Download Which one of the following statements is correct? An

1
Which one of the following statements is correct?
An electron follows a circular path when it is moving at right angles to
A
a uniform magnetic field.
B
a uniform electric field.
C
uniform electric and magnetic fields which are perpendicular.
D
uniform electric and magnetic fields which are in opposite directions.
(Total 1 mark)
2
Two electrons, X and Y, travel at right angles to a uniform magnetic field.
X experiences a magnetic force, FX, and Y experiences a magnetic force, FY.
What is the ratio
if the kinetic energy of X is half that of Y?
A
B
C
D
1
(Total 1 mark)
3
Charged particles, each of mass m and charge Q, travel at a constant speed in a circle of radius
r in a uniform magnetic field of flux density B.
Which expression gives the frequency of rotation of a particle in the beam?
A
B
C
D
(Total 1 mark)
Page 1 of 59
4
Protons and pions are produced in a beam from a target in an accelerator. The two types of
particles can be separated using a magnetic field.
(a)
State the quark composition of
(i)
a proton,
...............................................................................................................
(ii)
a positive pion, π+
...............................................................................................................
(2)
(b)
A narrow beam consisting of protons and positive pions, all travelling at a speed of
1.5 × 107 m s–1 , is directed into a uniform magnetic field of flux density 0.16 T, as shown in
the diagram.
(i)
Calculate the radius of curvature of the path of the protons in the field.
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(ii)
Sketch, on the diagram above, the path of the pions from the point of entry into the
field to the point of exit from the field.
Page 2 of 59
(iii)
If the magnetic field were increased, how would this affect the paths of the particles?
...............................................................................................................
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(7)
(Total 9 marks)
5
The diagram below shows a diagram of a mass spectrometer.
(a)
The magnetic field strength in the velocity selector is 0.14 T and the electric field strength is
20 000 V m–1.
(i)
Define the unit for magnetic flux density, the tesla.
...............................................................................................................
...............................................................................................................
(2)
(ii)
Show that the velocity selected is independent of the charge on an ion.
(2)
(iii)
Show that the velocity selected is about 140 km s–1.
(1)
Page 3 of 59
(b)
A sample of nickel is analysed in the spectrometer. The two most abundant isotopes of
nickel are
Ni and
Ni. Each ion carries a single charge of +1.6 × 10–19 C.
mass of a proton or neutron = 1.7 × 10–27 kg
The Ni ion strikes the photographic plate 0.28 m from the point P at which the ion beam
enters the ion separator.
Calculate:
(i)
the magnetic flux density of the field in the ion separator;
(3)
(ii)
the separation of the positions where the two isotopes hit the photographic plate.
(2)
(Total 10 marks)
Page 4 of 59
6
An electron moves into a region of uniform magnetic flux density between the poles of a magnet
as shown in the diagram.
The deflection of the electron will be
A
towards the pole marked S
B
towards the pole marked N
C
perpendicular to the plane of the paper towards you
D
perpendicular to the plane of the paper away from you
(Total 1 mark)
7
An alpha particle moves at one-tenth the velocity of a beta particle. They both move through the
same uniform magnetic field at right angles to their motion.
The magnitude of the ratio
is
A
B
C
D
(Total 1 mark)
Page 5 of 59
8
Figure 1 shows the plan view of a cyclotron in which protons are emitted in between the dees.
The protons are deflected into a circular path by the application of a magnetic field. Figure 2
shows a view from in front of the cyclotron.
Figure 1
Figure 2
(a)
(i)
Mark on Figure 2 the direction of the magnetic field in the region of the dees such
that it will deflect the proton beam in the direction shown in Figure 1.
(2)
(ii)
Show that the velocity of the proton, v, at some instant is given by:
v=
where m is the proton mass, r the radius of its circular path, B the magnetic flux
density acting on the proton and +e the proton charge.
(2)
Page 6 of 59
(iii)
Write down an equation for the time T for a proton to make a complete circular path
in this magnetic field.
(2)
(iv)
Explain how your equation leads to the conclusion that T is independent of the speed
with which the proton is moving.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(b)
In addition to this magnetic field there is an electric field provided between the dees. This
accelerates the proton towards whichever dee is negatively charged. An alternating
potential difference causes each dee to become alternately negative and then positive.
This causes the proton to accelerate each time it crosses the gap between the dees.
(i)
Describe and explain the effect the acceleration has on the path in which the proton
moves.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
In terms of T, write down the frequency with which the p.d. must alternate to match
the period of motion of the proton.
(1)
Page 7 of 59
(c)
(i)
Calculate the velocity of a proton of energy 0.12 keV.
the proton mass, m = 1.7 × 10–27 kg
the magnitude of the electronic charge, e = 1.6 × 10–19 C
(3)
(ii)
Calculate the de Broglie wavelength of the 0.12 keV proton.
the Planck constant, h = 6.6 × 10–34 J s
(3)
(iii)
Name the region of the electromagnetic spectrum which has an equivalent
wavelength to that of the proton.
...............................................................................................................
(1)
(Total 17 marks)
9
(a)
Explain why a particle is accelerating even when it is moving with a uniform speed in a
circular path.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 8 of 59
(b)
Figure 1 shows a schematic diagram of a proton synchrotron. This is a device for
accelerating protons to high speeds in a horizontal circular path.
