Magnetic Fields
2014
134 minutes
104 marks
Q1.
An α particle and a β– particle both enter the same uniform magnetic field, which is
perpendicular to their direction of motion. If the β– particle has a speed 15 times that of
theα particle, what is the value of the ratio
?
A
3.7
B
7.5
C
60
D
112.5
(Total 1 mark)
Q2.
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)
Q3.
Which line, A to D, correctly describes the trajectory of charged particles which enter, at
right angles, (a) a uniform electric field, and (b) a uniform magnetic field?
A
B
C
D
(a) uniform electric field
(b) uniform magnetic field
circular
circular
parabolic
parabolic
circular
parabolic
circular
parabolic
(Total 1 mark)
Q4.
The diagram shows a square coil with its plane parallel to a uniform magnetic field. Which one
of the following would induce an emf in the coil?
A
movement of the coil slightly to the left
B
movement of the coil slightly downwards
C
rotation of the coil about an axis through XY
D
rotation of the coil about an axis perpendicular to the planeof the coil through Z
(Total 1 mark)
Q5.
Protons, each of mass m and charge e, follow a circular path when travelling perpendicular
to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
A
B
C
D
(Total 1 mark)
Q6.
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)
Q7.
Particles of mass m carrying a charge Q travel in a circular path of radius r in a magnetic
field of flux density B with a speed v. How many of the following quantities, if changed one at a
time, would change the radius of the path?
•
m
•
Q
•
B
•
v
A
one
B
two
C
three
D
four
(Total 1 mark)
Q8.
Why, when transporting electricity on the National Grid, are high voltages and low currents
used?
A
The energy lost by radiation from electromagnetic waves is reduced.
B
The electrons move more rapidly.
C
The heat losses are reduced.
D
The resistance of the power lines is reduced.
(Total 1 mark)
Q9.
The primary winding of a perfectly efficient transformer has 200 turns and the secondary
has 1000 turns. When a sinusoidal pd of rms value 10 V is applied to the input, there is a
primary current of rms value 0.10 A rms. Which line in the following table, A to D, gives correct
rms output values obtainable from the secondary when the primary is supplied in this way?
A
rms output emf/V
rms output current/A
50
0.10
B
50
0.02
C
10
0.10
D
10
0.02
(Total 1 mark)
Q10.
The diagram shows a square coil with its plane parallel to a uniform magnetic field.Which one of
the following would induce an emf in the coil?
A
movement of the coil slightly to the left
B
movement of the coil slightly downwards
C
rotation of the coil about an axis through XY
D
rotation of the coil about an axis perpendicular to the plane of the coil through Z
(Total 1 mark)
Q11.
Protons, each of mass m and charge e, follow a circular path when travelling perpendicular
to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
A
B
C
D
(Total 1 mark)
Q12.
The magnetic flux, Ф, through a coil varies with time, t, as shown by the first graph. Which
one of the following graphs, A to D, best represents how the magnitude, , of the induced emf
varies in this same period of time?
(Total 1 mark)
Q13.
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)
Q14.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)
Q15.
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)
Q16.A transformer with 3000 turns in its primary coil is used to change an alternating pd from an rms
value of 240 V to an rms value of 12 V.
When a 60 W, 12 V lamp is connected to the secondary coil, the lamp lights at normal
brightness and a rms current of 0.26 A passes through the primary coil.
Which line, A to D, in the table gives correct values for the number of turns on the secondary
coil and for the transformer efficiency?
number of turns on the
secondary coil
efficiency
A
150
96%
B
60 000
96%
C
150
90%
D
60 000
90%
(Total 1 mark)
Q17.A transformer, which is not perfectly efficient, is connected to a 230 V rms mains supply and is
used to operate a 12 V rms, 60 W lamp at normal brightness. The secondary coil of the
transformer has 24 turns.
Which line, A to D, in the table is correct?
number of turns on primary coil
rms current in primary coil
A
92
less than 0.26 A
B
92
more than 0.26 A
C
460
less than 0.26 A
D
460
more than 0.26 A
(Total 1 mark)
Q18.A vertical conducting rod of length l is moved at a constant velocity v through a uniform
horizontal magnetic field of flux density B.
Which line, A to D, in the table gives a correct expression for the induced emf for the stated
direction of the motion of the rod?
direction of motion
induced emf
A
vertical
B
horizontal at right angles to the field
Blv
C
vertical
Blv
D
horizontal at right angles to the field
(Total 1 mark)
Q19.Which one of the following statements concerning power losses in a transformer isincorrect?
