Electric and Magnetic

TEST 2 MARK SCHEME

Surname

Name

American Academy Larnaca

Year 7 Advanced Physics

Semester 1Unit 4Topic 2Test 2Physics on the MoveElectric and Magnetic Fields

Tuesday 29 October 2013Time: 1 hour 30 min

You must have:Scientific calculatorRulerTotal 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 each question 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 qualityof your written communication will be assessed

you should take particular care with your spelling, punctuation and grammar, aswell 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

SEMESTER 1 TEST 2

YEAR 7 A2 PHYSICSTurn over

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.A 1500 F capacitor is connected to a 6 V battery to be charged.

6 VC = 1500 F

The total energy that can be stored in the capacitor is equal to

A0.0045 J

B0.027 J

C27 J

D54 J

[Total for Question 1 = 1 mark]_______________________________________________________________________________________

2.When the above capacitor is half-charged, the energy stored is

A1/16 of the total

B1/8 of the total

C1/4 of the total

D1/2 of the total


_______________________________________________________________________________________

3.A 75 cm wire is carrying a current of 2.3 A through a uniform magnetic field of flux density 0.4 T at an angle of 40 to the field.

B = 0.4 TI = 2.3 A4075 cm

The force acting on the wire is equal to

A0.44 N

B0.53 N

C0.69 N

D44.4 N


4.Here is the situation again. Which is the correct direction of the force on the wire?

BIDout of the paperCinto the paperBA

ABCD


_______________________________________________________________________________________

5.An equivalent unit for Tesla is

AN A m

BN A1 m1

CN A m1

DN A1 m


6.Two wires close to each other are carrying current in opposite directions.

I1I2

The direction of force on the left wire carrying current I1 is

Ato the left

Bto the right

Cinto the paper

Dout of the paper


_______________________________________________________________________________________

7.A 100% efficient transformer is shown below with the characteristics given in the diagram.

Primary coil = 1200 turnsPrimary voltage = 1600 VPrimary power = 2400 WSecondary coil = 150 turnsSecondary voltage = ?Secondary power = ?

Which row gives the correct values of secondary voltage and secondary power?

Secondary voltage / VSecondary power / W

A200300

B2002400

C12 8002400

D12 800300


8.An equivalent unit for Farad would be

AV C

BV A s1

CV C1

DC V1


_______________________________________________________________________________________

9.Below you can see the main elements of a simple DC electrical motor.

Which of the graphs below show correctly how current I in the motor changes with time t?Current is switched on at time T.

ttttIIIITTTT

ABCD


_______________________________________________________________________________________

10.An electron beam travelling at 4 105 m s1 enters a region of uniform magnetic field of flux density equal to 0.08 T.

B = 0.08 Telectron beamspeed = 4 105 m s1

Which row gives the correct magnitude and direction of the force on an electron.

Magnitude / NDirection

A5.1 1015to the left

B5.1 1015to the right

C3.2 104into the paper

D3.2 104out of the paper


TOTAL FOR SECTION A = 10 MARKS_______________________________________________________________________________________

SECTION B

Answer all questions in the spaces provided_______________________________________________________________________________________

11.Below you can see an RC circuit and a graph of how current changes with time during charging.

RCV = 3 V

(a)The power supply provides 3 V.Calculate the resistance R in the circuit.

[2 marks]

Resistance =

(b)Determine the time constant of the circuit.Hence calculate the capacitance C of the capacitor in the circuit.

[3 marks]

Time constant =

Capacitance =

(c)Describe what happens as a capacitor charges.

[2 marks]

(d)Calculate the charge of the capacitor after a time interval equivalent to three time constants as determined in (b).

[3 marks]

Charge =

[Total for Question 11 = 10 marks]_______________________________________________________________________________________

_______________________________________________________________________________________

12.Buzzers are used in many cases, for example as fire alarms or to indicate that a door or gate is opening or closing. These usually sound for a specific time interval that can be controlled with the help of a capacitor.Two such circuits, A and B, built by research students are shown below.18 mF capacitors are charged by a 12 V battery. When the alarm sounds they discharge through a buzzer that has 100 resistance.

buzzerR = 100 C = 18 mFbuzzerR = 100 C = 18 mF12 V12 V

AB

(a)Describe two disadvantages of circuit A with respect to circuit B.

