84 free, high-quality past-paper-style questions on Induction (HL) for IB Physics HL, each with a full mark scheme. Covers: Magnetic flux and flux linkage; Faraday's law; The a.c. generator; and more. For when you have finished the past papers, or want to practise one topic at a time.Read the D.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.396 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1D-1A-56Magnetic flux and flux linkagePaper 1AEasy1 mark
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?
Side view: the coil is seen edge-on; not to scale
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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.
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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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).
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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.
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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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.
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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.
4IBP2H-034-00-01Magnetic flux and flux linkagePaper 1AEasy1 mark
The antenna of a contactless payment card is a flat rectangular loop of area 3.0 × 10−3 m2. Near a card reader there is a uniform magnetic field of flux density 2.0 mT. The field makes an angle of 60° with the plane of the loop. What is the magnetic flux through the loop?
Side view: the loop is seen edge-on; not to scale
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Flux is Φ = BA cos θ, where θ is the angle between B and the normal to the loop. The field is at 60° to the plane, so it is at 30° to the normal. Φ = 2.0 × 10−3 × 3.0 × 10−3 × cos 30° = 5.2 × 10−6 Wb. Option A uses the 60° angle in the cosine. Option B ignores the angle. Option D divides B by A.
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Answer C
Answer: C
Common error: Check which angle you are given. If it is measured from the plane of the loop, use sin θ, or convert it to the angle from the normal first.
5IBP2H-034-01-02The a.c. generatorPaper 1AEasy1 mark
Which combination of units is equivalent to the weber, the unit of magnetic flux?
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Faraday's law gives emf = rate of change of flux linkage, so flux = emf × time. The weber is therefore equivalent to V s. It is also equivalent to T m2, because Φ = BA, not T m−2. V s−1 comes from inverting the relationship. A m2 is the unit of magnetic dipole moment, not flux.
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Answer B
Answer: B
Exam tip: Check flux units in two independent ways: Φ = BA gives T m², and ε = ΔΦ/Δt gives V s. Both must agree.
6IBP2H-034-01-05Magnetic flux and flux linkagePaper 1AEasy1 mark
A flat coil of N turns, each of area A, lies in a uniform magnetic field of flux density B. At first the normal to the coil is parallel to the field. The coil is then rotated through 180° about an axis perpendicular to the field. What is the magnitude of the change in flux linkage?
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The initial flux linkage is NBA cos 0° = +NBA. The final flux linkage is NBA cos 180° = −NBA. The change therefore has magnitude 2NBA. Choosing 0 ignores the reversal of the flux relative to the coil. Choosing NBA assumes the flux simply falls to zero. Choosing 2BA omits the number of turns.
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Answer C
Answer: C
Common error: Flux through a coil has a sign relative to the coil's normal. Turning a coil over reverses the flux, so the change is double the initial value.
7IBP3H-D4-20r9-01The a.c. generatorPaper 1AEasy1 mark
The rectangular coil of a wind-turbine generator rotates at constant angular speed in a uniform magnetic field. At the instant when the magnitude of the induced emf is greatest, the plane of the coil is
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Emf = rate of change of flux linkage. The flux is zero when the plane is parallel to the field, and there the flux is changing fastest (the gradient of the cosine curve is steepest), so the emf is at its peak: B is correct. A confuses maximum flux with maximum emf: there the gradient is zero and the emf is zero. C has no physical basis; the equality of flux and gradient is not a condition for maximum emf. D is wrong because the angle for peak emf is fixed; the speed only changes the size of the peak.
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Answer B
Answer: B
Common error: Maximum flux means zero emf; maximum emf occurs when the flux is passing through zero.
8IBP3H-D4-20r9-02The a.c. generatorPaper 1AEasy1 mark
A rectangular coil of a hand-cranked generator in a library display rotates in a uniform magnetic field. The crank is turned twice as fast. All other quantities are unchanged. The peak emf induced in the coil
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The peak emf is \(NBA\omega\), directly proportional to the angular speed, so doubling the speed doubles the peak emf: B is correct. A treats the emf as dependent only on field and area. C would be the factor for the mean power, which depends on the square of the emf. D confuses the peak emf with the rms emf, which involves a factor \(\sqrt2\) but is not a speed factor.
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Answer B
Answer: B
Exam tip: Write down \(\varepsilon_0 = NBA\omega\) and see which quantities change before comparing results.
9IBP3H-D4-20r9-06Magnetic flux and flux linkagePaper 1AEasy1 mark
A rectangular coil of 50 turns and area 0.020 m2 is placed in a uniform magnetic field of 0.30 T. The normal to the coil makes an angle of 60° with the field. The magnetic flux linkage of the coil is
Side view: the coil is seen edge-on; not to scale
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\(N\Phi = NBA\cos\theta = 50\times0.30\times0.020\times\cos60^\circ=0.15\) Wb, so B is correct. A omits the number of turns. C uses sin 60° instead of cos 60°. D omits the angle.
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Answer B
Answer: B
Exam tip: The angle in \(\Phi=BA\cos\theta\) is between the field and the normal to the coil, not the plane of the coil.
10IBP3H-D4-20r21-01A magnet oscillating above a coilPaper 1AEasy1 mark
The south pole of a bar magnet is moved towards the face of a flat circular coil that is connected to a resistor. Which row gives the direction of the induced current, as seen by an observer looking at the coil from the side of the magnet, and the force that the coil exerts on the magnet?
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The flux from the approaching south pole through the coil is increasing, so by Lenz's law the coil face nearest the magnet becomes a south pole to repel it, opposing the approach. A south-pole face is produced by a clockwise current when viewed from that face, so clockwise with repulsion is correct. Anticlockwise with repulsion is inconsistent, as anticlockwise would make that face a north pole. The two options with attraction would assist the motion, contradicting Lenz's law.
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Answer A
Answer: A
Exam tip: Decide first whether the induced effect must repel or attract, then use the clockwise-south, anticlockwise-north rule to get the current direction.
11IBP3H-D4-20r21-02Magnetic flux and flux linkagePaper 1AEasy1 mark
A flat coil of area 200 cm2 is in a uniform magnetic field of magnitude 0.50 T. The normal to the coil makes an angle of 60° with the field. What is the magnetic flux through the coil?
Side view: the coil is seen edge-on; not to scale
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Φ = BA cos θ = 0.50 × 200×10−4 × cos 60° = 5.0×10−3 Wb. The 8.7×10−3 Wb value uses sin 60° instead of cos 60°. The 1.0×10−2 Wb value omits the cosine. The 5.0×10−5 Wb value converts the area using 1 cm2 = 10−6 m2 instead of 10−4 m2.
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Answer B
Answer: B
Common error: The angle in Φ = BA cos θ is between the field and the normal to the coil, not between the field and the plane of the coil.
12IBP3H-D4-20r21-03Motional emf and Lenz's lawPaper 1AEasy1 mark
A straight metal rod of length 0.40 m moves at a constant speed of 3.0 m s−1 in a uniform magnetic field of flux density 0.25 T. The rod, its velocity and the field are mutually perpendicular. What is the emf induced between the ends of the rod?
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ε = BLv = 0.25 × 0.40 × 3.0 = 0.30 V. The 0.10 V value is BL, omitting the speed. The 0.75 V value is Bv, omitting the length. The 1.9 V value is Bv/L, dividing by the length instead of multiplying.
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Answer D
Answer: D
Exam tip: Motional emf is BLv only when B, L and v are mutually perpendicular; check the geometry before substituting.
13IBP3H-D4-20r21-04A magnet falling through a coilPaper 1AEasy1 mark
A bar magnet is released from rest above a coil and falls vertically along the axis of the coil, passing completely through it. The coil is connected to a data logger that records the induced emf. The magnet is moving faster as it leaves the coil than when it entered. Compared with the emf peak recorded as the magnet enters, the emf peak recorded as it leaves is
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On entry the flux linkage increases and on exit it decreases, so the emf changes sign. The magnet moves faster on exit, so the flux changes more rapidly and the magnitude of the emf is larger. Equal magnitude wrongly assumes that the same total flux change gives the same emf regardless of the time taken. The same sign ignores the reversal of the flux change. Smaller magnitude contradicts the stated greater speed on exit.
