Download (a) Figure 1 shows two coils, P and Q, linked by an iron bar

1
(a)
Figure 1 shows two coils, P and Q, linked by an iron bar. Coil P is connected to a battery
through a variable resistor and a switch S. Coil Q is connected to a centre-zero ammeter.
Figure 1
(i)
Initially the variable resistor is set to its minimum resistance and S is open.
Describe and explain what is observed on the ammeter when S is closed.
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(3)
(ii)
With S still closed, the resistance of the variable resistor is suddenly increased.
Compare what is now observed on the ammeter with what was observed in part (i).
Explain why this differs from what was observed in part (i).
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(2)
Page 1 of 25
(b)
Figure 2 shows a 40-turn coil of cross-sectional area 3.6 × 10–3 m2 with its plane set at
right angles to a uniform magnetic field of flux density 0.42 T.
Figure 2
(i)
Calculate the magnitude of the magnetic flux linkage for the coil.
State an appropriate unit for your answer.
flux linkage ....................................................... unit .....................
(2)
(ii)
The coil is rotated through 90° in a time of 0.50 s.
Determine the mean emf in the coil.
mean emf ................................................. V
(2)
(Total 9 marks)
Page 2 of 25
2
A rectangular coil of area A has N turns of wire. The coil is in a uniform magnetic field, as shown
in the diagram.
When the coil is rotated at a constant frequency f about its axis XY, an alternating emf of peak
value ε0 is induced in it.
What is the maximum value of the magnetic flux linkage through the coil?
A
B
C
π f ε0
D
2π f ε0
(Total 1 mark)
3
A rectangular coil is rotating anticlockwise at constant angular speed with its axle at right angles
to a uniform magnetic field. Figure 1 shows an end-on view of the coil at a particular instant.
Figure 1
(a)
At the instant shown in Figure 1, the angle between the normal to the plane of the coil and
the direction of the magnetic field is 30°.
(i)
State the minimum angle, in degrees, through which the coil must rotate from its
position in Figure 1 for the emf to reach its maximum value.
angle ................................. degrees
(1)
Page 3 of 25
(ii)
Calculate the minimum angle, in radians, through which the coil must rotate from its
position in Figure 1 for the flux linkage to reach its maximum value.
angle ................................. radians
(2)
(b)
Figure 2 shows how, starting in a different position, the flux linkage through the coil varies
with time.
(i)
What physical quantity is represented by the gradient of the graph shown in Figure
2?
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(1)
(ii)
Calculate the number of revolutions per minute made by the coil.
revolutions per minute ..............................................
(2)
Page 4 of 25
Figure 2
Figure 3
(iii)
Calculate the peak value of the emf generated.
peak emf ......................................... V
(3)
(c)
Sketch a graph on the axes shown in Figure 3 above to show how the induced emf varies
with time over the time interval shown in Figure 2.
(2)
Page 5 of 25
(d)
The coil has 550 turns and a cross-sectional area of 4.0 × 10–3m2.
Calculate the flux density of the uniform magnetic field.
flux density .......................................... T
(2)
(Total 13 marks)
4
The graph shows how the magnetic flux, Φ, passing through a coil changes with time, t.
Which one of the following graphs could show how the magnitude of the emf, V, induced in the
coil varies with t?
(Total 1 mark)
Page 6 of 25
5
A coil of 50 turns has a cross-sectional area of 4.2 × 10–3 m2. It is placed at an angle to a uniform
magnetic field of flux density 2.8 × 10–2 T, as shown in the diagram, so that angle θ = 50°.
What is the change in flux linkage when the coil is rotated anticlockwise until θ = 0°?
A
The flux linkage decreases by 2.1 × 10–3 Wb turns.
B
The flux linkage increases by 2.1 × 10–3 Wb turns.
C
The flux linkage decreases by 3.8 × 10–3 Wb turns.
D
The flux linkage increases by 3.8 × 10–3 Wb turns.
