Edexcel GCE A2 Physics Unit 4 Electric and magnetic fields test 14_15 with MS
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Transcript of Edexcel GCE A2 Physics Unit 4 Electric and magnetic fields test 14_15 with MS
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS Turn over
Surname Name
American Academy Larnaca
Year 7 Advanced Physics
Semester 1
Unit 4
Topic 2
Test 2
Physics on the Move
Electric and Magnetic Fields
Monday 8th
December 2014 Time: 1 hour 30 min
You must have:
Scientific calculator
Ruler
Total Marks
Instructions
Use black ink or ball-point pen
Write your name at the top of this page
Answer all questions in the spaces provided
there may be more space than you need
Information
The total mark for this paper is 80
The marks for eachquestion are shown in square brackets
use this as a guide as to how much time to spend on each question
Questions labelled with an asterisk (*) are ones where the quality
of your written communication will be assessed
you should take particular care with your spelling, punctuation and grammar, as
well as the clarity of expression, on these questions
The list of data, formulae and relationships is printed at the end of this paper
Candidates may use a scientific calculator
Advice
Read each question carefully before you start to answer it
Keep an eye on the time
Try to answer every question
Check your answers if you have time at the end
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 2
SECTION A
Answer all questions
For questions 110, in Section A, select one answer from A to D and put a cross in the box ( )
If you change your mind, put a line through the box ( ) and then mark your new answer with a cross_______________________________________________________________________________________
1. Which of the following is not a vector quantity?
A Electric field strength
B Magnetic flux density
C Magnetic flux linkage
D Momentum
[Total for Question 1 = 1 mark]
_______________________________________________________________________________________
2. A beam of electrons, moving with a constant velocity v in a vacuum, enters a uniform electric field
between two metal plates.
Which line, A to D, in the table describes the components of the acceleration of the electrons in the x
and y directions as they move through the field?
Acceleration in x direction Acceleration in y direction
A zero zero
B zero constant
C constant zero
D constant constant
[Total for Question 2 = 1 mark]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 3
_______________________________________________________________________________________
3. Electric field strength can have units of
A Nm
B N C
C VC-1
D Vm-1
[Total for Question 3 = 1 mark]
_______________________________________________________________________________________
4. Two small charged objects, a distance d apart, exert an attractive forceF on each other.
The charge on each object is triple and the distance increased to 4d.
The force of attraction would be
A 0.1875 F
B 0.5625 F
C 0.75 F
D 2.25 F
[Total for Question 4 = 1 mark]
_______________________________________________________________________________________
5. The voltage across a capacitor falls from 19.48 V to 9.0 V in 20 ms as it discharges through a
resistor.
What is the time constant of the circuit?
A 25.9 ms
B 40 ms
C 10 ms
D 38.6 ms
[Total for Question 5 = 1 mark]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 4
6. A vertical conducting rod of length l is moved at a constant velocity v through a
uniform horizontal magnetic field of flux densityB.
Which line, A to D, in the table gives a correct expression for the induced emf and current
for the stated direction of the motion of the rod?
Direction of motion Induced Emf Induced current
A Vertical
No current
B
Horizontal at right
angles to the field
Blv No current
C Vertical
Blv Clockwise
D Horizontal at right
angles to the field
Clockwise
[Total for Question 6 = 1 mark]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 5
_______________________________________________________________________________________
7. Two horizontal parallel plate conductors are separated by a distance of 5.0 mm in air.
The lower plate is earthed (0 V) and the potential of the upper plate is + 100 V.
Which line, A to D, in the table gives correctly the electric field strength, E, and thepotential, V, at a point midway between the plates?
Electric field strength E / Vm-1
Potential V / V
A 40000 upwards 100
B 20000 upwards 100
C 40000 downwards 50
D 20000 downwards 50
[Total for Question 7 = 1 mark]_______________________________________________________________________________________
8. A current of 8.0 A is passed through a conductor of length 0.40 m and cross-sectional area
1.0 106m2. The conductor contains 8.0 1028free electrons per m3. When the conductor
is at right angles to a magnetic field of flux density 0.20T, it experiences a magnetic force.
What is the average magnetic force that acts on one of the free electrons in the wire?
A 2.0 x 10-23
N
B 8.0 x 10-24
N
C 5.0 x 10-29
N
D 8.0 x 10-30
N
[Total for Question 8 = 1 mark]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 6
_______________________________________________________________________________________
9. The graph shows how the potential difference across a capacitor varies with the charge
stored by it.