[2 marks]

1.

2.

(b)Compare the current through the buzzer in circuit A during charging and discharging.

[2 marks]

(c)During discharge the buzzer will sound while the voltage is above 2 V.Calculate the time for which the buzzer sounds.

[3 marks]

Time =

(d)What could be done to the circuit to increase the time for which the buzzer sounds?

[1 mark]


_______________________________________________________________________________________

13.Old television sets contain an electron gun that produces, accelerates and deflects electron beams.The screen is luminescent i.e. has a coating that emits light when hit by fast moving electrons.

(a)The electron gun releases electrons horizontally at a speed of 4.2 106 m s1.The beam is directed midway between a positive and negative plate as shown below.

u = 4.2 106 m s1+d = 22 mmL = 4 cmA

(i)Continue the line to show the trajectory of an electron that just misses hitting the horizontal plate leaving at point A.

[1 mark]

(ii)The vertical deflection h of an electron will be given by

where a is the vertical acceleration and t is the time of travel through the plates.

Show that t is about 10 ns.Hence calculate the potential difference V between the two plates that with which the beam just misses hitting the charged plates.

[5 marks]

Potential difference, V =

(iii)Show the trajectory of an electron that is deflected by the same potential difference you found in (ii) but enters the plates at a lower speed.

[1 mark]

(b)After leaving the plates at point A the electrons hit the luminescent screen at point B.

screenBA

Draw the trajectory of the electrons as they move from A to B outside the plates.

[1 mark]

(c)The deflection could have been achieved using electromagnets instead of charged plates.Describe two differences between the two ways of deflecting an electron beam.

[2 marks]

1.

2.


_______________________________________________________________________________________

14.Electromagnetic brakes are now used in all trains and many trucks.They employ electromagnetic induction instead of friction to reduce their speed.

electromagnetrotor wheelmain axeltrain wheel85 cm

*(a)Explain how these brakes can slow down a train.Explain the relationship between force and speed of train.

[4 marks]

(b)Below you can see the electromagnet and the current flowing through it.Draw the magnetic field through the rotor wheel.

[1 mark]rotor wheel

(c)The diameter of the train wheel is 85 cm.Show that it performs about 160 revolutions per minute when the train moves at 7 m s1.

[2 marks]

(d)The rotor wheel has a diameter of 70 cm.30% of that area is covered by a magnetic field of effective flux density equal to 0.2 T.Calculate the induced emf in the rotor wheel.

[3 marks]

Emf =

(e)We can think of V2 / R as the energy per second changed from kinetic energy to heat when the brakes are applied, where V is the induced emf and R is the resistance in the rotor.Taking into account that the rotor conducts electricity as a typical metal, explain why for high speeds, the brakes are less effective than expected.

[2 marks]


_______________________________________________________________________________________

15.Below you can see the main principles of a mass spectrometer, used to measure masses of atoms or molecules that make up a chemical substance.

detectorionizing and accelerating chambersorting chamberdeflectorVd

A sample of the substance is vaporized and enters the ionizing and accelerating chamber, where the atoms or molecules are bombarded by electrons and become positive ions.Next they are accelerated by a potential difference V and then sorted out in the sorting chamber.Finally they are deflected and detected on a luminescent screen.

(a)Deflection is achieved by a magnetic field.Circle below the direction of the magnetic field that provides such deflection.

[1 mark]

into the paperout of the paperto the leftto the right

(b)The diameter of the trajectory d for a specific molecule is found to be 38.4 cm.The ions are singly charged and the magnetic flux density is equal to B = 0.13 T.

(i)Show that the momentum of the ions is about 4 1021 kg m s1.

[2 marks]

(ii)The detector gives two bright spots on the screen, as shown on the diagram.Explain what indication this gives about the masses of ions present in the sample.

[2 marks]

(c)The diagram below shows the sorting chamber in greater detail.The same magnetic flux density of 0.13 T acts in the same direction as in the deflector.Additionally there is an electric field of strength equal to 3100 N C1.

B = 0.13 TE

(i)Write + and in the boxes above in such a way so that the positive ions move through the chamber in a straight line.

[1 mark]

(ii)Show that, for ions moving in a straight line through the chamber at a fixed speed, the electric field strength and magnetic flux density are proportional.Hence or otherwise calculate the speed of the ions passing through undeflected.

