Download The diagram shows a rigidly-clamped straight horizontal current

1
The diagram shows a rigidly-clamped straight horizontal current-carrying wire held mid-way
between the poles of a magnet on a top-pan balance. The wire is perpendicular to the magnetic
field direction.
The balance, which was zeroed before the switch was closed, read 161 g after the switch was
closed. When the current is reversed and doubled, what would be the new reading on the
balance?
A
−322 g
B
−161 g
C
zero
D
322 g
(Total 1 mark)
2
Four rectangular loops of wire A, B, C and D are each placed in a uniform magnetic field of the
same flux density B. The direction of the magnetic field is parallel to the plane of the loops as
shown.
When a current of 1 A is passed through each of the loops, magnetic forces act on them. The
lengths of the sides of the loops are as shown.
Which loop experiences the largest couple?
A
B
C
D
(Total 1 mark)
Page 1 of 34
3
A horizontal straight wire of length 0.30 m carries a current of 2.0 A perpendicular to a horizontal
uniform magnetic field of flux density 5.0 × 10–2 T. The wire ‘floats’ in equilibrium in the field.
What is the mass of the wire?
A
8.0 × 10–4 kg
B
3.1 × 10–3 kg
C
3.0 × 10–2 kg
D
8.2 × 10–1 kg
(Total 1 mark)
Page 2 of 34
4
A horizontal straight wire of length 40 mm is in an east-west direction as shown in the diagram. A
uniform magnetic field of flux density 50 mT is directed downwards into the plane of the diagram.
When a current of 5.0 A passes through the wire from west to east, a horizontal force acts on the
wire. Which line, A to D, in the table gives the magnitude and direction of this force?
magnitude / mN
direction
A
2.0
north
B
10.0
north
C
2.0
south
D
10.0
south
(Total 1 mark)
Page 3 of 34
5
The diagram shows a rigidly-clamped straight horizontal current-carrying wire held mid-way
between the poles of a magnet on a top pan balance. The wire is perpendicular to the magnetic
field direction.
The balance, which was zeroed before the switch was closed, reads 112 g after the switch is
closed. If the current is reversed and doubled, what will be the new reading on the balance?
A
–224 g
B
–112 g
C
zero
D
224 g
(Total 1 mark)
Page 4 of 34
6
A coil, mounted on an axle, has its plane parallel to the flux lines of a uniform magnetic field B, as
shown. When a current I is switched on, and before the coil is allowed to move,
A
there are no forces due to B on the sides PQ and RS.
B
there are no forces due to B on the sides SP and QR.
C
sides SP and QR attract each other.
D
sides PQ and RS attract each other.
(Total 1 mark)
7
The diagram shows a magnet placed on a top-pan balance. A fixed horizontal wire, through
which a current can flow, passes centrally through the magnetic field parallel to the pole-pieces.
With no current flowing, the balance records a mass of 201.32 g. When a current of 5.0 A flows,
the reading on the balance is 202.86 g.
(a)
(i)
Explain why the reading on the balance increased when the current was switched on.
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 5 of 34
(ii)
State the direction of current flow and explain your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(iii)
If the length of the wire in the magnetic field is 60 mm, estimate the flux density of the
magnetic field.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(b)
Sketch a graph to show how you would expect the balance reading to change if the current
through the wire was changed.
(2)
(Total 8 marks)
Page 6 of 34
8
The diagram shows a metal rod suspended in a magnetic field by two vertical conducting
springs. The cell and rod have negligible resistance. When the switch S is closed the effect of the
magnetic field is to displace the rod vertically a distance y.
When both the spring constant and electrical resistance of each spring is doubled, closing the
switch would now cause the rod to be displaced a distance
A
B
C
y
D
4y
(Total 1 mark)
Page 7 of 34
9
The diagram shows a wire carrying a current, I, in the plane of the paper and in the south
direction. A magnetic field is applied perpendicularly to the paper and acts into the paper. What is
the direction of the force acting on the wire?