(Total 1 mark)
6
Which line, A to D, gives correct units for both magnetic flux and magnetic flux density?
magnetic flux
magnetic flux density
A
Wb m−2
Wb
B
Wb
T
C
Wb m−2
T m−2
D
T m−2
Wb m−2
(Total 1 mark)
7
(a)
Explain what is meant by the term magnetic flux linkage. State its unit.
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(2)
Page 7 of 25
(b)
Explain, in terms of electromagnetic induction, how a transformer may be used to step
down voltage.
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(4)
(c)
A minidisc player is provided with a mains adapter. The adapter uses a transformer with a
turns ratio of 15:1 to step down the mains voltage from 230 V.
(i)
Calculate the output voltage of the transformer.
(2)
(ii)
State two reasons why the transformer may be less than 100% efficient.
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(2)
(Total 10 marks)
8
An aircraft, of wing span 60 m, flies horizontally at a speed of 150 m s–1. If the vertical
component of the Earth’s magnetic field in the region of the plane is 1.0 × 10 –5 T, what is the
magnitude of the magnetic flux cut by the wings in 10 s?
A
1.0 × 10–5 Wb
B
1.0 × 10–4 Wb
C
9.0 × 10–2 Wb
D
9.0 × 10–1 Wb
(Total 1 mark)
Page 8 of 25
9
A rectangular coil measuring 20 mm by 35 mm and having 650 turns is rotating about a
horizontal axis which is at right angles to a uniform magnetic field of flux density 2.5 × 10–3 T.
The plane of the coil makes an angle θ with the vertical, as shown in the diagrams.
(a)
State the value of θ when the magnetic flux through the coil is a minimum.
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(1)
(b)
Calculate the magnetic flux passing through the coil when θ is 30°.
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(2)
(c)
What is the maximum flux linkage through the coil as it rotates?
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(2)
(Total 5 marks)
Page 9 of 25
10
Figure 1
A circular coil of diameter 140 mm has 850 turns. It is placed so that its plane is perpendicular to
a horizontal magnetic field of uniform flux density 45 mT, as shown in Figure 1.
(a)
Calculate the magnetic flux passing through the coil when in this position.
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(2)
(b)
The coil is rotated through 90° about a vertical axis in a time of 120 ms.
Calculate
(i)
the change of magnetic flux linkage produced by this rotation,
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(ii)
the average emf induced in the coil when it is rotated.
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(4)
(Total 6 marks)
Page 10 of 25
11
The magnetic flux through a coil of N turns is increased uniformly from zero to a maximum value
in a time t. An emf, E, is induced across the coil.
What is the maximum value of the magnetic flux through the coil?
A
B
C
EtN
D
(Total 1 mark)
12
The graph shows how the magnetic flux passing through a loop of wire changes with time.
What feature of the graph represents the magnitude of the emf induced in the coil?
A
the area enclosed between the graph line and the time axis
B
the area enclosed between the graph line and the magnetic flux axis
C
the inverse of the gradient of the graph
D
the gradient of the graph
(Total 1 mark)
Page 11 of 25
13
The Large Hadron Collider (LHC) uses magnetic fields to confine fast-moving charged particles
travelling repeatedly around a circular path. The LHC is installed in an underground circular
tunnel of circumference 27 km.
(a)
In the presence of a suitably directed uniform magnetic field, charged particles move at
constant speed in a circular path of constant radius. By reference to the force acting on the
particles, explain how this is achieved and why it happens.
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(4)
(b)
(i)
The charged particles travelling around the LHC may be protons. Calculate the
centripetal force acting on a proton when travelling in a circular path of circumference
27 km at one-tenth of the speed of light. Ignore relativistic effects.
answer = ................................ N
(3)
(ii)
Calculate the flux density of the uniform magnetic field that would be required to
produce this force. State an appropriate unit.
answer = ...................................... unit ........................
(3)
Page 12 of 25
(c)
The speed of the protons gradually increases as their energy is increased by the LHC.
State and explain how the magnetic field in the LHC must change as the speed of the
protons is increased.
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(2)
(Total 12 marks)
14
A 500 turn coil of cross-sectional area 4.0 × 10–3 m2 is placed with its plane perpendicular to a
magnetic field of flux density 7.5 × 10–4 T. What is the value of the flux linkage for this coil?