Which one of the following statements are correct?
Capacitance is given by: Energy stored in the capacitor is given by:
A
Area between the line and the charge axis
B
Gradient Area between the line and the charge axis
C
Area between the line and the potential difference axis
DGradient Area between the line and the potential difference axis
[Total for Question 9 = 1 mark]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 7
_______________________________________________________________________________________
10. The capacitor shown in the circuit below is initially charged to a potential difference (p.d.) V
by closing the switch.The power supply has negligible internal resistance.
The switch is opened and the p.d. across the capacitor allowed to fall. A
short time laterthe switch is closed again. Select the graph that shows how
the p.d. across the capacitorvaries with time, after the switch is opened.
A
B
C
D
[Total for Question 10 = 1 mark]
_______________________________________________________________________________________
TOTAL FOR SECTION A = 10 MARKS
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 8
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 9
SECTION B
Answer all questions in the spaces provided
_______________________________________________________________________________________
11. The magnetic forceF that acts on a current-carrying conductor in a magnetic field is
given by the equation
F = BIL
(a) State the condition under which this equation applies
(1)
(b) The unit for magnetic flux densityB is the tesla.
Express the tesla in base units.(2)
(Turn over)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 10
(c) The diagram shows a rectangular bar of aluminium which has a current of 5.0 A
through it.
The bar is placed in a magnetic field so that its weight is supported by the magnetic
field.
Calculate the minimum value of the magnetic flux densityB needed for this to occur.
density of aluminium = 2.7 103kg m3(3)
Minimum B = ........................................
(d) State the condition under which this equation applies
(1)
[Total for Question 11 = 7 marks]_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 11
______________________________________________________________________________________
12. The charge on an electron was originally measured in an experiment called the Millikan
Oil Drop experiment.
In a simplified version of this experiment, an oil drop with a small electric charge is
placed between two horizontal, parallel plates with a large potential difference (p.d.) acrossthem. The p.d. is adjusted until the oil drop is stationary.
For a particular experiment, a p.d. of 5100 V was required to hold a drop
of mass 2.50 1014kg stationary.
(a)Add to the diagram to show the electric field lines between the plates.
(3)
(b)State whether the charge on the oil drop is positive or negative.
(1)
(c)Complete the free-body force diagram to show the forces acting on the oil drop.
You should ignore upthrust.
(2)
(d)Calculate the magnitude of the charge on the oil drop.
(4)
(Turn over)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 12
(e)Calculate the number of electrons that would have to be removed or added to a neutral
oil drop for it to acquire this charge.
(2)
Number of electrons = .....................................
[Total for Question 12 = 12 marks]_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 13
_______________________________________________________________________________________
13. A particular experiment requires a very large current to be provided for a short time.
(a) An average current of 2.0 10 A is to be supplied to a coil of wire for a time of
1.4 103s. The resistance of the coil is 0.50 .
(i) Show that the charge that flows through the coil during this time is about 3 C.
(2)
(ii) The circuit shows how a capacitor could be charged and then discharged
through the coil to provide the current.
The circuit contains a capacitor of capacitance 600 F. This capacitor is
suitable to provide the current for 1.4 103s.
Explain why the capacitor is suitable.
(3)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 14
(b) It can be assumed that the 600 F capacitor completely discharges in 1.4 10 s.
(i) Calculate the potential difference of the power supply.
(2)
Potential difference = .................................
(i) Calculate the average power delivered to the coil in this time.
(3)
Average power = .................................
[Total for Question 13 = 10 marks]_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 15
_______________________________________________________________________________________
14.
(a) Explain what is meant by electric field.
(2)
(b) Calculate the force between an alpha particle, , and a gold nucleus,
, when
the distance between them is 1.9 x 10-12m.
(3)
Force = .......................
(c) The diagram shows the positions of the two protons.
Calculate the resultant electric field (size and direction) at position A, 3.0cm above
the midpoint of BC that is 8.0 cm long.
(5)
Electric field strength at A = ..............................
Direction:................................
[Total for Question 14 = 10 marks]_______________________________________________________________________________________
8.0 cm
3.0 cm
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 16
_______________________________________________________________________________________
15. Figure 4 shows two small, solid metal cylinders, P and Q.
P is made from aluminium. Q is made from a steel alloy.
