Grav & Electrical Fields Part II
98 minutes
69 marks
Q1.
What would the period of rotation of the Earth need to be if objects at the equator were to
appear weightless?
radius of Earth = 6.4 × 106 m
A
4.5 × 10–2 hours
B
1.4 hours
C
24 hours
D
160 hours
(Total 1 mark)
Q2.
The diagram shows two positions, X and Y, on the Earth’s surface.
Which line, A to D, in the table gives correct comparisons at X and Y for gravitational potential
and angular velocity?
gravitational potential at
Xcompared with Y
angular velocity at
Xcompared with Y
A
greater
greater
B
greater
same
C
greater
smaller
D
same
same
(Total 1 mark)
Q3.
The diagram shows two point masses each of mass m separated by a distance 2r.
What is the value of the gravitational field strength at the mid-point, P, between the two
masses?
A
B
C
D
zero
(Total 1 mark)
Q4.
The gravitational force between two uniform spheres is 3.1 × 10–9 N when the distance
between their centres is 150 mm. If the mass of one sphere is 2.5 kg, what is the mass of the
other?
A
0.043 kg
B
0.42 kg
C
2.8 kg
D
4.1 kg
(Total 1 mark)
Q5.
(a)
(i)
The equation F = BQv may be used to calculate magnetic forces.
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)
(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 thehorizontal,
so that they have no overall deflection, by applying a magnetic field over thesame 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
requiredstrength has been applied.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
(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
beenestablished.
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
(2)
(Total 11 marks)
Q6.
Figure 1 shows a parcel on the floor of a delivery van that is passing over a hump-backed
bridge on a straight section of road. The radius of curvature of the path of the parcel is r and the
van is travelling at a constant speed v. The mass of the parcel is m.
Figure 1
(a)
(i)
Draw arrows on Figure 2 below to show the forces that act on the parcel as it
passes over the highest point of the bridge. Label these forces.
Figure 2
(1)
(ii)
Write down an equation that relates the contact force, R, between the parcel and
thefloor of the van to m, v, r and the gravitational field strength, g.
.............................................................................................................
.............................................................................................................
(1)
(iii)
Calculate R if m = 12 kg, r = 23 m, and v = 11ms–1.
answer = .................................N
(2)
(b)
Explain what would happen to the magnitude of R if the van passed over the bridge at a
higher speed. What would be the significance of any van speed greater than 15ms–1?
Support your answer with a calculation.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 7 marks)
Q7.
Which one of the following is not a cause of energy loss in a transformer?
A
good insulation between the primary and secondary coil
B
induced currents in the soft iron core
C
reversal of magnetism in the soft iron core
D
resistances in the primary and secondary coil
(Total 1 mark)
Q8.
The above graph shows how the output emf, ε, varies with time, t, for a coil rotating at angular
speed ω in a uniform magnetic field of flux density B. Which one of the following graphs shows
how ε varies with t when the same coil is rotated at angular speed 2ω in a uniform magnetic
field of flux density 0.5 B?
A
B
C
D
(Total 1 mark)
Q9.
A bar magnet is pushed into a coil connected to a sensitive ammeter, as shown in
thediagram, until it comes to rest inside the coil.
Why does the ammeter briefly show a non-zero reading?
A
The magnetic flux linkage in the coil increases then decreases.
B
The magnetic flux linkage in the coil increases then becomes constant.
C
The magnetic flux linkage in the coil decreases then increases.
D
The magnetic flux linkage in the coil decreases then becomes constant.
(Total 1 mark)
Q10.
The graph shows how the flux linkage, N
into a magnetic field.
, through a coil changes when the coil is moved
The emf induced in the coil
A
increases then becomes constant after time t0.
B
is constant then becomes zero after time t0.
C
is zero then increases after time t0.
D
decreases then becomes zero after time t0.
(Total 1 mark)
Q11.
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
TP > TQ
C
TP < TQ
D
TP – TQ = 1
(Total 1 mark)
Q12.