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Answer A
Answer: A
Common error: The emf depends on the rate of change of flux linkage, not just on the total change in flux.
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)?
Top view; not to scale
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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.
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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.
15D-1A-16Peak emf of a rotating coilPaper 1AMedium1 mark
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
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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.
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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.
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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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).
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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.
17D-1A-25A magnet oscillating above a coilPaper 1AMedium1 mark
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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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.
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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.
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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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.
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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.
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?
Circuit diagram
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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.
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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.
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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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.
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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.
21IBP2H-034-00-02Self-inductionPaper 1AMedium1 mark
An electric bicycle uses regenerative braking. When the rider brakes, the hub motor acts as a generator and charges the battery. Which statement best explains why the bicycle slows down?
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By Lenz's law the induced current opposes the change that causes it, which here is the rotation of the wheel. The force on the current-carrying windings therefore opposes the motion. Work done against this force transfers the bicycle's kinetic energy to electrical energy, which is stored in the battery. Option B would break conservation of energy. Option C describes motor action, which drains the battery. Option D is wrong: a rotating generator has changing flux linkage.
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Answer A
Answer: A
Exam tip: Whenever an induced current flows, link Lenz's law to energy. The opposing force means work must be done, and that work is the source of the electrical energy.
22IBP2H-034-00-03The a.c. generatorPaper 1AMedium1 mark
A flat square coil rotates at a constant rate in a uniform magnetic field and produces a sinusoidal emf. Which single change doubles the peak emf but leaves the period of the emf unchanged?
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The peak emf is ε0 = BANω. Doubling N doubles ε0 and leaves ω unchanged, so the period stays the same. Option A doubles ε0 but halves the period. Option B multiplies the area, and so ε0, by 4. Option C changes the current for a given load but not the emf.
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Answer D
Answer: D
Common error: Doubling the side of a square quadruples its area. Area scales with the square of the length.
23IBP2H-034-00-05Self-inductionPaper 1AMedium1 mark
A circuit contains a battery, a switch and a coil with many turns wound on an iron core. When the switch is opened, a spark jumps across the switch contacts. What is the best explanation?
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When the switch opens, the current, and so the coil's own flux linkage, falls in a very short time. By Faraday's law a large emf is induced in the coil itself (self-induction). By Lenz's law this emf acts to oppose the decrease, so it tries to keep the current flowing. The emf is large enough to ionise the air between the contacts. Option B contradicts Lenz's law. Options A and D misdescribe the mechanism.
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Answer C
Answer: C
Exam tip: For self-induction, state both laws. Faraday gives the size of the emf (it is large because Δt is small). Lenz gives its direction (it opposes the change in current).
24IBP2H-034-00-06The a.c. generatorPaper 1AMedium1 mark
A coil rotates at a constant rate in a uniform magnetic field. At one instant the plane of the coil is parallel to the magnetic field lines. What are the flux linkage and the induced emf at this instant?
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When the plane of the coil is parallel to B, no field lines pass through the coil, so the flux linkage NΦ is zero. However, NΦ = BAN cos ωt is changing most rapidly at this point, where its gradient is greatest. The emf, which equals the rate of change of flux linkage, is therefore at its maximum. Flux linkage and emf are 90° out of phase.
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Answer D
Answer: D
Common error: Do not assume maximum flux means maximum emf. The emf follows the gradient of the flux-linkage graph, not its value.
An aircraft with a wingspan of 40 m flies horizontally, with its wings level, at 250 m s−1. The Earth's magnetic field there has magnitude 5.0 × 10−5 T and points at 53° below the horizontal (sin 53° ≈ 0.80, cos 53° ≈ 0.60). What is the magnitude of the emf induced between the wing tips?
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The wing and its velocity are both horizontal. The horizontal component of B gives a force v × Bh that is vertical, so it has no component along the wing. Only the vertical component of the field contributes: Bvertical = 5.0 × 10−5 × 0.80 = 4.0 × 10−5 T. The emf is then ε = BLv = 4.0 × 10−5 × 40 × 250 = 0.40 V. Option A wrongly uses the horizontal component (cos 53°). Option C wrongly uses the full field strength. Option D is wrong because an emf exists across a conductor moving through a field even when no current flows.
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Answer B
Answer: B
Common error: For motional emf, use the component of B that is perpendicular to both the conductor and its velocity. For a level wing moving horizontally, that is the vertical component of B, whatever the heading.
A small, strong cylindrical magnet is released from rest at the top of a long vertical copper tube. It soon falls at a constant terminal speed. Air resistance is negligible. Which statement is correct while the magnet falls at terminal speed?
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At terminal speed the kinetic energy of the magnet is constant. The flux through each section of tube still changes as the magnet passes, so eddy currents are still induced. By Lenz's law these currents oppose the relative motion, and the upward magnetic force balances the weight. The gravitational potential energy lost each second is therefore all converted to thermal energy in the tube. Below the magnet the flux is increasing and above it the flux is decreasing, so the two current loops circulate in opposite directions.
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Answer B
Answer: B
Common error: An induced emf needs a changing flux linkage, not a changing speed. A magnet moving at constant velocity past a conductor still changes the flux through it.
A maglev test vehicle travels horizontally at 120 m s−1. A straight metal antenna rod of length 2.0 m is mounted horizontally on the vehicle, perpendicular to the direction of travel. The vertical component of the Earth's magnetic field is 5.0 × 10−5 T. What is the emf between the ends of the rod?
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The rod cuts the vertical field component, so ε = BvL = 5.0 × 10−5 × 120 × 2.0 = 1.2 × 10−2 V. An emf exists across the rod even though there is no closed circuit. The value 6.0 × 10−3 V wrongly uses ½BvL, which applies only to a rod rotating about one end. The value 2.4 × 10−2 V wrongly doubles the emf.
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Answer C
Answer: C
Common error: An emf is induced across a moving conductor whether or not a current can flow. A closed circuit is needed only for a current.
28IBP2H-034-01-04A magnet oscillating above a coilPaper 1AMedium1 mark
A bar magnet hangs from a spring with its north pole pointing down. It oscillates vertically with frequency f above a fixed flat coil whose axis is vertical. The magnet never enters the coil. Which statement about the emf induced in the coil is correct?
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The emf depends on the rate of change of flux linkage. At either extreme of the oscillation the magnet is momentarily at rest, so the flux is momentarily not changing and the emf is zero. This is true at the lowest point even though the flux there is a maximum. The emf is largest near the equilibrium position, where the speed is greatest. The emf reverses sign when the direction of motion reverses, so it has one positive and one negative peak per cycle. Its frequency is therefore f, not 2f.
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Answer D
Answer: D
Exam tip: Always separate 'flux is large' from 'flux is changing quickly'. Induced emf depends only on the second.
29IBP2H-034-01-07The a.c. generatorPaper 1AMedium1 mark
A simple ac generator is connected to a fixed resistor. The internal resistance of the generator is negligible. The rotation frequency of the coil is doubled while the magnetic field is unchanged. By what factors do the peak emf and the mean power in the resistor change?
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The peak emf is ε0 = NBAω, so doubling ω doubles ε0. The mean power in a resistor is ε02/2R, so it increases by a factor of 22 = 4. The frequency of the ac also doubles. Option C wrongly assumes that only the frequency changes.
✓ 1
Answer B
Answer: B
Common error: Rotating faster changes the amplitude of the emf as well as its frequency, because the flux linkage changes more rapidly.
