From the graph: peak flux linkage NΦ0 = 0.040 Wb and period T = 20 ms.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The emf is the rate of change of flux linkage. For a sinusoidal variation the greatest rate is ωNΦ0, with ω = 2π/T = 314 rad s−1.
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Step 3ε0 = 314 × 0.040 = 12.6 ≈ 13 V, when NΦ = 0, where the graph is steepest.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses NΦ0/T = NΦ0f and leaves out the factor 2π in ω = 2πf.
BThis is the average emf over a quarter of a cycle (0.040 Wb in 5.0 ms), not the peak emf. The graph is steepest where it crosses the time axis.
CCorrect: ε0 = (2π/T) × NΦ0 = 13 V.
DThis reads the period as 10 ms, the time from a maximum to the next minimum, which is only half a cycle.
Syllabus understandingD.4 — that a uniform magnetic field induces a sinusoidal varying emf in a coil rotating within it; Faraday's law ε = −NΔΦ/Δt (the emf as the gradient of the flux linkage–time graph) Command term: Determine
5D-1A-21
The a.c. generator·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A simple generator coil rotates in a uniform magnetic field and produces a sinusoidal emf of peak value ε0.
The magnetic field strength is doubled and the angular speed of rotation is halved. What is the new peak emf?
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Marking point
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Notes
Step 1The emf is the rate of change of the flux linkage of the coil, ε = −NΔΦ/Δt.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Doubling B doubles the maximum flux linkage NBA; halving the rate of rotation halves how quickly the flux linkage changes. So the peak emf is proportional to B × (rate of rotation) (for a rotating coil ε0 = NBAω).
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Step 3The two factors 2 and ½ cancel, so the peak emf is unchanged (but the frequency of the emf halves).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis would be the result if the peak emf were proportional to (rate of rotation)/B.
BThis is the effect of halving ω alone.
CCorrect: ε0 ∝ Bω, and 2 × ½ = 1.
DThis is the effect of doubling B alone.
Syllabus understandingD.4 — that a uniform magnetic field induces a sinusoidal varying emf in a coil rotating within it; the effect on induced emf caused by changing the frequency of rotation; Faraday's law ε = −NΔΦ/ΔtCommand term: Determine
6D-1A-25
A magnet oscillating above a coil·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A bar magnet hangs from a spring and oscillates vertically with frequency f above a fixed coil. The coil is connected in series with a resistor and a switch.
Which statement is correct?
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Marking point
Mark
Notes
Step 1The emf depends on the rate of change of flux linkage. It is greatest when the magnet moves fastest (through the equilibrium position) and zero at the turning points.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The emf reverses when the direction of motion reverses, so it completes one cycle per oscillation: its frequency is f.
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Step 3With the switch closed there is a current. By Lenz's law it opposes the relative motion, and the energy of the oscillation is transferred to thermal energy in the circuit, so the oscillations die away faster. With the switch open there is an emf but no current.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: only with a complete circuit does an induced current flow; it opposes the motion and removes energy from the oscillator.
BThis confuses flux linkage with its rate of change. At the lowest point the magnet is momentarily at rest and the emf is zero.
CThe emf is in one direction while the magnet moves down and in the opposite direction while it moves up; together these make one cycle, so the frequency is f.
DAn emf is induced whenever the flux linkage changes. An open circuit prevents a current, not an emf.
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation; Faraday's law ε = −NΔΦ/Δt (relative motion: a magnet oscillating on a spring above a conducting coil) Command term: Deduce
7D-1A-56
Magnetic flux and flux linkage·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
A square coil of 50 turns and side 4.0 cm is in a uniform magnetic field of flux density 0.20 T. The normal to the plane of the coil makes an angle of 60° with the field. The coil is turned until its plane is parallel to the field.
What is the magnitude of the change in flux linkage NΦ?
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Notes
Step 1Φ = BA cos θ, where θ is the angle between the field and the normal to the coil. A = 0.040² = 1.6 × 10−3 m².
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All 3 steps must be completed — there is no mark for a part-answer.
CThis uses sin 60° in place of cos 60° (angle measured from the plane, not the normal): 50 × 0.20 × 1.6 × 10−3 × 0.866 = 1.4 × 10−2 Wb.
DThis leaves out the angle factor, as if the coil started face-on to the field: 50 × 0.20 × 1.6 × 10−3 = 1.6 × 10−2 Wb.
Syllabus understandingD.4 — magnetic flux Φ as given by Φ = BA cos θCommand term: Calculate
8D-1A-57
Motional emf·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A yacht has a vertical aluminium mast 16 m tall. In the harbour the Earth's magnetic field has a horizontal component of 1.8 × 10−5 T directed due north and a vertical component of 4.5 × 10−5 T directed downwards. The yacht sails at a constant speed of 5.0 m s−1.
Which row gives the magnitude of the emf induced between the top and the bottom of the mast when the yacht sails due east and when it sails due north?
Sailing due eastSailing due north
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Notes
Step 1An emf ε = BvL is induced along a straight conductor only by the field component perpendicular to both the conductor and its velocity.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The mast is vertical, so the vertical component never contributes. Sailing east, the mast moves perpendicular to the northward horizontal component: ε = 1.8 × 10−5 × 5.0 × 16 = 1.4 × 10−3 V.
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Step 3Sailing north, the velocity is parallel to the horizontal component, so the mast cuts no field lines: ε = 0.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis swaps the two cases: moving along the field lines (north) produces no emf.
BCorrect: 1.4 mV sailing east and zero sailing north.
CThis assumes the emf depends only on the speed, not on the direction of motion relative to the field.
DThis uses the vertical component, which is parallel to the mast: 4.5 × 10−5 × 5.0 × 16 = 3.6 × 10−3 V.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvLCommand term: Deduce
9D-1A-58
Faraday's law·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A coil on a laboratory carriage is moved into, through and out of a region of uniform magnetic field. The graph shows how the flux linkage NΦ of the coil varies with time t.
Which row gives the magnitude of the average emf induced in the coil between 0 and 0.20 s and between 0.30 s and 0.40 s?
Flux linkage of the coil against time.
0 to 0.20 s0.30 s to 0.40 s
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Notes
Step 1Faraday's law: ε = −NΔΦ/Δt, so the emf magnitude is the magnitude of the gradient of the NΦ–t graph.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 20 to 0.20 s: 0.060/0.20 = 0.30 V.
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Step 30.30 s to 0.40 s: 0.060/0.10 = 0.60 V (opposite sign, because the flux linkage is now decreasing).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis divides the second change by the time read from the axis, 0.40 s, instead of the interval 0.10 s: 0.060/0.40 = 0.15 V.
BThis assumes the emf is the same whenever the flux linkage changes by the same amount, ignoring the shorter time.
CCorrect: the gradients are 0.30 Wb s−1 and 0.60 Wb s−1.
DThis inverts the two intervals (0.060/0.10 and 0.060/0.20), attributing the steeper gradient to the first stage.
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday's law of induction ε = −NΔΦ/ΔtCommand term: Determine
10D-1A-59
Self-induction·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksExplain
Two identical lamps P and Q are connected in parallel branches across a battery through a single switch. P is in series with a resistor; Q is in series with a coil of many turns wound on an iron core. The resistor and the coil have equal resistance. The switch is closed.
Which statement describes and explains what is observed?
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Notes
Step 1When the switch is closed the current in the coil starts to grow, so the magnetic flux linking the coil's own turns grows.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2By Faraday's law this changing flux induces an emf in the coil; by Lenz's law it opposes the change that causes it — the growth of the current.
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Step 3The current in Q's branch therefore rises gradually, so Q lights after P; once the current is steady there is no induced emf and both lamps are equally bright.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis considers only the final steady state, where equal resistances do give equal currents; it ignores the self-induced emf while the current is changing.
BCorrect: self-induction opposes the change in current (Lenz's law), delaying the rise of current in Q's branch.
CThe self-induced emf opposes the change (Lenz's law); an emf that aided the growth would create energy from nothing.
DThe observation is right but the reason is wrong: the delay is not due to a slow drift of charge through the wire but to the induced emf.
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation; self-induction (qualitative treatment only) Command term: Explain
11D-1A-69
The a.c. generator·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A flat coil rotates at a constant frequency in a uniform magnetic field. The maximum magnetic flux linkage of the coil is NΦmax and the peak emf induced in it is ε0.
The frequency of rotation is doubled. Which row gives the new maximum flux linkage and the new peak emf?
maximum flux linkagepeak emf
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Notes
Step 1The maximum flux linkage NBA occurs when the plane of the coil is perpendicular to the field; it depends on N, B and A but not on how fast the coil turns.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Doubling the frequency halves the time for each change of flux linkage, so by Faraday's law (ε = −NΔΦ/Δt) the peak emf doubles.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes the flux linkage grows with the rate of rotation.
BThis assumes the emf depends only on N, B and A, ignoring the rate of change of flux linkage.
CThis doubles the flux linkage and then doubles the emf again for the higher frequency.
DCorrect: the same maximum flux linkage is swept through in half the time.
Syllabus understandingD.4 — magnetic flux Φ = BA cos θ; that a uniform magnetic field induces a sinusoidally varying emf in a coil rotating within it; the effect on induced emf caused by changing the frequency of rotation Command term: Deduce
12D-1A-70
Lenz's law·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The bob of a pendulum is a flat copper plate that swings in its own vertical plane. The plate swings through the narrow gap between the poles of a strong magnet, with the magnetic field perpendicular to the plate. The field occupies only a small region around the lowest point of the swing. The oscillations die away quickly.
I. The kinetic energy lost by the pendulum is transferred to internal energy of the plate.
II. If the uniform magnetic field extended over the whole region through which the plate swings, the oscillations would die away just as quickly.
III. With a stronger magnet, the oscillations would die away more quickly.
Which statements are correct?
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Notes
Step 1As parts of the plate enter or leave the field, the magnetic flux through loops in the plate changes, so emfs are induced and currents circulate in the plate. By Lenz's law the magnetic forces on these currents oppose the motion; the work done against them is dissipated in the resistance of the copper. I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2If the plate stayed inside a uniform field throughout its swing, the flux through every loop in the plate would remain constant, so no currents would be induced and there would be no magnetic damping. II is incorrect.
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Step 3A larger B gives a larger induced emf and so a larger current, and the force on that current is also proportional to B: the opposing force increases (roughly as B²) and the damping increases. III is correct: I and III only.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis accepts II, assuming that motion through a magnetic field is enough to induce currents; currents need a changing flux through loops in the plate. It also rejects III.
BCorrect: the eddy currents dissipate the kinetic energy, they need a changing flux, and a stronger field gives larger currents and larger forces.
CThis rejects I, as if the energy were stored in the magnet, and accepts II, ignoring that a constant flux induces no current.
DThis accepts II: in a field covering the whole swing the flux through each loop of the plate does not change, so there is no damping.
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday's law of induction; that the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation (relative motion between a conductor and a magnetic field) Command term: Deduce
13D-1A-71
Peak emf of a rotating coil·D.4 Induction (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A flat coil of N turns and area A rotates with constant angular velocity ω in a uniform magnetic field of flux density B. The peak emf induced in the coil is ε0 = NBAω.
The coil turns through 90° from the position in which its plane is perpendicular to the field. What is (magnitude of the average emf induced during this quarter turn)/ε0?
Show mark scheme
Marking point
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Notes
Step 1The flux linkage falls from NBA (plane perpendicular to the field) to zero (plane parallel to the field): Δ(NΦ) = NBA.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3Average emf = NBA/(π/(2ω)) = (2/π)NBAω, so the ratio is 2/π.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses half a period, π/ω, for the time of a quarter turn.
BThis assumes the emf rises linearly from zero to ε0, so that its average is half the peak.
CCorrect: Faraday's law gives the average emf as the change in flux linkage divided by the time.
DThis takes the change in flux linkage as 2NBA, as if the flux reversed during a quarter turn.
Syllabus understandingD.4 — Faraday's law of induction ε = −NΔΦ/Δt; magnetic flux Φ = BA cos θ; that a uniform magnetic field induces a sinusoidally varying emf in a coil rotating within it Command term: Determine
14D-1A-110
Magnetic flux in the Earth’s field·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two identical flat square loops of wire, each of side 0.50 m, are at a place where the Earth’s magnetic field has a magnitude of 5.0 × 10−5 T and is directed northwards at 65° below the horizontal. Loop X lies flat on the horizontal ground. Loop Y stands vertically, with its plane along the east–west direction.
Which row gives the magnetic flux through X and through Y?
flux through Xflux through Y
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Notes
Step 1Φ = BA cos θ, where θ is the angle between the field and the normal to the loop; A = 0.50² = 0.25 m².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2X: the normal is vertical and the field is 65° below the horizontal, so θ = 25°: Φ = 5.0 × 10−5 × 0.25 × cos 25° = 1.1 × 10−5 Wb.
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Step 3Y: the plane is east–west, so the normal is horizontal and points north–south; the angle between the field and the normal is 65°: Φ = 5.0 × 10−5 × 0.25 × cos 65° = 5.3 × 10−6 Wb.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis measures each angle from the plane of the loop instead of from its normal, so the two values are swapped.
BCorrect: X is threaded mainly by the large vertical component (cos 25°) and Y by the smaller horizontal component (cos 65°).