The electric potential at a distance r from a positive point charge is 45 V. The
potentialincreases to 50 V when the distance from the charge decreases by 1.5 m. What is the
value ofr?
A
1.3 m
B
1.5 m
C
7.9 m
D
15 m
(Total 1 mark)
Q13.
An ion carrying a charge of +4.8 × 10–19C travels horizontally at a speed of 8.0 × 105ms–1. It
enters a uniform vertical electric field of strength 4200 V m–1, which is directed downwards and
acts over a horizontal distance of 0.16m. Which one of the following statements is not correct?
A
The ion passes through the field in 2.0 × 10–7s.
B
The force on the ion acts vertically downwards at all points in the field.
C
The magnitude of the force exerted on the ion by the field is 1.6 × 10–9 N.
D
The horizontal component of the velocity of the ion is unaffected by the electric field.
(Total 1 mark)
Q14.
An electron and a proton are 1.0 × 10–10 m apart. In the absence of any other charges,what
is the electric potential energy of the electron?
A
+2.3 × 10–18J
B
–2.3 × 10–18J
C
+2.3 × 10–18J
D
–2.3 × 10–18J
(Total 1 mark)
Q15.
A satellite of mass m travels in a circular orbit of radius r around a planet of mass M.Which
one of the following expressions gives the angular speed of the satellite?
A
B
C
D
(Total 1 mark)
Q16.
At the surface of the Earth the gravitational field strength is g, and the
gravitationalpotential is V. The radius of the Earth is R. An object, whose weight on the surface
of the Earth is W, is moved to a height 3R above the surface. Which line, A to D, in the table
gives the weight of the object and the gravitational potential at this height?
weight
A
B
C
gravitational
potential
D
(Total 1 mark)
Q17.
A planet has a radius half the Earth’s radius and a mass a quarter of the Earth’s
mass.What is the approximate gravitational field strength on the surface of the planet?
A
1.6 N kg–1
B
5.0 N kg–1
C
10 N kg–1
D
20 N kg–1
(Total 1 mark)
Q18.
Masses of M and 2M exert a gravitational force F on each other when the
distancebetween their centres is r. What is the gravitational force between masses of 2M and
4M when the distance between their centres is 4r?
A
0.25 F
B
0.50 F
C
0.75 F
D
1.00 F
(Total 1 mark)
Q19.
(a) A transformer operating on a 230 V mains supply provides a 12 V output. There are
1150 turns on the primary coil.
(i)
Calculate the number of turns on the secondary coil.
answer = ........................... turns
(1)
(ii)
A number of identical lamps rated at 12 V, 24 W are connected in parallel across the
secondary coil. The primary circuit of the transformer includes a 630 mA fuse.
Calculate the maximum number of lamps that can be supplied by the transformer if
its efficiency is 85%.
answer = .......................... lamps
(2)
(iii)
The transformer circuit includes a fuse. Explain why this is necessary.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(iv)
Why is the fuse placed in the primary circuit rather than in the secondary circuit?
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(b)
The figure below shows an experimental arrangement that can be used to demonstrate
magnetic levitation. The iron rod is fixed vertically inside a large coil of wire. When the
alternating current supply to the coil is switched on, the aluminium ring moves up the rod
until it reaches a stable position ‘floating’ above the coil.
(i)
By reference to the laws of electromagnetic induction explain
•
why a current will be induced in the ring,
•
why the ring experiences a force that moves it upwards,
•
why the ring reaches a stable position.
The quality of your written communication will be assessed in your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(ii)
What would happen to the ring if the alternating current in the coil was increased
without changing the frequency? Explain your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 13 marks)
Q20.
(a)
Define the electric potential at a point in an electric field.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
Figure 1 shows part of the region around a small positive charge.
Figure 1
(b)
(i)
The electric potential at point L due to this charge is + 3.0 V. Calculate the
magnitude Q of the charge. Express your answer to an appropriate number of
significant figures.
answer = ................................. C
(3)
(ii)
Show that the electric potential at point N, due to the charge, is +1.0 V.
