Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
Electromagnetic induction
40 free, high-quality past-paper-style questions on Electromagnetic induction for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 145 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U4-6-A-01Unit 4 · A2Easy1 mark
A bar magnet is pushed into a coil that is connected to a sensitive meter. Which change would not increase the maximum e.m.f. induced in the coil?
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Answer
Mark
Notes
The e.m.f. depends on the rate of change of flux linkage. Speed, magnet strength and number of turns all change that rate. Thicker wire lowers the resistance of the coil, which would increase the current, but the flux linkage and its rate of change are unchanged, so the e.m.f. is the same.
✓ 1
Answer D
Answer: D
Common error: Mixing up e.m.f. and current. Resistance affects the current that an e.m.f. drives, not the e.m.f. itself.
A coil has 400 turns. The magnetic flux through each turn falls steadily from 3.0 × 10−4 Wb to zero in 0.060 s. What is the e.m.f. induced in the coil?
Two aluminium plates of the same size are made to swing as pendulums through the gap between the poles of a strong magnet. One plate is solid; the other has a series of narrow slots cut into it. Which statement is correct?
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Answer
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Notes
Moving through the field induces eddy currents in the plates, and by Lenz's law these oppose the motion. The slots break up the current loops, so the eddy currents and the braking force are much smaller in the slotted plate: the solid plate stops first.
✓ 1
Answer B
Answer: B
Exam tip: Magnetic braking does not depend on the metal being magnetic. Any conductor moving through a field has eddy currents induced in it.
A student connects a coil to a sensitive centre-zero ammeter. When she pushes the north pole of a bar magnet into the coil, the needle deflects briefly to the right.
(a)
State what the needle shows while the magnet is held stationary inside the coil.
(1)
(b)
State what the needle shows when the magnet is pulled out of the coil at the same speed.
(1)
(c)
State two changes that would give a larger deflection when the magnet is pushed in.
(2)
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Answer
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Notes
Part (a)
no deflection / reads zero
✓ 1
Part (b)
a brief deflection to the left, of about the same size
✓ 1
Part (c)
any two of: push the magnet in faster / use a stronger magnet / use a coil with more turns
✓ 2
Exam tip: “State” needs a short answer only. For (c), give two different changes; two ways of saying “faster” score once.
A small search coil has 60 turns, each of area 1.5 × 10−4 m2. It is held with its plane perpendicular to a uniform magnetic field of flux density 0.12 T. It is then pulled out of the field in 0.20 s.
(a)
Calculate the flux linkage of the coil while it is in the field.
(2)
(b)
Calculate the average e.m.f. induced while the coil is pulled out of the field.
(2)
(c)
Explain why the average e.m.f. would be larger if the coil were pulled out more quickly.
(1)
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Answer
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Notes
Part (a)
Nφ = NBA = 60 × 0.12 × 1.5 × 10−4
✓ 1
= 1.1 × 10−3 Wb (1.08 × 10−3 Wb)
✓ 1
Part (b)
ε = Δ(Nφ)/Δt = 1.08 × 10−3 ÷ 0.20
✓ 1
= 5.4 × 10−3 V — allow error carried forward from (a)
✓ 1
Part (c)
the same change in flux linkage happens in less time, so the rate of change of flux linkage is greater
✓ 1
Common error: Leaving out the number of turns. Faraday's law uses the flux linkage Nφ, not the flux through one turn.
The graph shows how the flux linkage of a coil changes during an experiment. What is the greatest magnitude of the e.m.f. induced in the coil between t = 0 and t = 60 ms?
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Answer
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Notes
The magnitude of the e.m.f. is the magnitude of the gradient. 0–20 ms: 3.0 × 10−3 ÷ 0.020 = 0.15 V. 20–30 ms: (3.0 − (−1.0)) × 10−3 ÷ 0.010 = 0.40 V. 30–60 ms: 3.0 × 10−3 ÷ 0.030 = 0.10 V. The greatest is 0.40 V, on the steepest section.
✓ 1
Answer D
Answer: D
Common error: Stopping a change at the time axis. The flux linkage falls from +3.0 to −1.0 × 10⁻³ Wb, a change of 4.0 × 10⁻³ Wb, not 3.0 × 10⁻³ Wb.
The diagram shows four arrangements of a pair of coils. In each, the same alternating current flows in the primary coil, and the secondary coil is connected to a voltmeter. In which arrangement is the e.m.f. induced in the secondary coil largest?
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Answer
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Notes
The e.m.f. in the secondary depends on the rate of change of its flux linkage. In X the coils are close together, share an axis, and the soft-iron rod concentrates the alternating flux of the primary and carries almost all of it through the secondary, so its flux linkage, and the rate at which that changes, is largest.