CThis confuses the plane of Y with its normal, taking the normal as east–west, at 90° to the field, so that no flux would pass through Y.
DThis ignores the angle between the field and each loop, using Φ = BA for both: 5.0 × 10−5 × 0.25 = 1.3 × 10−5 Wb.
Syllabus understandingD.4 — magnetic flux Φ as given by Φ = BA cos θCommand term: Determine
15D-1A-111
Average emf when a coil is turned over·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
A rectangular search coil of 150 turns measures 2.0 cm × 1.0 cm. It is held with its plane perpendicular to a uniform magnetic field of flux density 0.050 T, and is then turned over, through 180° about one of its sides, in 0.10 s.
What is the magnitude of the average emf induced in the coil while it is turned over?
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Notes
Step 1Initially NΦ = +NBA = 150 × 0.050 × 2.0 × 10−4 = 1.5 × 10−3 Wb; after the half turn the same field threads the coil from the other face, so NΦ = −1.5 × 10−3 Wb (cos 180° = −1).
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All 3 steps must be completed — there is no mark for a part-answer.
AThis compares only the magnitudes of the flux linkage at the start and the end, which are equal, ignoring that the flux has reversed relative to the coil.
BThis leaves out the number of turns: 2 × 0.050 × 2.0 × 10−4/0.10 = 2.0 × 10−4 V.
CThis takes the change in flux linkage as NBA, as for a quarter turn from maximum to zero, instead of 2NBA.
DCorrect: the flux linkage changes from +NBA to −NBA, a change of 3.0 × 10−3 Wb in 0.10 s.
Syllabus understandingD.4 — magnetic flux Φ as given by Φ = BA cos θ; that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/ΔtCommand term: Calculate
16D-1A-112
Units of magnetic flux·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
Which unit is equivalent to the weber (Wb), the unit of magnetic flux?
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Notes
Step 1Faraday’s law, ε = −NΔΦ/Δt, gives the unit of Φ as (unit of emf) × (unit of time) = V s (N is a number).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Check: Φ = BA cos θ gives T m²; with T = N A−1 m−1 (from F = BIL) this is N m A−1 = J A−1 = (J C−1)(s) = V s.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: flux linkage changes at a rate measured in volts, so flux is measured in volt seconds.
BThis inverts Faraday’s law, as if flux were emf divided by time.
CThis divides the flux density by the area instead of multiplying: Φ = BA, not B/A.
DThis is the tesla itself (from F = BIL); the area factor m² has been left out.
Syllabus understandingD.4 — magnetic flux Φ as given by Φ = BA cos θ; that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/ΔtCommand term: Deduce
17D-1A-113
Conditions for an induced emf·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A flat coil is held with its plane perpendicular to a uniform magnetic field. The field region is much larger than the coil.
Which action does not induce an emf in the coil?
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Marking point
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Notes
Step 1An emf is induced only while the flux linkage NBA cos θ of the coil changes.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Turning the coil over changes cos θ from +1 to −1; reversing the field changes the sign of B; squashing the coil reduces A. Each changes the flux linkage.
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Step 3Moving the coil without rotating it inside a uniform field leaves B, A and θ unchanged, so the flux linkage is constant and no emf is induced (the emfs in opposite sides of each turn cancel).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ATurning the coil reverses the flux relative to the coil (cos θ goes from +1 to −1), so the flux linkage changes and an emf is induced.
BReversing the field changes the flux linkage from +NBA to −NBA, so an emf is induced while the field changes.
CReducing the area reduces the flux linkage, so an emf is induced while the coil is being squashed.
DCorrect: in a uniform field a coil that moves without turning keeps the same flux linkage, so there is no induced emf, even though its sides move through the field.
Syllabus understandingD.4 — magnetic flux Φ as given by Φ = BA cos θ; that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; (guidance) relative motion between a conductor and a magnetic field Command term: Deduce
18D-1A-114
Emf from a changing current·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A small coil is placed inside a long solenoid, with its plane perpendicular to the axis of the solenoid. The magnetic field inside the solenoid is proportional to the current I in the solenoid. The graph shows how I varies with time t. The intervals W, X, Y and Z are labelled.
In which interval is the magnitude of the emf induced in the small coil greatest?
Current in the solenoid against time (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The flux linkage of the small coil is proportional to I, so by Faraday’s law the emf is proportional to the rate of change of I: the magnitude of the gradient of the graph.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradients: W 4/2 = 2.0 A s−1; X 0; Y 3/1 = 3.0 A s−1; Z 5/4 = 1.25 A s−1.
—
Step 3The steepest segment is Y.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis picks the interval in which the current increases, as if only an increasing flux could induce an emf. A decreasing flux also induces an emf; Y is steeper than W.
BThis picks the largest steady current. A constant current gives a constant flux, so the emf in X is zero.
CCorrect: the current changes fastest in Y (3.0 A s−1), so the flux linkage changes fastest there.
DThis picks the largest change in current (5 A, including the reversal) and ignores that it takes 4 s.
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; (guidance) time-varying magnetic fields Command term: Deduce
19D-1A-115
Energy transfer for a pulled rod·D.4 Induction (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A metal rod is pulled at constant speed by a force F along two horizontal rails that are joined by a resistor. A uniform vertical magnetic field acts on the rod. Friction and the resistance of the rod and the rails are negligible.
The rate at which F does work is equal to
Show mark scheme
Marking point
Mark
Notes
Step 1The current in the rod experiences a magnetic force BIL that, by Lenz’s law, opposes the motion; at constant speed F balances it.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2F does work at the rate Fv = BILv = εI, the electrical power generated in the rod.
—
Step 3All of this electrical energy is transferred to thermal energy in the resistor: εI = I2R.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThe speed is constant, so the kinetic energy of the rod does not change.
BThe resultant force is zero, but F itself acts along the direction of motion, so it does positive work.
CCorrect: the work done against the opposing magnetic force becomes electrical energy, which is dissipated in the resistor.
DThis counts the same energy twice: εI is the rate of electrical energy production, and it is this energy that becomes the thermal energy in the resistor.
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvLCommand term: Deduce
20D-1A-116
Terminal speed on rails·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A rod rests on horizontal, frictionless rails joined by a resistor, in a uniform vertical magnetic field. A constant horizontal force F is applied to the rod, which reaches a terminal speed.
The force is doubled and the magnetic flux density is doubled. What is (new terminal speed)/(original terminal speed)?
Show mark scheme
Marking point
Mark
Notes
Step 1At speed v the induced current is I = BLv/R, so the opposing force is BIL = B2L2v/R.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the terminal speed this equals F: vt = FR/(B2L2).
—
Step 3Ratio = 2/2² = ½.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis applies the factor 1/B2 but ignores the doubling of F.
BCorrect: vt ∝ F/B2, so the factor is 2/4 = ½.
CThis takes the opposing force as proportional to B rather than B2, forgetting that B affects both the current and the force on it.
DThis doubles the speed for the doubled force and ignores the stronger braking caused by the larger field.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θCommand term: Deduce
21D-1A-117
Emf from a flux-linkage–position graph·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A rectangular coil of 80 turns is pushed at a constant speed of 0.40 m s−1 into a region of uniform magnetic field. The graph shows how the flux linkage NΦ of the coil varies with the position x of the coil.
What is the maximum emf induced in the coil?
Flux linkage of the coil against its position (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1ε = Δ(NΦ)/Δt = (Δ(NΦ)/Δx) × (Δx/Δt) = (gradient of the graph) × v.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient of the rising section = 0.030/(0.070 − 0.020) = 0.60 Wb m−1.
—
Step 3ε = 0.60 × 0.40 = 0.24 V. (The graph already shows the flux linkage, so N must not be applied again.)
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the emf is the gradient of NΦ against x multiplied by the speed: 0.60 × 0.40 = 0.24 V.
BThis is the gradient of the graph, in Wb m−1, not multiplied by the speed: a rate of change with distance, not with time.
CThis divides the gradient by the speed (0.60/0.40) instead of multiplying.
DThis multiplies by the 80 turns again, although the graph already shows the flux linkage NΦ: 0.24 × 80 = 19 V.
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; magnetic flux Φ as given by Φ = BA cos θ; (guidance) rectangular coils moving in and out of fields Command term: Determine
22D-1A-118
Lenz’s law near a long straight wire·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A long straight wire carries a constant current I upwards in the plane of the page. A rectangular loop of wire, in the same plane, moves directly away from the wire at constant velocity, as shown.
Which row gives the direction of the current induced in the loop, as seen in the diagram, and the direction of the resultant magnetic force on the loop?
A rectangular loop moving away from a long straight wire, in the plane of the page (not to scale).
induced currentresultant force on the loop
Show mark scheme
Marking point
Mark
Notes
Step 1By the right-hand grip rule, the field of the wire to the right of it is into the page; it is weaker farther from the wire.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2As the loop moves away, the flux into the page through it decreases. By Lenz’s law the induced current opposes this decrease, producing a field into the page inside the loop: the current is clockwise.
—
Step 3In the side nearer the wire the current is upwards, parallel to I: attraction. In the far side it is downwards: repulsion, but the field there is weaker. The resultant force is towards the wire, opposing the motion, as energy conservation requires.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis makes the induced field oppose the field of the wire rather than the change in flux; an anticlockwise current would also make the near side antiparallel to I and repelled.
BThis gives a current that opposes the field instead of the change in flux, and a force that would help the loop move away, gaining energy from nowhere.
CCorrect: the clockwise current opposes the decrease of flux into the page, and the force on the loop opposes its motion.
DThe current direction is right, but the force is taken from the far side only; the near side is in the stronger field, so the attraction there dominates.
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; (guidance) relative motion between a conductor and a magnetic field; D.3 — the magnetic field patterns of a long straight wire; the force between parallel currents Command term: Deduce
23D-1A-119
A rod coasting on rails·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A metal rod lies across two horizontal, frictionless rails that are joined by a resistor of resistance R. A uniform vertical magnetic field acts on the rod. The rod is given a short push and then moves freely along the rails.
I. The speed of the rod decreases.
II. The magnitude of the deceleration of the rod decreases as it slows down.
III. The distance that the rod travels before it stops does not depend on R.
Which statements are correct?
Show mark scheme
Marking point
Mark
Notes
Step 1The moving rod has an emf BvL and a current BvL/R; the magnetic force on the current, B2L2v/R, opposes the motion (Lenz’s law), so the rod slows down: I is correct.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The force is proportional to v, so as the rod slows the force and the deceleration decrease: II is correct.
—
Step 3A larger R gives a smaller current and a smaller braking force at every speed, so the rod travels further: the distance depends on R and III is incorrect.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the braking force is proportional to the speed and inversely proportional to R.
BThis rejects II, treating the braking force as constant, and accepts III, overlooking that R sets the size of the current.
CThis rejects I, as if the rod could keep moving at constant speed on frictionless rails while energy is dissipated in the resistor.
DThis accepts III. The kinetic energy is always dissipated in the end, but with a larger R the braking force is smaller and the rod travels further.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θCommand term: Deduce
24D-1A-120
Energy dissipated as a coil leaves a field·D.4 Induction (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A rectangular coil of N turns has sides of length a, parallel to its velocity, and b, perpendicular to its velocity. It is pulled at constant speed v out of a region of uniform magnetic field of flux density B that is perpendicular to the plane of the coil. The ends of the coil are joined and its total resistance is R.
What is the thermal energy transferred in the coil while it leaves the field?
Show mark scheme
Marking point
Mark
Notes
Step 1While one side of length b is inside the field and the other outside, the emf is ε = NBbv.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The thermal power is ε2/R = N2B2b2v2/R, and it lasts for the time a/v taken to move a distance a.
—
Step 3Energy = N2B2b2av/R: pulling the coil out faster dissipates more energy.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses N rather than N2: the emf is proportional to N, and the power to the square of the emf.
BThis assumes that the energy, like the charge NBab/R that flows, does not depend on the speed.
CCorrect: power ∝ v2 and time ∝ 1/v, so the energy ∝ v.
DThis multiplies the power by the distance a instead of by the time a/v.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; (guidance) rectangular coils moving in and out of fields; B.5 — electrical power P = V2/RCommand term: Deduce
25D-1A-121
Terminal speed on inclined rails·D.4 Induction (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two parallel conducting rails, a distance L apart, are inclined at an angle θ to the horizontal and joined at the bottom by a resistor of resistance R. A uniform magnetic field of flux density B acts vertically. A horizontal rod of mass m slides down the rails without friction and reaches a terminal speed. The resistance of the rod and the rails is negligible.
What is the terminal speed?
Side view of the inclined rails (not to scale). The field is vertical.Show mark scheme
Marking point
Mark
Notes
Step 1The velocity is along the incline; only its horizontal component v cos θ is perpendicular to the vertical field, so ε = BLv cos θ and I = BLv cos θ/R.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The magnetic force BIL on the rod is horizontal; its component up the incline is BIL cos θ = B2L2v cos2θ/R.
—
Step 3At the terminal speed this balances mg sin θ, so v = mgR sin θ/(B2L2 cos2θ).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis applies the factor cos θ only once, either to the emf or to the force, but not to both.
BThis treats the field as perpendicular to the plane of the rails, so that neither the emf nor the force is reduced.