The steel cylinder Q is a strong permanent magnet. P and Q are released separately
from the top of a long, vertical copper tube so that they pass down the center of the
tube, as shown in Figure 5.
(a) The time taken for Q to pass through the tube is much longer than that taken by P.
Explain why would expect an emf to be induced in the tube as Q passes through it.(2)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 17
(b) State the consequences of this induced emf, and hence explain why Q takes longer
than P to pass through the tube.
(3)
(c) The copper tube is replaced by a tube of the same dimensions made from brass. The
resistivity of brass is much greater than that of copper. Describe and explain how, if
at all, the times taken by P and Q to pass through the tube would be affected.
(3)
P:
Q:
[Total for Question 15 = 8 marks]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 18
_______________________________________________________________________________________
16. Figure 1.1 shows a circuit with a capacitor of capacitance C.
The switch S is closed. The resistance of the variable resistor is manually adjusted so that
the current in the circuit is kept constant.
(a) Explain in terms of movement of electrons how the capacitor plates X and Y acquirean equal but opposite charge.
(2)
(b) The initial charge on the capacitor is zero.After 100s, the potential difference across the capacitor is 1.6 V.
The constant current in the circuit is 40A.
Calculate the capacitance C of the capacitor.
(3)
Capacitance =.........................
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 19
(c) On figure 1.2 sketch a graph to show the variation of potential difference V across the
capacitor with time t.
(2)
[Total for Question 16 = 7 marks]_______________________________________________________________________________________
17.
(a)Define Faradays law.
(2)
(Turn over)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 20
The diagram represents two identical coils X and Y. The planes of both coils are parallel
and their centres lie on a common axis.
Coil Y is connected to a cell, a variable resistor and a closed switch.
Under which of the following circumstances would a current be induced in coil X in the same
direction as the current shown in coil Y?
A The coils are moved closer together.
B The switch is opened.
C The resistance of the variable resistor is decreased.
D No change is made to the arrangement.
Explain your answer.
(4)
[Total for Question 17 = 6 marks]
_______________________________________________________________________________________
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 21
18.
(a)Describe how a beam of fast moving electrons is produced in the cathode ray tube of an
oscilloscope.
(3)
(b)Figure 1 shows the cathode ray tube of an oscilloscope. The details of how the beam of
electrons is produced are not shown.
The electron beam passes between two horizontal metal plates and goes on to strike a
fluorescent screen at the end of the tube. The plates are 0.040 m long and are separated by a
gap of 0.015 m. A potential difference of 270V is maintained between the plates.
An individual electron takes 1.5 109s to pass between the plates. The distance between
the right-hand edge of the plates and the fluorescent screen is 0.20 m.
(Turn over)
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SEMESTER 1TEST 2
YEAR 7 A2 PHYSICS 22
(b)Show that the vertical distance travelled by an electron as it passes between the
horizontal metal plates is approximately 3.6 mm.
(3)
(c)Calculate the vertical component of velocity achieved by an electron in the beam by the
time it reaches the end of the plates.
(2)
Vertical velocity = .........................
(d)Calculate the vertical displacement,y, of the electron beam from the centre of the screen.
Give your answer in m.
(3)
Vertical displacement y = ...........