[3 marks]

Speed =

(d)Calculate the mass of ions that are deflected to the mark d = 38.4 cm.

[1 mark]

Mass =

(e)Explain the significance of the sorting chamber.

[1 mark]


_______________________________________________________________________________________

16.Photocopiers work by attracting negatively charged toner (ink) grains (particles) to a positively charged drum and then releasing them onto a paper. When we zoom in down to the size of the toner particles the cylindrical drum will look like a plane surface.The main concepts of one such design are shown below.

drumpapercontainer with charged tonercharged tonerNOT TO SCALEdrumcharged toner particlesrotation+++++

Throughout the question you may assume that all toner grains are negatively charged, identical, spherical and have the following properties:

Diameter = 20 mCharge = 1.3 1015 CVolume = 4.2 1015 m3

(a)The toner has a density of 1500 kg m3.Show that the weight of each grain is about 6 1011 N.

[2 marks]

(b)Below you can see a toner grain.Draw the electric field formed around the particle.

[2 marks]

(c)Calculate the number of electrons on each toner grain.

[1 mark]

Number of electrons =

(d)Below you can see two toner grains next to each other.Calculate the electrostatic force between the two grains.

[2 marks]20 m

Force =

(e)The drum is connected to a positive voltage and creates a uniform electric field.Draw below the field created by the drum.

[2 marks]

+++++++++

YEAR 7 A2 PHYSICS1Turn over

(f)The electric field created by the drum has field strength equal to 1.2 105 N C1.Show that this is strong enough to hold a double layer of toner grains.

[4 marks]+ Vtoner grainsdrum

*(g)Explain how the drum can be used to control the darkness of a photocopy.

[5 marks]


TOTAL FOR SECTION B = 70 MARKS_______________________________________________________________________________________

END

TOTAL FOR PAPER = 80 MARKS

YEAR 7 A2 PHYSICS19

List of data, formulae and relationships

Acceleration of free fallg = 9.81 m s2(close to Earths surface)

Boltzmann constantk = 1.38 1023 J K1

Coulombs law constantk = 1 / 4 0 = 8.99 109 N m2 C2

Electron chargee = 1.60 1019 C

Electron massme = 9.11 1031 kg

Electronvolt1 eV = 1.60 1019 J

Gravitational constantG = 6.67 1011 N m2 kg2

Gravitational field strengthg = 9.81 N kg1(close to Earths surface)