Power losses can be reduced by
A
laminating the core.
B
using high resistance windings.
C
using thick wire.
D
using a core made of special iron alloys which are easily magnetised.
(Total 1 mark)
Q20.The magnetic flux through a coil of 5 turns changes uniformly from 15 × 10−3 Wb to 7.0 × 10−3 Wb
in 0.50 s. What is the magnitude of the emf induced in the coil due to this change in flux?
A
14 m V
B
16 m V
C
30 m V
D
80 m V
(Total 1 mark)
Q21.A rectangular coil is rotated in a uniform magnetic field.
When the coil is rotated at a constant rate, an alternating emf ε is induced in it. The variation of
emf ε, in volts, with time t, in seconds, is given by
ε = 20 sin (100 πt)
Which line, A to D, in the table gives the peak value ε0 and the frequency f of the induced emf?
A
ε0 / V
f / Hz
10
50
B
10
100
C
20
50
D
20
100
(Total 1 mark)
Q22.Which line, A to D, in the table correctly describes the trajectory of charged particles which enter
separately, at right angles, a uniform electric field, and a uniform magnetic field?
uniform electric field
uniform magnetic field
A
parabolic
circular
B
circular
parabolic
C
circular
circular
D
parabolic
parabolic
(Total 1 mark)
Q23.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)
Q24.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)
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)
(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)
Q25.(a)
(i)
State two situations in which a charged particle will experience no magnetic force
when placed in a magnetic field.
first situation...........................................................................................
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second situation.....................................................................................
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(2)
(ii)
A charged particle moves in a circular path when travelling perpendicular to a
uniform magnetic field. By considering the force acting on the charged particle,
show that the radius of the path is proportional to the momentum of the particle.
(2)
(b)
In a cyclotron designed to produce high energy protons, the protons pass repeatedly
between two hollow D-shaped containers called ‘dees’. The protons are acted on by a
uniform magnetic field over the whole area of the dees. Each proton therefore moves in a
semi-circular path at constant speed when inside a dee. Every time a proton crosses the
gap between the dees it is accelerated by an alternating electric field applied between the
dees. The diagram below shows a plan view of this arrangement.
(i)
State the direction in which the magnetic field should be applied in order for the
protons to travel along the semicircular paths inside each of the dees as shown in
the diagram above.
...............................................................................................................
(1)
(ii)
In a particular cyclotron the flux density of the uniform magnetic field is 0.48 T.
Calculate the speed of a proton when the radius of its path inside the dee is 190
mm.
speed .................................... ms–1
(2)
(iii)
Calculate the time taken for this proton to travel at constant speed in a semicircular
path of radius 190 mm inside the dee.
time .......................................... s
(2)
(iv)
As the protons gain energy, the radius of the path they follow increases steadily, as
shown in the diagram above. Show that your answer to part (b)(iii) does not depend
on the radius of the proton’s path.
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(2)
(c)
The protons leave the cyclotron when the radius of their path is equal to the outer radius
of the dees. Calculate the maximum kinetic energy, in Me V, of the protons accelerated by
the cyclotron if the outer radius of the dees is 470 mm.
maximum kinetic energy .................................... Me V
(3)
(Total 14 marks)
Q26.
A coil is connected to a centre zero ammeter, as shown. A student drops a magnet so that
it falls vertically and completely through the coil.
(a)
Describe what the student would observe on the ammeter as the magnet falls through the
coil.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
If the coil were not present the magnet would accelerate downwards at the acceleration
due to gravity. State and explain how its acceleration in the student’s experiment would be
affected, if at all,
(i)
as it entered the coil,
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.............................................................................................................
.............................................................................................................
(ii)
as it left the coil.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(c)
Suppose the student forgot to connect the ammeter to the coil, therefore leaving the circuit
incomplete, before carrying out the experiment. Describe and explain what difference this
would make to your conclusions in part (b).
You may be awarded marks for the quality of written communication provided in your
answer.
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(3)
(Total 9 marks)
Q27.
(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)
(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)
Q28.
(a)
The diagram above shows a doubly-charged positive ion of the copper isotope
that
is projected into a vertical magnetic field of flux density 0.28 T, with the field directed
upwards. The ion enters the field at a speed of 7.8 × 105 m s–1.