(1)
(iii)
Show that the electric field strength at point M, which is mid-way between L and N,
is 2.5 Vm–1.
(1)
(c)
R and S are two charged parallel plates, 0.60 m apart, as shown in Figure 2.They are at
potentials of + 3.0 V and + 1.0 V respectively.
Figure 2
(i)
On Figure 2, sketch the electric field between R and S, showing its direction.
(2)
(ii)
Point T is mid-way between R and S.
Calculate the electric field strength at T.
answer = .......................... Vm–1
(1)
(iii)
Parts (b)(iii) and (c)(ii) both involve the electric field strength at a point mid-way
between potentials of + 1.0 V and + 3.0 V. Explain why the magnitudes of these
electric field strengths are different.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(Total 12 marks)
Q21.
Which one of the following would not reduce the energy losses in a transformer?
A
B
C
D
coil
using thinner wire for the windings
using a laminated core instead of a solid core
using a core made from iron instead of steel
using a core that allows all the flux due to the primary coil to be linked to the secondary
(Total 1 mark)
Q22.
The output electromotive force (emf) of a simple ac generator can be increased by any of
the four factors listed.
Which one of these factors should not be changed if the frequency of the output is to remain
unaffected when the emf is increased?
A
the area of the coilB
the number of turns on the coilC
rotationD
the strength of the magnetic field
the speed of
(Total 1 mark)
Q23.
A 500 turn coil of cross-sectional area 4.0 × 10–3 m2 is placed with its plane perpendicular
to a magnetic field of flux density 7.5 × 10–4 T. What is the value of the flux linkage for this coil?
A
B
C
D
3.0 × 10–6 Wb turns
1.5 × 10–3 Wb turns
0.19 Wb turns
94 Wb turns
(Total 1 mark)
Q24.
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)
Q25.
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
0.2 FB
0.4 FC
0.8 FD
4.0 F
(Total 1 mark)
Q26.
A horizontal straight wire of length 0.30 m carries a current of 2.0 A perpendicular to a
horizontal uniform magnetic field of flux density 5.0 × 10–2 T. The wire ‘floats’ in equilibrium in the
field.
What is the mass of the wire?
A
8.0 × 10–4 kg
B
3.1 × 10–3 kg
C
3.0 × 10–2 kg
D
8.2 × 10–1 kg
(Total 1 mark)
Q27.
Which one of the following cannot be used as a unit for electric field strength?
A
J m–1 C–1
B
J A–1 s–1m–1
C
N A–1 s–1
D
J C m–1
(Total 1 mark)
Q28.
The distance between two point charges of + 8.0 nC and + 2.0 nC is 60 mm.
At a point between the charges, on the line joining them, the resultant electric field strength is
zero. How far is this point from the + 8.0 nC charge?
A
20 mm
B
25 mm
C
40 mm
D
45 mm
(Total 1 mark)
Q29.
A repulsive force F acts between two positive point charges separated by a distance r.
What will be the force between them if each charge is doubled and the distance between them
is halved?
A
F
B
2F
C
4F
D
16F
(Total 1 mark)
Q30.
A satellite is in orbit at a height h above the surface of a planet of mass M and
radius R.What is the velocity of the satellite?
A
B
C
D
(Total 1 mark)
Q31.
The gravitational potential at the surface of the Earth, of radius R, is V. What is the
gravitational potential at a point at a height R above the Earth’s surface?
A
B
C
V
D
2V
(Total 1 mark)
A spherical planet of uniform density ρ has radius R.
Q32.
Which line, A to D, in the table gives correct expressions for the mass of the planet and the
gravitational field strength at its surface?
mass of planet
gravitational fieldstrength
at surface
A
B
C
D
(Total 1 mark)
Q33.
If an electron and proton are separated by a distance of 5 × 10–11 m, what is the
approximate gravitational force of attraction between them?
A
2 × 10–57 N
B
3 × 10–47 N
C
4 × 10–47 N
D
5 × 10–37 N
(Total 1 mark)
M1.