A
north
B
south
C
east
D
west
(Total 1 mark)
10
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)
11
Page 8 of 34
A coil, mounted on an axle, has its plane parallel to the flux lines of a uniform magnetic field B, as
shown. When a current I is switched on, and before the coil is allowed to move,
A
there are no forces due to B on the sides SP and QR.
B
there are no forces due to B on the sides PQ and RS.
C
sides SP and QR tend to attract each other.
D
sides PQ and RS tend to attract each other.
(Total 1 mark)
12
(a)
A satellite moves in a circular orbit at constant speed. Explain why its speed does not
change even though it is acted on by a force.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
At a certain point along the orbit of a satellite in uniform circular motion, the Earth’s
magnetic flux density has a component of 56 μT towards the centre of the Earth and a
component of 17 μT in a direction perpendicular to the plane of the orbit.
(i)
Calculate the magnitude of the resultant magnetic flux density at this point.
...............................................................................................................
...............................................................................................................
Page 9 of 34
(ii)
The satellite has an external metal rod pointing towards the centre of the Earth.
Calculate the angle between the direction of the resultant magnetic field and the rod.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(iii)
Explain why an emf is induced in the rod in this position.
...............................................................................................................
...............................................................................................................
(4)
(Total 7 marks)
13
(a)
Complete the table of quantities related to fields. In the second column, write an SI unit for
each quantity. In the third column indicate whether the quantity is a scalar or a vector.
quantity
SI unit
scalar or vector
gravitational potential
electric field strength
magnetic flux density
(3)
(b)
(i)
A charged particle is held in equilibrium by the force resulting from a vertical electric
field. The mass of the particle is 4.3 × 10–9 kg and it carries a charge of magnitude
3.2 × 10–12 C. Calculate the strength of the electric field.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
If the electric field acts upwards, state the sign of the charge carried by the particle
.............................................................................................................
(3)
(Total 6 marks)
Page 10 of 34
14
A wire lies perpendicularly across a horizontal uniform magnetic field of flux density 20 × 10−3 T
so that 0.30 m of the wire is effectively subjected to the field. If the force exerted on this length of
wire due to a current in it is 30 × 10−3 N downward, what is the current in the wire?
A
0.45 A from P to Q
B
0.45 A from Q to P
C
5.0 A from P to Q
D
5.0 A from Q to P
(Total 1 mark)
15
(a)
The equation F = BIl, where the symbols have their usual meanings, gives the magnetic
force that acts on a conductor in a magnetic field.
Given the unit of each of the quantities in the equation.
F ..............................
B ..............................
I ..............................
l ...............................
State the condition under which the equation applies.
......................................................................................................................
......................................................................................................................
(2)
Page 11 of 34
(b)
The diagram shows a horizontal copper bar of 25 mm × 25 mm square cross-section and
length l carrying a current of 65 A.
(i)
Calculate the minimum value of the flux density of the magnetic field in which it
should be placed if its weight is to be supported by the magnetic force that acts upon
it.
density of copper = 8.9 × 103 kg m–3
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Draw an arrow on the diagram above to show the direction in which the magnetic field
should be applied if your calculation in part (i) is to be valid. Label this arrow M.
(5)
(Total 7 marks)
Page 12 of 34
16
The diagram shows a vertical square coil whose plane is at right angles to a horizontal uniform
magnetic field B. A current, I, flows in the coil, which can rotate about a vertical axis OO’.
Which one of the following statements is correct?
A
The forces on the two vertical sides of the coil are equal and opposite.
B
A couple acts on the coil.
C
No forces act on the horizontal sides of the coil.
D
If the coil is turned through a small angle about OO’, it will remain in position.
(Total 1 mark)
17
Which one of the following could not be used as a unit of force?
A
ATm
B
W s–2
C
kg m s–2
D
J m–1
(Total 1 mark)
Page 13 of 34
18
A section of current-carrying wire is placed at right angles to a uniform magnetic field of flux
density B. When the current in the wire is I, the magnetic force that acts on this section is F.