(i)
State the initial direction of the magnetic force that acts on the ion.
.............................................................................................................
(ii)
Describe the subsequent path of the ion as fully as you can.Your answer should
include both a qualitative description and a calculation.
mass of
ion = 1.05 × 10–25 kg
.............................................................................................................
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.............................................................................................................
(5)
(b)
State the effect on the path in part (a) if the following changes are made separately.
(i)
The strength of the magnetic field is doubled.
.............................................................................................................
.............................................................................................................
(ii)
A singly-charged positive
ion replaces the original one.
.............................................................................................................
.............................................................................................................
(3)
(Total 8 marks)
Q29.
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.
......................................................................................................................
......................................................................................................................
(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,
.............................................................................................................
.............................................................................................................
(ii)
the average emf induced in the coil when it is rotated.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 6 marks)
Q30.
(a) In an experiment to illustrate electromagnetic induction, a permanent magnet is
moved towards a coil, as shown in Figure 1, causing an emf to be induced across the
coil.
Figure 1
Using Faraday’s law, explain why a larger emf would be induced in this experiment if a
stronger magnet were moved at the same speed.
You may be awarded additional marks to those shown in brackets for the quality of written
communication in your answer.
......................................................................................................................
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(3)
(b)
A conductor of length l is moved at a constant speed v so that it passes perpendicularly
through a uniform magnetic field of flux density B, as shown in Figure 2.
Figure 2
uniform
magnetic field
(perpendicular
to the plane of
the diagram)
over this
region
(i)
Give an expression for the area of the magnetic field swept out by the conductor in
time Δt.
.............................................................................................................
.............................................................................................................
(ii)
Show that the induced emf,
, across the ends of the conductor is given by
.............................................................................................................
.............................................................................................................
= Blv.
.............................................................................................................
(3)
(c)
A simple electrical generator can be made from a copper disc, which is rotated at right
angles to a uniform magnetic field, directed into the plane of the diagram (Figure 3). An
emf is developed across terminals P (connected to the axle) and Q (connected to a
contact on the edge of the disc).
Figure 3
The radius of the disc is 64 mm and it is rotated at 16 revolutions per second in a uniform
magnetic field of flux density 28 mT.
(i)
Calculate the angular speed of the disc.
.............................................................................................................
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.............................................................................................................
(ii)
Calculate the linear speed of the mid-point M of a radius of the disc.
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(iii)
Hence, or otherwise, calculate the emf induced across terminals P and Q.
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(5)
(Total 11 marks)
Q31.(a)
State Lenz’s law.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
Figure 1 shows two small, solid metal cylinders, P and Q. P is made from aluminium. Q is
made from a steel alloy.
Figure 1
(i)
The dimensions of P and Q are identical but Q has a greater mass than P. Explain
what material property is responsible for this difference.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(ii)
When P and Q are released from rest and allowed to fall freely through a vertical
distance of 1.0 m, they each take 0.45 s to do so. Justify this time value and explain
why the times are the same.
...............................................................................................................
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...............................................................................................................
(2)
(c)
The steel cylinder Q is a strong permanent magnet. P and Q are released separately from
the top of a long, vertical copper tube so that they pass down the centre of the tube, as
shown in Figure 2.
Figure 2
The time taken for Q to pass through the tube is much longer than that taken by P.
(i)
Explain why you would expect an emf to be induced in the tube as Q passes
through it.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
State the consequences of this induced emf, and hence explain why Q takes longer
than P to pass through the tube.
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(3)
(d)
The copper tube is replaced by a tube of the same dimensions made from brass. The
resistivity of brass is much greater than that of copper. Describe and explain how, if at all,
the times taken by P and Q to pass through the tube would be affected.
P: ...................................................................................................................
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........................................................................................................................
........................................................................................................................
Q: ...................................................................................................................
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(3)
(Total 13 marks)
M1.
B
[1]
M2.
A
[1]
M3.
C
[1]
M4.
C
[1]
M5.
B
[1]
M6.
A
[1]
M7.
D
[1]
M8.
C
[1]
M9.
B
[1]
M10.
C
[1]
M11.
D
[1]
M12.
C
[1]
M13.
A
[1]
M14.A
[1]
M15.