Figure 1
In the synchrotron the protons of mass 1.7 × 10–27 kg are injected at point A at a speed of
8.0 × 106 m s–1. The diameter of the path taken by the protons is 400 m.
(i)
Show on Figure 1 the direction of the force required to make a proton move in the
circular path when the proton is at the position marked P.
(1)
(ii)
Calculate the force that has to be provided to produce the circular path when the
speed of a proton is 8.0 × 106 m s–1.
(2)
(iii)
Sketch on Figure 2 a graph to show how this force will have to change as the speed
of the proton increases over the range shown on the x-axis. Insert an appropriate
scale on the force axis.
Figure 2
Page 9 of 59
(c)
Before reaching their final energy the protons in the synchrotron in part (b) travel around
the accelerator 420 000 times in 2.0 s.
acceleration of free fall, g = 9.8 m s–2
(i)
Calculate the total distance travelled by a proton in the 2.0 s time interval.
(2)
(ii)
Unless a vertical force is applied the protons wold fall as they move through the
horizontal channel.
Calculate the distance a proton would fall in two seconds.
(2)
(iii)
Determine the force necessary to prevent the vertical movement.
(1)
(Total 12 marks)
10
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)
Page 10 of 59
11
The diagram shows an arrangement in a vacuum to deflect protons into a detector using a
magnetic field, which can be assumed to be uniform within the square shown and zero outside it.
The motion of the protons is in the plane of the paper.
(a)
Sketch the path of a proton through the magnetic deflector. At any point on this path draw
an arrow to represent the magnetic force on the proton. Label this arrow F.
(2)
(b)
State the direction of the uniform magnetic field causing this motion.
........................................................................................................................
(1)
(c)
The speed of a proton as it enters the deflector is 5.0 × 106 m s–1. If the flux density of the
magnetic field is 0.50 T, calculate the magnitude of the magnetic force on the proton.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(d)
If the path were that of an electron with the same velocity, what two changes would need to
be made to the magnetic field for the electron to enter the detector along the same path?
........................................................................................................................
........................................................................................................................
(2)
(Total 7 marks)
Page 11 of 59
12
A narrow beam of electrons is directed into a uniform electric field created by two oppositelycharged parallel metal plates at right angles to the field lines. A fluorescent screen is used to
make the beam give a visible trace.
(a)
(i)
Explain why the beam curves towards the positive plate.
...............................................................................................................
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(ii)
How does the trace show that, on entry to the electric field, all the electrons have the
same speed?
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(b)
The beam is produced as a result of accelerating electrons between the filament and a
metal anode.
(i)
Explain why the wire filament must be hot.
...............................................................................................................
...............................................................................................................
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Page 12 of 59
(ii)
Write down an equation relating the speed of the electrons, υ, to the potential
difference, VA, between the anode and the filament.
...............................................................................................................
...............................................................................................................
(2)
(c)
The deflection of the beam due to the electric field can be cancelled by applying a suitable
uniform magnetic field in the same region as the electric field.
(i)
What direction should the magnetic field be in to do this?
...............................................................................................................
...............................................................................................................
(ii)
Write down an equation relating the speed of the electrons υ to the plate voltage Vp,
the plate separation d, and the magnetic flux density B necessary to make the beam
pass undeflected between the plates.
...............................................................................................................
...............................................................................................................
Page 13 of 59
(iii)
The following measurements were made when the beam was undeflected.
VA = 3700 V
Vp = 4500 V
d = 50 mm
B = 2.5 mT
Use the two equations you have written down and the given data to calculate the
specific charge, e/m, of the electron.
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(5)
(Total 10 marks)
13
(a)
(i)
State two differences between a proton and a positron.
difference 1 ...........................................................................................
...............................................................................................................
difference 2 ...........................................................................................
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Page 14 of 59
(ii)
A narrow beam of protons and positrons travelling at the same speed enters a
uniform magnetic field. The path of the positrons through the field is shown in Figure
1.
Sketch on Figure 1 the path you would expect the protons to take.
Figure 1
(iii)
Explain why protons take a different path to that of the positrons.
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(5)
Page 15 of 59
(b)
Figure 2 shows five isotopes of carbon plotted on a grid in which the vertical axis
represents the neutron number N and the horizontal axis represents the proton number Z.
Two of the isotopes are stable, one is a beta minus emitter and two are positron emitters.
Figure 2
(i)
Which isotope is a beta minus emitter?
...............................................................................................................
(ii)
Which of the two positron emitters has the shorter half-life? Give a reason for your
choice.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(c)
A positron with kinetic energy 2.2 MeV and an electron at rest annihilate each other.
Calculate the average energy of each of the two gamma photons produced as a result of
this annihilation.
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........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 10 marks)
Page 16 of 59
14
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)
15
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)
16
An electron moves due North in a horizontal plane with uniform speed. It enters a uniform
magnetic field directed due South in the same plane. Which one of the following statements
concerning the motion of the electron in the magnetic field is correct?
A
It continues to move North with its original speed.
B
It slows down to zero speed and then accelerates due South.
C
It is accelerated due West.
D
It is accelerated due North.