✓ 1
Answer B
Answer: B
Exam tip: For linked coils, anything that sends more of the primary's changing flux through the secondary — closeness, a shared axis, an iron core — increases the induced e.m.f.
The diagram shows two identical coils, each with a soft-iron rod and an aluminium ring resting on top. Ring X is complete; ring Y is identical except for a narrow cut. Each coil is connected to an alternating supply through a switch, and both switches are closed at the same moment. What happens?
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Answer
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Notes
The alternating current produces an alternating flux in the rod, which links each ring. An e.m.f. is induced in both rings, but only in the complete ring X can it drive a current round the ring. By Lenz's law that current opposes the change in flux, so the ring is repelled and jumps. The cut in ring Y means no current flows, so there is no force on it.
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Answer A
Answer: A
Exam tip: An e.m.f. is induced whenever the flux linkage changes, but a force needs a current, and a current needs a complete path.
Two coils, P and S, are wound on a soft-iron ring as shown. When the switch in the circuit of coil P is closed, the needle of the centre-zero meter connected to coil S flicks briefly to the right and then returns to zero. The switch is left closed for a minute and is then opened. What does the meter show when the switch is opened?
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Answer
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Notes
Opening the switch makes the current in P, and so the flux in the ring, fall quickly to zero. The flux linkage of S changes for a short time in the opposite sense to when the switch was closed, so the e.m.f. and current in S are reversed: a brief deflection to the left.
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Answer D
Answer: D
Common error: Thinking an e.m.f. needs a current in the primary. It needs a changing flux: the switching-off moment gives a pulse just as switching on does, but in the opposite direction.
A student drops a small magnet through a series of coils of the same diameter but with different numbers of turns N. A data logger records the peak e.m.f. induced each time. Her results and a line of best fit are shown on the graph.
(a)
State two variables that she should keep the same.
(2)
(b)
Determine the gradient of the line of best fit.
(2)
(c)
Explain whether the results show that the peak e.m.f. is proportional to N.
(1)
(d)
Use the line to predict the peak e.m.f. for a coil of 400 turns.
(1)
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Answer
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Notes
Part (a)
any two of: the height from which the magnet is dropped / the magnet used (its strength) / the diameter (area) of the coils / dropping the same pole first, along the axis of the coil
✓ 2
Part (b)
use a large triangle, e.g. 60 mV ÷ 250 turns
✓ 1
= 0.24 mV per turn (allow 0.23–0.25)
✓ 1
Part (c)
yes: the points lie close to a straight line that passes through the origin
✓ 1
Part (d)
0.24 × 400 = 96 mV (allow 92–100 mV)
✓ 1
Exam tip: For a gradient, draw a large triangle on the line itself, not between two data points, and read its corners at gridlines.
The diagram shows cross-sections of two transformer cores, X and Y, of the same size and material. The same alternating magnetic flux passes along each core, at right angles to the page. The red loops show the paths of the induced currents in each core.
(a)
Explain why currents are induced in core X.
(2)
(b)
State which core wastes less energy, and explain why using the diagram.
(2)
(c)
State the energy transfer caused by the induced currents.
(1)
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Answer
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Notes
Part (a)
the alternating flux means the flux linking loops within the iron is changing, so e.m.f.s are induced in the core
✓ 1
iron is a conductor, so these e.m.f.s drive (eddy) currents round closed loops in it
✓ 1
Part (b)
core Y
✓ 1
the insulating layers stop the large current loop; each loop is confined to a thin sheet, so it encloses less flux (smaller e.m.f.) and has a higher resistance, so the eddy currents are much smaller
✓ 1
Part (c)
electrical energy is transferred to thermal energy, heating the core
✓ 1
Exam tip: Eddy currents need both a changing flux and a closed conducting path. Laminating a core attacks the conducting path, not the flux.
The graph of flux linkage against time for a coil is a straight line from 0.050 Wb at t = 0 to 0.010 Wb at t = 0.25 s. What is the magnitude of the e.m.f. induced in the coil during this time?
Figure: flux linkage of the coil against time
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Answer
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Notes
The e.m.f. is minus the gradient: ε = −(0.010 − 0.050) ÷ 0.25 = +0.16 V. Magnitude 0.16 V.
✓ 1
Answer B
Answer: B
Exam tip: A gradient uses the change in flux linkage between two points, not a single reading divided by a time.
A straight copper wire 0.60 m long is moved at 3.0 m s−1 through a uniform magnetic field of flux density 0.15 T. The wire, its velocity and the field are all at right angles to each other. What is the e.m.f. induced between the ends of the wire?
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Answer
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Notes
ε = Blv = 0.15 × 0.60 × 3.0 = 0.27 V.
