Test - A2 Physics
Primary focus Magnetic Fields - Secondary focus electric fields (including circular motion and
SHM elements)
Time allocation 40 minutes
These questions were ALL taken from the June 2010 Module 4 paper.
You are advised to spend no more than 15 minutes on Section A
Multiple Choice Solutions to section A: C - B - B - D - C - C
A20 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
B
C
D
A straight line in the direction of the field.
A straight line in a direction opposite to that of the field.
A circular arc in a plane perpendicular to the direction of the field.
An elliptical arc in a plane perpendicular to the direction of the field.
A21 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?
Use FLHR
A
B upwards downwards
C towards the N pole of the
magnet
D towards the S pole of the
magnet
A22 Which one of the following could not be used as a unit of force?
A
B
C
D
ATm F = BI l therefore N = TAm - so its corrrect
Ws –2 kg ms –2 F = ma - so its correct
Jm –1 work done = Fd so F=W/d - so its correct
A23 The graph shows how the magnetic flux passing through a loop of wire changes with time.
What feature of the graph represents the magnitude of the emf induced in the coil? so  = (  /N)  t
N =1 as it is a single loop so  =  t
Compare to Y=mx +c
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A
B
C
D the area enclosed between the graph line and the time axis the area enclosed between the graph line and the magnetic the inverse of the gradient of the graph the gradient of the graph
A24 A coil rotating in a magnetic field produces the voltage waveform below when connected to an oscilloscope.
With the same oscilloscope settings, which one of the voltage waveforms on the right would be produced if the coil were rotated at twice the original speed?
Twice the speed would result in doubling the frequency, also twice as many flux lines would be cut per second so the amplitude would double. - Choice C flux axis
A25 A 230V, 60 W lamp is connected to the output terminals of a transformer which has a 200 turn primary coil and a 2000 turn secondary coil. The primary coil is connected to an ac source with a variable output pd. The lamp lights at its normal brightness when the primary coil is supplied with an alternating current of 2.7 A.
What is the percentage efficiency of the transformer?
P = IV
I = 60/230 = 0.261A the output voltage must be 0.261A for the lamp to light with normal brightness.
Input current is 2.7A - output should be 0.27A as it is a step up transformer x 10 for voltage meaning a reduction in current of a factor of 10.
Efficiency = 0.261/0.27 x 100 = 97%
A
B
C
D
3%
10%
97%
100%
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Section B
B2 A small negatively charged sphere is suspended from a fine glass spring between parallel horizontal metal plates, as shown in the diagram
(a) Initially the plates are uncharged. When switch S is set to position X , a high voltage dc supply is connected across the plates. This causes the sphere to move vertically upwards so that eventually it comes to rest 18 mm higher than its original position.
(i) State the direction of the electric field between the plates.
(1 mark)
Sphere attracted to upper plate - so that must be positive.
Field is direction positive charge would move so it must be vertically down
(ii) The spring constant of the glass spring is 0.24Nm
–1 . Show that the force exerted on the sphere by the electric field is 4.3 × 10 –3 N. (1 mark) compression  L = 18 x 10 -3 m and k = 0.24N/m so F = x 0.24 = 4.3 × 10 –3 N
(iii) The pd applied across the plates is 5.0kV. If the charge on the sphere is –4.1 × 10 –8 determine the separation of the plates in metres. (3 marks)
C,
= 4.8 x 10 -2 d = V/E but E = F/Q so ldl = VQ/F = 5.0 × 10 3 x 4.1 × 10 –8 /4.3 × 10 –3 N m (3 marks: 1 for each correct use of the equations and one for the final answer).
(b) Switch S is now moved to position Y .
(i) State and explain the effect of this on the electric field between the plates. (2 marks)
The electric field becomes zero (1 mark) because the plates are connected to each other and therefore there will be a flow of charge (or electrons) from one plate to the other [ or plates discharge] (until) pd across plates becomes zero - they are at the same potential.
(ii) With reference to the forces acting on the sphere, explain why it starts to move with simple harmonic motion.