(Total 1 mark)
Page 17 of 59
17
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 the α particle, what is
the value of the ratio
?
A
3.7
B
7.5
C
60
D
112.5
(Total 1 mark)
18
The path followed by an electron of momentum p, carrying charge −e, which enters a magnetic
field at right angles, is a circular arc of radius r.
What would be the radius of the circular arc followed by an α particle of momentum 2p, carrying
charge +2e, which entered the same field at right angles?
A
B
r
C
2r
D
4r
(Total 1 mark)
19
(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.
.............................................................................................................
Page 18 of 59
(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)
Page 19 of 59
20
A negatively charged particle moves at right angles to a uniform magnetic field. The magnetic
force on the particle acts
A
in the direction of the field.
B
in the opposite direction to that of the field.
C
at an angle between 0° and 90° to the field.
D
at right angles to the field.
(Total 1 mark)
21
When travelling in a vacuum through a uniform magnetic field of flux density 0.43 m T, an
electron moves at constant speed in a horizontal circle of radius 74 mm, as shown in the figure
below.
(a)
When viewed from vertically above, the electron moves clockwise around the
horizontal circle. In which one of the six directions shown on the figure above,
+x, –x, +y, –y, +z or –z, is the magnetic field directed?
direction of magnetic field ......................................
(1)
Page 20 of 59
(b)
Explain why the electron is accelerating even though it is travelling at constant speed.
......................................................................................................................
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......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
(i)
By considering the centripetal force acting on the electron, show that its speed is
5.6 × 106 m s–1.
(2)
(ii)
Calculate the angular speed of the electron, giving an appropriate unit.
answer = ......................................
(2)
Page 21 of 59
(iii)
How many times does the electron travel around the circle in one minute?
answer = ......................................
(2)
(Total 9 marks)
22
Particles of mass m, each carrying charge Q and travelling with speed v, enter a magnetic field
of flux density B at right angles. Which one of the following changes would produce an increase
in the radius of the path of the particles?
A
an increase in Q
B
an increase in m
C
a decrease in v
D
an increase in B
(Total 1 mark)
23
An electron moving with a constant speed enters a uniform magnetic field in a direction
perpendicular to the magnetic field. What is the shape of the path that the electron would follow?
A
parabolic
B
circular
C
elliptical
D
a line parallel to the magnetic field
(Total 1 mark)
Page 22 of 59
24
An electron moves due North in a horizontal plane with uniform speed. It enters a uniform
magnetic field directed due South in the same plane. Which one of the following statements
concerning the motion of the electron in the magnetic field is correct?
A
It accelerated due West.
B
It slows down to zero speed and then accelerates due South.
C
It continues to move North with its original speed.
D
It is accelerated due North.
(Total 1 mark)
25
A jet of air carrying positively charged particles is directed horizontally between the poles of a
strong magnet, as shown in the diagram.
In which direction are the charged particles deflected?
A
upwards
B
downwards
C
towards the N pole of the magnet
D
towards the S pole of the magnet
(Total 1 mark)
Page 23 of 59
26
An electron moving with a constant speed enters a uniform magnetic field in a direction at right
angles to the field. What is the subsequent path of the electron?
A
A straight line in the direction of the field.
B
A straight line in a direction opposite to that of the field.
C
A circular arc in a plane perpendicular to the direction of the field.
D
An elliptical arc in a plane perpendicular to the direction of the field.
(Total 1 mark)
27
The path followed by an electron of momentum p, carrying charge –e, which enters a magnetic
field at right angles, is a circular arc of radius r.
What would be the radius of the circular arc followed by an α particle of momentum 2p, carrying
charge +2e, which entered the same field at right angles?
A
B
r
C
2r
D
4r
(Total 1 mark)
28
Charged particles, each of mass m and charge Q, travel at a constant speed in a circle of radius
r in a uniform magnetic field of flux density B. Which expression gives the frequency of rotation
of a particle in the beam?
A
B
C
D
(Total 1 mark)
Page 24 of 59
29
Two charged particles, P and Q, move in circular orbits in a magnetic field of uniform flux density.
The particles have the same charge but the mass of P is less than the mass of Q. TP is the time
taken for particle P to complete one orbit and TQ the time for particle Q to complete one orbit.
Which one of the following is correct?
A
TP = TQ
B
T P> T Q
C
T P< T Q
D
T P – T Q= 1
(Total 1 mark)
30
(a)
The equation F = BQv may be used to calculate magnetic forces.
(i)
State the condition under which this equation applies.
.............................................................................................................
.............................................................................................................
(1)
(ii)
Identify the physical quantities that are represented by the four symbols in the
equation.
F .........................................................................................................
B .........................................................................................................
Q..........................................................................................................
v...........................................................................................................
(1)
Page 25 of 59
(b)
The figure below shows the path followed by a stream of identical positively charged ions,
of the same kinetic energy, as they pass through the region between two charged plates.
Initially the ions are travelling horizontally and they are then deflected downwards by the
electric field between the plates.
While the electric field is still applied, the path of the ions may be restored to the horizontal,
so that they have no overall deflection, by applying a magnetic field over the same region
as the electric field. The magnetic field must be of suitable strength and has to be applied
in a particular direction.