30IBP3H-D4-20r9-03Motional emfPaper 1AMedium1 mark
A metal axle of length 0.80 m belongs to a library book trolley rolling at 1.5 m s−1 along a level floor. The axle is horizontal and perpendicular to the trolley's direction of motion. The vertical component of the Earth's magnetic field is 50 μT. The potential difference induced between the ends of the axle is
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Mark
Notes
\(\varepsilon = Blv = 50\times10^{-6}\times0.80\times1.5 = 6.0\times10^{-5}\) V, so D is correct. A uses half the length, as if the axle rotated about its centre. B comes from \(Bv/l\). C comes from \(Bl/v\). Both B and C misplace the quantities in the formula.
✓ 1
Answer D
Answer: D
Common error: Only the field component perpendicular to both the velocity and the conductor counts, so use the vertical component here, not the total field.
A strong magnet is dropped down a long vertical aluminium tube and takes much longer to emerge than an identical magnet dropped through a plastic tube of the same size. The best explanation is that
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Notes
As the magnet falls, the flux through each section of tube changes, inducing eddy currents. By Lenz's law, their field opposes the change, giving an upward force on the magnet: C is correct. A is wrong because aluminium is not ferromagnetic and would not attract a magnet steadily. B has no basis, because air flow is similar in both tubes. D is wrong because the magnet's own flux is not changing, so no emf is induced in it.
✓ 1
Answer C
Answer: C
Exam tip: For Lenz's law questions, state what is changing, what the induced effect opposes, and the resulting force direction.
32IBP3H-D4-20r9-05Peak emf of a rotating coilPaper 1AMedium1 mark
A coil in a wind-turbine generator produces a sinusoidal emf with a peak value of 200 V. Its resistance is negligible. It is connected across a 40 Ω resistor. The mean power dissipated in the resistor is
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Notes
Mean power \(=\frac{\varepsilon_0^2}{2R}=\frac{200^2}{2\times40}=500\) W, so D is correct. C uses \(\varepsilon_0^2/R\), the power at peak emf, forgetting the factor of one half. B uses \(\varepsilon_0^2/(\sqrt2 R)\), applying \(\sqrt2\) instead of squaring it. A halves twice, using \(V_{rms}\) and then dividing by 2 again.
✓ 1
Answer D
Answer: D
Common error: Mean power uses \(V_{rms}^2/R\); using the peak emf in \(V^2/R\) gives the maximum, not the average.
33IBP3H-D4-20r9-07The a.c. generatorPaper 1AMedium1 mark
A generator coil starts with its plane perpendicular to a uniform magnetic field and rotates at constant speed with period \(T\). The emf has peak value \(\varepsilon_0\). What is the first time at which the emf equals \(\frac12\varepsilon_0\)?
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Notes
Starting from maximum flux, \(\varepsilon=\varepsilon_0\sin\omega t\). Setting \(\sin\omega t = 0.5\) gives \(\omega t=30^\circ=\frac{\pi}{6}\), which is \(\frac{1}{12}\) of a period: C is correct. A uses 60°, as if the emf were \(\varepsilon_0\cos\omega t\). B uses 45°, which gives \(0.71\varepsilon_0\). D is the time of the peak emf.
✓ 1
Answer C
Answer: C
Common error: Check which function matches the start position: with the plane perpendicular to the field, the emf starts at zero and follows a sine.
34IBP3H-D4-20r21-05Electromagnetic flowmeterPaper 1AMedium1 mark
An electromagnetic flowmeter measures the speed of blood in a vessel of internal diameter 4.0 mm. A uniform magnetic field of 0.080 T is applied perpendicular to the vessel, and electrodes on opposite sides of the vessel record a potential difference of 0.24 mV. What is the speed of the blood?
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Notes
The electrodes are separated by the diameter, so ε = Bvd and v = ε/(Bd) = 0.24×10−3 / (0.080 × 4.0×10−3) = 0.75 m s−1. The 1.5 m s−1 value uses the radius 2.0 mm as the separation. The 0.375 m s−1 value divides by an extra factor of 2. The 3.0×10−3 m s−1 value is ε/B, omitting the diameter.
✓ 1
Answer A
Answer: A
Exam tip: Convert mV and mm to V and m before substituting into ε = Bvd.
35IBP3H-D4-20r21-06The a.c. generatorPaper 1AMedium1 mark
A wind-turbine generator contains a coil rotating at a constant frequency of 40 Hz in a uniform magnetic field, producing a peak emf of 15 V. The turbine speeds up so that the coil rotates at 60 Hz. The coil and field are unchanged. What are the new peak emf and the new period of the alternating emf?
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Notes
The peak emf is NBAω, proportional to the frequency, so it becomes 15 × 60/40 = 22.5 V, and the new period is 1/60 s = 16.7 ms. The 25 ms period is the original period (1/40 s). Leaving the peak emf at 15 V ignores its dependence on rotation rate. The 34 V value squares the frequency ratio (15 × 1.52), wrongly treating the emf as proportional to f2.
✓ 1
Answer B
Answer: B
Common error: Doubling the rotation frequency doubles the peak emf and halves the period; do not square the ratio.
36IBP3H-D4-20r21-07Motional emfPaper 1AMedium1 mark
A metal rod of length 0.50 m and negligible resistance slides at a constant speed of 2.0 m s−1 along horizontal conducting rails connected through a resistor, so that the total resistance of the circuit is 0.80 Ω. A uniform vertical magnetic field of 0.40 T acts on the rod. Ignoring friction, what horizontal force must be applied to keep the rod at constant speed?
Top view; not to scale
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Notes
ε = BLv = 0.40 V, I = ε/R = 0.50 A, and F = BIL = 0.40 × 0.50 × 0.50 = 0.10 N (equivalently B2L2v/R). The 0.20 N value uses only one factor of L in B2L2v/R. The 0.050 N value omits the speed. The 0.40 N value uses the emf in volts as if it were a force.
✓ 1
Answer D
Answer: D
Exam tip: At constant speed the applied force equals the magnetic force on the current-carrying rod, so F = BIL with I found from the emf and resistance.
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?
Square coil of side L moving to the right; not to scale
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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.
—
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.)
—
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.
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.
—
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.
—
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.
39D-1A-71Peak emf of a rotating coilPaper 1AHard1 mark
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?
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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.
—
All 3 steps must be completed — there is no mark for a part-answer.
40IBP2H-034-00-04Rectangular coil moving through a fieldPaper 1AHard1 mark
A square conducting loop of side 0.10 m moves at constant velocity through a region of uniform magnetic field 0.30 m wide. The field is perpendicular to the plane of the loop. The loop enters the region, passes through it, and leaves completely. Which describes the emf induced in the loop?
Top view; not to scale
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Notes
An emf is induced only when the flux linkage changes. While the loop enters, the flux increases at a constant rate. While it is wholly inside the field, the flux is constant, so the emf is zero. While it leaves, the flux decreases at the same rate. The emf therefore has the same magnitude but opposite sign, and by Lenz's law the induced current reverses. Option D confuses maximum flux with maximum emf.
✓ 1
Answer B
Answer: B
Exam tip: Emf depends on the rate of change of flux linkage, not on the flux itself. Constant flux means zero emf, however large the flux is.
On a linear test track, a metal bar slides without friction along two parallel conducting rails that are joined by a fixed resistor. A uniform magnetic field is perpendicular to the plane of the rails. An external force F keeps the bar moving at constant speed v, and the power dissipated in the resistor is P. The speed is changed to 2v. What are the new force and the new power?
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Notes
The emf is ε = BLv, so the current I = BLv/R doubles. The magnetic force on the bar is BIL, which is proportional to v, so it doubles. At constant speed the external force equals this magnetic force, so it becomes 2F. The power is P = Fv, or equally I2R, which is proportional to v2, so it becomes 4P. Option A forgets that power also depends on v. Option D assumes the force does not change.
✓ 1
Answer C
Answer: C
Exam tip: Build a proportionality chain: ε ∝ v, I ∝ v, F ∝ v, P = Fv ∝ v². Checking that Fv = I²R confirms that energy is conserved.