CThis uses the component of the weight perpendicular to the incline, mg cos θ, instead of the component along it.
DCorrect: the vertical field reduces both the emf and the useful component of the magnetic force by cos θ.
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; A.2 — resolving forces, terminal speed Command term: Deduce
26D-1A-122
Emf for a tilted square coil entering a field·D.4 Induction (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A flat square coil moves at constant velocity into a region of uniform magnetic field directed into the page. One diagonal of the square is parallel to the velocity, and the straight boundary of the field region is perpendicular to the velocity. The field region is larger than the coil. Time t = 0 when the leading corner reaches the boundary, and the coil is completely inside the field at time T.
Which graph shows how the emf ε induced in the coil varies with time?
Upper: the coil approaching the field (top view, not to scale). Lower: four possible graphs of the emf against time; at time T the coil is completely inside the field.Show mark scheme
Marking point
Mark
Notes
Step 1The emf is the rate of change of flux linkage. In a short time Δt the area inside the field grows by (width of the coil along the boundary) × vΔt, so ε ∝ (width of the coil at the boundary).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a square with a diagonal along the motion, this width grows linearly from zero to the full diagonal at T/2 and then shrinks linearly to zero at T.
—
Step 3After T the whole coil is inside the uniform field, the flux linkage is constant and ε = 0: graph X.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis treats the coil as if a straight side were parallel to the boundary, so that a constant length cuts the field lines.
BCorrect: the emf is proportional to the width of the coil along the boundary, which rises and falls linearly.
CThis makes the emf proportional to the total length of the coil’s edges inside the field, which keeps increasing until the coil is fully inside.
DThis is the shape of the flux linkage against time, not of its rate of change.
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; (guidance) rectangular coils moving in and out of fields Command term: Deduce
27D-1B-03
Faraday's law·D.4 Induction (HL)
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
A bar magnet is released from rest above a vertical coil that is connected to a data logger, which records the induced emf. The height h of the lower end of the magnet above the top of the coil is measured with a metre rule. The first peak of the emf, εpeak, occurs when the magnet has fallen a further, unknown distance h0 into the coil.
The student assumes that εpeak = kv, where v is the speed of the magnet at that instant and k is a constant, and that the magnet falls freely until the first peak. The graph shows five of the six values of εpeak² against h.
h / cm
εpeak / mV
εpeak² / mV²
5.0
17.2
296
10.0
22.2
493
15.0
26.3
692
20.0
29.9
30.0
35.9
1289
40.0
40.8
1665
Graph drawn to scale. The vertical dashed line marks h = 0.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient, including its unit.
(2)
(b)
Interpreting the line.
(i)
Explain why the line is expected to cut the h axis at h = −h0. Hence use the graph to determine h0.
(2)
(ii)
Determine k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
29.9² = 894 mV²
✓ 1
Accept 890–895.
Part (a)(ii)
Point plotted correctly and a straight line of best fit drawn and extended to cross the h axis
✓ 1
Allow ECF from (a)(i).
Gradient = 39 mV² cm−1 (3.9 × 103 mV² m−1)
✓ 1
Accept 37–41 mV² cm⁻¹. Unit required for this mark.
Part (b)(i)
The magnet falls freely through h + h0, so v² = 2g(h + h0) and εpeak² = 2gk²(h + h0), which is zero when h = −h0
✓ 1
Award for the kinematic relation leading to the linear form.
Line extrapolated to the h axis: intercept ≈ −2.6 cm, so h0 ≈ 2.6 cm
✓ 1
Accept 2.0–3.2 cm. Allow ECF from the candidate's line in (a)(ii).
Part (b)(ii)
Gradient = 2gk², with the gradient in mV² m−1: k² = 3.9 × 103/(2 × 9.81)
✓ 1
Allow ECF from (a)(ii). A gradient left per cm (giving k ≈ 1.4 mV s m⁻¹) scores this mark only.
k = 14 mV s m−1 (1.4 × 10−2 V s m−1)
✓ 1
Accept 13.7–14.5 mV s m⁻¹; unit not required.
Answers: (a)(i) 894 mV² · (a)(ii) 39 mV² cm−1 · (b)(i) h0 ≈ 2.6 cm · (b)(ii) k ≈ 14 mV s m−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — Faraday's law of induction ε = −NΔΦ/Δt (relative motion between a magnet and a coil); A.1 — the equations of motion for uniformly accelerated motion; Tools 3 — linearising a relationship, gradient with its unit, extrapolation to an intercept Command term: Determine
28D-1B-09
Peak emf of a rotating coil·D.4 Induction (HL)
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A small generator consists of a rectangular coil of 500 turns and area 2.0 × 10−3 m², rotating at a constant rate in a uniform magnetic field of flux density 0.050 T. The coil has a resistance r and is connected to a 30 Ω resistor. A data logger connected across the resistor records the potential difference V across it; the trace is shown.
The peak emf of a coil of N turns and area A rotating with frequency f in a field B is ε0 = 2πfNBA.
The potential difference across the 30 Ω resistor against time.
(a)
The emf of the coil.
(i)
Determine the frequency of rotation of the coil, using as much of the trace as possible.
(2)
(ii)
Calculate the peak emf ε0 of the coil.
(2)
(b)
The pd across the resistor.
(i)
Determine the ratio (peak value of V)/ε0.
(1)
(ii)
Explain why this ratio is less than 1 and hence determine r.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Time for four complete cycles read from the trace, e.g. between the upward zero crossings at t = 0 and t ≈ 182 ms
✓ 1
Accept 180–184 ms. Using a single cycle scores [1 max] for the part.
f = 4/0.182 s = 22 Hz
✓ 1
Accept 21.7–22.2 Hz.
Part (a)(ii)
ε0 = 2π × 22.0 × 500 × 0.050 × 2.0 × 10−3
✓ 1
Allow ECF from (a)(i).
ε0 = 6.9 V
✓ 1
Accept 6.8–7.0 V.
Part (b)(i)
Peak V from the trace = 5.8 V, so the ratio = 5.8/6.9 = 0.84
✓ 1
Accept 5.7–5.9 V and 0.82–0.86. Allow ECF from (a)(ii).
Part (b)(ii)
The current in the coil produces a potential difference across the resistance of the coil itself, so the pd across the resistor is less than the emf: ε0 = I0(R + r), so peak V = ε0 × 30/(30 + r)
✓ 1
Award for the idea of a pd across the coil's own resistance linked to ε = I(R + r).
r = 30 × (6.9/5.8 − 1) = 5.7 Ω ≈ 6 Ω
✓ 1
Allow ECF from (b)(i). Accept 5–7 Ω.
Answers: (a)(i) 22 Hz · (a)(ii) 6.9 V · (b)(i) 0.84 · (b)(ii) ≈ 6 Ω (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — a uniform magnetic field induces a sinusoidal varying emf in a coil rotating within it; the effect on induced emf caused by changing the frequency of rotation; B.5 — emf and internal resistance ε = I(R + r); Tools 3 — determining a frequency from many cycles (using the full range of the data); Inquiry 3 — comparing a result with theory and explaining the discrepancy Command term: Determine
29D-1B-19
Flux and charge in a rotated coil·D.4 Induction (HL)
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
A flat rectangular coil of 200 turns measures 5.0 cm × 4.0 cm. It is placed in a uniform magnetic field of flux density B with the normal to its plane at an angle θ to the field, and connected to a charge meter; the total resistance of the circuit is 25 Ω.
The coil is quickly turned until its plane is parallel to the field, and the total charge Q that flows through the meter is recorded. This is repeated for different starting angles θ.
θ / °
cos θ
Q / μC
0
1.000
813
20
0.940
767
40
0.766
627
50
0.643
530
60
0.500
416
70
0.342
285
80
0.174
150
Graph drawn to scale.
(a)
Theory.
(i)
Show that the charge is Q = NBA cos θ/R, where A is the area of the coil and R the resistance of the circuit.
(2)
(b)
Analysis.
(i)
Draw the line of best fit and, using the full range of the data, determine its gradient.
(2)
(ii)
Hence determine B.
(1)
(iii)
The line does not pass through the origin. Use it to determine the zero error of the charge meter, and the percentage error that this causes in the reading for θ = 80°.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Average emf ε = NΔΦ/Δt, so the average current is I = NΔΦ/(RΔt)
✓ 1
Q = IΔt = NΔΦ/R, with ΔΦ = BA cos θ − 0, giving Q = NBA cos θ/R
✓ 1
Part (b)(i)
Straight line of best fit; large triangle, e.g. (0.174, 150) to (1.000, 813)
✓ 1
Gradient = 8.0 × 102 μC (8.0 × 10−4 C)
✓ 1
Accept 780–820 μC.
Part (b)(ii)
B = gradient × R/(NA) = 8.0 × 10−4 × 25/(200 × 2.0 × 10−3) = 0.050 T
✓ 1
Allow ECF from (b)(i). Accept 0.049–0.051 T.
Part (b)(iii)
Intercept on the Q axis ≈ +12 μC: the meter reads about 12 μC too high (a zero error)
✓ 1
Accept 8–16 μC. Allow ECF from the candidate's line.
12/150 × 100 ≈ 8 % for θ = 80° (compared with about 1.5 % at 0°); the zero error does not change the gradient, so B in (b)(ii) is unaffected
✓ 1
Allow ECF from the intercept.
Answers: (b)(i) 8.0 × 102 μC · (b)(ii) 0.050 T · (b)(iii) ≈ 12 μC; ≈ 8 % (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — magnetic flux Φ = BA cos θ; Faraday's law of induction ε = −NΔΦ/Δt; B.5 — current as the rate of flow of charge I = Δq/Δt; Tools 3 — gradient using the full range, a zero error read from an intercept Command term: Determine
30D-1B-20
Lenz's law and energy·D.4 Induction (HL)
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
A strong bar magnet hangs from a spring of spring constant 12 N m−1 so that one pole is just above a flat coil fixed to the bench. The magnet oscillates vertically. With the ends of the coil unconnected the amplitude decreases slowly; the ends are then connected to a resistor and the amplitude A is measured every 5 s.
With the coil unconnected, a graph of ln(A/mm) against t has a gradient of −0.017 s−1 (dashed line). Assume that the damping forces from the air and from the induced current are each proportional to the speed of the magnet, so that their effects on the decay constant λ of the amplitude add.
t / s
A / mm
ln(A/mm)
0
39.9
3.69
5
28.7
3.36
10
20.8
3.03
15
14.7
2.69
20
10.5
2.35
25
7.6
2.03
30
5.4
1.69
Points: coil connected to the resistor. Dashed line: coil unconnected.
(a)
The decay.
(i)
Test the hypothesis that, with the coil connected, the amplitude decreases exponentially. Use at least three equal time intervals.
(2)
(ii)
Using the full range of the data, determine λ with the coil connected, including its unit.
(2)
(b)
Energy.
(i)
Calculate the energy lost by the oscillating system in the first 10 s with the coil connected.
(2)
(ii)
Estimate the energy transferred by the induced current in the first 10 s.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ratios of amplitudes over equal 10 s intervals: 20.8/39.9 = 0.521, 10.5/20.8 = 0.505, 5.4/10.5 = 0.514
✓ 1
Any three equal intervals are acceptable (e.g. 5 s: 0.72, 0.72, 0.71).
The ratio is constant to within about 2 %, so the decrease is exponential
✓ 1
The conclusion must be consistent with the candidate's ratios.
Part (a)(ii)
Gradient of the ln graph from the end points (or a large triangle): (1.69 − 3.69)/(30 − 0)
✓ 1
ALT: λ = −ln(ratio)/10 s averaged over the intervals in (a)(i). A value from a single 5 s interval scores [1 max].
λ = 0.067 s−1
✓ 1
Accept 0.064–0.070 s⁻¹. Unit required.
Part (b)(i)
E = ½kA²: at t = 0, ½ × 12 × 0.0399² = 9.6 × 10−3 J; at t = 10 s, ½ × 12 × 0.0208² = 2.6 × 10−3 J
✓ 1
Energy lost = 7.0 × 10−3 J
✓ 1
Accept 6.9–7.1 × 10⁻³ J.
Part (b)(ii)
Fraction = (0.067 − 0.017)/0.067 = 0.75, so energy ≈ 0.75 × 7.0 × 10−3 = 5.2 × 10−3 J
✓ 1
Allow ECF from (a)(ii) and (b)(i). Accept 5.0–5.4 × 10⁻³ J.
Answers: (a)(i) ≈ 0.51 per 10 s (constant) · (a)(ii) 0.067 s−1 · (b)(i) 7.0 × 10−3 J · (b)(ii) 5.2 × 10−3 J (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — relative motion between a conductor and a magnetic field (a magnet oscillating on a spring above a conducting coil); the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation; C.1/A.3 — energy of an oscillator, elastic potential energy ½k(Δx)²; Tools 3 — testing an exponential hypothesis (constant ratio), linearising with logarithms Command term: Determine
31D-1B-21
Peak emf and frequency of rotation·D.4 Induction (HL)
Paper 1BEasy7 marks
Data-based question5 steps to full marksDetermine
A student investigates how the peak emf ε0 of a small generator depends on the frequency of rotation f of its coil. The coil has 250 turns and an area of 1.20 × 10−3 m², and it rotates in a uniform magnetic field of flux density B. The peak emf is read from a data-logger trace, and each value has an absolute uncertainty of ±0.10 V. The frequency is measured with a stroboscope; its uncertainty is negligible.