Permittivity of free space0 = 8.85 1012 F m1

Planck constanth = 6.63 1034 J s

Proton massmp = 1.67 1027 kg

Speed of light in a vacuumc = 3.00 108m s1

Stefan-Boltzmann constant = 5.67 108 W m2 K4

Unified atomic mass unitu = 1.66 1027 kg

Unit 1

MechanicsKinematic equations of motionv = u + a t

s = u t + 1/2 a t2

v2 = u2 + 2 a s

ForcesF = m a

g = F / m

W = m g

Work and energyW = F s

Ek = 1/2 m v2

Egrav = m g h

MaterialsStokes lawF = 6 r v

Hookes lawF = k x

Density = m / V

Pressurep = F / A

Young modulusE = / where

Stress = F / A

Strain = x / x

Elastic strain energyEel = 1/2 F x

Unit 2

WavesWave speedv = f

Refractive index12 = sin i / sin r = v1 / v2

ElectricityPotential differenceV = W / Q

ResistanceR = V / I

Electrical powerP = V IP = I2 RP = V2 / R

EnergyW = V I t

Efficiency

ResistivityR = L / A

CurrentI = Q / t

I = n q v A

Resistors in seriesR = R1 + R2 + R3

Resistors in parallel

Quantum physicsPhoton modelE = h f

Einsteins photoelectric equation1/2

Unit 4

MechanicsMomentump = m v

Kinetic energy of non-relativistic particleEK = p2 / 2 m

Motion in a circlev = r

T = 2 /

F = m a = m v2 / ra = v2 / ra = r 2

FieldsCoulombs LawF = k Q1 Q2 / r2 where k = 1 / 4 0

Electric FieldE = F / QE = k Q / r2E = V / d

CapacitanceC = Q / V

Energy stored in capacitorW = 1/2 Q V

Capacitor dischargeQ = Q0 et / R C

In a magnetic fieldF = B I l sin F = B q v sin

r = p / B Q

Faradays and Lenzs Laws = d(N ) / dt

Particle physicsMass-EnergyE = c2 m

de Broglie wavelength = h / p

DATA & FORMULAE

YEAR 7 A2 PHYSICS22

TEST 2

MARK SCHEME

PART A

p. 1

1.B

2.C

p. 2

3.A

4.B

p. 3

5.B

6.A

p. 4

7.B

8.D

p. 5

9.B

p. 6

10.B

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 => = 3.2 3.3 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 platePlates get same charge as the battery pole they are connected toReject: positive charges moving / charges moving across plates&Opposite charges across plates attract each other (or similar) [1 mark]

(d)Q = Q0 (1 et / ) OR V = V0 (1 et / ) [1 mark]Q0 = C V = 3 C = 6.3 6.6 mC [2 marks] itsQ = Q0 (1 e3) = 0.95 Q0 = 5.99 6.27 mC [3 marks] itsV = 0.95 V0 = 2.85 V [2 marks] itsQ = C V = 5.99 6.27 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 secV = V0 et / (R C)=> 2 = 12 et / 1.8 [1 mark]t / 1.8 = ln (1 / 6) [2 marks] itst = 1.8 ln (1 / 6) = 3.2 sec [3 marks] its

(d)Any one from [1 mark]Use a larger capacitorAdd a resistor in series (or similar)

p. 11 & 12

13.(a)(i)&(iii)+A(i) parabola [1 mark](iii) line similar to but left of (i) [1 mark]

(ii)t = L / u = 0.04 / (4.2 106) = 9.5 109 sec [1 mark]&a = 2 h / t2 = 2 0.011 / t2 = (2.2 2.44) 1014 m s2 [1 mark]F = m a = 9.11 1031 a = (2 2.22) 1016 N [2 marks] itsE = F / q = F / (1.6 1019) = 1250 1389 N C1 [3 marks] itsE = V / d => V = E d = 27.5 30.56 V [4 marks] its

(b)BAstraight line [1 mark]

(c)Any two from [2 marks]Trajectory is circular (not parabolic)Force / Acceleration always perpendicular to velocitySpeed / Kinetic energy does not change / increaseForce / Acceleration does not maintain a constant direction (or similar)

p. 13 & 14

14.*(a)Any one from [1 mark]Emf induced because rotating metal cuts magnetic field lines / fluxForce acts on electrons moving through magnetic field as wheel rotates 0 as B acts through different areas of the wheel / area of B changes as wheel rotates&Any one from [1 mark]Eddy currents formed in the wheelCurrents due to electron flow in the wheel because of Faradays force (or similar)&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 rotationEddy currents transform kinetic energy into heat (or similar)&Any one from [1 mark]Speed => force because greater flux cut per secondSpeed => force because greater change in areaSpeed => force because more revolutions => (or similar)

(b)NSSN

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 m2 [1 mark]Vind = N / t = N B Aeff = N 0.2 0.115 [1 mark]Vind = 0.06 0.061 V [2 marks] its

(e)Eddy currents => heat produced => temperature [1 mark]R => V2 / R less than expected / smaller rate of KE transferred to heat etc. [1 mark]

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 1019 0.192 = 3.99 1021 kg m s1 [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]ORHeavier / More massive ions will hit at larger radius [2 marks]

(c)(i)B = 0.13 TE+[1 mark]

(ii)FE = FM => E q = B q v => E = B v [1 mark]v = E / B = 3100 / 0.13 [1 mark]v = 23.8 103 m s1 [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]

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)At least 4 lines [1 mark]Arrows pointing onto grain [1 mark]

(c)N = q / e = 1.3 1015 / (1.6 1019) = 8125 electrons [1 mark]

(d)Fgr = k q2 / r2 = 9 109 (1.3 1015)2 / (20 106)2 [1 mark]Fgr = 3.8 1011 N [2 marks] its

(e)+++++++++

At least 3 equidistant (by eye) lines [1 mark]Arrows pointing away from surface [1 mark]

(f)FDr on grain = E q = 1.2 105 1.3 1015 = 1.56 1010 N [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]FDrW + FgrORFTOT = 1.56 1010 1 1010 N

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]

YEAR 7 A2 PHYSICS1

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Electric and Magnetic