42IBP2H-034-01-06The a.c. generatorPaper 1AHard1 mark
A coil connected to a fixed resistor is flipped through 180° in a uniform magnetic field, taking time t. The experiment is repeated with the flip taking time t/2. Assume the emf changes with time in the same way in both trials. Compared with the first trial, what happens to the total charge that flows and to the total thermal energy produced in the second trial?
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Notes
The charge is Q = Δ(NΦ)/R, which depends only on the change in flux linkage, so it is unchanged. Halving the time doubles the emf and the current at each stage of the flip. The power I2R is therefore four times larger, but it acts for half the time. The thermal energy is 4 × ½ = 2 times larger.
✓ 1
Answer B
Answer: B
Exam tip: Charge depends on the flux change, not on how fast it happens. Energy dissipated depends on I²t, so it does depend on the speed.
A square conducting loop of side L and resistance R lies fully inside a region of uniform magnetic field directed perpendicular to its plane. The loop is pulled out of the field at constant speed v. The experiment is repeated at speed 2v. By what factor does the total energy dissipated in the loop change?
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Notes
While the loop is leaving the field, the emf is BLv and the current is BLv/R. The magnetic force on the loop is B2L2v/R, which must be balanced by the pulling force. This force acts over a distance L, so the work done is B2L3v/R. All of this work is dissipated as thermal energy, and it is proportional to v. Doubling the speed therefore doubles the energy. The power quadruples, but it acts for half the time.
✓ 1
Answer B
Answer: B
Exam tip: For constant-speed motion, find the force from the induced current, then use work = force × distance. Do not confuse the factor for power with the factor for energy.
44IBP3H-D4-20r9-08The a.c. generatorPaper 1AHard1 mark
A conducting tether of length \(L\) and resistance \(R\), with \(R\) proportional to \(L\), is dragged at constant speed perpendicular to a uniform magnetic field and to its own length, as part of a circuit whose resistance is entirely that of the tether. It is replaced by a tether of length \(2L\) made of the same wire. The speed and field are unchanged. Compared with the original tether, the current \(I\) in the new tether and the magnetic force \(F\) on it are
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Notes
The emf \(BLv\) doubles but the resistance also doubles, so \(I=\varepsilon/R\) is unchanged. The force is \(F=BIL\), so with \(I\) unchanged and \(L\) doubled, \(F\) doubles: C is correct. A and B forget that the resistance grows with length, so the current would not double. D forgets that the force depends on the length of conductor in the field.
✓ 1
Answer C
Answer: C
Exam tip: When one quantity changes, track every formula it appears in: here \(L\) appears in the emf, in \(R\) and in the force.
45IBP3H-D4-20r21-08A rectangular coil rotating in a uniform fieldPaper 1AHard1 mark
A metal rod of length 0.60 m is pivoted at one end and rotates at 5.0 revolutions per second in a plane perpendicular to a uniform magnetic field of 0.20 T. What is the emf between the ends of the rod?
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Notes
Each element of the rod has a speed proportional to its distance from the pivot, so the mean speed is ωL/2 and ε = ½BωL2 = 0.5 × 0.20 × (2π × 5.0) × 0.602 = 1.13 V. The 2.26 V value omits the factor ½. The 0.18 V value uses the frequency 5.0 Hz instead of ω. The 3.77 V value uses the speed ωL of the free end rather than the mean speed.
✓ 1
Answer D
Answer: D
Common error: Different points on a rotating rod move at different speeds, so use the average speed ωL/2 in ε = BLv.
46D-1B-19Flux and charge in a rotated coilPaper 1BEasy7 marks
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)
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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)
47D-1B-21Peak emf and frequency of rotationPaper 1BEasy7 marks
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)
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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)
48D-1B-09Peak emf of a rotating coilPaper 1BMedium7 marks
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)
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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)
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)
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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)
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)
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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)
51IBP2H-034-00-09Peak emf of a rotating coilPaper 2Easy5 marks
An electric toothbrush is recharged without any electrical contacts. A coil in the charging base produces a magnetic field that passes through a 200-turn coil inside the toothbrush handle. The handle coil has a cross-sectional area of 3.0 × 10−4 m2. In one part of a cycle, the magnetic flux density through the handle coil rises uniformly from 0 to 12 mT in 2.5 ms.
(a)
State Faraday's law of electromagnetic induction.
(1)
(b)
Calculate the average emf induced in the handle coil during the 2.5 ms interval.
(2)
(c)
Explain why the base coil must carry an alternating current rather than a steady direct current.
(2)
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Part (a)
The induced emf is equal to (proportional to) the rate of change of magnetic flux linkage
52IBP2H-034-00-10A rectangular coil rotating in a uniform fieldPaper 2Easy4 marks
A hand-cranked emergency radio contains a small a.c. generator. Its rectangular coil has 150 turns and an area of 4.0 × 10−3 m2. The coil rotates at 5.0 revolutions per second in a uniform magnetic field of flux density 0.080 T.
(a)
Calculate the peak emf produced by the generator.
(2)
(b)
Calculate the root-mean-square (rms) emf.
(1)
(c)
State the orientation of the plane of the coil relative to the magnetic field at the instant when the emf is zero.
A free-fall ride at a theme park is stopped by magnetic braking. Copper fins attached to the car pass at high speed between rows of strong permanent magnets fixed to the tower.
(a)
State Faraday's law of electromagnetic induction.
(1)
(b)
Explain, with reference to Lenz's law, why the car slows down as the fins pass between the magnets.
(3)
(c)
Suggest why the braking force decreases as the car slows, and why a separate mechanical brake is needed to hold the car at rest.
(2)
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Notes
Part (a)
The induced emf is equal to (proportional to) the rate of change of magnetic flux linkage .
✓ 1
Part (b)
As a fin moves through the field, the flux through parts of the fin changes, inducing emfs and therefore eddy currents in the copper . By Lenz's law, the induced currents produce effects that oppose the change causing them, which is the relative motion . The magnetic force on the currents acts opposite to the velocity of the car, and its kinetic energy is transferred to thermal energy in the fins .
✓ 3
Part (c)
The rate of change of flux, and so the emf, current and force, are proportional to the speed . At zero speed no current is induced and there is no braking force, so friction or a mechanical brake is needed to hold the car still .
✓ 2
Exam tip: In a Lenz's law explanation, name the change being opposed (here, the relative motion) and say where the energy goes.
At a science exhibit in a library, a student pulls a straight metal bar of length 0.45 m at a constant speed of 2.4 m s−1 along two horizontal parallel conducting rails. The rails are connected by a resistor of resistance 0.60 Ω, and the bar is in contact with the rails and perpendicular to them. A uniform vertical magnetic field of 0.35 T acts throughout. The resistances of the bar and rails are negligible.
Top view; not to scale
(a)
Calculate the emf induced in the bar.
(2)
(b)
Calculate the current in the resistor.
(2)
(c)
Calculate the horizontal force that must be applied to keep the bar moving at constant speed.
(2)
(d)
Show that the rate at which the applied force does work equals the electrical power dissipated in the resistor, and give this power.
(4)
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Part (a)
M1 \(\varepsilon=Blv\)
✓ 1
A1 \(0.35\times0.45\times2.4=0.378\) V
✓ 1
Part (b)
M1 \(I=\varepsilon/R\)
✓ 1
A1 \(0.63\) A
✓ 1
Part (c)
M1 \(F=BIL\) (ECF from (b))
✓ 1
A1 \(0.35\times0.63\times0.45=0.099\) N
✓ 1
Part (d)
M1 mechanical power \(=Fv=0.0992\times2.4\)
✓ 1
M1 electrical power \(=\varepsilon I=0.378\times0.63\)
✓ 1
A1 both equal \(0.238\) W
✓ 1
R1 energy conservation: the work done against the magnetic force is transferred to thermal energy in the resistor
✓ 1
Common error: The force needed to keep a bar at constant speed equals the magnetic force \(BIL\); students often write \(F=ma\) and find zero, forgetting that the magnetic force is balanced by the applied force.