Theory gives ε0 = 2πfNBA, where N is the number of turns and A the area of the coil. The graph shows the data with their error bars.
f / Hz
5
10
15
20
25
30
ε0 / V
0.80
1.44
2.32
2.98
3.84
4.46
Graph drawn to scale. Error bars show ±0.10 V.
(a)
The gradient.
(i)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(ii)
Draw the lines of maximum and minimum gradient consistent with the error bars, and hence determine the absolute uncertainty in the gradient.
(2)
(b)
The magnetic field.
(i)
Determine B.
(2)
(ii)
Determine the absolute uncertainty in B.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A single straight line passing through all the error bars
✓ 1
Gradient = 0.149 V Hz−1
✓ 1
Accept 0.145–0.155. The unit (V Hz⁻¹ or V s) is required for this mark.
Part (a)(ii)
Both lines drawn through all the error bars; maximum gradient ≈ 0.154 V Hz−1 and minimum gradient ≈ 0.147 V Hz−1
✓ 1
Accept maximum 0.152–0.158 and minimum 0.140–0.147. Lines through the first and last error bars only, which miss intermediate bars, score 0 for this mark.
Uncertainty = (maximum − minimum)/2 ≈ 0.004 V Hz−1
✓ 1
Allow ECF from the candidate's maximum and minimum gradients. Accept 0.003–0.008 V Hz⁻¹.
Part (b)(i)
Gradient = 2πNBA, so B = gradient/(2πNA)
✓ 1
B = 0.149/(2π × 250 × 1.20 × 10−3) = 0.079 T
✓ 1
Allow ECF from (a)(i). Accept 0.077–0.082 T.
Part (b)(ii)
ΔB/B = Δ(gradient)/gradient = 0.004/0.149 ≈ 3 %, so ΔB ≈ 0.002 T
✓ 1
Allow ECF from (a)(ii) and (b)(i). Accept 0.002–0.005 T. N and A have negligible uncertainty.
Answers: (a)(i) 0.149 V Hz−1 · (a)(ii) ±0.004 V Hz−1 · (b)(i) 0.079 T · (b)(ii) ±0.002 T (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces a sinusoidally varying emf in a coil rotating within it; the effect on induced emf caused by changing the frequency of rotation; Tools 3 — gradient with its unit, uncertainty bars, lines of maximum and minimum gradient, the uncertainty in a gradient Command term: Determine
32D-1B-33
Flux linkage from emf pulses·D.4 Induction (HL)
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine
A flat rectangular coil of 50 turns is mounted on a trolley. The coil is 4.0 cm long in the direction of motion and 3.0 cm wide. It starts completely between the large, flat poles of a magnet, where the field is uniform and perpendicular to the plane of the coil. The trolley is then pulled along a straight track so that the coil leaves the field at a constant speed. A data logger connected to the coil records a pulse of constant emf ε and its duration t. The experiment is repeated at different speeds.
The table shows the data. The graph shows five of the six values of ε against 1/t.
t / ms
ε / mV
(1/t) / s−1
200
36.3
5.0
133
54.3
7.5
101
72.4
74
97.7
13.5
53
133.6
18.9
40
179.5
25.0
Graph drawn to scale. The point for t = 101 ms is not plotted.
(a)
Theory.
(i)
Show that ε = NΦ/t, where NΦ is the flux linkage of the coil when it is completely in the field.
(2)
(b)
Analysis.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient, including its unit.
(2)
(iii)
Hence determine the magnetic flux density between the poles.
(1)
(iv)
The uncertainty in each value of t is ±1 ms. Identify the data point with the largest percentage uncertainty in 1/t and calculate that uncertainty.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
During the pulse one side is in the field and the other is outside: the flux linkage falls from NΦ to zero
✓ 1
ALT: ε = NBbv with v = a/t (3.0 cm side cutting the field lines).
At constant speed the rate of change is constant, so ε = Δ(NΦ)/Δt = NΦ/t
✓ 1
ALT: ε = NBab/t = NΦ/t since ab = A.
Part (b)(i)
1/0.101 = 9.9 s−1
✓ 1
Accept 9.9.
Part (b)(ii)
Point plotted correctly; straight line of best fit through (or very close to) the origin
✓ 1
Allow ECF from (b)(i).
Gradient = 7.1 mV s = 7.1 × 10−3 Wb (V s)
✓ 1
Accept 6.9–7.3 mV s. A correct unit (mV s, V s or Wb) is required for this mark.
Part (b)(iii)
B = gradient/(NA) = 7.1 × 10−3/(50 × 0.040 × 0.030) = 0.12 T
✓ 1
Allow ECF from (b)(ii). Accept 0.115–0.122 T. A gradient left in mV s gives 118 T: award 0.
Part (b)(iv)
The shortest time, t = 40 ms: 1/40 × 100 = 2.5 % (the percentage uncertainty in 1/t equals that in t)
✓ 1
Both the point and the value are needed.
Answers: (b)(i) 9.9 s−1 · (b)(ii) 7.1 mV s · (b)(iii) 0.12 T · (b)(iv) t = 40 ms, 2.5 % (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; (guidance) rectangular coils moving in and out of fields; Tools 3 — linearising a relationship, gradient with its unit, percentage uncertainty Command term: Determine
33D-1B-34
A magnet falling through copper tubes·D.4 Induction (HL)
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A small cylindrical magnet of mass 20 g is dropped down the axis of vertical copper tubes that have the same internal diameter but different wall thicknesses w. After a short distance the magnet falls at a constant speed v. A student uses a hand-held stopwatch to time the fall between two marks 0.800 m apart near the bottom of each tube. Each time t in the table is the mean of three trials; the uncertainty in each time is ±0.2 s.
The student’s hypothesis is that v is inversely proportional to w.
w / mm
t / s
v / m s−1
0.50
3.3
0.242
0.75
5.0
0.160
1.00
6.8
0.118
1.50
9.9
2.00
13.0
0.062
3.00
20.0
0.040
(a)
(i)
Explain, with reference to Lenz’s law, why the magnet reaches a constant speed.
(2)
(b)
(i)
Complete the table.
(1)
(ii)
Test the hypothesis, using at least three data points.
(2)
(iii)
Determine the absolute uncertainty in v for the thinnest tube.
(1)
(iv)
Calculate the rate at which thermal energy is transferred in the tube of thickness 1.00 mm while the magnet falls at constant speed.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The moving magnet changes the flux through rings of the copper tube, so emfs and (eddy) currents are induced in the copper; by Lenz’s law their magnetic field exerts an upward force on the magnet that opposes its motion
✓ 1
Do not accept "the magnet is attracted to the copper"; copper is not magnetic.
The faster the magnet moves, the faster the flux changes and the larger the current and the opposing force; the speed stops increasing when this force equals the weight of the magnet (resultant force zero)
✓ 1
Part (b)(i)
v = 0.800/9.9 = 0.081 m s−1
✓ 1
Accept 0.081.
Part (b)(ii)
wv calculated for at least three tubes, e.g. 0.50 × 0.242 = 0.121, 1.00 × 0.118 = 0.118, 3.00 × 0.040 = 0.120 (mm m s−1)
✓ 1
Any three points, preferably spread across the range. ALT: ratios of speeds compared with inverse ratios of thicknesses.
The product is constant to within about 3 %, so the data support the hypothesis
✓ 1
The conclusion must be consistent with the candidate’s values.
Part (b)(iii)
Δv/v = Δt/t = 0.2/3.3 = 6 %, so Δv = 0.06 × 0.242 = ±0.015 m s−1
✓ 1
Accept ±0.01–0.02 m s⁻¹. The distance has negligible uncertainty.
Part (b)(iv)
At constant speed the gravitational potential energy lost per second is transferred to the tube: P = mgv = 0.020 × 9.81 × 0.118 = 2.3 × 10−2 W
✓ 1
Accept 0.023 W.
Answers: (b)(i) 0.081 m s−1 · (b)(iii) ±0.015 m s−1 · (b)(iv) 2.3 × 10−2 W (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; (guidance) relative motion between a conductor and a magnetic field; A.2 — terminal speed; A.3 — power P = Fv; Tools 3 — testing a proportional relationship, absolute and percentage uncertainty Command term: Determine
34D-1B-35
Terminal speed of a rod pulled along rails·D.4 Induction (HL)
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
A copper rod rests across two horizontal rails 0.20 m apart in a uniform vertical magnetic field of flux density B. The rails are joined at one end by a resistor; the total resistance of the circuit is 0.040 Ω. A thread tied to the middle of the rod passes over a pulley and supports a hanging mass M. When the mass is released the rod quickly reaches a terminal speed v, which is measured with a motion sensor. A constant frictional force f acts on the rod whenever it moves.
The table and the graph show the results. The uncertainty in each value of v is ±0.010 m s−1.
M / g
5.0
10.0
15.0
20.0
25.0
30.0
v / m s−1
0.028
0.109
0.182
0.258
0.332
0.412
Graph drawn to scale. Error bars show ±0.010 m s−1.
(a)
(i)
Show that the terminal speed is v = (R/B2L2)(Mg − f), where L is the distance between the rails and R the resistance of the circuit.
(2)
(b)
(i)
Draw the line of best fit and determine its gradient, including its unit.
(2)
(ii)
Determine B.
(2)
(iii)
Use the graph to determine f.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ε = BvL, I = BvL/R, so the magnetic force on the rod is BIL = B2L2v/R
✓ 1
At the terminal speed nothing accelerates, so the tension is Mg and Mg = f + B2L2v/R; rearranging gives the result
✓ 1
Part (b)(i)
A single straight line through all the error bars
✓ 1
Gradient = 0.0152 m s−1 g−1 = 15.2 m s−1 kg−1
✓ 1
Accept 14.6–15.8 m s⁻¹ kg⁻¹ (0.0146–0.0158 m s⁻¹ g⁻¹). A correct unit is required for this mark.
Part (b)(ii)
Gradient = gR/(B2L2), so B = √(gR/(gradient × L2))
✓ 1
Allow ECF from (a)(i) and (b)(i). The gradient must be per kg.
B = √(9.81 × 0.040/(15.2 × 0.202)) = 0.80 T
✓ 1
Accept 0.78–0.82 T.
Part (b)(iii)
The line meets the M axis at M ≈ 3.0 g, where Mg = f: f = 3.0 × 10−3 × 9.81 = 0.030 N
✓ 1
Allow ECF from the candidate’s line. Accept 2.6–3.4 g, giving 0.026–0.033 N.
Answers: (b)(i) 15.2 m s−1 kg−1 · (b)(ii) 0.80 T · (b)(iii) 0.030 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; A.2 — Newton’s laws, terminal speed; Tools 3 — gradient and intercept of a linear graph with units Command term: Determine
35D-2-04
A rectangular coil rotating in a uniform field·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A small wind-turbine generator contains a rectangular coil of 150 turns measuring 6.0 cm × 4.0 cm. The coil rotates at a constant rate in a uniform magnetic field about an axis through its centre that is parallel to its 6.0 cm sides and perpendicular to the field. The graph shows how the emf ε across the ends of the coil varies with time t.
Graph drawn to scale.
(a)
The emf.
(i)
Use the graph to determine the frequency of rotation of the coil.
(1)
(ii)
By considering the speed of the 6.0 cm sides of the coil, deduce that the peak emf is ε0 = NBAω, where A is the area of the coil and ω is its angular velocity.
(3)
(iii)
Use the graph and your answer to (a)(ii) to determine the magnetic flux density.
(2)
(b)
Orientation and speed.
(i)
State the orientation of the plane of the coil relative to the magnetic field at t = 0, and explain your answer.
(1)
(ii)
The rate of rotation is halved. Describe how the graph changes.
(2)
(c)
The ends of the coil are connected to a lamp of resistance 12 Ω. The resistance of the coil is negligible.
(i)
Calculate the maximum instantaneous power delivered to the lamp.
(2)
(ii)
Explain why the turbine must now exert a larger torque on the coil to keep the rate of rotation constant.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Period = 50 ms, so f = 1/0.050 = 20 Hz
✓ 1
Part (a)(ii)
Each 6.0 cm side (length L) moves in a circle of radius b/2, where b = 4.0 cm, with speed v = ωb/2
✓ 1
When the sides move perpendicular to the field each has emf BvL, and the two emfs add round the turn: 2B(ωb/2)L = BωLb per turn
✓ 1
Lb = A and the N turns are in series, so ε0 = NBAω
✓ 1
Part (a)(iii)
ε0 = 5.4 V from the graph and ω = 2π × 20 = 126 rad s−1
✓ 1
Allow ECF from (a)(i). Accept a peak reading of 5.3–5.5 V.
B = ε0/(NAω) = 5.4/(150 × 2.4 × 10−3 × 126) = 0.12 T
✓ 1
Accept 0.11–0.13 T.
Part (b)(i)
Perpendicular to the field: the flux linkage is then a maximum and its rate of change, and so the emf, is zero (the sides move parallel to the field)
✓ 1
Both the orientation and the reason are needed.