✓ 1
Answer B
Answer: B
Exam tip: ε = Blv is Faraday's law for a moving rod: the rod sweeps an area lv every second, so it cuts a flux Blv every second.
The north pole of a bar magnet is pulled away from end P of a coil. The coil is part of a complete circuit. While the magnet is moving away, how does end P of the coil behave?
Figure: the north pole of the magnet moving away from end P of the coil (not to scale)
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Answer
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Notes
By Lenz's law the induced current opposes the change. The change is the magnet moving away, so the coil must attract it: end P becomes a south pole, attracting the north pole.
✓ 1
Answer A
Answer: A
Common error: Lenz's law does not always mean repulsion. The induced current opposes the change: approaching magnets are repelled, receding magnets are attracted.
Two coils are wound on the same iron core. A steady direct current flows in the first coil. A voltmeter connected across the second coil reads zero. Why is the reading zero?
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Answer
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Notes
A steady current produces a steady magnetic flux in the core. An e.m.f. is induced only when the flux linkage changes, so the constant flux linkage of the second coil induces nothing.
✓ 1
Answer D
Answer: D
Exam tip: In any linked-coil or transformer question, look for the word that tells you whether the primary current is changing. No change, no e.m.f.
A flat coil rotates at a constant rate about an axis at right angles to a uniform magnetic field. At which position is the e.m.f. induced in the coil zero?
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Answer
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Notes
With the plane of the coil perpendicular to the field, the flux linkage is at its maximum. At a maximum the flux linkage is momentarily not changing, so the e.m.f. (minus its gradient) is zero.
✓ 1
Answer A
Answer: A
Common error: Linking a large e.m.f. to a large flux linkage. The e.m.f. is zero when the flux linkage is largest and largest when the flux linkage is zero.
A car has a straight steel roof bar 1.2 m long, mounted horizontally and at right angles to the direction of travel. The car travels along a level road at 25 m s−1. At this place the vertical component of the Earth's magnetic field is 4.6 × 10−5 T.
(a)
Calculate the e.m.f. induced between the ends of the roof bar.
(2)
(b)
The bar is not part of a complete circuit. Explain why there is an e.m.f. between its ends but no continuing current in it.
(2)
(c)
The car turns round and travels in the opposite direction at the same speed. State the effect on the e.m.f.
(1)
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Answer
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Notes
Part (a)
ε = Blv = 4.6 × 10−5 × 1.2 × 25
✓ 1
= 1.4 × 10−3 V (1.38 × 10−3 V)
✓ 1
Part (b)
free electrons moving with the bar feel a magnetic force along it, so charge collects at one end
✓ 1
there is no complete path for charge to flow round, so once the charges have separated the current stops
✓ 1
Part (c)
same magnitude, opposite polarity (the other end of the bar becomes positive)
✓ 1
Exam tip: Only the component of the field at right angles to both the bar and its velocity counts. For horizontal motion of a horizontal bar, that is the vertical component.
A small transformer has a primary coil and a secondary coil wound on the same laminated iron core. The primary coil is connected to an alternating supply.
(a)
Explain how an alternating e.m.f. is produced across the secondary coil.
(3)
(b)
State why the transformer would give no output if the primary were connected to a steady d.c. supply.
(1)
(c)
State one change to the secondary coil that would increase the e.m.f. across it.
(1)
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Answer
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Notes
Part (a)
the alternating current in the primary produces a changing / alternating magnetic flux in the core
✓ 1
the core carries this flux through the secondary, so the flux linkage of the secondary changes
✓ 1
by Faraday's law an e.m.f. is induced in the secondary, alternating at the same frequency as the supply
✓ 1
Part (b)
a steady current gives a constant flux, so the flux linkage of the secondary does not change and no e.m.f. is induced
✓ 1
Part (c)
increase the number of turns on the secondary
✓ 1
Common error: Writing that current or voltage “passes through the core” to the secondary. The coils are linked only by the changing magnetic flux.
The current in an electromagnet is controlled by a computer. A small coil sits in the field of the electromagnet. The flux linkage of the coil rises steadily from zero at t = 0 to 4.5 × 10−3 Wb at t = 15 ms. It then stays constant until t = 40 ms, and then falls steadily to −4.5 × 10−3 Wb at t = 70 ms.
Figure: flux linkage of the small coil against time
(a)
Calculate the e.m.f. induced in the coil between t = 0 and t = 15 ms.
(2)
(b)
Calculate the e.m.f. induced between t = 40 ms and t = 70 ms.