55IBP3H-D4-20r21-09A magnet oscillating above a coilPaper 2Easy5 marks
In a demonstration of energy harvesting, the north pole of a bar magnet is withdrawn from one end of a coil. The coil is connected in a complete circuit to a sensitive ammeter.
(a)
State Lenz's law.
(2)
(b)
Deduce the polarity of the end of the coil nearest the magnet during the withdrawal, and explain why the student must do work to withdraw the magnet.
(3)
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Part (a)
A1 the direction of the induced emf (current) is such that it opposes
✓ 1
A1 the change in flux (flux linkage) that produces it
✓ 1
Part (b)
A1 the end nearest the magnet is a south pole
✓ 1
M1 this attracts the north pole, so the force on the magnet opposes the withdrawal
✓ 1
A1 work done by the student against this force is transferred to electrical energy in the circuit
✓ 1
Exam tip: Link Lenz's law to energy conservation: the opposing force is why mechanical work is needed to produce electrical energy.
56IBP3H-D4-20r21-10Magnetic flux and flux linkagePaper 2Easy5 marks
A contactless card reader in a shop contains a coil of 150 turns and cross-sectional area 4.0 cm2. A magnetic field, perpendicular to the coil, increases uniformly from 0 to 0.060 T in 0.020 s and then remains steady.
(a)
Calculate the change in magnetic flux through one turn of the coil during the increase.
(2)
(b)
State the emf induced in the coil while the field is steady at 0.060 T and explain your answer.
(1)
(c)
Calculate the magnitude of the average emf induced in the coil during the increase.
(2)
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Notes
Part (a)
M1 converts area to 4.0×10−4 m2 and uses ΔΦ = AΔB
✓ 1
A1 ΔΦ = 2.4×10−5 Wb
✓ 1
Part (b)
A1 zero, because there is no change in flux linkage
57D-2-04A rectangular coil rotating in a uniform fieldPaper 2Medium13 marks
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)
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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)
58D-2-13A magnet oscillating above a coilPaper 2Medium13 marks
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)
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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)
59D-2-16A magnet falling through a coilPaper 2Medium13 marks
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)
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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)
60D-2-29Rectangular coil moving through a fieldPaper 2Medium13 marks
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)
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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)
61IBP2H-034-00-11Rectangular coil moving through a fieldPaper 2Medium6 marks
In transcranial magnetic stimulation (TMS), a coil held against the scalp produces a brief, intense magnetic field in the brain. In a simple model, a circular path of conducting brain tissue of radius 1.0 cm lies perpendicular to the field. The magnetic flux density through this path falls uniformly from 1.5 T to zero in 0.10 ms.
(a)
Calculate the magnetic flux through the circular path at the start of the pulse.
(1)
(b)
Calculate the average emf induced around the path.
(2)
(c)
Use Lenz's law to explain the direction of the magnetic field produced by the current induced in the tissue.
(2)
(d)
Suggest why the field is made to change in a very short time.
(1)
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Part (a)
Φ = BA = 1.5 × π × (0.010)2 = 4.7 × 10−4 Wb
✓ 1
Part (b)
ε = ΔΦ/Δt with N = 1
✓ 1
ε = 4.7 × 10−4 / 1.0 × 10−4 = 4.7 V; ECF from (a)
✓ 1
Part (c)
The induced current opposes the change in flux, which here is a decrease
✓ 1
So the induced current produces a magnetic field in the same direction as the original field, tending to maintain the flux
✓ 1
Part (d)
A shorter Δt gives a greater rate of change of flux, so a larger induced emf and current in the tissue for the same change in B
✓ 1
Exam tip: When the flux is decreasing, the induced field acts in the same direction as the original field. Lenz's law opposes the change, not the field itself.
62IBP2H-034-00-13Rectangular coil moving through a fieldPaper 2Medium7 marks
A vehicle sensor buried under a traffic-light stop line contains a coil. As a car's steel body passes over, the magnetic flux linkage in the coil changes. The table gives a simplified model of the flux linkage NΦ during one event.
Between the tabulated values, NΦ changes linearly with time.
Simplified model from the table
(a)
Determine the magnitude of the emf induced in the coil between t = 0 and t = 40 ms.
(2)
(b)
State the emf between t = 40 ms and t = 60 ms, giving a reason.
(1)
(c)
Determine the magnitude of the emf between t = 60 ms and t = 80 ms. Compare its direction with that of the emf between t = 0 and t = 40 ms.
(2)
(d)
The coil is connected to a detector circuit with a total resistance of 15 Ω. Determine the charge that flows through the circuit between t = 60 ms and t = 80 ms.
(2)
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Part (a)
Δ(NΦ)/Δt = 12.0 × 10−3 / 40 × 10−3
✓ 1
ε = 0.30 V
✓ 1
Part (b)
Zero, because the flux linkage is constant, so its rate of change is zero
✓ 1
Part (c)
ε = 12.0 × 10−3 / 20 × 10−3 = 0.60 V
✓ 1
Opposite polarity (sign) to the emf between 0 and 40 ms, because the flux linkage is now decreasing rather than increasing
✓ 1
Part (d)
I = 0.60/15 = 0.040 A, or use Q = Δ(NΦ)/R; allow error carried forward from the emf found for 60–80 ms
✓ 1
Q = 0.040 × 0.020 = 8.0 × 10−4 C
✓ 1
Exam tip: The charge that flows equals Δ(NΦ)/R. It depends only on the total change in flux linkage, not on how quickly the change happens.
63IBP2H-034-00-16A rectangular coil rotating in a uniform fieldPaper 2Medium7 marks
A gym exercise bike uses an eddy-current brake. An aluminium flywheel spins between the poles of strong permanent magnets, and no part of the brake touches the flywheel. Tests show that the braking torque is proportional to the angular speed of the flywheel. At an angular speed of 40 rad s−1 the braking torque is 3.0 N m.
(a)
Explain why eddy currents are induced in the flywheel and why they produce a braking torque.
(3)
(b)
Calculate the power dissipated in the flywheel at 40 rad s−1.
(1)
(c)
The rider doubles the angular speed of the flywheel. Determine the new power dissipated.
(2)
(d)
State the effect on the braking torque, at a fixed angular speed, of moving the magnets closer to the flywheel.
(1)
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Part (a)
As the flywheel rotates, parts of it move through the magnetic field, so the flux through regions of the metal changes
✓ 1
By Faraday's law emfs are induced, which drive circulating (eddy) currents in the conducting aluminium
✓ 1
By Lenz's law these currents experience forces in the field that oppose the relative motion, giving a torque against the rotation
✓ 1
Part (b)
P = τω = 3.0 × 40 = 120 W
✓ 1
Part (c)
The torque doubles to 6.0 N m because it is proportional to ω
✓ 1
P = 6.0 × 80 = 480 W, which is four times the power found in the previous part; allow error carried forward from the previous part
✓ 1
Part (d)
The torque increases, because the stronger field gives a greater rate of change of flux and so larger eddy currents
✓ 1
Common error: If torque is proportional to ω, then the power P = τω is proportional to ω². Doubling the speed quadruples the power; it does not just double it.
64IBP2H-034-01-10Magnetic flux and flux linkagePaper 2Medium8 marks
A small search coil of 20 turns, each of area 4.0 cm2, is used to monitor the field inside a medical MRI scanner. The uniform magnetic flux density is 1.5 T, and the normal to the plane of the coil makes an angle of 30° with the field. The coil is connected to a meter, and the total resistance of the circuit is 2.5 Ω.
Side view: the coil is seen edge-on; not to scale
(a)
Calculate the magnetic flux linkage of the coil.
(2)
(b)
For a service check, the field is reduced uniformly to zero in 12 s. Calculate the average emf induced in the coil.
(2)
(c)
Determine the total charge that flows through the meter while the field is reduced.
(2)
(d)
Use Lenz's law to describe the direction of the magnetic field produced by the induced current, and explain your answer.