What force acts when the same section of wire is placed at right angles to a uniform magnetic
field of flux density 2B when the current is 0.25 I?
A
B
C
F
D
2F
(Total 1 mark)
19
The figure below shows a horizontal wire, held in tension between fixed points at P and Q. A
short section of the wire is positioned between the pole pieces of a permanent magnet, which
applies a uniform horizontal magnetic field at right angles to the wire. Wires connected to a circuit
at P and Q allow an electric current to be passed through the wire.
(a)
(i)
State the direction of the force on the wire when there is a direct current from P to Q,
as shown in the figure above.
.............................................................................................................
(1)
Page 14 of 34
(ii)
In a second experiment, an alternating current is passed through the wire. Explain
why the wire will vibrate vertically.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
The permanent magnet produces a uniform magnetic field of flux density 220 mT over a 55
mm length of the wire. Show that the maximum force on the wire is about 40 mN when
there is an alternating current of rms value 2.4 A in it.
(3)
(c)
The length of PQ is 0.40 m. When the wire is vibrating, transverse waves are propagated
along the wire at a speed of 64 m s–1. Explain why the wire is set into large amplitude
vibration when the frequency of the a.c. supply is 80 Hz.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Page 15 of 34
20
(a)
Figure 1 shows a negative ion which has a charge of –3e and is free to move in a uniform
electric field. When the ion is accelerated by the field through a distance of 63 mm parallel
to the field lines its kinetic energy increases by 4.0 × 10sup class="xsmall">–16 J.
Figure 1
(i)
State and explain the direction of the electrostatic force on the ion.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Calculate the magnitude of the electrostatic force acting on the ion.
magnitude of electrostatic force ........................................ N
(2)
(iii)
Calculate the electric field strength.
electric field strength .................................. NC–1
(1)
Page 16 of 34
(b)
Figure 2 shows a section of a horizontal copper wire carrying a current of 0.38 A.
A horizontal uniform magnetic field of flux density B is applied at right angles to the wire in
the direction shown in the figure.
Figure 2
(i)
State the direction of the magnetic force that acts on the moving electrons in the wire
as a consequence of the current and explain how you arrive at your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Copper contains 8.4 × 1028 free electrons per cubic metre. The section of wire in
Figure 2 is 95 mm long and its cross-sectional area is 5.1 × 10–6 m2.
Show that there are about 4 × 1022 free electrons in this section of wire.
(1)
(iii)
With a current of 0.38 A, the average velocity of an electron in the wire is
5.5 × 10–6 m s–1 and the average magnetic force on one electron is 1.4 × 10–25 N.
Calculate the flux density B of the magnetic field.
flux density ......................................... T
(2)
(Total 10 marks)
Page 17 of 34
21
A horizontal straight wire of length 0.30 m carries a current of 2.0 A perpendicular to a horizontal
uniform magnetic field of flux density 5.0 × 10–2 T. The wire ‘floats’ in equilibrium in the field.
What is the mass of the wire?
A
8.0 × 10–4 kg
B
3.1 × 10–3 kg
C
3.0 × 10–2 kg
D
8.2 × 10–1 kg
(Total 1 mark)
Page 18 of 34
22
The diagram shows a vertical square coil whose plane is at right angles to a horizontal uniform
magnetic field B. A current, I, is passed through the coil, which is free to rotate about a vertical
axis OO'.
Which one of the following statements is correct?
A
The forces on the two vertical sides of the coil are equal and opposite.
B
A couple acts on the coil.
C
No forces act on the horizontal sides of the coil.
D
If the coil is turned through a small angle about OO' and released, it will remain in position.
(Total 1 mark)
Page 19 of 34
23
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)
(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?
...............................................................................................................
(1)
(ii)
Calculate the number of revolutions per minute made by the coil.
revolutions per minute ..............................................