(2)
(c)
Describe the graph of e.m.f. against time from t = 0 to t = 70 ms.
a constant e.m.f. of −0.30 V from 0 to 15 ms, then zero from 15 ms to 40 ms
✓ 1
then a constant +0.30 V from 40 ms to 70 ms, with no change as the flux linkage passes through zero at 55 ms
✓ 1
Model sketch
Figure: e.m.f. against time (taking ε = −d(NΦ)/dt); the size is 0.30 V in both changing sections and zero while the flux linkage is constant
Common error: Thinking the e.m.f. becomes zero when the flux linkage passes through zero. On a straight line the gradient, and so the e.m.f., is the same all the way along.
A student investigates how the e.m.f. induced in a coil depends on the speed of a magnet. She holds a small magnet a height h above a horizontal coil and releases it so that it falls through the coil. A data logger connected to the coil records the peak e.m.f. She repeats this for several values of h.
(a)
Explain why the peak e.m.f. increases as h increases.
(2)
(b)
Air resistance is negligible, so the speed of the magnet at the coil is proportional to √h. The student predicts that the peak e.m.f. is proportional to this speed. State the graph she should plot to test this prediction, and what it would look like if the prediction is correct.
(2)
(c)
For h = 0.20 m the peak e.m.f. is 36 mV. Use the prediction to calculate the peak e.m.f. for h = 0.80 m.
(1)
(d)
Explain why a data logger is used rather than a voltmeter that is read by eye.
(1)
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Answer
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Notes
Part (a)
the magnet is moving faster when it reaches the coil
✓ 1
so the flux linkage of the coil changes at a greater rate, and the e.m.f. is larger (Faraday's law)
✓ 1
Part (b)
plot peak e.m.f. against √h
✓ 1
a straight line through the origin
✓ 1
Part (c)
√(0.80 ÷ 0.20) = 2, so 2 × 36 = 72 mV
✓ 1
Part (d)
the e.m.f. lasts only a small fraction of a second, too short to read a peak value by eye
✓ 1
Model sketch
Figure: if the prediction is correct, peak e.m.f. against √h is a straight line through the origin
Exam tip: To test a proportionality, choose axes that should give a straight line through the origin. Here e.m.f. ∝ v ∝ √h, so plot e.m.f. against √h, not against h.
The diagram shows the north pole of a bar magnet moving towards the left end of a coil. The front of each turn of the coil is drawn solid and the back is dashed. The ends of the coil are connected to a resistor between terminals X and Y. While the magnet approaches, what is the direction of the current in the resistor, and what does the left end of the coil become?
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Answer
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Notes
By Lenz's law the induced current opposes the approach, so the left end of the coil must repel the north pole: it becomes a north pole. Seen from the magnet, the current at that end must therefore flow anticlockwise. Following the winding drawn, an anticlockwise current seen from the left runs through the coil from the right-hand lead to the left-hand lead, so it leaves the coil at the left lead and flows through the resistor from X to Y.
✓ 1
Answer A
Answer: A
Exam tip: Do it in two steps: first decide the pole from Lenz's law (approaching magnet is repelled), then use the right-hand grip rule on the winding as drawn to find which lead the current leaves by.
The top graph shows how the flux linkage of a coil varies with time. Which of the graphs A to D shows the e.m.f. induced in the coil over the same time?
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Answer
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Notes
The e.m.f. is minus the gradient of the flux-linkage graph. From 0 to 4 ms the gradient is 6.0 × 10−3 Wb ÷ 0.004 s = +1.5 Wb s−1, so ε = −1.5 V. From 4 to 16 ms the gradient is −6.0 × 10−3 Wb ÷ 0.012 s = −0.50 Wb s−1, so ε = +0.50 V. Two constant values, −1.5 V then +0.50 V: graph C.
✓ 1
Answer C
Answer: C
Common error: Drawing the e.m.f. graph the same shape as the flux-linkage graph. A straight-line section of flux linkage gives a constant e.m.f., and a steeper section gives a larger one.
The diagram shows a square coil moving at constant velocity out of a region of uniform magnetic field. The number of turns, the size of the coil, its speed and the flux density are given on the diagram. What is the e.m.f. induced in the coil at the instant shown?
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Answer
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Notes
Only the left side of the coil is in the field, so only one side cuts field lines. The rate of loss of flux linkage is NBLv = 20 × 0.30 × 0.050 × 0.40 = 0.12 V.
✓ 1
Answer B
Answer: B
Exam tip: For a coil leaving a field, ask which sides are cutting field lines. Here the right side is already outside, the top and bottom move along their own length, so only one side counts.
A metal rod slides to the right along two horizontal rails in a uniform magnetic field directed into the page, as shown. The rails are joined by a resistor PQ. What is the direction of the induced current in the resistor, and the direction of the magnetic force on the rod?
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Answer
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Notes
The force on the moving charges in the rod is along v × B for positive charge: velocity to the right, field into the page, so conventional current is driven up the rod (towards the top rail). It flows along the top rail to P and down through the resistor from P to Q. The current in the rod is upwards in a field into the page, so the force BIl on the rod is to the left, opposing the motion, as Lenz's law requires.