(2)
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Notes
Part (a)
NΦ = NBA cos θ = 20 × 1.5 × 4.0 × 10−4 × cos 30°
✓ 1
NΦ = 1.0 × 10−2 Wb (0.0104 Wb)
✓ 1
Part (b)
ε = Δ(NΦ)/Δt = 1.04 × 10−2/12 (ECF from (a))
✓ 1
ε = 8.7 × 10−4 V
✓ 1
Part (c)
Q = Δ(NΦ)/R, or Q = It with I = ε/R
✓ 1
Q = 1.04 × 10−2/2.5 = 4.2 × 10−3 C (ECF from (a))
✓ 1
Part (d)
The induced current's field inside the coil is in the same direction as the scanner's field . The scanner's flux is decreasing, so the induced current acts to oppose the decrease, tending to maintain the flux .
✓ 2
Common error: θ is measured from the normal to the coil, so use cos θ. Using sin θ with this angle gives the wrong flux.
65IBP2H-034-01-12Flux and charge in a rotated coilPaper 2Medium7 marks
An engineer measures the field between the poles of an industrial electromagnet with a 'flip coil'. The coil has 150 turns, each of area 2.0 cm2, and is connected to a charge-measuring meter. The total circuit resistance is 40 Ω. The coil starts with its plane perpendicular to the field and is quickly rotated through 180°. The meter records a charge of 3.6 × 10−5 C.
(a)
Show that the magnitude of the change in flux linkage is 2NBA.
(2)
(b)
Show that the magnitude of the charge flowing is given by Q = Δ(NΦ)/R.
(2)
(c)
A technician suggests that flipping the coil faster would give a larger charge reading. Evaluate this suggestion.
(1)
(d)
Determine the magnetic flux density of the electromagnet.
(2)
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Notes
Part (a)
Initially NΦ = +NBA. After rotation through 180° the field passes through the coil in the opposite sense, so NΦ = −NBA .
✓ 1
Magnitude of change = NBA − (−NBA) = 2NBA
✓ 1
Part (b)
Uses Faraday's law, average emf magnitude ε = Δ(NΦ)/Δt, so the average current is I = ε/R = Δ(NΦ)/(RΔt). Accept instantaneous form I = (1/R)·d(NΦ)/dt. Ignore sign .
✓ 1
Q = IΔt = Δ(NΦ)/R, or Q = ∫I dt = Δ(NΦ)/R
✓ 1
Part (c)
Incorrect: Q depends only on Δ(NΦ) and R, not on the time taken. A faster flip gives a larger current for a shorter time, so the charge is unchanged .
66IBP2H-034-01-13A rectangular coil rotating in a uniform fieldPaper 2Medium9 marks
A prototype generator for a small wind turbine has a coil of 200 turns, each of area 1.5 × 10−3 m2, rotating in a uniform magnetic field. The peak emf ε0 is measured at several rotation frequencies f:
67IBP2H-034-01-15Peak emf of a rotating coilPaper 2Medium6 marks
Each metal blade of a helicopter rotor has a length of 5.0 m from the hub to the tip. The rotor turns in a horizontal plane with a frequency of 6.0 Hz. The vertical component of the Earth's magnetic field is 5.0 × 10−5 T.
(a)
Show that the emf between the hub and the tip of one blade is ε = ½Bωr2, where ω is the angular speed and r is the length of the blade.
(2)
(b)
Calculate the emf between the hub and the tip of one blade.
(2)
(c)
The rotor has two blades pointing in opposite directions. State and explain the potential difference between the two blade tips.
(2)
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Part (a)
In time Δt the blade turns through ωΔt and sweeps out an area ½r2ωΔt . The flux cut is B × ½r2ωΔt, so ε = flux cut per unit time = ½Bωr2 .
✓ 2
Part (b)
ω = 2π × 6.0 = 37.7 rad s−1
✓ 1
ε = ½ × 5.0 × 10−5 × 37.7 × 5.02 = 2.4 × 10−2 V
✓ 1
Part (c)
The potential difference is zero . Each blade has the same emf between hub and tip, in the same sense, so both tips are at the same potential relative to the hub .
✓ 2
Common error: For a rotating rod, different points move at different speeds, so BvL needs the average speed ½ωr. Using the tip speed doubles the answer.
68IBP3H-D4-20r9-14Rectangular coil moving through a fieldPaper 2Medium9 marks
A library security gate contains a rectangular coil of 240 turns, 0.10 m by 0.15 m, with its plane perpendicular to a uniform magnetic field. The field strength decreases uniformly from 0.080 T to 0.020 T in 0.150 s.
(a)
State Faraday's law of electromagnetic induction.
(2)
(b)
Calculate the average emf induced in the coil.
(3)
(c)
The field points into the plane of the page. Explain the direction of the induced current seen from the viewer.
(2)
(d)
The same change in field now takes place in 0.075 s. Calculate the new average emf.
(2)
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Notes
Part (a)
R1 the induced emf is proportional to the rate of change of magnetic flux linkage
69IBP3H-D4-20r21-11Electromagnetic flowmeterPaper 2Medium8 marks
An electromagnetic flowmeter is used to measure the speed of blood in an artery. A uniform magnetic field of magnitude 0.10 T is applied perpendicular to the artery, whose internal diameter is 5.0 mm. Two electrodes touch opposite sides of the artery wall, along a line perpendicular to both the field and the flow. The blood contains positive and negative ions.
(a)
Show that the potential difference between the electrodes is ε = Bvd, where v is the speed of the blood and d the diameter of the artery.
(3)
(b)
The blood moves at 0.42 m s−1. Calculate the potential difference between the electrodes.
(2)
(c)
The artery narrows to a diameter of 3.0 mm. The volume flow rate is unchanged. Determine the potential difference measured by electrodes placed on the narrowed section.
(3)
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Notes
Part (a)
A1 ions of charge q moving with speed v experience a magnetic force Bqv, which pushes opposite charges to opposite sides
✓ 1
M1 charge separation continues until the electric force qE balances the magnetic force: qE = Bqv
70IBP3H-D4-20r21-12A magnet falling through a coilPaper 2Medium7 marks
A bar magnet of mass 0.045 kg is released from rest 0.80 m above the top of a coil of 120 turns and falls with its axis vertical along the axis of the coil. Air resistance is negligible before the magnet reaches the coil. The coil is connected to a data logger.
(a)
Calculate the speed of the magnet as it reaches the coil.
(2)
(b)
The data logger records an emf peak of opposite sign, and larger magnitude, as the magnet leaves the coil compared with when it entered. Explain both features.
(3)
(c)
As the magnet enters the coil, the flux through each turn increases from 0 to 1.8×10−4 Wb in 12 ms. Calculate the average emf induced.
(2)
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Part (a)
M1 v2 = 2gh (or mgh = ½mv2)
✓ 1
A1 v = 3.96 m s−1
✓ 1
Part (b)
A1 the flux linkage increases as the magnet enters and decreases as it leaves, so the emf has the opposite sign by Lenz's law
✓ 1
A1 the magnet is moving faster on leaving, because gravity exceeds the retarding magnetic force and it still accelerates
✓ 1
A1 faster motion means a greater rate of change of flux linkage, so a larger emf magnitude
✓ 1
Part (c)
M1 ε = NΔΦ/Δt = 120 × 1.8×10−4 / 0.012
✓ 1
A1 1.8 V
✓ 1
Common error: Do not equate the magnetic force with the weight while the magnet is still accelerating; the magnetic force is smaller than the weight in this situation.
71IBP3H-D4-20r21-13Motional emfPaper 2Medium8 marks
A conducting rod of length 0.35 m slides at a constant speed of 1.8 m s−1 along two horizontal parallel rails connected by a resistor of resistance 2.5 Ω. The rails and rod have negligible resistance. A uniform vertical magnetic field of magnitude 0.60 T acts over the whole circuit. Use unrounded values in later parts.
Top view; not to scale
(a)
Calculate the emf induced in the rod.