(2)
Page 20 of 34
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 21 of 34
(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)
Page 22 of 34
Mark schemes
1
2
3
4
5
6
7
A
[1]
B
[1]
B
[1]
B
[1]
A
[1]
A
[1]
(a)
(i)
interaction between current and B-field gives force on wire (1)
equal and opposite force on magnet (down) (1)
(ii)
force on wire must be up (1)
∴ current right to left (1)
by left hand rule (1)
(iii)
(force = BIl = mg = change in mass × 9.8)
B × 5.0 × 0.060 = 1.54 × 10–3 × 9.8 (1)
B = 0.050 T [50.3 mT] (1)
(max 6)
Page 23 of 34
(b)
straight line (1)
intercept, upward slope (1)
(2)
[8]
8
9
10
11
12
B
[1]
C
[1]
B
[1]
B
[1]
(a)
gravity or force acts towards centre (1)
force acts at right angles to velocity or direction of motion
[or velocity is tangential] (1)
no movement in direction of force (1)
no work done so no change of kinetic energy so no change in speed (1)
3
(b)
(i)
B = (562 + 172)½ = 59 μT (1)
(ii)
tanθ =
(1)
θ = 17° (1) (± 1°)
(iii)
rod sweeps out or cuts (magnetic) flux
[or rod cuts field] (1)
4
[7]
Page 24 of 34
13
(a)
quantity
SI unit
(gravitational potential)
J kg–1 or N m kg –1
scalar
(electric field strength)
N C–1 or V m–1
vector
(magnetic flux density
T or Wb m–2 or N A–1 m–1
vector
6 entries correct (1) (1) (1)
4 or 5 entries correct (1) (1)
2 or 3 entries correct (1)
3
(b)
(i)
mg = EQ (1)
= 1.32 × 104 (V m–1) (1)
(ii)
positive (1)
3
[6]
14
15
D
[1]
(a)
units: F - newton (N), B - tesla (T) or weber metre–2 (Wb m–2),
I - ampere (A), l - metre (m) (1)
condition: I must be perpendicular to B (1)
2
(b)
(i)
mass of bar, m = (25 × 10–3)2 × 8900 × l (1)
(= 5.56l) weight of bar (= mg) = 54.6l (1)
mg = BIl or weight = magnetic force (1)
54.6l = B × 65 × l gives B = 0.840 T (1)
(ii)
arrow in correct direction (at right angles to I, in plane of bar) (1)
5
[7]
Page 25 of 34
16
17
18
19
A
[1]
B
[1]
B
[1]
(a)
(i)
(vertically) downwards (1)
1
(ii)
force F is perpendicular to both B and I [or equivalent correct
explanation using Fleming LHR] (1)
magnitude of F changes as size of current changes (1)
force acts in opposite direction when current reverses
[or ac gives alternating force] (1)
continual reversal of ac means process is repeated (1)
max 3
(b)
appreciation that maximum force corresponds to peak current (1)
peak current = 2.4 ×
= 3.39 (A) (1)
Fmax (= B Ipk L) = 0.22 × 3.39 × 55 × 10–3 (1) (= 4.10 × 10–2 N)
3
Page 26 of 34
(c)
wavelength (λ) of waves =
= 0.80 (m) (1)
length of wire is λ/2 causing fundamental vibration (1)
[or λ of waves required for fundamental (= 2 × 0.40) = 0.80 m (1)
natural frequency of wire
= 80 (Hz) (1)]
wire resonates (at frequency of ac supply) [or a statement that
fundamental frequency (or a natural frequency) of the wire is
the same as applied
frequency] (1)
3
[10]
20
(a)
(i)
force acts towards left or in opposite direction to field lines ✓
because ion (or electron) has negative charge
(∴ experiences force in opposite direction to field) ✓
Mark sequentially.
Essential to refer to negative charge (or force on + charge is to
right) for 2nd mark.