✓ 1
Answer C
Answer: C
Exam tip: Check any answer with Lenz's law: the magnetic force on the rod must always oppose its motion. An answer with the force along the velocity cannot be right.
The diagram shows a rectangular coil moving at a constant velocity towards a region of uniform magnetic field. The field region has sharp edges and is taller than the coil. The plane of the coil stays perpendicular to the field.
(a)
Calculate the time it takes for the coil to move completely into the field.
(1)
(b)
Calculate the e.m.f. induced in the coil while it is moving into the field.
(2)
(c)
Explain why there is no e.m.f. while the coil is completely inside the field.
(1)
(d)
Describe the variation of e.m.f. with time from the moment the leading side reaches the field until the coil has completely left it. Give the times involved.
(2)
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Answer
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Notes
Part (a)
0.030 ÷ 0.050 = 0.60 s
✓ 1
Part (b)
only the leading side, of length 4.0 cm, cuts the field: ε = NBLv = 50 × 0.25 × 0.040 × 0.050
✓ 1
= 0.025 V
✓ 1
Part (c)
the whole coil is in a uniform field, so the flux linkage stays constant (both vertical sides cut field lines and their e.m.f.s cancel)
✓ 1
Part (d)
constant 0.025 V for 0.60 s, then zero for 1.4 s while the coil is completely inside
✓ 1
then 0.025 V in the opposite direction for 0.60 s as the coil leaves
✓ 1
Exam tip: Track the flux linkage first: increasing, constant, decreasing. The e.m.f. graph then follows from the gradient — constant, zero, constant with the opposite sign.
A metal rod is pulled to the right at a constant speed along two horizontal rails, as shown. The rails are joined at one end by a resistor PQ. A uniform magnetic field acts out of the page. The resistance of the rod and rails is negligible, and friction can be ignored.
(a)
Calculate the e.m.f. induced across the rod.
(2)
(b)
Calculate the current in the resistor and state its direction, from P to Q or from Q to P. Explain how you decided the direction.
(3)
(c)
Calculate the horizontal force needed to keep the rod moving at constant speed.
(2)
(d)
Show that the rate at which this force does work equals the rate at which energy is dissipated in the resistor.
(1)
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Answer
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Notes
Part (a)
ε = Blv = 0.40 × 0.25 × 1.5
✓ 1
= 0.15 V
✓ 1
Part (b)
I = 0.15 ÷ 0.50 = 0.30 A
✓ 1
from Q to P
✓ 1
by Lenz's law the force on the rod must oppose its motion (act to the left); with the field out of the page this needs a current down the rod, which returns up through the resistor (or by the right-hand dynamo rule)
✓ 1
Part (c)
F = BIl = 0.40 × 0.30 × 0.25
✓ 1
= 0.030 N
✓ 1
Part (d)
Fv = 0.030 × 1.5 = 0.045 W and I2R = 0.302 × 0.50 = 0.045 W, so they are equal
✓ 1
Exam tip: In a rod-on-rails question, the pulling force at constant speed equals the magnetic force BIl, and Fv must equal I²R. Use this as a check on your numbers.
Two coils are wound on the same soft-iron core. The current in the primary coil is controlled by a computer and varies as shown in the graph. The secondary coil is connected to a data logger. Between t = 0 and t = 20 ms the e.m.f. across the secondary coil is constant at 0.12 V.
(a)
Explain why there is an e.m.f. across the secondary coil between t = 0 and t = 20 ms.
(2)
(b)
State the e.m.f. across the secondary coil between t = 20 ms and t = 50 ms. Give a reason.
(1)
(c)
Determine the e.m.f. across the secondary coil between t = 50 ms and t = 60 ms, and compare its direction with that between 0 and 20 ms.
(2)
(d)
The iron core is replaced by a plastic tube of the same shape, and the primary current is the same. State the effect on the e.m.f.s.
(1)
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Answer
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Notes
Part (a)
the increasing current in the primary produces an increasing magnetic flux in the core
✓ 1
the core carries this flux through the secondary, so its flux linkage changes and an e.m.f. is induced (Faraday's law)
✓ 1
Part (b)
zero, because the primary current, and so the flux linkage of the secondary, is constant
✓ 1
Part (c)
the current changes by 2.0 A in 10 ms instead of 20 ms, so the rate of change is twice as great: 0.24 V
✓ 1
the current is decreasing, so the e.m.f. is in the opposite direction
✓ 1
Part (d)
all the e.m.f.s are smaller (much less flux links the secondary for the same current); the zero section stays zero
✓ 1
Exam tip: With linked coils, the secondary e.m.f. follows the gradient of the primary current–time graph: steeper means larger, flat means zero, falling means reversed.