Part (b)(ii)
The peak emf halves, to about 2.7 V
✓ 1
Allow ECF from the peak value read in (a)(iii).
The period doubles, to 100 ms (still a sinusoidal emf)
✓ 1
Part (c)(i)
I0 = ε0/R = 5.4/12 = 0.45 A
✓ 1
Allow ECF from the peak emf read in (a)(iii).
Pmax = ε0I0 = 5.4 × 0.45 = 2.4 W
✓ 1
Accept 2.3–2.6 W.
Part (c)(ii)
A current now flows in the coil; by Lenz's law the magnetic forces on the current-carrying sides oppose the rotation
✓ 1
The work done by the turbine against this opposing torque supplies the energy transferred in the lamp (energy is conserved)
✓ 1
Do not accept "more energy is needed" without reference to the opposing force or torque.
Answers: (a)(i) 20 Hz · (a)(iii) 0.12 T · (b)(i) perpendicular to the field · (b)(ii) 2.7 V peak, period 100 ms · (c)(i) 2.4 W (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — a uniform magnetic field induces a sinusoidally varying emf in a coil rotating within it; the effect on induced emf of changing the frequency of rotation; ε = BvL for a straight conductor; Lenz's law as a consequence of energy conservation; B.5 — electrical power P = IVCommand term: Determine
36D-2-10
Motional emf and Lenz's law·D.4 Induction (HL)
Paper 2Hard13 marks
Short answer & extended response9 steps to full marksDetermine
A rectangular loop of copper wire has a mass of 8.0 g and a total resistance of 0.020 Ω. Its horizontal sides are 0.10 m long and its vertical sides 0.25 m long. The loop is released from rest, with its plane vertical, above a deep region of uniform horizontal magnetic field that is perpendicular to the plane of the loop. The upper boundary of the field region is horizontal. The graph shows how the speed v of the loop varies with time t after its release. Air resistance is negligible.
Graph drawn to scale.
(a)
Entering the field.
(i)
Use the graph to determine the height of the lower side of the loop above the field when the loop was released.
(1)
(ii)
Explain why a current is induced in the loop as it enters the field, and state the direction of the magnetic force on the loop.
(2)
(b)
The terminal speed.
(i)
Show that, while only the lower side of the loop is in the field, the magnetic force on the loop is F = B²L²v/R, where L is the length of the lower side.
(2)
(ii)
Use the graph to determine the magnetic flux density.
(2)
(iii)
Determine the acceleration of the loop just after its lower side enters the field.
(2)
(c)
Energy and the final stage.
(i)
For the loop moving at its terminal speed, show that the rate of thermal energy transfer in the loop is equal to the rate of decrease of its gravitational potential energy.
(2)
(ii)
Explain the shape of the graph after t = 0.46 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Area under the first straight section = ½ × 0.143 × 1.40 = 0.10 m
✓ 1
Accept 0.09–0.11 m. ALT: v²/(2g) with v = 1.40 m s⁻¹.
Part (a)(ii)
The lower side cuts the field lines (the flux linkage of the loop increases), so an emf is induced and a current flows in the closed loop
✓ 1
By Lenz's law the force on the current in the lower side opposes the motion: it is upwards
✓ 1
Part (b)(i)
ε = BvL and I = ε/R = BvL/R
✓ 1
F = BIL = B²L²v/R (no force on the vertical sides: the forces on them cancel)
✓ 1
Part (b)(ii)
Terminal speed from the graph vt = 0.63 m s−1, where mg = B²L²vt/R
✓ 1
Allow ECF from (b)(i). Accept 0.60–0.66 m s⁻¹.
B = √(mgR/(L²vt)) = √(0.0080 × 9.81 × 0.020/(0.10² × 0.63)) = 0.50 T
✓ 1
Accept 0.49–0.51 T.
Part (b)(iii)
The magnetic force is proportional to v, so at entry it is mg × (1.40/0.63) = 2.2mg
✓ 1
Allow ECF from (b)(ii). ALT: F = B²L²v/R = 0.50² × 0.10² × 1.40/0.020.
a = g(1 − 1.40/0.63) = −12 m s−2: a deceleration of about 12 m s−2 (acceleration upwards)
✓ 1
Accept 11–13 m s⁻² upwards.
Part (c)(i)
I = BLv/R = 0.50 × 0.10 × 0.63/0.020 = 1.6 A, so I²R = 1.6² × 0.020 = 0.049 W
✓ 1
Allow ECF from (b)(ii).
mgv = 0.0080 × 9.81 × 0.63 = 0.049 W: equal
✓ 1
Part (c)(ii)
The whole loop is inside the field: the flux linkage is constant, so there is no induced emf, current or magnetic force
✓ 1
The loop accelerates at g: a straight line with the same gradient as the first section
✓ 1
Answers: (a)(i) 0.10 m · (b)(ii) 0.50 T · (b)(iii) 12 m s−2 upwards · (c)(i) 0.049 W (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — the emf induced in a straight conductor moving perpendicularly to a uniform field ε = BvL; rectangular coils moving in and out of fields; Lenz's law as a consequence of energy conservation; D.3 — F = BIL sin θ; A.1 — interpretation of a velocity–time graph; A.2 — terminal speed; A.3 — power and energy transfer Command term: Determine
37D-2-13
A magnet oscillating above a coil·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A bar magnet of mass 0.20 kg hangs from a spring with its lower pole just above a flat coil fixed to the bench. The coil is connected to a data logger of very high resistance, so that no current flows in the coil. The magnet is pulled up, then released at t = 0. The graph shows how the displacement x of the magnet from its equilibrium position varies with time t. Air resistance is negligible.
Graph drawn to scale. Upward displacements are positive.
(a)
The oscillation.
(i)
Use the graph to show that the spring constant is about 20 N m−1.
(2)
(ii)
Determine the maximum speed of the magnet.
(2)
(b)
The emf induced in the coil.
(i)
Identify the times during the first cycle at which the induced emf is zero, and explain your answer.
(2)
(ii)
State, in radians, the phase difference between the induced emf and the displacement of the magnet, and explain your answer.
(2)
(iii)
The magnet is replaced by one of the same strength with a quarter of the mass, released with the same amplitude. State and explain the effect on the peak emf.
(1)
(c)
The coil is now connected to a resistor.
(i)
Explain why the amplitude of the oscillation decreases.
(2)
(ii)
Suggest one change to the circuit that would make the amplitude decrease more rapidly, and explain your suggestion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two cycles take 1.26 s, so T = 0.63 s
✓ 1
Accept 0.62–0.64 s.
k = 4π²m/T² = 4π² × 0.20/0.63² = 19.9 N m−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
ω = 2π/T = 2π/0.63 = 10 rad s−1
✓ 1
Allow ECF from (a)(i).
vmax = ωx0 = 10 × 0.020 = 0.20 m s−1
✓ 1
Part (b)(i)
t = 0, 0.31 s and 0.63 s, at the maximum displacements
✓ 1
Allow ECF from the period read in (a)(i).
The magnet is momentarily at rest, so the flux linkage of the coil is not changing and ε = −NΔΦ/Δt = 0
✓ 1
Part (b)(ii)
π/2 rad
✓ 1
Unit (rad) required; do not accept 90° alone.
The emf is greatest when the magnet passes through the equilibrium position at its maximum speed (the flux linkage changes fastest) and zero at maximum displacement, a quarter of a cycle later
✓ 1
Part (b)(iii)
T ∝ √m, so the period halves and the maximum speed (2πx0/T) doubles: the peak emf approximately doubles
✓ 1
Allow ECF from (a)(ii).
Part (c)(i)
A current is induced in the coil; by Lenz's law its magnetic field exerts a force on the magnet that opposes its motion
✓ 1
The work done against this force is transferred to thermal energy in the resistor, so the energy of the oscillator decreases
✓ 1
Part (c)(ii)
Use a resistor of smaller resistance (or a coil with more turns)
✓ 1
Do not accept "a larger resistance".
For the same induced emf the current is larger (I = ε/R), so the opposing magnetic force on the magnet is larger and energy is transferred from the oscillator at a greater rate (P = ε²/R)
✓ 1
For more turns: a larger emf and current. The link to a larger opposing force or a greater rate of energy transfer is required.
Answers: (a)(ii) 0.20 m s−1 · (b)(ii) π/2 rad (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — relative motion between a conductor and a magnetic field (a magnet oscillating on a spring above a conducting coil); Faraday's law ε = −NΔΦ/Δt; Lenz's law as a consequence of energy conservation; C.1 — the time period of a mass–spring system; C.1 (HL) — phase angle, v = ωx0; B.5 — electrical power P = V²/RCommand term: Determine
38D-2-16
A magnet falling through a coil·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response9 steps to full marksEstimate
A bar magnet is released from rest, north pole downwards, with its lower end 18 cm above the top of a vertical coil of 500 turns. The coil is connected to a data logger. The graph shows how the induced emf varies with time as the magnet falls through the coil. Air resistance is negligible.
A bar magnet dropped through a coil, and the induced emf recorded against time.
(a)
The shape of the graph.
(i)
Explain why pulse 2 has the opposite sign to pulse 1.
(1)
(ii)
Explain why pulse 2 is taller and narrower than pulse 1.
(2)
(iii)
Show that the areas between each pulse and the time axis are equal in magnitude.
(2)
(b)
Numbers.
(i)
Calculate the speed of the magnet when it reaches the top of the coil.
(1)
(ii)
Pulse 1 has a peak emf of 2.0 V. Pulse 2 occurs when the magnet has fallen a further 12 cm. Assume that the peak emf is proportional to the speed of the magnet. Estimate the peak emf of pulse 2.
(3)
(iii)
Pulse 1 lasts 0.050 s. Estimate the duration of pulse 2.
(1)
(iv)
State one assumption made in your estimates in (b)(ii) and (b)(iii).
(1)
(c)
(i)
The coil is now connected to a resistor and the magnet is dropped again from the same height. State and explain the effect on the motion of the magnet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
As the magnet enters, the flux linkage of the coil increases; as it leaves, the flux linkage decreases, so the rate of change of flux linkage, and the emf, has the opposite sign
✓ 1
Part (a)(ii)
The magnet accelerates, so it is moving faster when it leaves the coil than when it enters
✓ 1
The same change of flux linkage takes place in a shorter time: a larger rate of change (taller pulse) over a shorter interval (narrower pulse)
✓ 1
Part (a)(iii)
ε = −NΔΦ/Δt, so the area under a pulse, ΣεΔt, is equal to the total change in flux linkage NΔΦ
✓ 1
The flux linkage rises from zero to its maximum and then falls back to zero, so both changes have the same magnitude
✓ 1
Part (b)(i)
v = √(2gh) = √(2 × 9.81 × 0.18) = 1.88 m s−1
✓ 1
Part (b)(ii)
v2 = √(1.88² + 2 × 9.81 × 0.12) = 2.43 m s−1
✓ 1
Allow ECF from (b)(i).
Peak emf ∝ speed, so ε2/ε1 = 2.43/1.88 = 1.29
✓ 1
ε2 = 2.0 × 1.29 = 2.6 V
✓ 1
Accept 2.5–2.7 V.
Part (b)(iii)
The same change of flux linkage takes place over the same distance of fall, so Δt2 = 0.050 × 1.88/2.43 = 0.039 s
✓ 1
Allow ECF from (b)(ii). Accept 0.038–0.040 s. (Peak × duration is then about the same for both pulses, as in (a)(iii).)
Part (b)(iv)
The speed is approximately constant during each pulse / the magnet falls freely: no current flows in the coil, so there is no magnetic braking
✓ 1
Part (c)(i)
A current flows; by Lenz's law its magnetic field repels the approaching magnet and attracts the departing magnet: the magnetic force opposes the motion
✓ 1
The acceleration of the magnet is less than g and it leaves the coil more slowly; kinetic energy is transferred to thermal energy in the resistor
✓ 1
Answers: (b)(i) 1.88 m s−1 · (b)(ii) 2.6 V · (b)(iii) 0.039 s (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday's law of induction ε = −NΔΦ/Δt; (guidance) relative motion between a conductor and a magnetic field; that the direction of induced emf is determined by Lenz's law and is a consequence of energy conservation; A.1 — the equations of motion for uniform acceleration Command term: Estimate
39D-2-29
Rectangular coil moving through a field·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response7 steps to full marksDetermine
A speed sensor on a roller-coaster car consists of a flat rectangular coil of 40 turns, 6.0 cm long in the direction of motion and 3.0 cm wide. As the car passes a checkpoint the coil moves horizontally at a constant 12 m s−1 through a region of uniform magnetic field of flux density 0.25 T, directed into the page. The field region is 4.0 cm long in the direction of motion — shorter than the coil — and wider than the coil, as shown.
Top view, not to scale. The coil is shown just before its leading edge enters the field.
(a)
The coil enters the field.
(i)
Calculate the maximum flux linkage of the coil.
(1)
(ii)
Show that the emf induced in the coil while its leading edge is in the field is 3.6 V.
(2)
(b)
The emf pulse.
(i)
After the leading edge leaves the field there is an interval during which no emf is induced, even though part of the coil is still in the field. Explain why.
(1)
(ii)
Taking t = 0 when the leading edge enters the field, describe how the emf varies from t = 0 until the coil has completely left the field. Give the times at which the emf changes.