(2)
(b)
Calculate the current in the resistor.
(1)
(c)
Calculate the horizontal force needed to keep the rod moving at constant speed.
(2)
(d)
Calculate the power dissipated in the resistor and use it to show that energy is conserved in the system.
(3)
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Notes
Part (a)
M1 ε = BLv = 0.60 × 0.35 × 1.8
✓ 1
A1 0.378 V (0.38 V)
✓ 1
Part (b)
A1 I = ε/R = 0.378 / 2.5 = 0.15 A (0.1512 A)
✓ 1
Part (c)
M1 F = BIL = 0.60 × 0.1512 × 0.35
✓ 1
A1 0.0318 N
✓ 1
Part (d)
M1 P = I2R = 0.15122 × 2.5 = 0.0572 W
✓ 1
M1 mechanical power supplied P = Fv = 0.03175 × 1.8 = 0.0572 W
✓ 1
A1 the two powers are equal, so the work done by the applied force is all transferred to thermal energy in the resistor
✓ 1
Exam tip: Check energy conservation by comparing Fv with I²R or ε²/R; they must agree.
72IBP3H-D4-20r21-14Peak emf of a rotating coilPaper 2Medium7 marks
A coil of 250 turns and area 0.040 m2 rotates at a constant frequency of 2.5 Hz in a uniform magnetic field of magnitude 0.30 T, as part of a small wind-turbine generator. At time t = 0 the plane of the coil is perpendicular to the field. Use unrounded values in later parts.
(a)
Calculate the peak emf generated.
(3)
(b)
Calculate the root mean square emf.
(2)
(c)
State the emf at t = 0 and determine the time at which the emf first reaches its peak value.
(2)
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Part (a)
M1 ω = 2πf = 15.7 rad s−1
✓ 1
M1 ε0 = NBAω = 250 × 0.30 × 0.040 × 15.7
✓ 1
A1 47 V (47.1 V)
✓ 1
Part (b)
M1 εrms = ε0/√2
✓ 1
A1 33 V (33.3 V)
✓ 1
Part (c)
A1 emf is zero at t = 0, because the flux is at its maximum and not changing
✓ 1
A1 a quarter of a period: t = 1/(4 × 2.5) = 0.10 s
✓ 1
Common error: When the coil plane is perpendicular to the field the flux is maximum but the emf is zero; the emf peaks when the coil plane is parallel to the field.
73D-2-10Motional emf and Lenz's lawPaper 2Hard13 marks
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)
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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)
74D-2-30Motional emf and Lenz's lawPaper 2Hard13 marks
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)
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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)
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)
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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)
76D-2-36Electromagnetic flowmeterPaper 2Hard14 marks
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)
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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)
77D-2-37An electrodynamic tetherPaper 2Hard19 marks
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)
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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)
78IBP2H-034-00-14A rectangular coil rotating in a uniform fieldPaper 2Hard8 marks
A small wind turbine drives an a.c. generator. The rectangular coil has an area of 0.060 m2 and rotates at 25 revolutions per second in a uniform magnetic field of flux density 0.25 T. The design must give a peak emf of 340 V. The resistance of the coil may be neglected.
(a)
Determine the minimum number of turns needed on the coil.
(2)
(b)
In calm conditions the coil rotates at 12.5 revolutions per second. Describe how the emf–time graph for these conditions differs from the graph for 25 revolutions per second, giving the new peak emf and the new period.
(2)
(c)
At 25 revolutions per second, the generator supplies a purely resistive load of 60 Ω. Calculate the average power delivered to the load.
(2)
(d)
Explain why the wind must exert a larger torque to keep the coil rotating at the same rate once the load is connected.
(2)
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Part (a)
ω = 2π × 25 = 157 rad s−1 and N = ε0/(BAω)
✓ 1
N = 340/(0.25 × 0.060 × 157) = 144.3, so a minimum of 145 turns, since fewer turns give a peak emf below 340 V
✓ 1
Part (b)
The graph is still sinusoidal, and the peak emf halves to 170 V because ε0 ∝ ω
✓ 1
The period doubles from 40 ms to 80 ms
✓ 1
Part (c)
Vrms = 340/√2 = 240 V
✓ 1
P = Vrms2/R = 2402/60 ≈ 963 W (accept 960 W)
✓ 1
Part (d)
Without the load no current flows, so there is no opposing magnetic torque (friction ignored). With the load connected, a current flows in the coil and, by Lenz's law, the magnetic force on the current-carrying coil produces a torque that opposes its rotation
✓ 1
Work must be done against this torque at the rate the load dissipates electrical energy (torque × ω = electrical power), so a larger driving torque is needed
✓ 1
Common error: Use the rms value, not the peak value, when calculating average power: P = ε₀²/(2R), not ε₀²/R.
In a car ignition system, a current of 4.0 A flows in the primary coil of an ignition coil. The current is switched off and falls to zero in 0.20 ms. The primary coil has a self-inductance L = 6.0 mH. The self-induced emf is given by ε = −L ΔI/Δt, and the energy stored in the coil is ½LI2.
(a)
Explain why an emf is induced in the primary coil itself when the current is switched off.
(2)
(b)
Calculate the magnitude of the average self-induced emf.
(2)
(c)
Calculate the energy stored in the primary coil just before the switch opens.
(1)
(d)
A secondary coil with 100 times as many turns as the primary is wound on the same core, so the same changing flux links every turn. Estimate the emf induced across the secondary coil.
(2)
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Notes
Part (a)
The current in the coil produces magnetic flux that links the coil's own turns, so when the current falls, the flux linkage of the coil changes
✓ 1
By Faraday's law a changing flux linkage induces an emf, which by Lenz's law opposes the fall in current
✓ 1
Part (b)
ε = LΔI/Δt = (6.0 × 10−3 × 4.0) / (0.20 × 10−3)
✓ 1
ε = 120 V
✓ 1
Part (c)
E = ½ × 6.0 × 10−3 × 4.02 = 0.048 J
✓ 1
Part (d)
The emf per turn is the same in both coils, so the secondary emf is 100 times the primary emf
✓ 1
ε ≈ 100 × 120 = 1.2 × 104 V. Allow an error carried forward from the average emf calculated earlier
✓ 1
Exam tip: The very large spark voltage comes from the very small Δt. When explaining why the emf is large, state explicitly that the current is switched off rapidly.
80IBP2H-034-01-11A magnet oscillating above a coilPaper 2Hard9 marks
A bar magnet of mass 0.20 kg hangs from a spring of spring constant 20 N m−1. It oscillates vertically with amplitude 3.0 cm just above a fixed coil whose axis is vertical. The coil is connected to a resistor, and the total resistance of the circuit is 15 Ω. The emf induced in the coil is approximately sinusoidal, with a peak value of 45 mV.
(a)
Calculate the frequency of the induced emf.
(2)
(b)
Explain why the emf is zero when the magnet is at its lowest point, even though the flux linkage of the coil is then a maximum.
(2)
(c)
Explain why the oscillations die away more quickly when the resistor is connected than when the coil is on open circuit.
(2)
(d)
Estimate the fraction of the oscillation energy converted to thermal energy during one cycle.
(3)
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Part (a)
T = 2π√(m/k) = 2π√(0.20/20) = 0.63 s
✓ 1
The emf has the same frequency as the oscillation: f = 1/T = 1.6 Hz
✓ 1
Part (b)
The emf is proportional to the rate of change of flux linkage (Faraday's law) . At the lowest point the magnet is momentarily at rest, so the flux is not changing and the emf is zero .
✓ 2
Part (c)
With a closed circuit a current flows, and its magnetic field opposes the motion of the magnet (Lenz's law), giving a damping force . Energy of oscillation is transferred to thermal energy in the resistance; on open circuit there is no current, so there is no such transfer .