A small magnet is released from rest just above a horizontal coil and falls vertically through it. The coil is connected to an oscilloscope, and the diagram shows the trace recorded. The Y-gain and time-base settings are given below the screen.
(a)
Determine the peak e.m.f. of each of the two pulses.
(2)
(b)
Determine the time between the two peaks.
(1)
(c)
Explain why the second pulse is in the opposite direction to the first.
(1)
(d)
Explain why the second pulse has a larger peak and is narrower than the first.
second pulse: 2.7 divisions → +27 mV (allow 26–28 mV)
✓ 1
Part (b)
about 2.1 divisions × 10 ms = 21 ms (allow 20–22 ms)
✓ 1
Part (c)
the flux linkage of the coil increases as the magnet enters and decreases as it leaves, so the rate of change has the opposite sign
✓ 1
Part (d)
the magnet is accelerating, so it is moving faster as it leaves the coil than as it enters
✓ 1
the flux linkage changes by a similar amount in a shorter time, so the rate of change, and the e.m.f., is larger
✓ 1
Exam tip: Read an oscilloscope in divisions first, using the small ticks for tenths, then multiply by the setting. Write both steps so a misread setting does not cost every mark.
A simple generator consists of a flat coil rotating at a constant rate about an axis at right angles to a uniform magnetic field. The graph shows how the e.m.f. induced in the coil varies with time. Point K is marked on the graph.
(a)
Use the graph to determine the rate of rotation of the coil in revolutions per second.
(2)
(b)
State the orientation of the plane of the coil relative to the magnetic field at point K, and explain your answer.
(2)
(c)
State the two times between t = 0 and t = 50 ms at which the flux linkage of the coil is zero.
(1)
(d)
The rate of rotation is reduced to 10 revolutions per second. State the new peak e.m.f. and the new period.
(2)
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Answer
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Notes
Part (a)
period = 50 ms
✓ 1
rate = 1 ÷ 0.050 = 20 rev s−1
✓ 1
Part (b)
the plane of the coil is perpendicular to the field lines
✓ 1
the flux linkage is at a maximum there and is momentarily not changing, so the e.m.f. is zero
✓ 1
Part (c)
12.5 ms and 37.5 ms, where the e.m.f. has its peaks
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Part (d)
peak e.m.f. 2.0 V
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period 100 ms
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Common error: Linking zero e.m.f. with zero flux. The e.m.f. is zero when the flux linkage is at a maximum, because the flux linkage is then not changing.
A coil of 150 turns and area 4.0 × 10−3 m2 rotates at 25 revolutions per second in a uniform magnetic field of flux density 0.050 T. What is the average e.m.f. induced while the coil turns from the position where its plane is perpendicular to the field to the position where its plane is parallel to the field?
Show mark scheme
Answer
Mark
Notes
The flux linkage falls from NBA = 150 × 0.050 × 4.0 × 10−3 = 0.030 Wb to zero. This takes a quarter of a turn: (1 ÷ 25) ÷ 4 = 0.010 s. Average ε = 0.030 ÷ 0.010 = 3.0 V.
✓ 1
Answer C
Answer: C
Exam tip: From face-on to edge-on is a quarter of a revolution, so it takes a quarter of the period.
A metal rod rests across two smooth horizontal rails 0.20 m apart. The rails are joined at one end by a resistor, and the total resistance of the circuit is 0.40 Ω. A uniform vertical magnetic field of flux density 0.50 T acts over the rails. A constant horizontal force of 0.050 N, at right angles to the rod, is applied to the rod, which starts from rest.
Figure: the rod on the rails, seen from above (not to scale)
(a)
Explain why the rod reaches a constant (terminal) speed.
(3)
(b)
Show that the terminal speed is given by v = FR/(B2l2) and calculate its value.
(3)
(c)
Calculate the rate at which energy is dissipated in the circuit when the rod moves at its terminal speed.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
as the speed increases, the induced e.m.f. Blv increases
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so the current increases and the force BIl on the rod, which opposes the motion (Lenz), increases
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when this force equals 0.050 N the resultant force is zero, so the speed stays constant
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Part (b)
I = ε/R = Blv/R
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opposing force = BIl = B2l2v/R; at terminal speed this equals F, so v = FR/(B2l2)
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v = 0.050 × 0.40 ÷ (0.502 × 0.202) = 2.0 m s−1
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Part (c)
P = Fv = 0.050 × 2.0 = 0.10 W (or I2R with I = 0.50 A)
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Exam tip: In a terminal-speed argument, say which force grows, why it grows, and what happens when it equals the driving force. Each of the three steps is worth a mark.