(3)
(c)
Current and force.
(i)
The coil is connected to a data logger; the total resistance of the circuit is 12 Ω. Calculate the current while the leading edge is in the field.
(1)
(ii)
State and explain, using Lenz's law, whether the current is clockwise or anticlockwise (as seen in the diagram) while the trailing edge is in the field.
(2)
(iii)
Calculate the magnitude of the magnetic force on the coil while the leading edge is in the field, and state its direction.
(2)
(d)
The sensor.
(i)
Outline one way in which the emf signal could be used to measure the speed of the car.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Maximum when the whole 4.0 cm field region is inside the coil: NBA = 40 × 0.25 × (0.030 × 0.040) = 1.2 × 10−2 Wb
✓ 1
Part (a)(ii)
Only the 3.0 cm leading edge cuts field lines: ε = NBvL
✓ 1
Or: flux linkage rises by 1.2 × 10−2 Wb in 0.040/12 = 3.33 ms.
ε = 40 × 0.25 × 12 × 0.030 = 3.6 V
✓ 1
Part (b)(i)
The whole field region is then inside the coil (both edges are outside the field), so the flux linkage is constant and, by Faraday's law, no emf is induced
✓ 1
Part (b)(ii)
Constant 3.6 V from 0 to 3.3 ms (leading edge crossing the 4.0 cm field: 0.040/12 s)
✓ 1
Zero from 3.3 ms to 5.0 ms (until the trailing edge reaches the field: 0.060/12 s)
✓ 1
Constant 3.6 V of opposite sign from 5.0 ms to 8.3 ms, then zero
✓ 1
Allow ECF for the magnitude from (a)(ii).
Part (c)(i)
I = 3.6/12 = 0.30 A
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
While the trailing edge crosses the field, the flux into the page through the coil is decreasing
✓ 1
The induced current opposes the decrease, producing a field into the page inside the coil: clockwise
✓ 1
Part (c)(iii)
F = NBIL = 40 × 0.25 × 0.30 × 0.030 = 0.090 N
✓ 1
Allow ECF from (c)(i).
Opposite to the velocity (to the left), opposing the motion as required by Lenz's law
✓ 1
Part (d)(i)
The duration of either pulse equals the field length divided by the speed (4.0 cm ÷ v), or the height of the pulse is proportional to the speed; so v can be calculated from the pulse duration or height
✓ 1
Answers: (a)(i) 1.2 × 10−2 Wb · (a)(ii) 3.6 V · (c)(i) 0.30 A · (c)(ii) clockwise · (c)(iii) 0.090 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — Faraday's law ε = −NΔΦ/Δt; the emf induced in a straight conductor moving perpendicularly to a uniform field ε = BvL; rectangular coils moving in and out of fields; Lenz's law; D.3 — the force on a current-carrying conductor F = BIL sin θCommand term: Determine
40D-2-30
Motional emf and Lenz's law·D.4 Induction (HL)
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksDetermine
A model electromagnetic launcher consists of two long horizontal rails 0.25 m apart, joined at one end through a switch to a battery of emf 6.0 V. A copper rod of mass 0.050 kg rests across the rails in a uniform vertical magnetic field of flux density 0.80 T. The total resistance of the circuit is 1.5 Ω and remains constant. A constant frictional force of 0.12 N acts on the rod whenever it moves. The switch is closed and the rod accelerates along the rails, as shown.
Top view, not to scale. The switch in series with the battery is not shown. The long thin plate of the cell is the positive terminal.
(a)
Starting.
(i)
Calculate the initial current and the initial magnetic force on the rod.
(2)
(ii)
Calculate the initial acceleration of the rod.
(1)
(b)
Speeding up.
(i)
Explain why the current decreases as the speed of the rod increases.
(2)
(ii)
Show that the maximum (terminal) speed of the rod is about 26 m s−1.
(2)
(c)
Energy at the terminal speed.
(i)
Determine the power delivered by the battery, the rate of work done against friction and the rate of thermal energy transfer in the resistance, and show that they are consistent.
(3)
(ii)
Explain why the rod never reaches the speed at which the induced emf equals the battery emf.
(1)
(d)
A stronger field.
(i)
The magnetic flux density is doubled. Determine the new terminal speed and comment on the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
I = 6.0/1.5 = 4.0 A
✓ 1
F = BIL = 0.80 × 4.0 × 0.25 = 0.80 N
✓ 1
Part (a)(ii)
a = (0.80 − 0.12)/0.050 = 14 m s−2
✓ 1
Accept 13.6. Award 0 for 16 m s−2 (friction ignored). Allow ECF from (a)(i).
Part (b)(i)
The moving rod cuts the field lines, so an emf BvL is induced in it, increasing with speed
✓ 1
By Lenz's law this emf opposes the current that drives the motion, so the net emf (6.0 − BvL) and hence I = (6.0 − BvL)/1.5 decrease
✓ 1
Part (b)(ii)
At terminal speed the magnetic force equals friction: BIL = 0.12, so I = 0.12/(0.80 × 0.25) = 0.60 A
✓ 1
BvL = 6.0 − 0.60 × 1.5 = 5.1 V, so v = 5.1/(0.80 × 0.25) = 25.5 m s−1 ≈ 26 m s−1
Allow ECF from the current and terminal speed in (b)(ii).
3.06 + 0.54 = 3.6 W: the battery output equals the total rate of energy transfer (energy conservation)
✓ 1
Part (c)(ii)
A current (0.60 A) is always needed to provide a magnetic force that balances friction, so the induced emf must stay below 6.0 V by IR = 0.9 V; only with no friction would it approach 6.0 V (at 30 m s−1, with zero current)
✓ 1
Part (d)(i)
I = 0.12/(1.6 × 0.25) = 0.30 A; v = (6.0 − 0.30 × 1.5)/(1.6 × 0.25) = 13.9 m s−1
✓ 1
Accept 13.8–13.9.
The terminal speed roughly halves: a stronger field produces the opposing (back) emf at a lower speed, although the initial force is doubled
✓ 1
Answers: (a)(i) 4.0 A, 0.80 N · (a)(ii) 14 m s−2 · (b)(ii) 25.5 m s−1 · (c)(i) 3.6 W = 3.06 W + 0.54 W · (d)(i) 13.9 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — the emf induced in a straight conductor moving perpendicularly to a uniform field ε = BvL; Lenz's law as a consequence of energy conservation; D.3 — the force on a current-carrying conductor F = BIL sin θ; B.5 — electrical power Command term: Determine
41D-2-31
Lenz's law·D.4 Induction (HL)
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksExplain
A vertical solenoid has an iron core that extends well above its top. An aluminium ring of mass 5.0 g rests on the top of the solenoid, around the core. The solenoid is connected through a switch to a d.c. supply. When the switch is closed the ring jumps vertically upwards to a height of 0.45 m, then falls back and comes to rest on the solenoid, although the current in the solenoid remains on. The graph shows how the current I in the solenoid varies with time t after the switch is closed.
Upper: the apparatus (not to scale; the ring is free to slide up the core). Lower: the current in the solenoid after the switch is closed (graph drawn to scale).
(a)
The jump.
(i)
Explain why the ring jumps upwards when the switch is closed.
(3)
(ii)
Explain why the ring falls back and stays at rest although the solenoid current is still on.
(1)
(b)
The current in the solenoid.
(i)
Explain, in terms of self-induction, why the current takes about 200 ms to reach its final value.
(2)
(ii)
Assume that the current in the ring is proportional to the rate of change of the solenoid current, and that the magnetic field at the ring is proportional to the solenoid current. Use the graph to explain why the upward force on the ring is zero at t = 0 and becomes very small after about 200 ms.
(2)
(c)
Numbers.
(i)
Calculate the speed with which the ring leaves the solenoid. Ignore air resistance.
(2)
(ii)
The ring reaches this speed in about 40 ms. Estimate the average magnetic force on the ring during this time.
(2)
(iii)
The original ring is cooled in liquid nitrogen, which greatly reduces its resistance, and the experiment is repeated. Suggest and explain the effect on the height reached by the ring.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
As the solenoid current grows, the magnetic flux through the ring increases; by Faraday's law an emf is induced in the ring, which drives a current round it (the ring is a closed conductor)
✓ 1
By Lenz's law the ring current opposes the increase in flux, so it circulates in the opposite sense to the solenoid current
✓ 1
Opposite currents repel (the ring acts as a magnet with a like pole facing the solenoid), so there is an upward force on the ring larger than its weight
✓ 1
Part (a)(ii)
Once the current is steady the flux through the ring is constant, so no emf or current is induced and there is no magnetic force; the ring falls under gravity
✓ 1
Part (b)(i)
The growing current produces a growing magnetic flux through the solenoid's own turns, which induces an emf in the solenoid
✓ 1
By Lenz's law this self-induced emf opposes the growth of the current, so the current rises gradually
✓ 1
Part (b)(ii)
At t = 0 the gradient of the graph (and so the ring current) is a maximum but the solenoid current, and hence the field at the ring, is zero: no force
✓ 1
After about 200 ms the field is large but the gradient is almost zero, so there is almost no ring current and no force; the force is greatest at an intermediate time
✓ 1
Part (c)(i)
½mv² = mgh, so v = √(2 × 9.81 × 0.45)
✓ 1
v = 3.0 m s−1
✓ 1
Accept 2.97 m s⁻¹.
Part (c)(ii)
Average resultant force = Δp/Δt = 5.0 × 10−3 × 2.97/0.040 = 0.37 N
✓ 1
Allow ECF from (c)(i).
Magnetic force = resultant force + weight = 0.37 + 0.049 = 0.42 N
✓ 1
Award [1 max] for 0.37 N (weight not added).
Part (c)(iii)
The induced emf is the same (the same rate of change of flux), but the smaller resistance gives a larger induced current in the ring
✓ 1
The magnetic force on the ring (which depends on the ring current and the field) is larger, so the impulse and the launch speed are larger and the ring rises higher
✓ 1
Do not accept "the ring is lighter when cold".
Answers: (c)(i) 3.0 m s−1 · (c)(ii) 0.42 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — Faraday's law ε = −NΔΦ/Δt; Lenz's law as a consequence of energy conservation; a qualitative treatment of self-induction; production of an induced emf by time-varying magnetic fields; A.2 — impulse and the change in momentum J = FΔt = Δp; A.3 — conservation of energy Command term: Explain
42D-2-36
Electromagnetic flowmeter·D.4 Induction (HL)
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksDetermine
An electromagnetic flowmeter measures the flow of seawater along a horizontal plastic pipe of internal diameter d = 0.050 m. The water flows due east. Coils outside the pipe produce a uniform magnetic field of flux density B, directed vertically downwards, across the pipe. Two small metal electrodes are set in the wall of the pipe at the ends of a horizontal diameter, one on the north side and one on the south side.
Seawater contains positive and negative ions. Assume that the water moves with the same speed v across the whole cross-section of the pipe.
Graph drawn to scale.
(a)
The potential difference between the electrodes.
(i)
Explain why a potential difference develops between the electrodes when the water flows.
(2)
(ii)
Show that the potential difference between the electrodes is V = Bvd.
(2)
(iii)
State and explain which electrode is at the higher potential.
(1)
(b)
The meter is calibrated by measuring V, with a voltmeter of very high resistance, for known volume flow rates Q. The graph shows the results.
(i)
Determine the gradient of the graph, including its unit.
(2)
(ii)
The volume flow rate is Q = v × (area of cross-section of the pipe). Use your answers to (a)(ii) and (b)(i) to determine B.
(3)
(iii)
The line does not pass through the origin. Explain whether this affects the value of B found in (b)(ii).
(1)
(c)
The electrodes are connected to a voltmeter of resistance 450 Ω. The seawater between the electrodes has a resistance of 50 Ω. Ignore the offset.
(i)
The water flows at 2.0 m s−1. Determine the reading on the voltmeter.
(2)
(ii)
While there is a current in the water, the magnetic field exerts a force on the current. State and explain the direction of this force.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The water is a conductor moving across the magnetic field; the ions in it, moving with the water, experience a magnetic force qvB, with positive and negative ions pushed towards opposite electrodes
✓ 1
Accept: an emf is induced across the moving water, as in a conductor moving through a magnetic field.
Charge builds up on opposite sides of the pipe, creating an electric field across it; the build-up stops when the electric force on an ion balances the magnetic force
✓ 1
Part (a)(ii)
The column of water between the electrodes acts as a straight conductor of length d moving at speed v perpendicular to B, so the induced emf is ε = BvL with L = d
✓ 1
ALT: in the steady state qE = qvB for an ion [1].
With no current drawn, V = ε: V = Bvd
✓ 1
ALT: E = V/d, so V = Bvd [1].
Part (a)(iii)
The north electrode: by Fleming's left-hand rule the force on a positive ion moving east in a downward field is towards the north, so positive charge collects on the north side
✓ 1
The explanation is required for the mark.
Part (b)(i)
Line of best fit drawn; gradient found from two widely separated points on the line
✓ 1
Gradient = 3.05 mV s L−1 (= 3.05 V s m−3)
✓ 1
Accept 2.9–3.2 mV s L⁻¹. A correct unit is required for this mark.