[Total for Question 17 = 11 marks]
TOTAL FOR SECTION B = 70 MARKS
_______________________________________________________________________________________
END
TOTAL FOR PAPER = 80 MARKS
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DATA & FORMULAE
YEAR 7 A2 PHYSICS 23
List of data, formulae and relationships
Acceleration of free fall g = 9.81 m s2
(close to Earths surface)
Boltzmann constant k= 1.38 1023
J K1
Coulombs law constant k= 1 / 4 0= 8.99 109N m
2C
2
Electron charge e = 1.60 1019C
Electron mass me= 9.11 1031
kg
Electronvolt 1 eV = 1.60 1019
J
Gravitational constant G= 6.67 1011
N m2kg
2
Gravitational field strength g = 9.81 N kg1
(close to Earths surface)
Permittivity of free space 0= 8.85 1012
F m1
Planck constant h = 6.63 1034
J s
Proton mass mp= 1.67 1027
kg
Speed of light in a vacuum c = 3.00 108m s1
Stefan-Boltzmann constant = 5.67 108
W m2
K4
Unified atomic mass unit u= 1.66 1027
kg
Unit 1
Mechanics
Kinematic equations of motion v= u+ a t
s= u t+ 1/2at2
v2= u
2+ 2 a s
Forces F= m a
g= F/ m
W= m g
Work and energy W= F s
Ek=1/2m v
2
Egrav= m g h
Materials
Stokes law F= 6 r v
Hookes law F= k x
Density = m/ V
Pressure p= F/A
Young modulus E= / where
Stress = F/A
Strain = x/x
Elastic strain energy Eel=1/2F x
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DATA & FORMULAE
YEAR 7 A2 PHYSICS 24
Unit 2
Waves
Wave speed v=f
Refractive index 12= sin i/ sin r= v1/ v2
Electricity
Potential difference V= W/ Q
Resistance R= V/ I
Electrical power P= VI
P= I2R
P= V2/ R
Energy W= V I t
Efficiency effcenc =
useful ener outut
ener nut 100
effcenc =useful ower outut
ower nut 100
Resistivity R=L/A
Current I= Q/ t
I= n q v A
Resistors in series R= R1+ R2+ R3
Resistors in parallel1
=1
1
1
1
Quantum physics
Photon model E= h f
Enstens hotoelectrc equaton = 1/2m vma
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DATA & FORMULAE
YEAR 7 A2 PHYSICS 25
Unit 4
Mechanics
Momentum p= m v
Kinetic energy of non-relativistic particle EK=p2/ 2 m
Motion in a circle v= r
T= 2 /
F= m a= m v2/ r
a= v2/ r
a= r 2
Fields
Coulombs Law F= k Q1Q2/ r2where k= 1 / 4 0
Electric Field E= F/ Q
E= k Q/ r2
E= V/ d
Capacitance C= Q/ V
Energy stored in capacitor W=1/2Q V
Capacitor discharge Q= Q0et/ R C
In a magnetic field F= B I lsin
F= B q vsin
r=p/ B Q
Faradas and Lenzs Laws = d(N ) / dt
Particle physics
Mass-Energy E= c2m
de Broglie wavelength = h/p
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 1
TEST 2
MARK SCHEME
PART A
p. 2
1. C
2. B
p. 3
3. D
4. B
5. A
p. 4
6. B
p. 5
7. D
8. A
p. 6
9. A and C
p. 6
10. B
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 2
PART B
p. 7 & 8
11. (a) R= V/ I= 3 / 0.002 [1 mark]
R= 1500 [2 marks] its
(b) 2 0.37 = 0.74 mA => = . . sec [1 mark]
C= / R= / 1500 [1 mark]
C= 2.1 2.2 mF [3 marks] its
(c) Any one from [1 mark]
Electrons move to positive battery end & from negative battery end to one plate
Plates get same charge as the battery pole they are connected to
Reject: positive charges moving / charges moving across plates
&
Opposite charges across plates attract each other (or similar) OR current decreases
exponentially [1 mark]
(d) Q= Q0(1 et/
) OR V= V0(1 et/
) [1 mark]
Q0= C V= 3 C= 6.36.6 mC [2 marks] its
Q= Q0(1 e3
) = 0.95 Q0= 5.996.27 mC [3 marks] its
V= 0.95 V0= 2.85 V [2 marks] its
Q= C V= 5.996.27 mC [3 mark] its
p. 9 & 10
12. (a) 1. Will buzz during charging (or similar) [1 mark]
2. Longer to charge (or similar) [1 mark]
(b) Same magnitude / rate of change / value at equal times (or similar) [1 mark]
Opposite direction / sign (or similar) [2 marks] its
(c) = 100 18 103
= 1.8 sec
V= V0et/ (R C)
=> 2 = 12 et/ 1.8
[1 mark]
t/ 1.8= ln (1 / 6) [2 marks] its
t= 1.8 ln (1 / 6) = 3.2 sec [3 marks] its
(d) Any one from [1 mark]
Use a larger capacitor
Add a resistor in series (or similar)
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 3
p. 11 & 12
13. (a) (i)
&
(iii)
(ii) t= L/ u= 0.04 / (4.2 106) = 9.5 10