B
[1]
M16.A
[1]
M17.D
[1]
M18.B
[1]
M19.B
[1]
M20.D
[1]
M21.C
[1]
M22.A
[1]
M23.B
[1]
M24.(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
(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)
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[ 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)
M25.(a)
(i)
Two examples (any order):
• when charged particle is at rest or not moving relative to field
• when charged particle moves parallel to magnetic field
2
(ii)
Acceptable answers must include correct force equation (1st point).
B and Q are constant so r ∝ momentum (mv)
.
Insist on a reference to B and Q constant for 2nd mark.
2
(b)
(i)
upwards (perpendicular to plane of diagram)
Accept “out of the page” etc.
1
(ii)
2
(iii)
length of path followed (= length of semi-circle) =
time taken
Allow ECF from incorrect v from (b)(ii).
Max 1 if path length is taken to be 2
(gives 1.37 × 10−7s).
2
(iv)
v ∝ r (and path length ∝ r )
t = (path length / v) or (
/v)
so r cancels (∴time doesn't depend on r)
[or BQv = mω2 r gives BQωr = mω2 r and BQ = mω = 2πfm
∴ frequency is independent of r
]
2
(c)
1st mark can be achieved by full substitution, as in (b)(ii), or by use
of data from (b)(i) and / or (b)(ii).
Ek (= ½ mvmax 2) = ½ × 1.67 × 10−27 × (2.16 × 107)2
( = 3.90 × 10−13 J)
.
Allow ECF from incorrect v from (b)(ii), or from incorrect t from
(b)(iii).
3
(Total 14 marks)
M26.
(a) deflects one way (1)
then the other way (1)
2
(b)
(i)
acceleration is less than g [or reduced] (1)
suitable argument (1) (e.g. correct use of Lenz’s law)
(ii)
acceleration is less than g [or reduced] (1)
suitable argument (1) (e.g. correct use of Lenz’s law)
4
(c)
magnet now falls at acceleration g (1)
emf induced (1)
but no current (1)
no energy lost from circuit (1)
[or no opposing force on magnet, or no force from
magnetic field or no magnetic field produced]
3
QWC 2
[9]
M27.
(a) units: F - newton (N), B - tesla (T) or weber metre–2 (Wb m–2),
metre (m) (1)condition: I must be perpendicular to B (1)
I - ampere (A), l 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]
M28.
(a)
(i)
out of plane of diagram (1)
(ii)
circular path (1)
in a horizontal plane [or out of the plane of the diagram] (1)
(1)
radius of path, r
(1)
= 0.91(4) m (1)
max 5
(b)
(i)
radius decreased (1)
halved (1)
[or radius is halved (1) (1)]
(ii)
radius increased (1)
doubled (1)
[or radius is doubled (1) (1)]
max 3
[8]
M29.
(= BA) = 45 × 10–3 × π × (70 × 10–3)2 (1)
(a)
4
Wb)
= 6.9 × 10–4 Wb (1) (6.93 × 10–
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]
M30.
(a) greater flux (linkage) or more flux lines (at same distance)
[or stronger magnet produces flux lines closer together] (1)
greater rate of change of flux (linkage)
[or more flux lines cut per unit time] (1)
emf
rate of change of flux (linkage) (1)
[or using
, where Δ
= A ΔB, v and Δt are the same (1)
ΔB is larger since magnet is stronger (1)
N and A are constant,
is larger (1)]
3
(b)
(i)
area swept out, ΔA = lvΔt (1)
Δ
(= BΔA) = Blv Δt (1)
gives result (1)
3
(c)
(i)
w(=2πf) = 2π × l6 (1) = 101 rads–1 (1)
(ii)
v(= rw) = 32 × 10–3 × 101 = 3.2(3)ms–1 (1)
(allow C.E. for value of w from (i))
(iii)
(= Blv) = 28 × 10–3 × 64 × 10–3 × 3.23 (1)= 5.7(9) × 10–3V (1)
(allow C.E. for values of v from (ii))
(solutions using = Bfπr2 to give 5.7(6) × 10–3 V acceptable)
5
[11]
M31.(a)
direction of induced emf (or current)
opposes change (of magnetic flux) that produces it
2
(b)
(i)
(volumes are equal and mass of Q is greater than that of P) density of steel > density
of aluminium
Allow density of Q greater (than density of P).
1
(ii)
use of s =½ g t2 gives t 2 =
(from which t = 0.45 s)
Backwards working is acceptable for 1st mark
(vertical) acceleration [or acceleration due to gravity] is independent of mass of
falling object
[or correct reference to F = mg = ma with m cancelling ]
2 nd mark must refer to mass.