B
[1]
M2.
B
[1]
M3.
D
[1]
M4.
B
[1]
M5.
(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
(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 isunchanged [or magnetic force now
exceeds electrostatic force]
2
[11]
M6.
(a)
(i)
arrows to show R (or N) vertically up and mg (or W)
vertically down and along the same line (within ± 2 mm)
1
(ii)
mg – R =
R = mg –
1
(iii)
use of R = m
gives R = 12(9.81 –
)
= 55 (54.6) (N)
2
(b)
R decreases (as v increases)
because mg is unchanged but
is larger
at higher speeds R becomes = 0 [or package is not in
contact with the floor]
supported by calculation eg when v = 15 m s–1,
R = 0.33 N (or ≈ 0)
max 3
[7]
M7.
A
[1]
M8.
B
[1]
M9.
B
[1]
M10.
B
[1]
M11.
C
[1]
M12.
D
[1]
M13.
C
[1]
M14.
B
[1]
M15.
D
[1]
M16.
A
[1]
M17.
C
[1]
M18.
A
[1]
M19.
(a)
(i)
use of
= gives NS =
= 60 (turns)
1
(ii)
max output power = 0.85 × 0.630 × 230
max number of lamps
[or efficiency =
max number of lamps
=5
(= 123 W)
(no mark for non-integer answer)
(and max IS = 10.3 (A))
gives 0.85 =
=5
]
2
(iii)
fuse prevents transformer from overheating [or prevents transformer from
supplying excessive currents]
1
(iv)
(all of) transformer is disconnected from supply when fuse fails [or fuse in
secondary circuit would leave primary circuit live]
1
(b)
(i)
The candidate’s writing should be legible and the spelling, punctuation
and grammar should be sufficiently accurate for the meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be
assigned to one of three levels according to the following criteria.
High level (good to excellent) 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and coherent,
using appropriate specialist vocabulary correctly. The form and style of writing is
appropriate to answer the question.
The candidate states that the ac in the coil produces a constantly changing
magnetic field that passes through the ring, causing an emf to be induced according
to Faraday’s law.
The candidate recognises that the induced emf will cause a current to flow in the
ring, that the current is likely be large because the coil acts as a single conductor
with low resistance, and that this current also produces a magnetic field.
The candidate appreciates that Lenz’s law indicates that the direction of the induced
current is such as to produce a magnetic field that will oppose the existing field, and
that the two fields will interact.
The candidate refers to the force that acts on a current-carrying conductor when it is
in a magnetic field and that this force lifts the ring upwards (into an area where the
magnetic field is weaker) until the upwards magnetic force is equal to the
downwards weight of the ring.
Intermediate level (modest to adequate) 3 or 4 marks
The information conveyed by the answer may be less well organised and not fully
coherent. There is less use of specialist vocabulary, or specialist vocabulary may be
used incorrectly. The form and style of writing is less appropriate.
The candidate is familiar with either or both Faraday’s and Lenz’s laws but only
applies one of them to explain what happens in this demonstration. There are
correct references to the two forces that act on the ring, and a reasonable
explanation of why the ring reaches a stable position.
Low level (poor to limited) 1 or 2 marks
The information conveyed by the answer is poorly organised and may not be
relevant or coherent. There is little correct use of specialist vocabulary. The form
and style of writing may be only partly appropriate.
The candidate refers much more superficially to either Faraday’s or Lenz’s law (or to
both of them) but shows some understanding of why the forces acting on the ring
cause it to reach equilibrium.
The explanation expected in a competent answer should include a coherent
selection of the following points concerning the physical principles involved
and their consequences in this case.
Faraday’s law
•
An emf is induced whenever there is a change in the magnetic flux passing
through a conductor.
•
The magnitude of the emf is proportional to the rate of change of magnetic flux
linkage.
•
The induced emf will cause a current to flow in any complete circuit, such as a
single conducting ring.