(i)
State the direction in which the magnetic field should be applied.
.............................................................................................................
(1)
(ii)
Explain why the ions have no overall deflection when a magnetic field of the required
strength has been applied.
.............................................................................................................
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.............................................................................................................
(2)
Page 26 of 59
(iii)
A stream of ions passes between the plates at a velocity of 1.7 x 105ms–1. The
separation d of the plates is 65 mm and the pd across them is 48 V. Calculate the
value of B required so that there is no overall deflection of the ions, stating an
appropriate unit.
answer = ....................................
(4)
(c)
Explain what would happen to ions with a velocity higher than 1.7 x 105ms–1 when they
pass between the plates at a time when the conditions in part (b)(iii) have been
established.
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.....................................................................................................................
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(2)
(Total 11 marks)
Page 27 of 59
31
A beam of positive ions enters a region of uniform magnetic field, causing the beam to change
direction as shown in the diagram.
What is the direction of the magnetic field?
A
out of the page and perpendicular to it
B
into the page and perpendicular to it
C
in the direction indicated by +y
D
in the direction indicated by -y
(Total 1 mark)
32
The Large Hadron Collider (LHC) uses magnetic fields to confine fast-moving charged particles
travelling repeatedly around a circular path. The LHC is installed in an underground circular
tunnel of circumference 27 km.
(a)
In the presence of a suitably directed uniform magnetic field, charged particles move at
constant speed in a circular path of constant radius. By reference to the force acting on the
particles, explain how this is achieved and why it happens.
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(4)
Page 28 of 59
(b)
(i)
The charged particles travelling around the LHC may be protons. Calculate the
centripetal force acting on a proton when travelling in a circular path of circumference
27 km at one-tenth of the speed of light. Ignore relativistic effects.
answer = ................................ N
(3)
(ii)
Calculate the flux density of the uniform magnetic field that would be required to
produce this force. State an appropriate unit.
answer = ...................................... unit ........................
(3)
(c)
The speed of the protons gradually increases as their energy is increased by the LHC.
State and explain how the magnetic field in the LHC must change as the speed of the
protons is increased.
........................................................................................................................
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........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 12 marks)
Page 29 of 59
33
When a β particle moves at right angles through a uniform magnetic field it experiences a force
F. An α particle moves at right angles through a magnetic field of twice the magnetic flux density
with velocity one tenth the velocity of the β particle. What is the magnitude of the force on the α
particle?
A
B
C
D
0.2 F
0.4 F
0.8 F
4.0 F
(Total 1 mark)
34
(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.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
(2)
(ii)
Calculate the magnitude of the electrostatic force acting on the ion.
magnitude of electrostatic force ........................................ N
(2)
Page 30 of 59
(iii)
Calculate the electric field strength.
electric field strength .................................. NC–1
(1)
(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.
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...............................................................................................................
(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)
Page 31 of 59
(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)
35
(a)
(i)
State two situations in which a charged particle will experience no magnetic force
when placed in a magnetic field.
first situation...........................................................................................
...............................................................................................................
...............................................................................................................
second situation.....................................................................................
...............................................................................................................
...............................................................................................................
(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)
Page 32 of 59
(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)
Page 33 of 59
(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)
36
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)
Page 34 of 59
Mark schemes
1
2
3
4
A
[1]
C
[1]
A
[1]
(a)
(i)
uud (1)
(ii)
(1)
2
(b)
(i)
= Bev [or r =
] (1)
m = 1.67 × 10-27 (1)
(1)
= 0.98 m (1)
(ii)
pion path more curved than proton path (1)
(iii)
path more curved
[or radius (of path) smaller] (1)