✓ 2
Part (d)
Mean power P = ε02/2R = 0.0452/(2 × 15) = 6.8 × 10−5 W
✓ 1
Oscillation energy E = ½kA2 = ½ × 20 × 0.0302 = 9.0 × 10−3 J
81IBP2H-034-01-16A rectangular coil rotating in a uniform fieldPaper 2Hard10 marks
A portable emergency generator contains a coil that rotates at 1800 revolutions per minute in a uniform magnetic field. The maximum flux linkage of the coil is 0.12 Wb. The coil has a resistance of 3.0 Ω and is connected to an external resistor of 15 Ω.
(a)
Calculate the angular frequency of the coil.
(1)
(b)
Calculate the peak emf.
(2)
(c)
State the orientation of the plane of the coil relative to the field when the emf is a maximum, and explain your answer.
(2)
(d)
Determine the mean power dissipated in the external resistor.
(3)
(e)
Explain why the coil is harder to turn when the resistor is connected than when the generator is on open circuit.
(2)
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Part (a)
f = 1800/60 = 30 Hz, so ω = 2π × 30 = 1.9 × 102 rad s−1 (188 rad s−1)
✓ 1
Part (b)
ε0 = ω(NΦ)max = 188 × 0.12 (ECF from (a))
✓ 1
ε0 = 23 V (22.6 V)
✓ 1
Part (c)
The plane of the coil is parallel to the field, so the flux linkage is zero . At this orientation the flux linkage is changing at its greatest rate, so the emf is a maximum .
✓ 2
Part (d)
I0 = ε0/(R + r) = 22.6/18 = 1.26 A (ECF from (b))
✓ 1
Irms = 1.26/√2 = 0.889 A
✓ 1
P = Irms2R = 0.8892 × 15 = 12 W (11.8 W)
✓ 1
Part (e)
When connected, a current flows in the coil in the magnetic field, producing a torque that opposes the rotation (Lenz's law) . Work must be done against this torque to supply the electrical energy dissipated in the resistances; on open circuit there is no current and so no opposing torque .
✓ 2
Common error: Use the total resistance, including the coil's internal resistance, to find the current. Then use only the external resistance to find the power delivered to that resistor.
82IBP3H-D4-20r9-13A rectangular coil rotating in a uniform fieldPaper 2Hard12 marks
A generator coil of 80 turns and area 0.050 m2 rotates at 6.0 Hz in a uniform field of 0.60 T, driven by a wind turbine. At t = 0 the plane of the coil is perpendicular to the field and the emf is zero. The coil, whose resistance is negligible, is connected across a 15 Ω load. Give numerical answers to three significant figures.
(a)
Calculate the peak emf.
(3)
(b)
Calculate the mean power dissipated in the load.
(3)
(c)
Determine the time after \(t=0\) at which the emf first reaches 60 V.
(3)
(d)
In a gale the rotation frequency doubles. Determine the new mean power in the load.
(3)
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Part (a)
M1 \(\omega=2\pi\times6.0=37.7\) rad s−1
✓ 1
M1 \(\varepsilon_0=NBA\omega\)
✓ 1
A1 \(80\times0.60\times0.050\times37.7=90.5\) V
✓ 1
Part (b)
M1 \(\varepsilon_{rms}=\varepsilon_0/\sqrt2=64.0\) V (ECF from (a))
✓ 1
M1 \(P=\varepsilon_{rms}^2/R\)
✓ 1
A1 \(273\) W
✓ 1
Part (c)
M1 \(\varepsilon=\varepsilon_0\sin\omega t\)
✓ 1
M1 \(\sin\omega t=\frac{60}{90.5}\) giving \(\omega t=0.725\) rad
✓ 1
A1 \(t=0.0192\) s
✓ 1
Part (d)
M1 \(\varepsilon_0\) doubles because \(\varepsilon_0\propto\omega\)
✓ 1
M1 \(P\propto\varepsilon_0^2\), so the mean power increases by a factor of 4
✓ 1
A1 \(1090\) W (1.09 × 103 W)
✓ 1
Common error: Doubling the speed of a generator quadruples the mean power into a fixed resistor, because power depends on the square of the emf.
83IBP3H-D4-20r21-15Rectangular coil moving through a fieldPaper 2Hard12 marks
A rectangular conducting loop of width 0.20 m and length 0.30 m has resistance 0.050 Ω. It moves at a constant speed of 1.5 m s−1 in the direction of its length, towards a region of uniform magnetic field 0.40 T directed into the page. The field region has sharp edges and extends 0.50 m in the direction of motion. The loop is entirely in the plane of the page and its leading edge enters the field at t = 0.
Top view; not to scale
(a)
Calculate the emf induced in the loop while it is entering the field.
(2)
(b)
Calculate the current in the loop while it is entering the field and state its direction.
(2)
(c)
Determine the magnetic force on the loop while it is entering the field, including its direction.
(2)
(d)
Describe the graph of emf against time from t = 0 until the loop has completely left the field, taking the entering emf as positive.
(3)
(e)
Calculate the total thermal energy dissipated in the loop while it is entering the field.
(3)
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Notes
Part (a)
M1 ε = Bwv = 0.40 × 0.20 × 1.5
✓ 1
A1 0.12 V
✓ 1
Part (b)
A1 I = 0.12/0.050 = 2.4 A
✓ 1
A1 anticlockwise, since the induced field must point out of the page to oppose the increasing flux into the page
✓ 1
Part (c)
A1 F = BIw = 0.40 × 2.4 × 0.20 = 0.19 N (0.192 N)
✓ 1
A1 directed opposite to the velocity of the loop
✓ 1
Part (d)
A1 0.12 V from t = 0 to t = 0.20 s while the loop enters
✓ 1
A1 zero from t = 0.20 s to t = 0.33 s while the loop is entirely inside the field
✓ 1
A1 −0.12 V (opposite sign, same magnitude) from t = 0.33 s to t = 0.53 s while the loop leaves
✓ 1
Part (e)
M1 time to enter t = 0.30/1.5 = 0.20 s
✓ 1
M1 P = ε2/R = 0.122/0.050 = 0.288 W, so E = Pt
✓ 1
A1 E = 0.0576 J (equivalently F × 0.30 m)
✓ 1
Exam tip: Check your answer for energy by comparing the heat produced with the work done by the force that keeps the loop moving.
84IBP3H-D4-20r21-16Motional emf and Lenz's lawPaper 2Hard9 marks
A magnet of mass 0.050 kg falls vertically through a long solenoid whose ends are connected to a resistor, so that the total resistance of the circuit is 4.0 Ω. After falling a certain distance the magnet moves through the solenoid at a constant terminal speed of 0.80 m s−1. Air resistance is negligible and all electrical energy is transferred to thermal energy in the circuit.
(a)
Determine the magnitude of the magnetic force that the solenoid exerts on the magnet at terminal speed.
(2)
(b)
Calculate the rate at which thermal energy is produced in the circuit.
(2)
(c)
Determine the emf induced in the solenoid at terminal speed.
(2)
(d)
The total resistance of the circuit is halved. Assuming that the induced emf is proportional to the speed of the magnet, determine the new terminal speed.
(3)
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Part (a)
A1 constant speed, so the resultant force is zero
✓ 1
A1 magnetic force = weight = 0.050 × 9.8 = 0.49 N upward
✓ 1
Part (b)
M1 power = rate of loss of gravitational potential energy = mgv
✓ 1
A1 0.49 × 0.80 = 0.39 W (0.392 W)
✓ 1
Part (c)
M1 P = ε2/R so ε = √(PR) = √(0.392 × 4.0) (ECF from (b))
✓ 1
A1 1.25 V
✓ 1
Part (d)
M1 ε = kv, and at terminal speed mgv = ε2/R = k2v2/R
✓ 1
A1 so v = mgR/k2, which is proportional to R
✓ 1
A1 new terminal speed = 0.80/2 = 0.40 m s−1
✓ 1
Common error: At terminal speed the magnet's kinetic energy is constant, so all the gravitational energy lost becomes thermal energy; do not include a kinetic energy term.
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