A long solenoid is connected to a signal generator that supplies an alternating current. The amplitude of the current is kept constant. A small search coil is placed at the centre of the solenoid with its axis along the axis of the solenoid, and is connected to an oscilloscope. At a frequency of 50 Hz the peak e.m.f. across the search coil is 12 mV.
(a)
Explain why an e.m.f. is induced in the search coil.
(2)
(b)
The frequency is increased to 200 Hz with the same current amplitude. Explain the effect on the peak e.m.f. and calculate its new value.
(2)
(c)
The frequency is returned to 50 Hz. The search coil is turned so that its axis makes an angle of 60° with the axis of the solenoid. Calculate the new peak e.m.f.
(2)
(d)
Suggest why the search coil is placed at the centre of the solenoid.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the alternating current in the solenoid produces a changing magnetic flux inside it
✓ 1
the flux linkage of the search coil changes, so an e.m.f. is induced (Faraday's law)
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Part (b)
the flux linkage changes through the same range in one-quarter of the time, so its rate of change is four times as great
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peak e.m.f. = 4 × 12 = 48 mV
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Part (c)
the area at right angles to the field, and so the flux linkage, is multiplied by cos 60°
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12 × cos 60° = 6.0 mV
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Part (d)
the field is uniform there, so small changes in the position of the coil do not change the reading
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Exam tip: When a frequency changes but the amplitude does not, the flux linkage goes through the same values faster. The e.m.f. scales with the frequency.
On a drop-tower fairground ride, a passenger car falls freely down a vertical tower. Near the bottom, thick copper fins fixed to the car pass between two rows of strong permanent magnets mounted on the tower. The car slows down to a low speed without touching the magnets.
(a)
Explain how the magnets slow the car down. Your answer should refer to the energy transfers involved.
(6)
Show mark scheme
Answer
Mark
Notes
Indicative content:
—
the fins move through the magnetic field, so the flux linking regions of each fin changes / the fins cut the field lines
✓ 1
an e.m.f. is induced in the fins (Faraday's law)
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copper is a good conductor, so the e.m.f.s drive eddy currents in the fins
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by Lenz's law the eddy currents oppose the relative motion: the field exerts an upward force on the currents in the fins
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the braking force is larger at higher speed, so the deceleration is greatest at first and decreases as the car slows
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kinetic (and gravitational potential) energy of the car is transferred to thermal energy in the fins by the eddy currents
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Exam tip: Build the answer as a chain: change of flux → e.m.f. → eddy currents → opposing force → effect on motion → energy. Each link is a separate point.
A simple generator has a flat coil of 80 turns, each of area 6.0 × 10−3 m2. The coil rotates at 12 revolutions per second about an axis at right angles to a uniform magnetic field of flux density 0.15 T.
(a)
Calculate the maximum flux linkage of the coil.
(1)
(b)
Calculate the average e.m.f. induced as the coil turns from the position of maximum flux linkage to the position of zero flux linkage.
(2)
(c)
State the position of the coil when the e.m.f. is a maximum, and explain why the e.m.f. is a maximum there.
(2)
(d)
The rate of rotation is increased to 24 revolutions per second. Describe two changes to the graph of e.m.f. against time.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
NBA = 80 × 0.15 × 6.0 × 10−3 = 0.072 Wb
✓ 1
Part (b)
time for a quarter turn = (1 ÷ 12) ÷ 4 = 0.0208 s
✓ 1
ε = 0.072 ÷ 0.0208 = 3.5 V (3.46 V)
✓ 1
Part (c)
the plane of the coil is parallel to the field lines (flux linkage zero)
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the flux linkage is changing at its greatest rate there / the sides of the coil cut the field lines at right angles
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Part (d)
the peak e.m.f. doubles
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the period halves / the frequency doubles
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Model sketch
Figure: e.m.f. against time; at 24 rev/s the peak e.m.f. doubles and the period halves
Common error: Using the whole period for a change of flux linkage that happens in a quarter of a turn. Sketch the rotation and count the quarter turns.
A coil rotates at a constant rate in a uniform magnetic field. The graph shows how its flux linkage varies with time. Four points, P, Q, R and S, are marked on the graph. At which point is the induced e.m.f. at its maximum positive value?
Show mark scheme
Answer
Mark
Notes
ε = −d(Nφ)/dt. The e.m.f. has its largest magnitude where the graph is steepest, which is where the flux linkage passes through zero (Q and S). At Q the flux linkage is falling, so the gradient is negative and the e.m.f. is positive: the answer is Q. At S the gradient is positive, so the e.m.f. there is at its maximum negative value.
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Answer B
Answer: B
Common error: Picking the point where the flux linkage is largest. At P and R the graph is momentarily flat, so the e.m.f. there is zero.