Do not allow “both in free fall” for 2nd mark.
2
(c)
(i)
moving magnet [or magnetic field] passes through tube
(linkage)(in the tube)
there is a change of flux
[or flux lines are cut or appropriate reference to ɛ = N (Δɸ / Δt)]
In this part marks can be awarded for answers which mix and
match these schemes.
[Alternative:
(conduction) electrons in copper (or tube) acted on by (moving)
magnetic field of Q
induced emf (or current) is produced by redistributed electrons
]
2
(ii)
emf produces current (in copper)
this current [allow emf] produces a magnetic field
this field opposes magnetic field (or motion) of Q
[or acts to reduce relative motion or produces upward force]
no emf is induced by P because it is not magnetised (or not magnet)
[or movement of P is not opposed by an induced emf or current]
Alternative to 3 rd mark:
current gives heating effect in copper and energy for this comes
from ke of Q
max 3
(d)
time for P is unaffected because there is still no (induced) emf[or because P is not
magnetisedor because there is no repulsive force on P] time for Q is shorter (than in
(c)) current induced by Q would be smaller
because resistance of brass ∝ resistivity
and is therefore higher[or resistance of brass is higher because resistivity is greater]
giving weaker (opposing) magnetic field[or less opposition to Q’s movement]
Condone “will pass through faster” for 2nd mark.If emf is stated to
be smaller for Q, mark (d) to max 2.
max 3
[13]
E1.
This question required candidates to recall the charges of α and β-particles, as well as to be
familiar with F = BQv. Both the facility (54%) and the discrimination were an improvement on the
pre-examination values. 24% of the candidates chose distractor C (presumably because 15 × 4
= 60). This suggests these candidates had difficulty with the physics as well as the arithmetic.
E2.
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.
E3.
Trajectories of charged particles as they pass through electric and magnetic fields ought to
be a fairly simple topic, but the facility of this question improved only slightly, from 55% to 57%,
between pre-test and examination. Candidates who did not understand these topics were
attracted in almost equal numbers to distractors B and D.
E4.
This question was answered correctly by 60% of the candidates. Electromagnetic induction
involves three-dimensional thinking, and it is likely that the 24% who chose distractor D
experienced difficulty in visualising the meaning of the words in the statement for this distractor.
E5.
Difficulties over rearrangement of the algebra no doubt caused 20% of the candidates to
choose distractor A and 13% distractor C in this question, but the facility was still 55%. This was
another two-stage calculation, where Bev, mv2/r and 2πr/v were all to be combined to find the
time for one orbit of a proton in a magnetic field.
E6.
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.
E7.
This question required candidates to be familiar with the factors that affect the radius of the
path followed by a charged particle in a magnetic field. 58% of the candidates spotted that all
four factors would change the radius, but 27% thought that only three would matter (distractor
C). This question was the best discriminator in the test.
E13.
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.
E14.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.
E15.
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
E16.This question tested both the turns ratio equation and efficiency; again there were few problems
and the facility was 72%.
E17.The question showed that most of the candidates were able to apply the turns ratio equation
correctly, because over 80% of them selected Np = 460. A simple conservation of power
calculation would also show them that the primary current would be 0.26 A if the transformer
was perfectly efficient. Since this transformer has an efficiency of less than 100%, the better
candidates (51%) realised that the primary current had to be greater than 0.26 A.
E18.By far the majority of the students realised that the induced emf would be B / v. It was the
direction of motion of the rod that caused a problem for the 30% who chose distractor C;
evidently they did not realise that the rod would not cut field lines if it were to move vertically.
E19.This qestion was about transformers. Causes of power loss were well known in this question,
where three quarters of the candidates evidently saw that using windings with higher resistance
would have a detrimental effect.
E20.This question, which involved the direct application of ε = ΔNΦ / Δt, probably required less
thought before committing a response. The facility of the question was 74%, an improvement of
10% over the last occasion when it appeared in an examination. Thinking that the induced emf
is equal to the rate of change of flux, instead of flux linkage, probably caused 16% of the
candidates to select distractor B.
E21.This question tested the candidates’ understanding of the peak voltage and frequency terms in
the equation ε = ε0 sin (2πf t) for a coil rotating at constant speed in a uniform magnetic field.
