chapter 12 test question answers

8/2/2019 Chapter 12 Test Question Answers

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1

1. B[1]

2. C[1]

3. A[1]

4. (a) (i) lines parallel and normal to plates; (ignoreanyedgeeffect)

equally spaced;

direction from (+) to (); 3

(ii) curved path between plates and no curvature outside;

in downward direction; 2

(b) (i) change = qV;

= 1.6 1019

750

= 1.2 1016

J; 2

Or 750 eV.

(ii) mv2

= 1.2 1016

;

9.1 1031

v2

= 1.2 1016

/v2

= 2.64 1014

to give v = 1.67 107

m s1

; 2

(c) (i) inside solenoid, lines parallel to axis;

line spacing about double at ends / lines equally spaced in solenoid;

reasonable shape (symmetry and curving);

correct direction (to left); 4(ii) path with no deviation along axis; 1

(d) (i) velocity component normal to field = 1.6 107

sin 35;

= 9.2 106

m s1

; 2

(ii) circular motion;

in plane normal to paper;

r

m v2

=Bqv /r=)106.1100.4(

)102.9101.9(193

631

;

radius of circle = 1.3 102

m; 4

(iii) velocity component along field = 1.6 107 cos 35

= 1.3 107

m s1

; 1

(iv) force is zero;

because F=Bqv sinand = 0 orinwords; 2

(e) helical shape (allowspiralshape);

any further detail e.g. constant pitch etc.; 2

Award[2]for a good diagram.

[25]

5.(a) situation1: EE/GG;situation2: EG/GE; 2

(b) equal amounts of each type are produced in electrification by friction;


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2

and the normal state of matter is neutral; 2 max

OWTTE; Accept each cancels the other out.

(c) 6 max

Hypothesis /theory Explanation

Franklin

all matter contains an

electrical fluid;

fluid is transferred from one

object to another by friction;two objects with excess fluid

or less fluid will repel and

excess and less will attract;

Modernatomic

theory

protons and electrons

carry equal and

opposite charges;

electrons are transferred during

friction;

two objects with excess or less

electrons will repel and excess

and less will attract;

or electron transfer leaves on +

the other object;

two like charges repel, unlike

attract;

Award[1]each for sensibly worded hypothesis and 2 each for

an explanation in terms of the hypothesis which shows that they

have an understanding of what is going on up to[6 max].

[10]

6. B[1]

7. C[1]

8. D[1]

9. B[1]

10.A [1]

11. C[1]

12. (a) (i) M shown at peak or trough; 1

(ii) Z shown on t-axis; 1


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3

(b) by Lenzs law, e.m.f. (or current) must change direction as flux cutting

changes direction;

as magnet oscillates, flux is cut in opposite directions; 2[4]

13. (a) (i) if independent of charge;

must be gravitational;

(ii) depends on charge so electric or magnetic;

independent of velocity so electric;

(iii) (depends on velocity and charge, so) magnetic; 5 max

(b) idea of change in electric potential energy = gain in kinetic energy;

qV= mv2;

(1.6 1019

2.1 103) = ( 9.1 10

31v

2) /v = 7.38 10

14; 3

Award[0]for v = 2.7 107

m s1

.

(c) (i) correct force direction (upwards); 1

(ii) force =d

qV;

=)102.2(

)95106.1(2

19

= 6.9 1016

N; 3

(d) (i) time to cross plates =)107.2(

)1012(7

2

= 4.44 10

9s; 1

(ii) vertical acceleration =

)101.9(

)109.6(31

16

(= 7.58 10

14m s

2);

distance = a t2;

= 7.58 1014

(4.44 109

)2allowecf;

= 7.5 103

m; 3

(e) gravitational force is very small;

small in comparison with electric or magnetic force; 2

(f) (i) force due to B-field must be downwards;

mention of Flemings left-hand rule / right-hand palm rule;

hence field into paper; 3

(ii) Bqv = 6.9 1016

N allowecf

B =)107.2106.1(

)109.6(719

16

;

= 1.6 104

T; 2 max


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4

(g) (i) electric force unchanged;

magnetic force is greater;

hence deflection downwards;

(ii) both forces reversed in direction;

but not changed in magnitude;

hence undeflected;

(iii) undeflected; 7 max[30]

14. C[1]

15. A[1]

16. D[1]

17.D[1]

18. B[1]

19. B[1]

20.C[1]

21. (a) [1]foranyvalidandrelevantpointe.g.