9sec [1 mark]
&
a= 2 h/ t2= 2 0.011 / t
2= (. .44) 10
14m s
2[1 mark]
F= m a= 9.11 1031
a= (22.22) 1016
N [2 marks] its
E= F/ q= F/ (1.6 1019
) = 12501389 N C1
[3 marks] its
E= V/ d=> V= E d= 27.530.56 V [4 marks] its
(b)
(c) Any two from [2 marks]
Trajectory is circular (not parabolic)
Force / Acceleration always perpendicular to velocity
Speed / Kinetic energy does not change / increase
Force / Acceleration does not maintain a constant direction (or similar)
+ A
(i)parabola [1 mark]
(iii)line similar to but left of (i)[1 mark]
B
A
straight line [1 mark]
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 4
p. 13 & 14
14. *(a) Any one from [1 mark]
Emf induced because rotating metal cuts magnetic field lines / flux
Force acts on electrons moving through magnetic field as wheel rotates 0 as Bacts through different areas of the wheel / area of Bchanges as wheel rotates
&
Any one from [1 mark]
Eddy currents formed in the wheel
Currents due to electron flow n the wheel because of Faradas force (or smlar)
&
Any one from [1 mark]
Current turns wheel into a magnet of such polarity as to oppose motion (or similar)
Lenzs law causes current flow to stop rotation
Eddy currents transform kinetic energy into heat (or similar)
&
Any one from [1 mark]
Seed => force because reater flu cut er second
Seed => force because reater chane n area
Seed => force because more revolutons => (or smlar)
(b)
Marking polarity on magnets or field lines with arrows [1 mark]
(c) N = v/ ( D) = 7 / ( 0.85) = 2.62 rev / sec [1 mark]
2.62 60 =157.3 rpm [1 mark]
(d) Aeff= r2 0.3 = (0.35)
2 0.3 = 0.115 m
2[1 mark]
Vind= N/ t= N B Aeff= N 0.2 0.115 [1 mark]
Vind= 0.060.061 V [2 marks] its
(e) Eddy currents => heat produced => temperature [1 mark]R=> V
2/ Rless than expected / smaller rate of KE transferred to heat etc. [1 mark]
N SS
N
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 5
p. 15 & 16
15. (a) out of the paper[1 mark]
(b) (i) p= B q r=> r= 0.384 / 2 = 0.192 m [1 mark]
p=0.13 1.6 10
19
0.192 = 3.99 10
21
kg m s
1
[2 marks] its
(ii) Not all ions have the same mass / Ions with different masses present etc. [1 mark]
Any connection between mass and radius e.g. Radius as mass [1 mark]
OR
Heavier / More massive ions will hit at larger radius [2 marks]
(c) (i)
(ii) FE= FM=> E q= B q v=> E= B v[1 mark]
v= E/ B= 3100 / 0.13 [1 mark]
v= 23.8 103m s
1[2 marks]its
(d) m=p/ v = 4 1021
/ 23800 = 1.677 1025
kg [1 mark]
(e) Allows accurate determination of speed (to calculate mass) (or similar) [1 mark]
B= 0.13 TE
+ [1 mark]
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TEST MARK SCHEME
YEAR 7 A2 PHYSICS 6
p. 17, 18 & 19
16. (a) m=V= 1500 4.2 1015
= 6.3 1012
kg [1 mark]
W= m g= 6.3 1012
9.81 = 6.18 1011
N [1 mark]
(b)
(c) N= q/ e= 1.3 1015
/ (1.6 1019
) = 8125 electrons [1 mark]
(d) Fgr= k q2/ r
2= 9 10
9 (1.3 10
15)
2/ (20 10
6)
2[1 mark]
Fgr= 3.8 1011N [2 marks] its
(e)
At least 3 equidistant (by eye) lines [1 mark]
Arrows pointing away from surface [1 mark]
(f) FDron grain = E q= 1.2 105 1.3 10
15= 1.56 10
10N [1 mark]
Downward force on bottom grain = Fgr+ W= 6.2 1011
+ 3.8 1011
= 1 1010
N [1 mark]
Any indication of correct appreciation of forces by diagram or calculation e.g. [1 mark]
FTOT= 5.6 1011
upwards / FDr> Fgr+ W (or similar) [1 mark]
(g) Language should contain at least some technical terms for 5th
mark
&
Darker => more layers of grains / toner, lighter => fewer layers of grains / toner [1 mark]
More layers => greater repulsion on lower layer / grain by negative grains above [1 mark]
Greater upward attraction needed by drum [1 mark]
=> drum must be more positive / at higher voltage [1 mark]
=> drum produces a stronger electric field [1 mark]
At least 4 lines [1 mark]
Arrows pointing onto grain [1 mark]
+ + + + + + + + +
FDr
W+ Fgr
OR
FTOT =1.56 1010
1 1010
N
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Mark scheme
Part A
1. C
2. B
3. D
4. B
5.