This equation was not understood as well as might have been expected, because the facility
was less than 60%. Common misapprehensions seem to have been that ε0 represents the peakto-peak voltage (because distractors A and B each attracted more than 10% of the responses)
and that 2f represents the frequency (because distractor D attracted 17% of responses).
E22.This question had appeared in an examination previously; it tested the fairly familiar knowledge
of the trajectory of charged particles in electric and magnetic fields and this time had a facility of
71%.
E23.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.
E24.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.
E25.The two situations expected to be given in part (a)(i) were when the charge is at rest, or when it
is moving parallel to the direction of the magnetic field. These answers were given by a high
proportion of the candidates. Inexact expressions such as “when the charge is placedparallel to
the field” were viewed with suspicion and went unrewarded. Also unsuccessful were attempts
such as “when it is not moving perpendicular to the field” and “when it does not cut any flux
lines”. Some candidates thought they could answer by subjecting the moving charge to an
electric field over the same region (as in an ion velocity selector) so that there would be no
resultant force on the charge. This was not acceptable because the magnetic force would still
be acting.
In part (a)(ii) most candidates gained the first mark by quoting BQv = mv2 / r. Cancelling one v
then gives mv = BQr. However, to show that mv ∝ r it is necessary to point out
that B and Qmust be constant. The large number of answers which failed to do this did not
receive the second mark.
Examiners were surprised by the large number of incorrect answers to part (b)(i), on a topic that
has usually been well understood. Perhaps this was because the question is set in the context
of a device being used to accelerate protons (rather than electrons). Consequently many
candidates could not see that the magnetic field has to act upwards, out of the plane of the
diagram.
Errors in part (b)(ii) included using the wrong mass and/or charge for a proton, but the majority
of answers were correct. The frequent slip of using 2πr instead of πr for the path length incurred
a one mark penalty in part (b)(iii); many candidates got around this problem by dividing their
answer for time by 2. Part (b)(iv) was often rewarding, but it also defeated many candidates.
The expected approaches included using algebraic equations for the time, or an argument
based on the proportionality of the speed and radius. Less precise attempts, such as “gwhen
the speed increases the radius increases so the time is the same’ were not credited. A few
candidates repeated the calculation in part (b)(iii) for a different radius to show that the time was
unaltered. In part (c) the candidates who thought that the protons would still be travelling at the
speed they had calculated in part (b)(ii) were under a serious misapprehension and gained no
marks. Surprisingly few used v ∝ r to find the new velocity, most preferring to repeat their earlier
calculation in full but using r = 0.47m. The conversion of the kinetic energy unit from J to MeV .
which is an AS topic . defeated many.
E26.
This question was intended as a straightforward test of the “simple experimental
phenomena” of electromagnetic induction and Lenz’s law, as required by Section 13.4.4 of the
Specification. It is recognised that most A level candidates have difficulty with these topics and
examiners were not very surprised by the many relatively weak answers that were written.
Partial (or superficial) understanding of the phenomena appeared to be the main obstacle to
progress. For example, in part (a) almost all candidates appreciated that the ammeter needle
would deflect, but relatively few saw that it would move one way, and then the other way, before
returning to zero. In this part, examiners sometimes wondered what was going through the
minds of candidates who wrote things such as “the current through the ammeter would
increase, and then return to its normal value”. Perhaps this suggests that these students had
never previously encountered a centre zero instrument. Inappropriate use of English also
handicapped some candidates in part (a) – typical of which were answers that began with “the
ammeter moves to the right”.
Failure to address the question was the main difficulty encountered in most answers to part (b).
Instead of stating clearly that the acceleration of the magnet decreased, candidates usually
preferred to resort to woolly descriptions of the effect on the motion of the magnet. Responses
such as “the magnet slows down” and “it decelerates” were rejected. “The acceleration slows
down” was not a preferred response but it was accepted. The major problem in part (b)(ii) was
the failure of candidates to read the question properly: this was about the effect on the
acceleration of the magnet as it left the coil, not after it had left the coil. Consequently a large
number of candidates followed a broadly correct deduction in (i) by an incorrect one in (ii): they
thought that the acceleration would increase. The two explanation marks in part (b) escaped all
but the most knowledgeable candidates. Some understanding of what was induced and why,
was almost a prerequisite to progress here. Bald reference to Lenz’s law was not considered to
be adequate.