for both paths (1)
7
[9]
Page 35 of 59
5
(a)
(i)
1 N per A per m
or 1 Wb m–2
or quotes: B = F/IL with terms defined
or induced EMF = ΔBAN/t with terms defined
or a slightly flawed attempt at the definition in
statement form
C1
It is the flux density (perpendicular to a wire) that
produces a force of 1N per m on the wire when
the current is 1A
or
B = F/IL and 1 T is flux density when F = 1N; I = 1A
and L = 1 m
or induced EMF = ΔBAN /t and 1 T is the flux change
when emf = 1V for A=1 N =1 and t =1 or similar
A1
2
(ii)
force on charge due to E field , FE= Eq or Vq/d
and
force due to B field, FB = Bqv
or Eq=Bqv
B1
= Bqv; cancels q and states explicitly v =
or v =
B1
2
(iii)
v = 20000/0.14 (seen) or 143 × 103 m s–1
B1
1
Page 36 of 59
(b)
(i)
Bqv =mv2/r or r = mv/Bq ( allow e instead of q)
mass of ion = 1.7 × 10–27 × 58 (may be in equation)
or (9.86 × 10–26 kg seen)
C1
or
radius = 0.14 m (may be in equation)
C1
Substitutes and arrives at 0.62 to 0.63 T
A1
3
(ii)
Calculates new radius (0.145 m) or diameter (0.288 m)
using r
m or otherwise allowing ecf
C1
0.010 m (condone 0.01 m) or 0.0096 – 0.0097 m
(Allow 0.0079 m or 0.008 m due to use of different
sfs for B and v )
A1
2
[10]
6
7
8
D
[1]
B
[1]
(a)
(i)
vertical field line(s)
B1
directed downwards
B1
2
(ii)
mv2/r and Bev seen
M1
equated and correctly rearranged
A1
2
Page 37 of 59
(iii)
v=
or equivalent
M1
T=
A1
2
(iv)
no v in the equation for T (m, B and e
all independent of v)
B1
1
(b)
(i)
proton spirals outwards/suitable diagram
B1
as v ↑ r ↑
B1
2
(ii)
f =1/T
B1
1
(c)
(i)
conversion of keV to J (1.92 × 10–17)
C1
use of ½ mv2
1.50 × 105 ms–1
A1
3
Page 38 of 59
(ii)
λ=
p = mv or substituted values
C1
2.6 × 10–12 m
A1
3
(iii)
y-rays or X-rays or answer consistent with
candidate’s λ
B1
1
[17]
9
(a)
acceleration is (rate of) change of velocity
or velocity is a vector
or velocity has magnitude and direction
B1
velocity is changing since direction is changing
(must be clear that it is the velocity that is changing direction)
B1
allow 1 mark for ‘it would move in a straight line at constant speed if it were not
accelerating’
do not allow
‘because there is a force acting’
‘because direction is changing’
(2)
(b)
(i)
arrow toward centre of circle at P
B1
(1)
(ii)
F = mv2 / r or mrω2
or numerical equivalent (r must be 200 m)
C1
5.4 × 10–16 N
A1
(2)
Page 39 of 59
(iii)
graph showing correct curvature with F plotted correctly (e.c.f. for F)
(should be between 5 × 10–14 and 6 × 10–16 N
B1
double v, quadruple F
(should be possible to do these tasks to ±½ a square)
B1
(2)
(c)
(i)
circumference = 1256 m or 2πr × 420 000
(allow e.c.f. for incorrect r from (b)(ii))
C1
distance travelled = 5.3 × 108 m
A1
(2)
(ii)
s = ½ gt2 or ut + ½at2
C1
19.6m (20m)
A1
(2)
(iii)
mg = 1.7 (1.67) × 10–26 N
B1
(1)
[12]
10
11
D
[1]
(a)
(uniformly) curved path continuous with linear paths at entry and exit points (1)
arrow marked F towards top left-hand corner (1)
2
(b)
into (the plane of) the diagram (1)
(not accept “downwards”)
1
Page 40 of 59
(c)
F(= BQυ) = 0.50 × 1.60 × 10-19 × 5.0 × 106 (1)
= 4.0 × 10-13 N (1)
2
(d)
B must be in opposite direction (1)
(much) smaller magnitude
(1)
2
[7]
12
(a)
(i)
electrons are negatively charged so beam is attracted to positive plate
[or repelled by negative plate or electron experiences force towards positive plate]
(1)
(ii)
beam does not spread out (1)
if speeds varied, faster electrons would be
deflected less than slower electrons (1)
3
(b)
(i)
(ii)
to give conduction electrons sufficient
k.e. to leave metal [or to cause thermionic
emission or electrons have insufficient
ke. in a cold filament to leave filament] (1)
mυ2 = eVA [or υ =
] (1)
2
(c)
(i)
into the plane of the diagram (1)
perpendicular to the diagram [or the electric field] (1)
(ii)
Beυ =
(iii)
combine the two equations to give
(1)
(1)
(1)
1.75 × 1011 Ckg–1 (1)
max 5
[10]
Page 41 of 59
13
(a)
(i)
including, for example:
positron is an antimatter particle; proton is a matter particle (*)
positron is a lepton; proton is a hadron (*)
positron has a smaller rest mass than a proton (*)
positron is not composed of other particles; proton is made up of quarks (*)
(*) any two [1] [1]
(ii)
proton path has greater radius of curvature than positron (1)
(iii)
radius of curvature r =
and υ, B and e are constants (1)
therefore r proportional to m (1)
mass of proton is (much) greater than mass
of positron (at same speed) (1)
5
(b)
(i)
C - 14 (1)
(ii)
C - 10 (1)
as this is furthest from stability (1)
3
(c)
rest mass of electron = 0.51 MeV therefore total energy available
= (2.2 +2 × 0.51)= 3.22 (MeV) (1)
gamma photons produced have average energy =
= 1.6 MeV(1)
2
[10]
14
15
16
17
C
[1]
B
[1]
A
[1]
B
[1]
Page 42 of 59
18
19
B
[1]
(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]
20
21
D
[1]
(a)
magnetic field direction: −z (1)
1
(b)
direction changes meaning that velocity is not constant (1)
acceleration involves change in velocity
(or acceleration is rate of change of velocity) (1)
[alternatively
magnetic force on electron acts perpendicular to its velocity (1)
force changes direction of movement causing acceleration (1)]
2
Page 43 of 59
(c)
(i)
BQv =
(1) gives v
(1) (= 5.59 × 106 m s−1)
2
(ii)
angular speed ω
= 7.5(5) × 107 (1)
unit: rad s−1 (1) (accept s−1)
2
(iii)
frequency of electron’s orbit f
(1)
(= 1.20 × 107 s−1)
number of transits min−1 = 1.20 × 107 × 60 = 7.2 × 108 (1)
[alternatively
orbital period
[or
]
(= 8.32 × 10−8 s)
number of transits min−1 =
(1)]
2
[9]
22
23
B
[1]
B
[1]
Page 44 of 59
24
25
26
27
28
29
30
C
[1]
B
[1]
C
[1]
B
[1]
A
[1]
C
[1]
(a)
(i)