(3 marks)
For shm to be occurring, acceleration
(1 mark)
∝ − ) displacement [ or for shm, force ∝ − ) displacement]
There is now no electric field therefore there is a net downward force on sphere due to its weight (or the force of gravity) (1 mark). This force extends spring (1 mark) accelerating as it moves to its equilibrium position. The acceleration in down and the extension is decreasing. (1 mark) so the acceleration is proportional to (change in) extension of spring (1 mark) acceleration is in opposite direction to displacement (or towards equilibrium) (1 mark) MAX 3
(10 MARKS)
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B4 When travelling in a vacuum through a uniform magnetic field of flux density 0.43mT, an electron moves at constant speed in a horizontal circle of radius 74 mm, as shown in the diagram.
(a) When viewed from vertically above, the electron moves clockwise around the horizontal circle. In which one of the six directions shown on the diagram ,
+ x, –x, +y, –y, +z or –z, is the magnetic field directed?
FLHR
-z
- movement is direction of force - into centre of the circle
- current in opposite direction to electron remember the electron's direction is opposite to the direction of the middle (current) finger!
(1 mark)
(b) Explain why the electron is accelerating even though it is travelling at constant speed.
(2 marks)
Its direction and therefore velocity is constantly changing (1 mark) - acceleration is a change in velocity with time. (1 mark)
(i) By considering the centripetal force acting on the electron, show that its speed is 5.6 x
10 6 ms –1 (2 marks)
= 0.43 x 10 -3
BQv = mv 2 /r ( 1 mark) v = BQr/m
x 1.60 x 10 -19 x 74 x 10
= 5.59 x 10 -6
-3 /9.11 x 10 -31 ( 1 mark) m/s
(ii) Calculate the angular speed of the electron, giving an appropriate unit.
(2 marks)
= 5.6 x 10 6 /74 x 10 -3 = 7.6 x 10 7 (1 mark) s -1
(radians per second or per second for the unit mark)
(iii) How many times does the electron travel around the circle in one minute?
(2 marks)
Frequency is the number of times per second f =  /(2  ) = 7.6 x 107/(2p) = 1.2 x 10
= 60 x 1.2 x 10 7
7 times per second (1 mark)
times per minute = 7.2 x 10 8 (1 mark)
OR reason as follows: circumference = 2  r
In one second it travels 5.6 x 10
This would be 5.6 x 10 6
6 m
/2  x 74 x 10 -3
= 60 x 1.2 x 10 7
= 1.2 x 10
times per minute = 7.2 x 10 8
7 times per second (1 mark)
(1 mark)
(9 MARKS)
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B5 The diagram on the right shows an end view of a simple electrical generator. A rectangular coil is rotated in a uniform magnetic field with the axle at right angles to the field direction.
When in the position shown the angle between the direction of the magnetic field and the normal to the plane of the coil is θ .
(a) The coil has 50 turns and an area of 1.9 x10 –3 m 2 . The flux density of the magnetic field is 2.8 x 10 –2 T.
Calculate the flux linkage (in Wb) for the coil when θ is 35°, expressing your answer to an appropriate number of significant figures.
(3 marks)
= 2.8 × 10 − 2 × 1.9 × 10 − 3
= 2.2 × 10 − 3
× 50 × cos 35° (1 mark)
Wb (1 mark) answer must be to 2 sf only (1 mark)
(b) The coil is rotated at constant speed, causing an emf to be induced.
(i) Sketch a graph on a sketch of diagram below to show how the induced emf varies with angle θ during one complete rotation of the coil, starting when θ = 0. Values are not required on the emf axis of the graph.
(1 mark)
Reasonable sine curve drawn on axes, showing just one cycle, starting at emf = 0
(ii) Give the value of the flux linkage for the coil at the positions where the emf has its greatest values.
(1 mark)
EMF will be greatest when  = 90 or 270 - cos 90 is zero therefore flux linkage is zero.
(iii) Explain why the magnitude of the emf is greatest at the values of θ shown in your
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answer to part (b)(i).
(3 marks) induced emf ∝ flux (linkage) through the coil changes as it is rotated (1 mark) from maximum at θ = 0, 180° to zero at 90° and 270° (1 mark) rate of change is greatest when plane of coil is parallel to B [ or reference to
ε = BAN ω sin  t, or ε = BAN  sin θ ] (1 mark) because coil then cuts flux lines perpendicularly [ or ε = BAN  sin  t shows
ε is greatest when  = 90° or 270°] (1 mark) (3 MAX)
Section A total - 12 marks
Section B total - 26 marks
Overall - 38 marks
(9 MARKS)
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