2
(ii)
✓
(use of W = F s gives) force F =
= 6.3(5) × 10–15 (N) ✓
If mass of ion m is used correctly using algebra with F = ma, allow
both marks (since m will cancel). If numerical value for m is used,
max 1.
2
(iii)
= 1.3(2) ✓ 104 (N C-1) ✓
electric field strength
[or
(833 V)
= 1.3(2) ✓ 104 (V m-1) ✓ ]
Allow ECF from wrong F value in (ii).
1
Page 27 of 34
(b)
(i)
(vertically) downwards on diagram ✓
reference to Fleming’s LH rule or equivalent statement ✓
Mark sequentially.
1st point: allow “into the page”.
2
(ii)
number of free electrons in wire = A × l × number density
= 5.1 × 10–6 × 95 × 10–3 × 8.4 × 1028 = 4.1 (4.07) × 1022 ✓
Provided it is shown correctly to at least 2SF, final answer alone is
sufficient for the mark. (Otherwise working is mandatory).
1
✓ = 0.16 (0.159) (T) ✓
(iii)
✓ = 0.16 (0.158) (T) ✓ ]
[or
In 2nd method allow ECF from wrong number value in (ii).
2
[10]
21
22
23
B
[1]
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
(b)
(i)
(magnitude of the induced) emf
Accept “induced voltage” or “rate of
change of flux linkage”, but not
“voltage” alone.
1
Page 28 of 34
(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 29 of 34
Examiner reports
1
2
4
5
6
7
In this question the students needed to know that reversing the current in a wire placed in a
magnetic field would reverse the direction of the force on it, and that doubling the current would
double the force. 60% of the responses were correct, up from 41% the last time this question
appeared in an examination. The most common incorrect answer was distractor D (22%), where
the force would be doubled but not reversed.
This question gave the most surprising outcome because, although its facility of 54% was
satisfactory, it was a very poor discriminator between the strongest and weakest students. The
physics of the question is clear enough: the couple on the coil is proportional to its crosssectional area. The puzzling outcome may have arisen because 39% of the answers were for
distractor A. The ratios of the areas are actually 0.20, 0.25, 0.15 and 0.16 respectively, so why so
many students selected 0.20 remains a mystery. Maybe they were put off by the greater length of
the long side of the rectangle.
This question dealt with various aspects of electromagnetism. The question required the
application of F = B I L together with Fleming’s left hand rule. The facility was 75% and the most
common incorrect response was distractor D – from those who could not decide the correct
direction.
This question involved a basic current balance and assumed familiarity with F = B I l. The simple
idea here was that reversing the current and doubling it would produce a force in the opposite
direction that would be twice the original. This eluded so many of the candidates that only 41% of
them could select the correct response, whilst many of them chose distractors C or D.
This question was correctly answered by 61% of the candidates. It involved the topic, familiar
from previous tests, of the forces acting on the sides of a current-carrying coil when in a uniform
magnetic field. This is another question that had been re-banked after use in an earlier PA04 test,
when the facility had been only 53%. Evidently, a greater majority were able this time to spot that
sides PQ and RS would experience no forces when I and B are parallel, but distractors C and B
were chosen by 18% and 15% respectively.
There were some very vague statements in answer to part (a). The majority of candidates said
that a downward force was acting, either on the wire, pushing it down on to the balance, or by
some other mysterious means producing an increase in the balance reading. A surprising
number of candidates stated that the increase in mass was due to “the moving electrons”. Only a
small proportion of those who said clearly that there was a force on a current-carrying conductor
in a magnetic field went on to deduce that there would be an equal opposite force on the magnet
and hence the balance. Thus only a few candidates gained all the marks in part (a)(ii) for the
correct application of Fleming’s left hand rule to a wire experiencing an upward force in the field.
One mark was awarded to those candidates who considered a downward force correctly,
provided they gave their reasoning. The majority of candidates knew how to do the calculation,
but some omitted to change kg to N, did not take the difference in the balance readings, made
mistakes in arithmetic or gave the wrong unit. Similarly. The majority of candidates knew that the
graph in part (b) was a straight line, but many forgot the intercept.