magnetic field (or B) must be at right angles to velocity (or v)
1
(ii)
F = (magnetic) force (on a charged particle or ion)
B = flux density (of a magnetic field)
Q = charge (of particle or ion)
v = velocity [or speed] (of particle or ion)
all four correct
1
Page 45 of 59
(b)
(i)
into plane of diagram
1
(ii)
magnetic force = electric force [or BQv = EQ]
these forces act in opposite directions [or are balanced
or resultant vertical force is zero]
2
(iii)
BQv = EQ gives flux density B =
(= 738 V m–1)
= 4.3 × 10–3
T
4
(c)
ions would be deflected upwards
magnetic force increases but electrostatic force is
unchanged [or magnetic force now exceeds electrostatic force]
2
[11]
31
32
A
[1]
(a)
(magnetic) field is applied perpendicular to path
or direction or velocity of charged particles
(magnetic) force acts perpendicular to path
or direction or velocity of charged particles
force depends on speed of particle or on B [or F ∞ v or F = BQv explained]
force provides (centripetal) acceleration towards centre of circle
[or (magnetic) force is a centripetal force]
shows that r is constant when B and v are constant
4
Page 46 of 59
(b)
(i)
= 4.30 × 103 (m)
radius r of path =
(allow 4.3km)
= 3.50 × 10–16(N)
centripetal force
3
(ii)
magnetic flux density
= 7.29 × 10-5
T
3
(c)
magnetic field must be increased
to increase (centripetal) force or in order to keep r constant
[or otherwise protons would attempt to travel in a path of larger radius]
[or, referring to
, B must increase when v increases to keep r constant ]
2
[12]
33
34
B
[1]
(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
Page 47 of 59
(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
(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]
35
(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
Page 48 of 59
(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
(iv)
(gives 1.37 × 10−7s).
2
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
Page 49 of 59
(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)
36
A
[1]
Page 50 of 59
Examiner reports
1
2
4
This question required students to understand the trajectory of an electron moving in electric and
/ or magnetic fields. 73% gave a correct response.
Calculation of the force acting on electrons moving in magnetic fields in relation to their kinetic
energies was the basis of this question. Because the kinetic energy of X is half that of Y, it follows
that vx = vy /
and that the ratio of the forces is 1 / . The facility of this question was 54%;
18% of the students gave distractor A and 23% gave B.
The majority of candidates were able to give the correct quark composition of the proton in part
(a), but many were unable to give the correct quark composition of the positive pion, usually as a
result of stating that the antiquark was strange.
In part (b) the calculation of the radius of curvature was usually correct although some
candidates did not use the correct mass value for the proton. In part (ii) the majority of candidates
knew the pion path was more curved and were thus able to score full marks. In the final part
candidates often failed to state that the radius of curvature would be less for both types of
particles.
5
(a)
(i)
Only a small proportion of the candidates was able to provide a correct definition of
the telsa. Most gave another form of the unit Wb m–2 or less commonly N A–1 m–1.
(b)
(ii)
It would have been useful to see some words to support the algebraic argument but
this was very rare. Responses were usually a number of formulae with cancellations
(including some irrelevant formulae amongst the relevant ones in the poorest
answers). In view of the poor explanations candidates were at least expected to
make v the subject of the final formula to gain both marks.
(iii)
This was often correct but whether the answers were based on an understanding of
the physics or just number crunching was often difficult to follow. Candidates were
given the benefit of the doubt.
(i)
Many knew the equation r = mv/Bq or were able to derive it from first principles. A
common error was to use the diameter instead of the radius so arriving at half the
answer. Some did not calculate the mass of the ion and others used 28 × 1.6 × 10–19
C for the charge on the ion.
(ii)
Many were able to gain credit in this part using the error carried forward for an
incorrect flux density. However, the majority subtracted the given diameter from their
new radius so losing the second mark.
Page 51 of 59
8
(a)
(b)
(c)
9
(i)
Surprisingly, several candidates failed to answer this part. Of those who did most
recognised that the magnetic field would be vertically downwards.
(ii)
Most candidates were able to complete this part successfully.
(iii)
This was found to be more difficult and only the best candidates were able to arrive
at the expected equation.
(iv)
Of those arriving at an appropriate equation in (iii) most recognised that all the
quantities were independent of v.
(i)
Few candidates recognised that the proton would spiral outwards as its speed
increases. Several candidates did discuss an increase in the radius of the proton
path.
(ii)
Most candidates correctly recognised the relationship between f and T.
(i)
Weaker candidates had difficulty in converting the energy from keV into J but
otherwise this part was well answered.
(ii)
The de Broglie relationship was well known although, even with error carried forwards
from (i), several candidates had difficulty in calculating a consistent wavelength.
(iii)
Few candidates could relate the wavelength calculated in (ii) to its appropriate area
in the electromagnetic spectrum.