Part (b)(ii)
v = 4Q/(πd²), so V = Bvd = 4BQ/(πd)
✓ 1
Allow ECF from (a)(ii).
Gradient = 4B/(πd), so B = gradient × πd/4
✓ 1
B = 3.05 × π × 0.050/4 = 0.12 T
✓ 1
Allow ECF from (b)(i); the gradient must be in V s m⁻³. Accept 0.11–0.13 T.
Part (b)(iii)
No: the intercept is a constant (systematic) offset added to every reading, which does not change the gradient used to find B
✓ 1
Allow ECF from the candidate's graph.
Part (c)(i)
emf = Bvd = 0.12 × 2.0 × 0.050 = 12 mV
✓ 1
Allow ECF from (b)(ii).
The water acts as the internal resistance: reading = 12 × 450/(450 + 50) = 10.8 mV
✓ 1
Award [1 max] for 12 mV, which ignores the resistance of the water. Accept 10–11 mV.
Part (c)(ii)
Inside the water the current flows from the south electrode to the north electrode, so the force is towards the west, opposing the flow; it must oppose the flow because the electrical energy delivered to the meter comes from work done on the water (Lenz's law / energy conservation)
✓ 1
Both the direction and a reason (Fleming's rule or energy) are required.
Answers: (a)(iii) the north electrode · (b)(i) 3.05 mV s L−1 · (b)(ii) 0.12 T · (c)(i) 10.8 mV · (c)(ii) towards the west (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it, ε = BvL; Lenz's law as a consequence of energy conservation; D.3 — the force on a charge moving in a magnetic field F = qvB sin θ; the motion of charged particles in perpendicular electric and magnetic fields; D.2 — E = V/d; B.5 — emf and internal resistance Command term: Determine
43D-2-52
Speed of a train from the Earth’s field·D.4 Induction (HL)
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksDetermine
A test train runs along a straight, level track. The two rails are 1.435 m apart and are insulated from each other except through the steel wheels and axles of the train. A voltmeter of very high resistance is connected between the rails far ahead of the train. At this location the vertical component of the Earth’s magnetic field is 4.0 × 10−5 T.
The graph shows how the reading ε of the voltmeter varies with time t. The train starts to brake uniformly at t = 10 s.
Voltmeter reading against time (drawn to scale).
(a)
(i)
Explain why there is a potential difference between the rails while the train moves.
(2)
(ii)
Explain why the horizontal component of the Earth’s field makes no contribution to the reading.
(1)
(b)
(i)
Show that the speed of the train before it brakes is about 50 m s−1.
(2)
(ii)
Determine the magnitude of the deceleration of the train.
(2)
(iii)
Determine the distance travelled by the train from t = 0 until it stops.
(3)
(c)
(i)
An identical train approaches the same voltmeter from the opposite direction at the same speed. State and explain how the reading differs.
(1)
(ii)
A second train travels at the same speed but has twice as many axles. Explain whether the reading of the voltmeter is different.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Each axle, with its wheels, is a conductor moving perpendicular to the vertical component of the field, so an emf BvL is induced across it (the free electrons experience a magnetic force along the axle)
✓ 1
The rails connect the ends of the axles to the voltmeter; it draws almost no current, so it reads this emf
✓ 1
Part (a)(ii)
The velocity and the horizontal component are both horizontal, so the magnetic force on the charges (perpendicular to both) is vertical, perpendicular to the horizontal axle; it drives no charge along the axle
✓ 1
Accept: the axle does not cut the field lines of the horizontal component.
Part (b)(i)
ε = 2.87 mV read from the graph
✓ 1
Accept 2.8–2.9 mV.
v = ε/(BL) = 2.87 × 10−3/(4.0 × 10−5 × 1.435) = 50.0 m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(ii)
The emf is proportional to the speed, so the speed falls uniformly from 50 m s−1 to zero between t = 10 s and t = 60 s
✓ 1
Allow ECF from (b)(i).
a = 50/50 = 1.0 m s−2
✓ 1
ALT: gradient of the graph 0.0574 mV s−1 ÷ (BL).
Part (b)(iii)
At constant speed: 50 × 10 = 500 m
✓ 1
Allow ECF from (b)(i).
While braking: ½ × 50 × 50 = 1250 m
✓ 1
ALT: v2/(2a) using (b)(ii).
Total = 1750 m
✓ 1
Accept 1.7–1.8 km.
Part (c)(i)
Same magnitude but opposite polarity: the velocity is reversed, so the magnetic force on the charges in each axle is reversed
✓ 1
Part (c)(ii)
Each axle has the same emf and the axles are connected in parallel by the rails
✓ 1
The voltmeter draws (almost) no current, so it reads the emf of one axle: the reading is unchanged
✓ 1
Do not accept "twice as many conductors, so twice the emf".
Answers: (b)(i) 50.0 m s−1 · (b)(ii) 1.0 m s−2 · (b)(iii) 1750 m (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; (guidance) relative motion between a conductor and a magnetic field; A.1 — the equations of motion for uniform acceleration; displacement as the area under a velocity–time graph Command term: Determine
44D-2-53
An electric guitar pickup·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
An electric guitar uses a pickup to detect the vibration of a steel string. The pickup is a coil of many turns wound around a small permanent magnet placed just below the string, near one end. The magnet magnetises the part of the string directly above it. The coil is connected to an amplifier of very high input resistance.
A string with a vibrating length of 0.650 m is plucked so that it vibrates in its first harmonic. The graph shows how the emf ε across the coil varies with time t.
Emf of the pickup against time (drawn to scale).
(a)
(i)
Explain why an emf is induced in the coil while the string vibrates.
(2)
(ii)
Use the graph to determine the frequency of the vibration. Use as many cycles as possible.
(2)
(iii)
Determine the speed of the transverse waves on the string.
(2)
(b)
The peak emf is proportional to the maximum speed of the part of the string directly above the pickup. When the graph was recorded, the amplitude of the vibration of that part of the string was 1.2 mm.
(i)
Calculate the maximum speed of the part of the string above the pickup.
(2)
(ii)
As the note fades, the frequency stays the same and the amplitude above the pickup falls to 0.40 mm. Determine the new peak emf.
(1)
(iii)
The string is now made to vibrate in its second harmonic, with the same amplitude of 1.2 mm above the pickup. Deduce the peak emf and the period of the emf.
(2)
(c)
(i)
The ends of the coil are now connected together. Explain, with reference to Lenz’s law, the effect on the vibration of the string.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The magnetised part of the string moves towards and away from the coil, so the magnetic flux linking the coil changes
✓ 1
Accept "the field at the coil changes as the string moves".
By Faraday’s law an emf equal to the rate of change of flux linkage is induced; it alternates as the string moves up and down
✓ 1
Do not accept "the string cuts the field lines" without reference to a change of flux linkage of the coil.
Part (a)(ii)
Three cycles take 15.3 ms (from the zero crossings at 0 and 15.3 ms)
✓ 1
Accept 15.1–15.5 ms. Using one cycle: [1 max].
f = 3/0.0153 = 196 Hz
✓ 1
Accept 194–199 Hz.
Part (a)(iii)
First harmonic: λ = 2L = 1.30 m
✓ 1
v = fλ = 196 × 1.30 = 255 m s−1
✓ 1
Allow ECF from (a)(ii). Accept 252–259 m s⁻¹.
Part (b)(i)
vmax = ωx0 = 2πfx0
✓ 1
= 2π × 196 × 1.2 × 10−3 = 1.48 m s−1
✓ 1
Allow ECF from (a)(ii). Accept 1.4–1.5 m s⁻¹.
Part (b)(ii)
vmax ∝ amplitude at constant frequency, so the peak emf = 36 × 0.40/1.2 = 12 mV
✓ 1
Allow ECF from the peak value read from the graph (accept 35–37 mV).
Part (b)(iii)
The frequency doubles, so vmax = 2πfx0 doubles: peak emf = 72 mV
✓ 1
Allow ECF from (b)(i) and the graph.
Period = 5.1/2 = 2.6 ms (2.55 ms)
✓ 1
Allow ECF from (a)(ii).
Part (c)(i)
A current now flows in the coil; by Lenz’s law its magnetic field exerts a force on the magnetised string that opposes the string’s motion
✓ 1
Energy of the vibration is transferred to thermal energy in the coil, so the amplitude of the vibration decreases more rapidly (the note dies away sooner)
✓ 1
Do not accept "the string stops vibrating" without reference to energy or a faster decrease in amplitude.
Answers: (a)(ii) 196 Hz · (a)(iii) 255 m s−1 · (b)(i) 1.48 m s−1 · (b)(ii) 12 mV · (b)(iii) 72 mV; 2.6 ms (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; (guidance) relative motion between a conductor and a magnetic field; C.4 — standing waves on strings, harmonics; C.1 — vmax = ωx0 for simple harmonic motion Command term: Determine
45D-2-54
A coil on a trolley entering a field·D.4 Induction (HL)
Paper 2Medium13 marks
Short answer & extended response7 steps to full marksDetermine
A trolley of total mass 0.25 kg carries a flat, horizontal, rectangular coil of 120 turns. The front and back sides of the coil are 5.0 cm long and perpendicular to the track; the other two sides are 8.0 cm long. The ends of the coil are joined together and its total resistance is 2.4 Ω. The trolley runs along a level, frictionless track into a region of uniform vertical magnetic field of flux density B. The field region is longer than the coil and its boundary is perpendicular to the track.
A motion sensor records the speed v of the trolley. The graph shows v against the distance x of the front side of the coil beyond the boundary of the field.
Speed of the trolley against the position of the front side of the coil (drawn to scale).
(a)
(i)
Explain why the trolley slows down between x = 0 and x = 8.0 cm, but moves at constant speed once x is greater than 8.0 cm.
(2)
(ii)
Show that, while only the front side of the coil is in the field, the magnetic force on the coil is F = N2B2b2v/R, where b is the length of the front side and R the resistance of the coil.
(2)
(iii)
By considering the impulse of this force in a short time Δt, show that the decrease in speed Δv of the trolley is proportional to the distance Δx it moves. Hence explain the shape of the graph between x = 0 and x = 8.0 cm.
(2)
(b)
(i)
Use the graph to determine B.
(3)
(c)
(i)
Determine the thermal energy transferred in the coil while the trolley enters the field.
(2)
(ii)
The experiment is repeated with the trolley entering the field at 0.30 m s−1. Deduce whether the coil enters the field completely.
(2)
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Marking point
Mark
Notes
Part (a)(i)
From 0 to 8.0 cm the front side is in the field and the back side is not: the flux linkage of the coil increases, so an emf and a current are induced; by Lenz’s law the force on the current in the front side opposes the motion
✓ 1
Beyond 8.0 cm the whole coil is in the uniform field: the flux linkage is constant, so there is no emf, no current and no magnetic force (no friction, so constant speed)
✓ 1
Part (a)(ii)
ε = NBbv (the N front sides in series) and I = ε/R = NBbv/R
✓ 1
Force on the N front sides: F = NBIb = N2B2b2v/R (the forces on the two long sides cancel)
✓ 1
Part (a)(iii)
FΔt = mΔv with vΔt = Δx gives mΔv = (N2B2b2/R)Δx
✓ 1
Allow ECF from (a)(ii).
N2B2b2/(mR) is constant, so v decreases linearly with x: a straight line of constant negative gradient
✓ 1
Part (b)(i)
Speed at x = 8.0 cm read as 0.41 m s−1
✓ 1
Accept 0.40–0.42 m s⁻¹.
Gradient = (0.41 − 0.60)/0.080 = −2.4 s−1, and its magnitude equals N2B2b2/(mR)
✓ 1
Allow ECF from (a)(iii).
B = √(2.4 × 0.25 × 2.4/(1202 × 0.0502)) = 0.20 T
✓ 1
Accept 0.19–0.21 T.
Part (c)(i)
Loss of kinetic energy = ½ × 0.25 × (0.602 − 0.412)
✓ 1
Allow ECF from the speed read in (b).
= 0.024 J
✓ 1
Accept 0.023–0.025 J.
Part (c)(ii)
By (a)(iii) the decrease in speed over the 8.0 cm is the same whatever the initial speed: 0.19 m s−1
✓ 1
Allow ECF from (a)(iii) and (b).
0.30 − 0.19 = 0.11 m s−1 > 0, so the coil enters completely and then moves on at about 0.11 m s−1
✓ 1
Award [1 max] for a kinetic-energy argument that assumes the same energy loss as in (c)(i) (it predicts that the trolley stops).
Answers: (b)(i) 0.20 T · (c)(i) 0.024 J · (c)(ii) yes; about 0.11 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; (guidance) rectangular coils moving in and out of fields; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; A.2 — impulse J = FΔt = Δp; A.3 — kinetic energy Command term: Determine
46D-2-55
A bicycle generator·D.4 Induction (HL)
Paper 2Medium14 marks
Short answer & extended response9 steps to full marksDetermine
A bicycle lamp is powered by a small generator fixed to the frame. A rubber drive wheel of radius 1.0 cm on the generator shaft presses against the outer edge of the tyre and turns without slipping. The tyre has a radius of 0.34 m. Inside the generator a rectangular coil of 400 turns and area 6.0 × 10−4 m2 turns with the shaft in a uniform magnetic field of flux density 0.050 T.