A
B
C
D
3.0 × 10–6 Wb turns
1.5 × 10–3 Wb turns
0.19 Wb turns
94 Wb turns
(Total 1 mark)
Page 13 of 25
15
A bar magnet is pushed into a coil connected to a sensitive ammeter, as shown in the diagram,
until it comes to rest inside the coil.
Why does the ammeter briefly show a non-zero reading?
A
The magnetic flux linkage in the coil increases then decreases.
B
The magnetic flux linkage in the coil increases then becomes constant.
C
The magnetic flux linkage in the coil decreases then increases.
D
The magnetic flux linkage in the coil decreases then becomes constant.
(Total 1 mark)
Page 14 of 25
Mark schemes
1
(a)
(i)
meter deflects then returns to zero ✓
current produces (magnetic) field / flux ✓
change in field / flux through Q induces emf ✓
induced emf causes current in Q (and meter) ✓
Deflection to right (condone left) then zero is equivalent to 1st mark.
Accept momentary deflection for 1st point.
“Change in field / flux induces current in Q” is just ✓ from the last
two marking points.
max 3
(ii)
meter deflects in opposite direction (or to left, or ecf) ✓
field / flux through P is reduced ✓
induces emf / current in opposite direction ✓
Ignore references to magnitude of deflection.
max 2
(b)
(i)
flux linkage (= nΦ = nBA) = 40 × 0.42 × 3.6 × 10 −3
= 6.0(5) × 10−2 ✓
Unit mark is independent.
Allow 6 × 10−2.
Wb turns ✓
Accept 60 mWb turns if this unit is made clear.
Unit: allow Wb.
2
(ii)
change in flux linkage = Δ(nΦ)= 6.05 × 10−2 (Wb turns) ✓
induced emf
=
= 0.12(1) (V) ✓
Essential to appreciate that 6.05 × 10−2 is change in flux linkage for
1st mark. Otherwise mark to max 1.
2
[9]
2
3
A
[1]
(a)
(i)
60 (degrees)
1
(ii)
angle required is 150°
which is 5
/ 6 [or 2.6(2)] (radians)
Correct answer in radians scores both marks.
2
Page 15 of 25
(b)
(i)
(magnitude of the induced) emf
Accept “induced voltage” or “rate of
change of flux linkage”, but not
“voltage” alone.
1
(ii)
no of revolutions per minute = 25 × 60 = 1500
1500 scores both marks.
Award 1 mark for 40s → 1.5 rev min−1.
2
(iii)
maximum flux linkage (=BAN) = 0.55 (Wb turns)
peak emf (= BANω) = 0.55 × 157 = 86(.4) (V)
[ or, less accurately, use of gradient method
(V)
(max 2 for (iii) for values between 63 and 72 V or 94 and 103V) ]
3
(c)
sinusoidal shape of constant period 40 ms
Mark sequentially.
Graph must cover at least 80ms.
correct phase (i.e. starts as a minus sin curve)
For 2nd mark, accept + sin curve.
Perfect sin curves are not expected.
2
(d)
= 0.25(0) (T)
OR by use of ε from (b)(iii) and f from
(b)(ii) substituted in ε = BAN
.