•
Because the ring is made from aluminium, which is a good conductor, a large
initial current will be induced in it.
Lenz’s law
•
The induced current flows in such a direction as to oppose the increase in
magnetic flux when the current is switched on in the coil.
•
The current produces a magnetic field in the opposite direction to that
produced by the coil.
•
These two (alternating) fields interact like the fields between two facing like
magnetic poles, giving repulsion.
Forces
•
The ring is a current-carrying conductor in a magnetic field, and
consequently it experiences a force.
•
This magnetic force acts upwards, in the opposite direction to the
weight of the ring.
•
As the ring rises, the magnetic field to which it is exposed becomes
weaker as it moves away from the coil.
•
This reduces the induced current, reducing also the magnetic force
on the ring.
•
The ring reaches a stable height when the magnetic force has
decreased to the point where it is equal to the weight of the ring.
6
(ii)
ring would ‘float’ higher [or be expelled upwards]
because (initial) current or emf (induced) in ring is greater
or ring moves into weaker field until magnetic force balances weight
[or (initially) magnetic force exceeds weight]
2
[13]
M20.
(a) work done [or energy needed] per unit charge
[or (change in) electric pe per unit charge]
on [or of] a (small) positive (test) charge
in moving the charge from infinity (to the point)
[not from the point to infinity]
3
(b)
gives Q (= 4πε0rV) = 4π × 8.85 × 10–12 × 0.30 × 3.0
(i)
= 1.0 × 10–10 (C)
to 2 sf only
3
(ii)
use of V ∞
gives VM =
(= (+) 1.0 V)
1
(iii)
=
(= 2.50 V m–1)
1
(c)
(i)
uniformly spaced vertical parallel lines which start and end on plates
relevant lines with arrow(s) pointing only downwards
2
(ii)
= 3.3(3) (V m–1)
1
(iii)
part (b) is a radial field whilst part (c) is a uniform field
[or field lines become further apart between L and M but are equally spaced
between R and S]
1
[12]
M21.
A
[1]
M22.
C
[1]
M23.
B
[1]
M24.
A
[1]
M25.
B
[1]
M26.
B
[1]
M27.
D
[1]
M28.
C
[1]
M29.
D
[1]
M30.
A
[1]
M31.
B
[1]
M32.
B
[1]
M33.
C
[1]
E1.
This question required familiarity with the idea that a body appears to become weightless
when its centripetal acceleration is just equal to the local value of the acceleration due to
gravity. Hence, if this were to happen at the surface of the Earth, ω2R would have to equal 9.81
m s–2. The question had a facility of 55%, but one in five candidates selected distractor A.
E2.
This question was a re-banked question about the gravitational potential and angular
velocity at two points whose height above the Earth’s surface was different. The outcome was a
very similar facility to that obtained on the previous occasion, with half of the candidates
appreciating that the point at greater height would have greater V but the same ω. More than a
quarter of responses were for distractor C (greater V, smaller ω) and almost a fifth for distractor
A (both V and ω greater).
E3.
This question was about the value of the gravitational field strength at the mid-point
between two equal masses; surprisingly, only 60% of the candidates knew that this would be
zero.
E4.
This question, involving a rearrangement of the force equation from Newton’s law, had a
facility of 77%.
E5.
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.
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.
E6.
The context of this question may have been unfamiliar to many candidates. Although most
had some awareness of the ‘lift off’ sensation when a vehicle passes over a hump-backed
bridge, relatively few were able to give good answers to explain the mechanics involved.
Attempts at part (a)(i) were often muddled by the introduction of arrows marked ‘centripetal
force’, ‘driving force’, ‘momentum’ and so on. In a correct answer, two labelled vertical arrows
acting through the same point on the parcel were all that was expected; the weight downwards
and the reaction upwards. Frictional forces were not expected but their inclusion did not nullify
the answer. Poor understanding was revealed by labels such as ‘upthrust’ and ‘gravity’. The
principal error in part (a)(ii) was to assume that the resultant (centripetal) force acts outwards,
resulting in the incorrect equation R – mg = mv2/r. Candidates doing this evidently did not
understand that, for a body to move in a circular path, it has to experience a resultant force that
acts towards the centre of the circle. This kind of incorrect response in part (ii) almost invariably
meant that part (a)(iii) – where the calculation is based on the equation – would also be
incorrect.