(a)
A number of candidates referred to the fact that there was a force and therefore
acceleration but ignored the reference to uniform speed. Candidates were expected to refer
to change in direction resulting in a change in velocity, owing to its vector nature, and to
state the link between change in velocity and acceleration.
(b)
(i)
Most candidates completed this successfully.
(ii)
The majority of candidates did this part correctly but some used the given diameter as
the radius. Many who started with mrω2 had difficulty determining ω.
(iii)
Many candidates drew careful graphs, plotting the data from (ii) correctly and using a
scale that covered the whole range. The quadrupling of force for a doubling of the
velocity was also very clear in the best graphs. There were, however, many
candidates whose skills in graphical communication left much to be desired. There
were many instances where, for example, the value from (ii), 5.4 × 10–16 N, was
plotted on the 20 mm grid line. These candidates rarely showed other values
correctly.
(i)
Most candidates were able to complete this successfully but there were a significant
number who used πr2 as the circumference.
(c)
Page 52 of 59
10
11
(ii)
Most appreciated the need to use s = ut + ½ at2 and the majority obtained the correct
answer. There were a significant proportion of candidates who did not distinguish
between horizontal and vertical motion and used 8 × 106 m s–1 for u. This led to a silly
answer for distance fallen in 2 s that usually passed without comment.
(iii)
Most completed this part successfully. Some candidates simply stated ‘gravitation’
instead of a value. A few gave 9.8 N as the answer.
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.
Apart from the very occasional script in which the candidate showed the proton to be banging
around inside the ‘box’ like a gas molecule, the path of the proton was well recognised in part (a).
The majority of candidates made an acceptable attempt at a freehand circular arc to join the
entry and exit points. The direction of the magnetic force caused greater difficulty for some
candidates, with arrows shown in all kinds of directions instead of towards the top left-hand
corner. An arrow tangential to the path was a common wrong response.
Parts (b) and (c) were generally done well, but the charge of the proton was not always extracted
correctly from the Data booklet as e. A value of +1 (or 9.58 × 107 which is e / mp) was frequently
encountered.
Reversal of the direction of B was usually correct as the answer for one of the marks in part (d),
but many fewer candidates were able to understand that the less massive electron would require
a field of much smaller magnitude in order to follow the same path as the proton. As has
happened often in the past, weaker candidates tended to give incomplete responses such as
“change” the field direction and “alter” the strength of the field; these answers gained no marks.
12
Most candidates explained correctly in part (a) why the beam bends towards the positive plate
and why the trace shows that the electrons have the same speed. Relatively few candidates
went on to state that faster-moving electrons would deflect less and slower electrons more.
Many candidates scored both marks in part (b), often with a good account of thermionic emission
which indicated clear understanding of the process involved.
Few candidates scored both marks in part (c)(i), although many did know that the magnetic field
direction needs to be into the plane of the diagram. Weaker candidates generally failed to score
here, often confusing the force direction and the field direction, Most candidates knew the correct
equation in part (c)(ii) and many realised that the equation was to be used in part (c)(iii), although
some lost a mark through failure to state the correct unit.
13
Parts (a)(i) and (a)(ii) were found to be straightforward by most candidates, but only the best
explained the ideas behind the trajectory with any rigour. This required the essential statement
that B, v, and e are constants in the equation R = mʋ / Be.
Page 53 of 59
Part (b) discriminated well at the bottom end of the ability range.
Many more candidates than expected failed to include the rest mass of both particles in the total
energy calculation in part (c).
14
15
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.
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.
17
19
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.
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.
20
This question, with facilities of 67%, was about charged particles moving at right angles to a
magnetic field. Relatively few candidates chose any one of the incorrect responses in the
question.
Page 54 of 59
21
In part (a) the correct application of Fleming’s left hand rule to moving electrons was a much
sterner test than it ought to have been for A2 candidates. It seemed that responses were
distributed almost randomly between the six alternative directions. Part (b) was the familiar test
of whether candidates understood the significance of the directional nature of velocity for a
particle moving in a circle. The expected approach was to point out that a change in direction
shows that velocity is changing, and that acceleration involves a change in velocity.
Alternatively, it could be argued that the force on the electron always acts at right angles to its
velocity, thus changing the electron’s direction of travel and causing it to accelerate.
Candidates with a superficial acquaintance with this situation tended to refer to centripetal force
in their answers, without conveying any proper understanding of the directional nature of velocity.
As suggested by the question, the starting point for successful answers to part (c) (i) was the
equation BQv = mv2/r. Most candidates arrived at a correct result for the speed of the electron, by
substituting either the separate values for e and me, or for the specific charge e/me, from the
Data and Formulae Booklet. Calculation of the angular speed in part (c) (ii) usually caused little
difficulty, but its unit was not always known: m s–1 was often written down. Those who had quoted
m s–1 usually then got into difficulty in part (c) (iii), because they tried to find the orbital period by
dividing the circumference of the circle by their value for ω/m s–1. A particularly worrying error by
many candidates in this part was a calculator error when trying to divide ω (= 7.55 × 107 rad s–1)
by 2π. This led to a final incorrect answer of 7.1 × 109 revolutions per minute, which appeared in
a large number of scripts. The error appears to have been caused by an incorrect sequence of
division and multiplication operations on calculators.