Even after presenting indifferent answers to the earlier parts of this question, many candidates
salvaged most of the three marks in part (c). Most appreciated that an incomplete circuit meant
that no current could flow, but many candidates wrongly thought that the missing ammeter
would also prevent the induction of an emf.
E27.
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.
E28.
Students are much more accustomed to diagrams which show magnetic fields acting at
right angles to the plane of a diagram, than magnetic fields acting in the plane of a diagram.
Consequently the seeds of confusion were sown at the start of part (a) for a large proportion of
the candidates, many evidently treating the question as though it referred to an electric field.
Therefore the path of the ion in part (ii) was stated to be parabolic, and not circular, in a large
number of the scripts. Perhaps the aim of the required calculation was a little obscure, but a
question about a circular path ought to have triggered ‘radius’ in the minds of the candidates.
Many calculated this radius very successfully, the principal error being a wrong value for the
charge of the doubly-charged ion.
In part (b) it was not possible to award any marks to candidates who were convinced that the
path was parabolic; they tended to write about curves that were ‘steeper’ or ‘with a bigger
slope’, etc.
E29.
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.
E30.
Most candidates gained at least one mark in part (a), and many went on to score all three.
Ability to use the accepted terminology for electromagnetism at this level contributed to well
written answers: the fact that a stronger magnet produces a greater magnetic flux (or flux
linkage) was preferred to ‘bigger magnet gives a stronger magnetic field’. Some candidates
even referred to the magnet producing an electric field. Faraday’s law was quite well known,
although some candidates made confusing references to Lenz’s law. For all three marks to be
awarded, it was necessary for candidates to make it clear that moving a stronger magnet at the
same speed would cause a greater change of flux in the same time and hence a greater
induced emf. Many answers fell short of this by failing to mention the time factor (or the rate of
change).
Part (b) was found to be straightforward by many candidates, but totally impossible by others.
Examiners insisted on seeing Δt in the first part; t was not accepted. In the second part, the
application of the equation ΔA = lvΔt to Faraday’s law easily led to the required equation. A
large number of unsuccessful attempts seemed to rely on extracting as many electromagnetic
equations as possible from the Data Sheet and trying to rearrange them into = B/v. The main
difficulty in part (b) (ii) arose from using the wrong value for the radius. Part (b) (iii) was most
easily answered by substitution of the data into = Blv, but the traditional approach leading
to = Bfπr2 also rewarded several candidates with full marks.
Despite the relatively unfamiliar device involved, part (c) of the question proved to be accessible
to a large proportion of the candidates, and full marks were frequently awarded.
E31.Acceptable statements really needed to refer to both the direction of the induced emf (or current)
and to the change (in magnetic flux) that produces the effect. In part (b)(i) an explanation of the
greater mass of Q was required, so a simple statement that density was involved was
inadequate; candidates had to state that steel (or Q) has a greater density than aluminium (or
P). In part (b)(ii) the time of 0.45s was usually justified through the application of s = ut + ½ at2,
although some candidates made no attempt to justify this value. Backwards working, such as
showing that the distance fallen is approximately 1.0m when the time of fall is 0.45s, was
accepted. Explanations of why the two times are equal were expected to refer to acceleration
due to gravity being independent of the mass of a falling body.
There was widespread misunderstanding in candidates’ attempts to answer part (c). In part (i),
clearly Q is a moving magnet passing through a conducting tube and so the magnet’s flux lines
are cut by the tube – hence an emf is induced. A significant number of responses stated that Q
would be cutting through the flux lines of the tube. The tube was regularly referred to as a
magnet. A very common misapprehension was that when a current is induced in the tube, it is
the current that causes the emf. In part (c)(ii) many answers were too trivial, such as ones which
referred to the repulsion of poles, or were simply wrong, such as attributing the effect to
inducedcharges. Some responses even suggested that the induced electromotive force acts as
a mechanical force to oppose the falling magnet. Examiners were pleased to encounter logical
answers stating that the induced emf caused a current to flow in the copper, which then
produced a magnetic field to oppose the movement of the falling magnet Q by opposing the
magnet’s own field. Relatively few answers made any reference as to why cylinder P would fall
without opposition.
Full marks were regularly awarded in part (d), where it was usually seen that the time for P
would be unaffected (an explanation was needed for the mark) but that for Q would be shorter.
Some candidates thought that the increased resistance of the tube would cause a reduced emf;
these answers were subjected to a two mark maximum.