2
(Total 13 marks)
Page 16 of 25
4
5
6
7
D
[1]
B
[1]
B
[1]
(a)
product of flux and number of turns
B1
Wb or equivalent
C1
(2)
(b)
changing primary magnetic field due to alternating voltage
B1
(applied to primary)
varying flux links with secondary
B1
induced emf ∑ rate of change of flux linkage
B1
NS < NP so less voltage on secondary
C1
(4)
(c)
(i)
equation or correct substitution
C1
15.3 V
A1
(2)
(ii)
<100% flux linkage / flux leakage / copper losses / iron losses / hysterysis
losses not just “heating” or “heat loss”
B2
(2)
[10]
8
9
D
[1]
(a)
θ = 90° (or 270° or
or
) (1)
1
Page 17 of 25
(b)
Φ = BA cosθ (1)
= 2.5 × 10-3 × 35 × 10-3 × 20 × 10-3 × cos 30° = 1.5 × 10-6 Wb (1)
2
(c)
Φmax = 2.5 × 10-3 × 35 × 10-3 × 20 × 10-3 (Wb) (1) (= 1.75 × 10-6)
flux linkage = 650 × 1.75 × 10-6 = 1.1(4) × 10-3 (Wb turns) (1)
2
[5]
10
(= BA) = 45 × 10–3 × π × (70 × 10–3)2 (1)
(a)
= 6.9 × 10–4 Wb (1) (6.93 × 10–4 Wb)
2
(b)
(i)
NΔ
( = NBA – 0) = 850 × 6.93 × 10–4 (1)
= 0.59 (Wb turns) (1) (0.589 (Wb turns))
(if
= 6.9 × 10–4, then 0.587 (Wb turns))
(allow C.E. for value of
(ii)
induced emf ( = N
= 4.9 V (1)
)=
from (a))
(1)
(4.91 V)
(allow C.E. for value of Wb turns from (ii)
4
[6]
11
12
A
[1]
D
[1]
Page 18 of 25
13
(a)
(magnetic) field is applied perpendicular to path
or direction or velocity of charged particles
(magnetic) force acts perpendicular to path
or direction or velocity of charged particles
force depends on speed of particle or on B [or F ∞ v or F = BQv explained]
force provides (centripetal) acceleration towards centre of circle
[or (magnetic) force is a centripetal force]
shows that r is constant when B and v are constant
4
(b)
(i)
= 4.30 × 103 (m)
radius r of path =
(allow 4.3km)
= 3.50 × 10–16(N)
centripetal force
3
(ii)
magnetic flux density
= 7.29 × 10-5
T
3
(c)
magnetic field must be increased
to increase (centripetal) force or in order to keep r constant
[or otherwise protons would attempt to travel in a path of larger radius]
[or, referring to
, B must increase when v increases to keep r constant ]
2
[12]
14
B
[1]
Page 19 of 25
15
B
[1]
Page 20 of 25
Examiner reports
1
The topic of electromagnetic induction continues to challenge the understanding of A level
students, as well as their ability to describe a sequence of processes systematically. Part (a) was
set in the context of two coils linked by an iron bar, where the first coil acts as an electromagnet
and the second is subject to magnetic flux changes produced by current changes in the
electromagnet. Relatively few students stated in part (i) that the centre-zero meter would deflect
and then return to zero when the current in coil P was switched on. There were frequent
references to current flowing through the iron bar from P to Q and also to “ac batteries” and
alternating currents. Only the best students described the processes sequentially and coherently:
current in P produces magnetic flux, change in flux induces emf in Q, emf causes current in Q
and meter, current falls to zero when flux becomes steady.
In part (a)(ii) more answers attempted to address the magnitude of the induced current than its
direction. The effect on the magnitude could not in fact be determined, because there is no
indication in the question of how rapidly the slider of the resistance is moved. What could be
deduced is that a reduction in the electromagnet’s current would reduce the flux linkage and that
this change would induce an emf in the opposite direction. This would cause a momentary
deflection of the ammeter in the opposite direction to that in part (i).
Most students found the calculation of flux linkage in part (b)(i) to be routine. Both marks were
usually awarded. The unit of flux linkage caused problems for some. The accepted unit for flux
linkage is Wb turns. Some text books omit “turns” (which anyway is a dimensionless quantity)
and quote flux linkage values in Wb. Either Wb turns or Wb were therefore considered to be
acceptable; derived units such as T m2 were not. Calculation of the emf induced when the coil
was rotated by 90° was required in part (b)(ii). This tempted many students to attempt their
solution by using the equation in the data booklet for a uniformly rotating coil, ε = BANω sinωt,
which does not apply in this case. Correct solutions should have started from ε = Δ(NΦ) / Δt, and
it should therefore have been clear that the induced emf is derived from the change in flux
linkage rather than just one value of flux linkage. Almost inevitably, a few students confused flux
with flux linkage.