The peak emf of a coil of N turns and area A rotating with angular velocity ω in a uniform field B is ε0 = NBAω. The graph shows the emf ε of the generator against time t when the bicycle moves at constant speed with the lamp disconnected.
Emf of the generator against time (drawn to scale).
(a)
(i)
Determine the frequency of rotation of the coil. Use as many cycles as possible.
(2)
(ii)
Determine the speed of the bicycle.
(2)
(iii)
Show that the peak emf on the graph is consistent with the data for the generator.
(1)
(b)
The lamp needs a peak emf of at least 3.0 V to give enough light.
(i)
Determine the minimum speed of the bicycle for which the lamp gives enough light.
(2)
(ii)
State the period of the emf at this speed.
(1)
(c)
The lamp is now connected and the bicycle again moves at the speed found in (a)(ii). At the instant when the emf has its peak value, the current in the coil is 0.40 A. Ignore the resistance of the coil.
(i)
State the orientation of the plane of the coil relative to the magnetic field at this instant.
(1)
(ii)
Deduce that at this instant the magnetic torque on the coil is NBIA, and calculate its value.
(3)
(iii)
Show that the rate at which work must be done against the magnetic forces on the two sides of the coil equals the electrical power delivered at this instant, and explain how this illustrates Lenz’s law.
(2)
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Marking point
Mark
Notes
Part (a)(i)
Three cycles take 37.7 ms
✓ 1
Accept 37–38.5 ms. Using one cycle: [1 max].
f = 3/0.0377 = 80 Hz
✓ 1
Accept 78–81 Hz.
Part (a)(ii)
The edge of the tyre moves relative to the frame at the speed of the bicycle, and the edge of the drive wheel moves at the same speed: v = ωr = 2πfr with r = 1.0 cm
✓ 1
Using the tyre radius (giving about 170 m s⁻¹): [1 max].
v = 2π × 79.6 × 0.010 = 5.0 m s−1
✓ 1
Allow ECF from (a)(i). Accept 4.9–5.1 m s⁻¹.
Part (a)(iii)
NBAω = 400 × 0.050 × 6.0 × 10−4 × (2π × 79.6) = 6.00 V, equal to the peak of 6.0 V on the graph
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
ε0 ∝ ω, and ω ∝ the speed of the bicycle
✓ 1
Minimum speed = 5.0 × 3.0/6.0 = 2.5 m s−1
✓ 1
Allow ECF from (a)(ii).
Part (b)(ii)
The period doubles: 2 × 12.6 = 25 ms
✓ 1
Allow ECF from (a)(i).
Part (c)(i)
Parallel to the field (the flux linkage is zero and changing fastest)
✓ 1
Part (c)(ii)
The two sides parallel to the axis, of length L, carry current perpendicular to the field: each has a force NBIL, perpendicular to the plane of the coil
✓ 1
Each force acts at a distance w/2 from the axis, where w is the width of the coil: τ = 2 × NBIL × w/2 = NBILw = NBIA
✓ 1
τ = 400 × 0.050 × 0.40 × 6.0 × 10−4 = 4.8 × 10−3 N m
✓ 1
Part (c)(iii)
Each side moves perpendicular to the field at speed ωw/2 against a force NBIL: P = 2 × NBIL × ωw/2 = NBIAω = 400 × 0.050 × 0.40 × 6.0 × 10−4 × 500 = 2.4 W, and ε0I = 6.0 × 0.40 = 2.4 W: equal
✓ 1
Allow ECF from (a)(i) and (c)(ii). ω = 2π × 79.6 = 500 rad s−1.
The magnetic forces on the induced current oppose the rotation (Lenz’s law), so mechanical work must be done against them at the rate at which electrical energy is delivered: energy is conserved
✓ 1
Do not accept "the lamp needs energy" without reference to the opposing forces.
Answers: (a)(i) 80 Hz · (a)(ii) 5.0 m s−1 · (b)(i) 2.5 m s−1 · (b)(ii) 25 ms · (c)(i) parallel to the field · (c)(ii) 4.8 × 10−3 N m (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces a sinusoidally varying emf in a coil rotating within it; the effect on induced emf caused by changing the frequency of rotation; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; A.2 — circular motion v = ωr; A.4 — torque; A.3 — power P = Fv; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θCommand term: Determine
47D-2-56
Two rods on rails·D.4 Induction (HL)
Paper 2Hard14 marks
Short answer & extended response10 steps to full marksDeduce
Two identical copper rods P and Q, each of mass 0.020 kg, rest across a pair of long, horizontal, parallel rails 0.20 m apart. A uniform vertical magnetic field of flux density 0.50 T acts over the whole region. The rods and rails form a closed circuit of total resistance 0.10 Ω, which may be taken as constant. Friction is negligible.
At t = 0, rod P is set moving and is then kept moving at a constant speed of 2.0 m s−1, directly away from Q, by an external force. Q is initially at rest. The graph shows how the speed of Q varies with time.
Speed of rod Q against time (drawn to scale). The dashed line shows the speed of P.
(a)
(i)
Explain why a current flows in Q although Q is at rest.
(1)
(ii)
Show that the initial current is 2.0 A.
(1)
(iii)
State and explain the direction of the magnetic force on Q.
(1)
(b)
(i)
Show that, when Q moves with speed vQ, the current is I = BL(2.0 − vQ)/R.
(2)
(ii)
Calculate the initial acceleration of Q and show that it is consistent with the graph.
(3)
(iii)
Explain why the speed of Q approaches 2.0 m s−1.
(1)
(c)
(i)
Explain why, at every instant, the external force on P has the same magnitude as the magnetic force on Q.
(1)
(ii)
After a long time Q moves at 2.0 m s−1. By considering the total impulse on Q, show that by then the total work done by the external force is 0.080 J.
(2)
(iii)
Determine the total thermal energy transferred in the circuit, and comment on your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
P moves, so the area of the circuit and its flux linkage increase (P cuts field lines): an emf is induced in P and drives a current round the closed circuit, which includes Q
✓ 1
Part (a)(ii)
I = BLv/R = 0.50 × 0.20 × 2.0/0.10 = 2.00 A
✓ 1
Part (a)(iii)
In the direction of motion of P (Q is pulled after P): by Lenz’s law the induced current opposes the increase of flux, and a force that moves Q after P reduces the rate at which the area of the circuit increases
✓ 1
Both the direction and a reason are required. Accept: the currents in P and Q are in opposite directions across the rails, so the force on P opposes its motion and the force on Q is in the opposite direction to that on P.
Part (b)(i)
The area of the circuit now increases at the rate L(2.0 − vQ), because Q also moves in the same direction
✓ 1
ALT: the emfs BL × 2.0 in P and BLvQ in Q oppose each other round the circuit.
ε = BL(2.0 − vQ), so I = ε/R = BL(2.0 − vQ)/R
✓ 1
Part (b)(ii)
F = BIL = 0.50 × 2.0 × 0.20 = 0.20 N
✓ 1
Allow ECF from (a)(ii).
a = 0.20/0.020 = 10 m s−2
✓ 1
The tangent to the graph at t = 0 has a gradient of about 10 m s−2 (it would reach 2.0 m s−1 at about 0.20 s)
✓ 1
Accept 9–11 m s⁻². A tangent drawn on the graph or a gradient from the curve near t = 0 is needed.
Part (b)(iii)
As vQ increases the current, and hence the force BIL on Q, decrease (by (b)(i)); when the speeds are equal the flux linkage is constant, so there is no current and no force
✓ 1
Allow ECF from (b)(i).
Part (c)(i)
The same current flows in P and Q, in the same field and with the same length, so the magnetic forces on them are equal in magnitude (opposite in direction); P moves at constant velocity, so the external force balances the magnetic force on P
✓ 1
Part (c)(ii)
Total impulse of the external force = total impulse on Q = 0.020 × 2.0 = 0.040 N s (equal forces at every instant)
✓ 1
Allow ECF from (c)(i).
P moves at a constant 2.0 m s−1, so the work done in each short interval is F × 2.0Δt; the total work = 2.0 × 0.040 = 0.0800 J
✓ 1
Part (c)(iii)
Kinetic energy gained by Q = ½ × 0.020 × 2.02 = 0.040 J, so thermal energy = 0.080 − 0.040 = 0.040 J
✓ 1
Allow ECF from (c)(ii).
Exactly half of the work done is dissipated: the result does not depend on B or R
✓ 1
Accept any valid comment, e.g. a stronger field makes Q reach 2.0 m s⁻¹ sooner but the same energy is dissipated.
Answers: (b)(ii) 10 m s−2 · (c)(iii) 0.040 J (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — that a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it as given by ε = BvL; that a time-changing magnetic flux induces an emf ε as given by Faraday’s law of induction ε = −NΔΦ/Δt; that the direction of induced emf is determined by Lenz’s law and is a consequence of energy conservation; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; A.2 — Newton’s second law, impulse; A.3 — work and kinetic energy Command term: Deduce
48D-2-57
An electrodynamic tether·D.4 Induction (HL)
Paper 2Hard19 marks
Short answer & extended response12 steps to full marksEstimate
An electrodynamic tether can be used to bring a disused satellite down from orbit. A straight conducting tether 5.0 km long hangs vertically below a satellite that is in a circular orbit at an altitude of 700 km. The total mass of the satellite and tether is 200 kg. The satellite moves eastwards; the component of the Earth's magnetic field perpendicular to both the tether and the velocity is 2.5 × 10−5 T, directed northwards, and may be treated as uniform along the tether.
The circuit is completed through the ionised gas around the satellite, and the total resistance of the circuit is 1.2 kΩ. ME = 5.97 × 1024 kg, RE = 6.37 × 106 m.
Not to scale. Viewed from the south, looking north.
(a)
The tether in orbit.
(i)
Show that the orbital speed of the satellite is about 7.5 km s−1.
(1)
(ii)
Explain, in terms of the forces on the free electrons in the tether, why an emf is induced between the ends of the tether, and state which end is at the higher potential.
(2)
(iii)
Calculate the emf induced in the tether.
(1)
(b)
Current and force.
(i)
Calculate the current in the tether.
(1)
(ii)
Determine the magnitude of the magnetic force on the tether and state its direction.
(2)
(iii)
Show that the rate at which this force removes energy from the orbiting satellite is P = B²L²v²/R.
(2)
(iv)
Deduce the fundamental SI units of B²L²/R.
(2)
(c)
Bringing the satellite down.
(i)
Determine the decrease in the total energy of the satellite when its orbit falls from an altitude of 700 km to an altitude of 300 km.
(2)
(ii)
Estimate the time taken for this descent.
(2)
(iii)
State one assumption made in your estimate, and explain whether it leads to an overestimate or an underestimate of the time.
(2)
(d)
(i)
A power supply on the satellite could instead drive a current through the tether in the opposite direction to the induced current. Explain how this could be used to raise the orbit of a satellite.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10−11 × 5.97 × 1024/7.07 × 106) = 7.50 × 103 m s−1
✓ 1
r = 6.37 × 10⁶ + 7.0 × 10⁵ m must be used; full substitution or 3 s.f. required.
Part (a)(ii)
The electrons move with the tether at speed v through the field, so a magnetic force evB acts on them along the tether, downwards (F = qv × B with q negative)
✓ 1
Do not accept "the tether cuts the field lines" alone.
Electrons gather at the lower end, so the upper end is at the higher potential
t = ΔE/P = 3.4 × 108/7.3 × 102 = 4.6 × 105 s (about 5 days)
✓ 1
Allow ECF from (c)(i). Accept 4.4–4.8 × 10⁵ s.
Part (c)(iii)
Assumption: the power stays at its value for the 700 km orbit (B, v and the resistance constant)
✓ 1
As the orbit falls the speed increases (to 7.73 × 103 m s−1 at 300 km) and P ∝ v², so the power increases: the time is overestimated
✓ 1
Allow ECF from (b)(iii). Accept a reasoned argument based on a different assumption, e.g. B increasing closer to the Earth.
Part (d)(i)
The current in the tether is reversed (down the tether) in the same field, so the force BIL on the tether is reversed: eastwards, in the direction of the velocity
✓ 1
Allow ECF from the direction in (b)(ii). Accept the observation that the supply must provide an emf greater than the induced emf (940 V).
This force does positive work on the satellite, so its total energy −GMm/(2r) increases and the orbital radius increases (the energy comes from the power supply, e.g. solar cells)
✓ 1
Do not accept "the force pushes it upwards".
Answers: (a)(i) 7.50 × 103 m s−1 · (a)(ii) upper end · (a)(iii) 940 V · (b)(i) 0.78 A · (b)(ii) 0.098 N, westwards · (b)(iv) kg s−1 · (c)(i) 3.4 × 108 J · (c)(ii) 4.6 × 105 s (the remaining parts are explanations — see the table above)
Syllabus understandingD.4 — a uniform magnetic field induces an emf in a straight conductor moving perpendicularly to it, ε = BvL; Lenz's law as a consequence of energy conservation; D.3 — the force on a moving charge and on a current-carrying conductor, F = BIL sin θ; D.1 (HL) — the orbital speed vorbital = √(GM/r); (guidance) changes in energy when a satellite changes orbit; B.5 — R = V/I; A.3 — power P = Fv; Tools 3 — fundamental SI units Command term: Estimate
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