A6. B
7. D
8. A
9. A and C
10.B
Part B
Question Answers Acceptable answers
11.(a) B perpendicular to I (1)
(b) kgs-2
A-1
(2) kgs-1
C-1
(2)
NA-1
m-1
(1)
(c) mg = BIL (1)
1x10-4xLx2.7x103x9.81 = Bx5.0xL (1)
B = 0.53T (1)
(d) B perpendicular to I (1)
12.(a)
equally spaced lines/at least
three (2)
downwards (1)
(b) negative (1)
(c) weight down (1)
electric force up (1)
(d) FE = W
EQ = mg (1)
VQ/d = mg (1)
Q = mgd/V
= 2.50x10-14
x9.81x2.00x10-2
/5100 (1)
= 9.61x10-19
C (1)
(e) 9.61x10-19
/1.60x10-19
(1)
=6 (1)
13.(a)(i) Q = It
= 2.0x103x1.4x10
-3(1)
= 2.8C (1)
(a)(ii) = RC = 0.50x600x10-6
= 3.00x10-4
s (1)
1.4x10-3
s /3.00x10-4
s = 4.7 (1)
Capacitor takes 4-5 to dischargesuitable (1)
(b)(i) V = Q/C
= 2.8/600x10-6 (1)
=4.7kV (1)(b)(ii) P = I2R (1)
= (2.0x103)
2x0.5 (1)
= 2.0x106W (1)
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Question Answers Acceptable answers
14.(a) region in which an electric chargewould experience a force(2)
(b) F = +kQ1Q2/x2
= 8.99x109x2x79x(1.6x10
-19)
2/(1.9x10
-12)
2(2)
= 0.010N (1)
(c) x = (8.02+3.0
2)
=5.0 (cm) (1)
EB,C= +kQ/x2
= 8.99x109x1.6x10
-19/(0.05)
2
= 5.75 x10
-7
(Nm
-1
) (1)= tan-1
(4.0/3.0) = 53.1 (O) (1)
E = 2xEB,Ccos = 2x5.75 x10-7
xcos53.1 = 6.90x10-7
NC-1
(1) upwards (1)
15.(a) changing flux in copper tube (1)
due to falling magnet (1)
(b) induced emf/current produces magnetic field (1)
that opposes field of falling magnet (1)
reducing resultant force/acceleration of magnet (1)
(c) P: no change (1)
Q: faster/less time (1)
induced currentinduced field(1)
16.(a) same number of es are pulled off X (1)as are pushed onto Y (1)
(b) C = Q/V
= It/V (1)
= (40x10-6
x100)/1.6 (1)
= 2500F (1)
(c)
straight line thru origin (1)
about 1.6V at 100s (1)
17.(a) inducedemf (1)
equals rate of changeof [magnetic] flux (1)
(b) B
(c) when switch is opened current in Y stops (1)
so there is a change in B/being caused by Y (1)
this change in induces emf in X and a current flows in same direction as current in
Y (1)so that due to X is opposing decreasing due to Y (1)
18.(a) [thermal] electrons boiled off cathode/heater (1)
acceleratedby electric field between cathode and anode/voltage across plates (1)
small number pass through hole in anode (1)
(b) E = V/d = 270/0.015 = 18x103
(NC-1
) (1)
a = F/m = Ee/me = 18x103x1.6x10-19/9.11x10-31= 3.16 x1015 (ms-2) (1)
s = ut+at2= x3.16 x10
15x(1.5 x10
-9)
2= 3.55mm (1)
(c) vY= u+at
= 3.16 x1015
x1.5 x10-9
(1)
= 4.74 x106ms
-1(1)
(d) x-dirn
:vx= 0.040/1.5 x10
-9= 2.67x10
7(ms
-1) (1)
time from plates to screen t = 0.20/2.67x107
= 7.5 x10-9
(s) (1)
y-dirn:
y = vYt= 4.74 x10
6x7.5 x10
-9= 36mm (1)
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