2
This question could be answered by knowing that the emf generated in a coil rotating in a
magnetic field is given by ε = BANω sin ωt, that and that ω = 2πf. The maximum emf ε0 = BANω,
which is (maximum flux linkage) × 2πf. 59% of the answers were correct.
Page 21 of 25
3
It has long been clear that electromagnetism is a challenging topic for quite a lot of A level
candidates, and that flux and flux linkage are regularly confused. Several parts of this question
revealed these weaknesses once again. In part (a) the initial orientation of the coil in relation to
the magnetic field has to be understood. It is also necessary to know that flux linkage is a
maximum when the plane of the coil is perpendicular to the field, and that the emf through a
rotating coil is a maximum when the flux linkage is a minimum. There were more correct answers
to part (a)(i) than to part (a)(ii), although a significant number of candidates gave the wrong
answer in (i) and the gave the right answer in (ii). Conversion of the angle from degrees to
radians was a problem for some in (ii), whilst others had not noticed that the answer was
required in radians in this part.
The majority of responses to part (b)(i) were correct, but part (b)(ii) caused more problems than
expected. Answers of 1.5 revolutions per minute that had been reached by misreading the time
axis of Figure 3 were allowed one of the two available marks. In part (b)(iii) the most accurate
value for the peak emf can be found by reading the maximum flux linkage (0.55 Wb turns) from
Figure 3 and then applying εmax = BAN × 2πf. Only a few candidates used this method, the
majority choosing a gradient evaluation instead. Different tolerances were applied, depending on
how far their answer for the gradient was away from the expected result of around 85 V. It was
very clear from their calculations that many candidates had not understood that the peak value of
the emf is represented by the maximum gradient, because a large number of the evaluations
were closer to an average emf. Almost inevitably, there were candidates who resorted to the
introduction of
into their calculation.
Plenty of good, carefully drawn -sine or sine curves were presented in part (c). These received
both marks provided the graph covered at least two full cycles and had periods of 40 ms and
consistent maxima. Good cosine curves were allowed one mark, but there was no credit for the
triangular or square waves that were occasionally drawn.
4
5
This question was a graphical test of the relationship between an induced emf and the rate of
change of magnetic flux causing it. 59% of the students saw that the increasing gradient of the
original graph had to imply that the emf would increase, and that therefore only graph D could be
correct. 24% of the responses were for distractor B, where the emf is shown to decrease at an
increasing rate.
This was the most demanding question in the test, about the change in flux linkage when a coil is
rotated in a magnetic field. 40% of the responses to this question were correct. Distractors C and
D were each selected by almost a quarter of the candidates; this is probably because they
considered the flux linkage to be zero when the plane of the coil was perpendicular to the
magnetic field.
Page 22 of 25
7
8
9
(a)
Few of the candidates were able to give a good explanation of magnetic flux linkage.
several confusing the term with flux density. Few correctly identified the Weber as the
correct unit.
(b)
Most candidates were unable to give a coherent explanation of how electromagnetic
induction underpinned the working of the step down transformer. The most common
observation gaining credit was that there are more turns on the primary than the secondary
coil.
(c)
(i)
Most candidates calculated the output voltage correctly.
(ii)
Few candidates’ answers gained credit for this part. It was common for candidates to
cite heating or eddy currents as the reason for less than 100% efficiency. Although
these answers are not incorrect they are too vague to allow credit at this level.
The most common incorrect response was distractor C, which accounted for 25% of the
answers. This wrong choice is likely to have been caused by trying to work out the emf across
the wing tips of a moving aircraft using the equation E = B L v, rather than finding the magnetic
flux cut by the wing of a moving aircraft, as required by the question. However, 59% of the
candidates chose the correct answer. This question discriminated well between the most able
and the least able candidates.
Attempts at part (a) revealed huge uncertainty in the minds of many of the candidates, with 0°
and 90° repeatedly crossed out before arriving at a considered decision. In the end, more than
half of the candidates seemed to get a correct answer.