Page 30 of 34
12
In part (a) the large majority of candidates knew that the force on the satellite acted towards the
centre of the Earth and that the direction of motion or velocity of the satellite was at right angles
to the force. However, few of them were able to use these facts to explain adequately why the
speed remained constant. Little or no reference was made to the absence of work done or zero
change in potential or kinetic energy.
There were many correct calculations in parts (b)(i) and (ii), but some candidates did confuse the
two components when calculating the angle in part (ii). In the final part, correct explanations for
the induced emf were usually given in terms of the rod cutting the non-radial component of the
Earth's magnetic field.
13
Units of the various physical quantities related to fields and the scalar/vector nature of them, are
generally not well known by the candidates. Part (a) showed that the 2004 cohort were no better
than their predecessors. Six correct entries in the table were required for three marks, and it was
very rare for all three to be awarded. The unit of N m kg–1 was accepted as an alternative to J
kg–1 for gravitational potential, but candidates regularly put N kg–1 in the table. The unit of electric
field strength was known better, and that of magnetic flux density was usually shown correctly.
Candidates often resorted to guesswork when completing the second column of the table. Many
did not appreciate that the concept of potential arises from energy considerations and that it is
therefore a scalar quantity, whilst the other two quantities are force-related and therefore vectors.
Completely correct answers to part (b) were encountered in many of the scripts. Since the unit of
E had already been tested in the table in part (a), no penalty was imposed for wrong or missing
units in the answer to part (b)(i). A worrying error, made by a significant minority of the
candidates, was to equate the electric force on the particle to its mass, rather than to its weight.
15
The lack of familiarity of candidates with the units of electromagnetic quantities continues to be a
cause for concern. All four units had to be correct for the first mark in part (a). It might have been
anticipated that candidates would make an incorrect choice for B, such as the regularly
encountered Wb. The many candidates who could not identify the SI unit of force (sometimes N
m–1 or J m–1 were given) came as a greater surprise. The most common error in the second
aspect of part (a) was to state that the force must be perpendicular to the magnetic field,
although some candidates confused the question with electromagnetic induction and thought that
the conductor had to be moving.
A large number of clear and succinct solutions were seen in the answers to part (b), although
many other candidates were stumped by the need to combine ideas about magnetic force and
weight. Equating mass with magnetic force was regarded as a serious error of physics for which
no further marks could be given. The final part of the question required the accurate application
of Fleming’s left-hand rule; this defeated far more candidates than it ought to have done.
Page 31 of 34
16
17
This question, on the forces which act on a current-carrying coil in a magnetic field, had a facility
of 63%, which was a substantial improvement on the pre-examination facility. It was not a
particularly effective discriminator.
This question contained a synoptic element, because two of the unit combinations quoted were
mechanical. The facility of the question was 44%.
Overlooking the word not in the stem of the question presumably caused 34% of the candidates
to choose distractor A, where F = B I l ought to have shown that A T m is a correct unit of force.
18
19
This question was a test of F = B // that, with a facility of 86%, proved to be the least demanding
question in this Section A. When the question was pre-tested just over half of the students
selected the correct response.Fleming’s left hand rule, as applied to a beam of positive ions
Most candidates were able to use Fleming’s left hand rule in order to give the correct force
direction in part (a) (i). Sometimes a candidate’s answer was contradictory and went unrewarded,
for example ‘downwards towards the S pole’. Most answers to part (a) (ii) were reasonably good
when explaining why the wire would vibrate, but rarely explained why these vibrations are
vertical. An explanation by reference to the mutually perpendicular field, current and force
directions was required in a complete answer. The reversal of force direction with change of
current direction was well understood.
Fewer candidates made reference to the continuous current reversals brought about by ac
causing the process to repeat, or to the fact that the size of the current affects the magnitude of
the magnetic force.