A coil of 400 turns is moved quickly away from a magnet. The graph shows the e.m.f. induced in the coil. What is the change in the magnetic flux through each turn of the coil?
Show mark scheme
Answer
Mark
Notes
The area under an e.m.f.–time graph is the change in flux linkage. Area = ½ × (0.060 − 0.010) s × 0.80 V = 0.020 Wb = NΔφ. So Δφ = 0.020 ÷ 400 = 5.0 × 10−5 Wb.
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Answer A
Answer: A
Exam tip: Area under ε–t gives Δ(Nφ), the change in flux linkage. Divide by N for the flux through one turn, and convert ms to s before multiplying.
A pick-up coil is fixed close to a large electromagnet. When the supply to the electromagnet is disconnected, the flux linkage of the pick-up coil decreases as shown. A tangent has been drawn to the curve at t = 25 ms.
(a)
State the flux linkage of the coil at t = 0.
(1)
(b)
Use the tangent to determine the magnitude of the e.m.f. induced in the coil at t = 25 ms.
(3)
(c)
Explain how the e.m.f. changes between t = 25 ms and t = 100 ms.
(2)
(d)
The experiment is repeated with the flux linkage falling twice as fast, so that every value of flux linkage is reached in half the time. State the e.m.f. when the flux linkage is the same as at t = 25 ms in the original experiment.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
8.0 × 10−3 Wb
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Part (b)
the tangent runs from about 5.9 × 10−3 Wb at t = 0 to zero at t = 50 ms
✓ 1
gradient = −5.9 × 10−3 ÷ 0.050 = −0.118 Wb s−1
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ε = −gradient, magnitude 0.12 V (allow 0.11–0.13 V)
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Part (c)
the gradient of the graph becomes less steep, so the rate of change of flux linkage decreases
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so the e.m.f. decreases, tending to zero as the flux linkage stops changing
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Part (d)
the graph is twice as steep at that flux linkage: 0.24 V (allow twice the value from (b))
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Common error: Dividing a single reading by a time, such as 2.9 × 10⁻³ ÷ 0.025. On a curve, the e.m.f. at an instant is minus the gradient of the tangent at that instant.
An exercise bike has a heavy flywheel with an aluminium rim. A strong magnet can be moved close to the rim, without touching it, to make pedalling harder. The graph shows how the rotation rate f of the flywheel changes after the rider stops pedalling: line R with the magnet removed, and curve M with the magnet close to the rim.
(a)
Explain why the flywheel slows down much more quickly with the magnet close to the rim, and why curve M becomes less steep as time goes on. Your answer should refer to the energy transfers involved.
(6)
(b)
Use curve M to find the time taken for the rotation rate to fall from 10 rev s−1 to 5.0 rev s−1.
(1)
(c)
Suggest why line R is straight.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
Indicative content:
—
the rim moves through the field of the magnet, so the flux linking regions of the rim changes / the rim cuts field lines
✓ 1
e.m.f.s are induced in the rim (Faraday's law)
✓ 1
aluminium is a good conductor, so these e.m.f.s drive eddy currents in the rim
✓ 1
by Lenz's law the eddy currents oppose the relative motion: the field exerts a force on the currents that opposes the rotation
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the faster the rim moves, the greater the rate of change of flux, so the larger the eddy currents and the braking force; as the flywheel slows the braking force falls, so the gradient of M decreases
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the kinetic energy of the flywheel is transferred to thermal energy in the rim by the eddy currents
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Part (b)
4.0 s (allow 3.8–4.2 s)
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Part (c)
with the magnet removed only friction acts, which is (roughly) constant, so the deceleration is constant
✓ 1
Exam tip: For a 6-mark explanation, build a chain: changing flux → e.m.f. → eddy currents → opposing force → effect on the motion → energy transfer. Each link is a separate point.
A coil of 150 turns, each of area 4.0 × 10−4 m2, is placed with its plane perpendicular to a uniform magnetic field. The flux density varies with time as shown in the graph. The coil is connected to a resistor, and the total resistance of the circuit is 5.0 Ω.
(a)
Calculate the maximum flux linkage of the coil.
(1)
(b)
Calculate the e.m.f. induced between t = 0 and t = 25 ms.
(2)
(c)
Calculate the e.m.f. and the current between t = 40 ms and t = 60 ms.
(3)
(d)
Calculate the charge that flows between t = 40 ms and t = 60 ms, and show that it is equal to the change in flux linkage divided by the resistance of the circuit.
Δ(Nφ)/R = 6.0 × 10−3 ÷ 5.0 = 1.2 × 10−3 C, the same
✓ 1
Common error: Taking the change in flux linkage between 40 ms and 60 ms as 3.0 × 10⁻³ Wb. The field reverses, so the flux linkage changes from +3.0 to −3.0 × 10⁻³ Wb.
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