After the examination it was decided that part (b) should be removed and the answers were not
marked by the examiners.
The amplification column of section 13.4.3 of Physics Specification A reads “ Φ = BA, B normal to
A ”. This same entry has appeared in the previous two NEAB syllabuses, stretching back over
ten years, out of which the current AQA specification was developed. It was intended by the
designers of the specification, and the previous syllabuses, as shorthand for “ Φ = BA, where B is
normal to A ”. “B normal to A” defines the condition under which the equation applies, rather than
implying that the 90° condition is the only case in which magnetic flux may be calculated.
However, it has been pointed out to the examiners that teachers at many centres have
interpreted “B normal to A” as defining the only condition under which candidates could be
expected to calculate magnetic flux. (This is consistent with the 90° case being a restriction
which definitely applies to the equations F = B11 and F = BQ v in sections 13.4.1 and 13.4.2 of
the specification ).
Therefore, in fairness to all candidates taking the test, the two marks for part (b) of Question 3
were discounted from the assessment and the whole paper was marked out of 28 rather than 30.
Part (c) was seldom answered correctly. Very few candidates knew the distinction between flux
and flux linkage, and there were considerable difficulties over powers of 10. Many candidates
obtained one mark by determining the maximum flux correctly but did not then multiply their
result by 650. The correct unit of flux linkage evaded almost all.
Page 23 of 25
10
The topic of electromagnetism continues to present greater difficulty than most of the remainder
of the Unit 4 content. Candidates who had mastered the distinction between magnetic flux and
flux linkage, and who appreciated that induced emf = (rate of change of NΦ), readily gained all
six marks. Only a small minority of the candidates came into this category, however. When
finding the cross-sectional area presented to the flux, there was evidence of the usual confusion
between diameter and radius, leading to the loss of one mark on the question. More worrying
were those candidates who wrote the area of a circle as 2πr, or as 2πr2. In part (b), examiners
took the view that candidates should know that an emf is measured in V – final answers
expressed in Wb turns s–1 were not accepted.
11
This question moved on to electromagnetic induction and tested E = N ΔΦ /Δt for a uniform rate
of change of magnetic flux. 72% of the responses were correct. Like questions 22 and 23, this
question was a very good discriminator.
12
This question was easy, with a facility of 73%. It was a direct test of ε = N (ΔФ/Δt) in a graphical
context. The most common incorrect answer was distractor A.
13
It was rare for all four marks to be awarded in part (a). The essence of this question was well
understood, but poor use of English and an inability to write logically limited the mark that could
be given. An alarming proportion of answers made no reference at all to the magnetic field; these
students appeared to be answering a more general question about circular motion. Many of the
students evidently thought that the purpose of the magnetic force (presumably acting outwards)
was to balance the centripetal force, rather than to provide it. Relatively few correct solutions
were seen that used r = mv / BQ to show that r is constant when B and v are constant.
The common error in part (b)(i) was failure to deduce the radius of the path of the protons from
the 27 km circumference of the LHC. This only meant the loss of one of the three marks,
however, provided the principles of the rest of the calculation were correct. Careless arithmetic
such as failure to square v, and/or forgetting to convert km to m, was also a frequent source of
loss of marks. F = BQv was usually applied successfully in part (b)(ii), where the unit of magnetic
flux density was quite well known. Almost inevitably, there was some confusion between flux
density and magnetic flux.
The fact that had to be appreciated in part (c) was that in the LHC the radius of the path of the
charged particles must remain constant as they are accelerated. A large proportion of students
thought that it was necessary to maintain a constant centripetal force for this to happen, whereas
it ought to have been clear to them that F must increase as v increases if r is to be constant.
Page 24 of 25
14
15
The remaining three questions all tested aspects of electromagnetism.This question which, with
a facility of 89% was the easiest in the test, was a straightforward calculation of flux linkage for a
coil in a magnetic field.
In this question a large proportion of candidates did not realise that the flux linkage increases to
a constant value once the magnet is at rest inside the coil, and therefore selected the incorrect
distractor A. 43% gave the correct response.
Page 25 of 25