It was evident that a large number of candidates had made a second, more enlightened, attempt
at part (b) once they had realised that direct substitution of I = 2.4 A into F = BIL did not lead to
the value of force (about 40 mN) they had been asked to show. Once they realised that the
maximum force is caused by the peak current, it became a straightforward matter to secure three
marks.
Page 32 of 34
The final part of the question, part (c), involved the resonance effect observed when the wire is
supplied with ac current at the frequency of its fundamental vibration. Resonance was usually
mentioned, but fewer candidates used the values provided in the question together with c = fλ to
give a wholly convincing account of why the wire would vibrate in its fundamental mode at 80 Hz.
A large number of candidates had forgotten that the fundamental condition would be L = λ/2 (this
should be studied in unit 2). After using c = fλ with λ = 0.40 m, they concluded that the frequency
of waves on the wire would be 160 Hz. These candidates then attempted to argue that
resonance would occur at 80 Hz because 80 is one half of 160, not understanding that if 160 Hz
was the fundamental frequency, no frequency lower than 160 Hz could possibly set the wire into
resonance.
20
The direction of the force on the negatively charged ion in part (a)(i) was mainly correct.
Explanations of the direction of the force were good and marks for this part were high.
At A2 level the students are expected to have retained a comprehensive knowledge of earlier
work, which is tested by synoptic components within the questions. Part (a)(ii) was an example of
this, because the simplest solution followed directly from “energy gained = work done = force ×
distance”. This eluded most students, many of whom were completely defeated. Many of them
became successful after taking a very roundabout route, involving calculation of the pd from V =
W / Q, the field strength from E = V / d and the force from F = EQ. Others produced answers
based on the uniform acceleration equations and F = ma. When this was done using algebra the
mass m of the ion could be cancelled and the answer was accepted. When a numerical value
was chosen for m the mark that could be awarded was limited to 1 out of 2. There were fewer
difficulties in part (a)(iii), where E = F / 3e gave the most direct answer but where E = V / d
offered an alternative method. Incorrect force values from part (a)(ii) were permitted for full credit
in part (a)(iii).
In part (b)(i) the successful application of Fleming’s left hand rule readily showed the majority that
the magnetic force on the wire would be downwards (or into the page). Most students gained
both marks. It seemed that some students who thought too deeply about the direction of this
force eventually got it wrong: their line of thought was that the force due to the current was
downwards, but electrons carry a negative charge so the force on them must act upwards. It had
not occurred to them that the force on the complete wire has to act in the same direction as the
force on all of the free electrons within it.
The simple calculation that led to the number of free electrons in the section of wire was almost
always worked out correctly in part (b)(ii). This was a “show that” question, so students should
have realised that a more precise answer than 4 × 1022 (such as 4.07 × 1022) would be expected
before the mark became available. Part (b)(iii) offered two approaches to the value of the flux
density B. By considering the average force on a single electron, F = BQv could be used.
Alternatively, by considering the force on the section of wire, F = BIl could be used. In the latter
method many got into difficulty by forgetting to consider the number of electrons in the wire.
Page 33 of 34
21
22
23
In this question the correct answer could be found by equating the weight of a section of wire to
the magnetic force that acts on it when the wire carries a current at right angles to magnetic field.
This was answered correctly by 75% if the students. Distractor C, where the mass was
approximately 9.8 times the expected mass, was the choice of almost one in five. This suggests
that the students involved equated the mass of the wire with the magnetic force, forgetting g.
This question has been used to test candidates in earlier years. The former one showed an
improvement in facility and the latter one a deterioration. This question was about the magnetic
forces that act on a coil which is free to rotate in a uniform magnetic field. 58% of the responses
were correct. The most popular incorrect answer was distractor B. Candidates who chose this
response had not spotted the fact that when the coil lies with its plane perpendicular to the field,
the two forces on its sides act in opposite directions along the same straight line. Therefore they
do not constitute a couple.
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.
Page 34 of 34