Some good answers were seen to part (b), but most only received partial credit. Good
understanding of the mechanics of part (a) allowed more able candidates to see that at higher
speeds, because mv2/r increases but mg remains constant, R must decrease. They could then
calculate R when v = 15 ms-1 and find it to be almost zero. The obvious deduction is that, with a
slight increase in speed, R would become zero and the parcel would lose contact with the floor
of the van. Very few candidates who carried out the calculation for speeds higher than 15 ms1
were able to give a correct interpretation of the negative value they calculated for R. Less able
candidates sometimes argued that the van would lift up vertically from the road surface at high
speed because the upwards reaction would be greater than its weight. Candidates would be
able to approach questions of this kind more successfully if more of them realised that
‘centripetal force’ (mv2/r) really is ‘mass × acceleration towards centre’ rather than being an
actual force, but that it is equal to the real resultant force (mg – R here) acting towards the
centre of the circle.
E7.
The facts about energy losses in transformers were well known in this question, where 73%
of candidates gave the correct answer. The strongest incorrect distractor was C, suggesting that
the energy losses caused by magnetic hysteresis are not always recognised.
E8.
Electromagnetic induction continued to cause difficulty in this question, which had a facility
of 40%. Candidates should have noticed that, although the speed of rotation of the coil was
doubled, the flux density was halved. This has the net effect of leaving the peak emf unchanged
whilst the frequency is doubled. 30% of the candidates realised that the frequency would be
doubled but thought the peak emf would also double (distractor D).
E9.
In this question a large proportion of candidates did not realise that the flux linkage
increases to a constant value once the magnet is at rest inside the coil, and therefore selected
the incorrect distractor A. 43% gave the correct response.
E10.
This question required candidates to know that an induced emf is proportional to the rate
of change of flux linkage. Almost one-third of the candidates considered it to be proportional to
the flux linkage, distractor A. 56% of the responses were correct, whilst very few chose
alternatives C or D.
E11.
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.
E12.
This question tested the relationship V
1/r for a point charge, but made appreciable
mathematical demands because it required candidates to deal with a change in V. Rather fewer
than half of the responses were correct, with distractor C as the most popular incorrect answer.
E13.
This question tested candidates’ understanding of the mechanics of the motion of a
positive ion as it passes through a uniform electric field. Quite a lot of calculation was needed to
arrive at the correct response, but 57% were successful. Incorrect answers were evenly
distributed amongst the other distractors.
E14.
Failure to realise that a negatively charged electron has an associated negative value of
electric potential caused many candidates to go wrong in this question. As in gravitation, the
force between an electron and a proton is attractive and so the potential is negative. 40% of the
candidates made the correct selection, B. 26% selected distractor A; the only difference
between A and B is the – sign in B.
E15.
This question, with a facility of 71%, required the angular speed of a satellite in circular
orbit to be found and appeared to cause little difficulty.
E16.
This question continued the theme of gravitation. At first sight, it should be easy. In fact it
was the most demanding question in the test, with a facility of only 33%. Marginally more
candidates chose the incorrect distractor D than the correct answer. This was a fairly simple test
of inverse square proportion for force and inverse proportion for potential. Candidates made
matters difficult by confusing the distance from an external point to the centre of the Earth with
the distance to the surface of the Earth.
E17.
This question, on the gravitational field strength at the surface of a planet, made similar
mathematical demands to the previous question but was answered more successfully. The
facility was 72%, an improvement of over 10% on the result when this question last appeared in
an examination. The question was also an effective discriminator.
E18.