22
23
24
This question looked at factors that might increase the radius of curvature of charged particles
following a circular arc in a B field. The facility of this question was 63%, and incorrect responses
were fairly equally spread.
This question, with facilities of 62%, was about charged particles moving at right angles to a
magnetic field. Distractor A (parabolic path) attracted 29% of the responses.
The magnetic force on a moving electron was tested in this question, but this time qualitatively
instead of quantitatively. Since this electron was moving anti-parallel to the magnetic field
through which it was travelling, it would experience no magnetic force. Therefore, its motion
would not be affected by the B field. 45% of candidates appreciated this (answer C), but 21%
thought it would grind to a halt and set off in the opposite direction (distractor B) whilst 18%
thought it would accelerate in its original direction (distractor D). Confusion with the effects of
electric fields is evident in these incorrect responses.
Page 55 of 59
25
26
27
28
29
This question could be answered by applying Fleming’s left hand rule to a beam of positive ions.
Around half of the responses were correct, but a quarter were for distractor A (which was
upwards, instead of downwards).
The principles of the magnetic deflection of an electron beam by a magnetic field were well
understood in this question, where 73% of the candidates made the correct choice. Almost one in
five candidates chose distractor D, an obvious confusion over the shape of the curved path.
This question, which had been used in an earlier examination, had a facility of 60% in 2014. On
the previous occasion its facility was 55%. It may be readily seen that the radius of the path of a
moving charged particle in a magnetic field is proportional to momentum p and inversely
proportional to charge Q. When both p and Q are doubled, the charge will continue in a path of
the same radius. Incorrect responses were evenly spread between distractors A, C and D. This
was the most discriminating question in the test.
This question was a fairly standard test of the algebraic equations which govern the motion of a
charged particle as it moves through a magnetic field at right angles, but involving the frequency
of rotation around the circle. Almost 70% of the students selected the correct response.
This question concerned the time period of charged particles moving in a circular orbit in a
magnetic field. Its facility was 55% but distractors A and B were each selected by more than 20%
of candidates. Successful solutions required mv2/r = BQv to be combined with T = 2πr Iv.
30
In part (a)(i) many candidates were unaware of the condition under which F = BQv applies,
which is given clearly in the specification. A common incorrect answer was to state that the force
has to be perpendicular to B, without any reference to v. In part (a)(ii) the main difficulty proved to
be the meaning of B; magnetic flux density was correct and the loose ‘magnetic field strength’
was not accepted. Some candidates thought that v represents voltage.
Part (b)(i) was a test of Fleming’ left hand rule when applied to a stream of positive ions. Together
with the figure, the first paragraph of part (b) defines ‘downwards’ as the direction towards the
lower (negative) plate. The correct answer in (b)(i) is ‘into the plane of the diagram’, not
downwards.
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In part (b)(ii) candidates were expected to consider the force conditions applying to the
undeflected ions. A common misconception was that the magnetic field is equal to the electric
field. The main errors in part (b)(iii), where the numerical value obtained was often correct, were
the omission of clear working and not knowing that the unit of B is T. Some candidates could only
quote F = BQv and were at a loss to make further progress without F = EQ and E = V/d.
Many candidates were totally lost in part (c). Others correctly explained that the ions would now
be the magnetic force (which is proportional to v) increases whilst the electrostatic force (which is
independent of v) remains constant.
31
32
This question 61% of the students could apply the rule correctly, but almost one quarter of them
chose distractor B, where the magnetic field would have acted “down” into the page instead of
“up” out of it.
It was rare for all four marks to be awarded in part (a). The essence of this question was well
understood, but poor use of English and an inability to write logically limited the mark that could
be given. An alarming proportion of answers made no reference at all to the magnetic field; these
students appeared to be answering a more general question about circular motion. Many of the
students evidently thought that the purpose of the magnetic force (presumably acting outwards)
was to balance the centripetal force, rather than to provide it. Relatively few correct solutions
were seen that used r = mv / BQ to show that r is constant when B and v are constant.
The common error in part (b)(i) was failure to deduce the radius of the path of the protons from
the 27 km circumference of the LHC. This only meant the loss of one of the three marks,
however, provided the principles of the rest of the calculation were correct. Careless arithmetic
such as failure to square v, and/or forgetting to convert km to m, was also a frequent source of
loss of marks. F = BQv was usually applied successfully in part (b)(ii), where the unit of magnetic
flux density was quite well known. Almost inevitably, there was some confusion between flux
density and magnetic flux.
The fact that had to be appreciated in part (c) was that in the LHC the radius of the path of the
charged particles must remain constant as they are accelerated. A large proportion of students
thought that it was necessary to maintain a constant centripetal force for this to happen, whereas
it ought to have been clear to them that F must increase as v increases if r is to be constant.
33
This question had the unusual outcome that one of the distractors (A) was slightly more popular
than the correct answer (B). This had not happened when the question was pre-tested. No doubt
the mistake made by students who chose distractor A was to forget that the charge of an α
particle is +2e, not +e. The question had a facility of 44%.
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34
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.
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35
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 placed parallel 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 Q must
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.
36
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%.
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