edexcel gce a2 physics unit 4 electric and magnetic fields test 14_15 with ms

8/21/2019 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


_______________________________________________________________________________________


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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


_______________________________________________________________________________________

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


_______________________________________________________________________________________

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


_______________________________________________________________________________________


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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


_______________________________________________________________________________________


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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


_______________________________________________________________________________________


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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


_______________________________________________________________________________________


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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 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


______________________________________________________________________________________

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


(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 = .....................................



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SEMESTER 1TEST 2


_______________________________________________________________________________________

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


(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 = .................................



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SEMESTER 1TEST 2


_______________________________________________________________________________________

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:................................


8.0 cm

3.0 cm


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SEMESTER 1TEST 2


_______________________________________________________________________________________

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


(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


_______________________________________________________________________________________

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


(c) On figure 1.2 sketch a graph to show the variation of potential difference V across the

capacitor with time t.

(2)


17.

(a)Define Faradays law.

(2)

(Turn over)


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SEMESTER 1TEST 2


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)


_______________________________________________________________________________________


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SEMESTER 1TEST 2


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


(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 SECTION B = 70 MARKS

_______________________________________________________________________________________

END

TOTAL FOR PAPER = 80 MARKS


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DATA & FORMULAE


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


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


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)

&


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)

&


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)


33/34

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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edexcel gce a2 physics unit 4 electric and magnetic fields test 14_15 with ms

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