This was a fairly demanding calculation on the inverse square law of gravitation, in which
candidates had to consider the effect of changing both the size of the attracting masses and
their separation. Just over half reached the correct conclusion. No doubt it was errors in
rearranging the arithmetic and/or algebra that caused 34% of candidates to opt for distractor B,
where the new force was double what it ought to be.
E19.
The transformer turns ratio equation was familiar territory for most in part (a) (i), but correct
application of the efficiency formula proved to be a greater challenge in part (a) (ii). Many
correct answers were seen, and almost all students knew that the number of lamps has to be an
integer. Most difficulties arose from mixing up data for the secondary coil with that for the
primary (for example, multiplying the primary current by the secondary voltage).
Parts (a) (iii) and (iv) proved to be an exacting test of whether students could think through to
the real reasons or had enough practical experience of transformers to know these
reasons.Many attempts at part (iii) were general answers about the reason for fitting a fuse in
any circuit rather than specifically in a transformer’s circuit. Very few students stated that
transformer coils can overheat and become damaged when they handle excessive currents and
that they therefore need to be protected.
Similarly, it was only a small minority of answers to part (iv) that were properly valid; that
stopping the primary current would isolate the whole transformer from the mains or that a failed
fuse in the secondary circuit would leave the primary live.
Most students find that electromagnetic induction is one of the most demanding topics in the
specification. In these circumstances perhaps it should not be surprising that many of the
attempts to answer part (b) (i) were very disappointing. Even when pointed at a logical and
sequential structured answer by three bullet points, many students could not construct a
coherent, ordered response. In assessing the Quality of Written Communication, one aspect
that must be taken into consideration is the appropriate use of technical terminology. This was
often absent from the responses seen. The term induction has a very special meaning in
physics; magnetic induction involves magnetising a material by applying a magnetic field,
electrostatic induction involves charge separation by applying an electric field, electromagnetic
induction involves producing an emf by applying a changing magnetic field. In all cases, direct
contact is unnecessary. Many answers contained the word ‘induced’ used much more
carelessly than its technical meaning in physics; a current was stated to be induced in the coil
because it was connected to the ac supply, for example. This current was then said to induce a
magnetic field. Many students seemed obsessed by effects in the iron rod, rather than in the
aluminium ring. The aluminium ring was confused with the coil for example ‘the coil is pushed
upwards by the magnetic field’. It was evident that a large proportion of students were familiar
with statements of the laws of electromagnetic induction but could not apply them meaningfully
to explain what happens in this demonstration. The field produced was regularly referred to as
an electric field. The repulsion of the ring was sometimes attributed to Coulomb’s law and the
repulsion between charges. Fleming’s left hand rule was confused with his right hand rule, or
with the right hand grasp rule.
Broadly, an outline plan of an appropriate answer to this question would be along the following
lines. The ac current in the coil produces an alternating magnetic field, which is concentrated in
the iron rod and passes through the ring. This changing magnetic field induces an emf in the
ring. Because the ring is aluminium it is a good conductor and the emf causes a large current in
it. A current-carrying conductor in a magnetic field experiences a force so this current produces
a magnetic field whose direction opposes the applied field. Interaction between these fields
gives a net upwards repulsion of the ring. As the ring moves upwards the magnetic field
becomes weaker and the force on the ring decreases. The ring’s position becomes stable when
the upwards magnetic force balances its weight. Answers written in this fashion were rare but
not difficult to identify and they were rewarded well.
Answers to part (b) (ii) suffered from the same lack of general understanding as the previous
part. It was often realised that the ring would move to a higher position, or be expelled upwards
from the rod. Reasons were less well presented. Reference to a larger induced current (or emf)
in the ring was considered a prerequisite for an acceptable explanation.
E20.
The definition of electric potential in part (a) was generally well known. Where students did
not score all three marks this was down to oversight; typically either omitting to mention that the
charge involved in the definition is positive or that the definition involves the work done per
unitcharge.
In part (b) (i), most students successfully applied V = Q /4πε0r in order to determine the
magnitude of the charge (1.0 × 10–10 C) from the value of V when r = 0.30 m. As the data in the
question is given to two significant figures, an answer was expected to two significant figures.
Some students need to appreciate that the number of significant figures they should quote in an
answer needs to be limited to the least precise data they are working with, not the most (in
theData Sheet (see Reference Material) ε0 is given to three significant figures). At the same
time, in these circumstances the answer should never be abbreviated to one significant figure(1
× 10–10 C), as was the case in many answers.
The mark in part (b) (ii) was gained easily, usually by applying V = Q/4πε0r, although more
perceptive students saw that V ∝ 1/r could lead to a more concise answer. Part (b) (iii) caused a
little more difficulty for some students. Application of E = Q/4πε0r2 with r = 0.60 m was the
obvious route. The pitfall for many was that, by first finding V at M (by the same method as
before), they then had to apply E = V/d to find the field strength. This last equation specifically
applies to a uniform field and it therefore cannot be used here. Surprisingly, there were many
students who, having obtained an incorrect charge in part (b) (i) as a result of an arithmetical
slip, did not revisit part (i) when they could not show either of the required values in parts (ii) and
(iii).
Many good attempts to represent the electric field between two plates were seen in part (c) (i),
but careless sketching, such as field lines stopping short of the plates, often meant that it was
not possible to award both marks. Because this was the field between two plates at different
positive potentials, some students were thrown off course both when sketching the field and
when the uniform field strength had to be found in part (c) (ii). In part (c) (iii) the respective radial
and uniform fields were usually recognised but a precise statement that identified which was
which was required to gain the mark.
E21.
This question which were each correctly answered by just over three-quarters of the
students, respectively tested the rotating coil and energy losses in a transformer.
E22.
This question which were each correctly answered by just over three-quarters of the
students, respectively tested the rotating coil and energy losses in a transformer.
E23.
The remaining three questions all tested aspects of electromagnetism.This question which,
with a facility of 89% was the easiest in the test, was a straightforward calculation of flux linkage
for a coil in a magnetic field.
E24.
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.
E25.
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%.
E26.
In this question the correct answer could be found by equating the weight of a section of
wire to the magnetic force that acts on it when the wire carries a current at right angles to
magnetic field. This was answered correctly by 75% if the students. Distractor C, where the
mass was approximately 9.8 times the expected mass, was the choice of almost one in five.
This suggests that the students involved equated the mass of the wire with the magnetic force,
forgetting g.
E27.
This question was the most demanding question on the paper, with only 39% of the
students giving the correct answer. In order to identify the correct combinations of units to give
V m-1, it was necessary to remember that 1 V = 1 J C-1 and that 1 C = 1 A s. Distractor C was the
choice of over a quarter of the students.
E28.
This question was also on electrostatics but, with a facility of 48%, was much more
demanding. At first sight it appears necessary to solve a quadratic equation to answer the
question, but this difficulty can be overcome by taking the square root of the expression
obtained. Incorrect distractor D was chosen by 35% of the students and consequently the
discrimination of the question was poor.
E29.
This question was a fairly direct test of Coulomb’s law for charges under changing
circumstances; 70% of the students had the correct answer.
E30.
This question three quarters of the students were successful when dealing with the algebra
giving the velocity motion of a satellite in stable orbit of radius (R + h). This question had
appeared in a 2002 examination, when the students found it marginally harder and it was
slightly less discriminating.
E31.
Appreciation that gravitational potential V is proportional to 1/r was all that was required to
arrive at the correct response in this question which had a facility of 71%. The most common
incorrect choice was distractor A, where the students may have thought V is proportional to 1/r2.
E32.
The algebra required to relate the density of a planet to its mass and gravitational field
strength in this question did not prove to be an obstacle to most students because 79% of them
gave the correct combination from the table.
E33.
Data for the gravitational constant and the masses of the electron and proton had to be
extracted from the Data Sheet (see Reference Material) for use in this question where the topic
was the gravitational force between two particles. Over four-fifths of the students succeeded
with this.