Geiger-Marsden experiment involved bombardment of gold foil byalpha particles;

most passed straight through / were deviated through small angles but,

some deflected through large angles;

these alpha particles were heading towards central nucleus; 3 max

(b) [1]foranyvalidandrelevantpointe.g.

protons in nucleus repel each other (seen or implied);

but are held together by the strong nuclear force / or neutrons are

involved keeping it bound together /OWTTE; 2 max


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5

(c) (i) attempted use ofF=2

0

21

4 r

qq

;

with q1 = q2 = 1029

e = 1.6 1010

C;

and r= 100 m;

to get F= 2.3 1026

N 1026

N; 4 max

(ii) people are overall electrically neutral;equal numbers of positive charges mean that overall the electrical

force is zero /OWTTE; 2 max[11]

22. (a) (i) correct substitution into power = p.d. current

to give power = 12 0.5 = 6 W; 1 max

(ii) correct substitution into V= I R

to giveR =5.0

12= 24; 1 max

(b) correct positioning of ammeter;correct positioning of voltmeter;

e.g.

12 V battery

A

V2 max

(c) (i) the battery (or the ammeter or the wires) must have some resistance;

some p.d. is used up so less available / OWTTE; 2 max

(ii) low voltage requires low current and thus large resistance;

max resistance of variable resistor not infinite /OWTTE; 2 max


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6

(d) (i) any circuit involving potentiometer or equivalent;

that correctly controls the p.d. across the bulb;

with meters still correctly connected;

A

V3 max

(ii) [1]foreachrelevantpointe.g.

the 12 V is shared by the two halves of the resistor;

if the LH half is zero resistance, the p.d. will be zero /OWTTE; 2 max

(e) (i) appropriate statement of Ohms law;

e.g. p.d. proportional to current of constant temperature.

temperature is not constant as current varies /OWTTE; 2 max

(ii) lamp B must have greater power dissipation;

since it has a greater current for the same p.d. /OWTTE;

so power dissipation (= V I) is greater; 3 max

(f) (i) current lamp A equals the current in lamp B /OWTTE; 1 max

(ii) any answer that is less than 0.5 A but above 0.3 A;

realization (seen or implied) that each lamp does not have the same p.d.;

explanation (or evidence from the graph) of trying to find the

current when the individual p.d.s sum to 12 V;

to give 0.4 A ( 0.1); 4 max

(iii) lamp A will have greater power dissipation;

since current the same, but it takes greater share of p.d.; 2 max[25]

23. (a) [1]foreachappropriateandvalidpointe.g.

thermal energy is the K.E. of the component particles of an object;

thus measured in joules;

the temperature of an object is a measure how hot something is

(it can be used to work out the direction of the natural flow of thermal

energy between two objects in thermal contact) / measure of the average

K.E. of molecules;

it is measured on a defined scale (Celsius, Kelvin etc.); 4 max

(b) (i) correct substitution: energy = power time;

= 1200 W (30 60) s;= 2.2 10 J 2 max

(ii) use of E = mc;


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7

to get = 2.2 106

/ (4200 70) K;

= 7.5 K; 3 max

(c) [1]namingeachprocessupto[3max].

convection;

conduction;

radiation;

[1]for an appropriate (matching) piece of information / outlinefor each process up to[3 max].

e.g. convection is the transfer of thermal energy via bulk movement of a gas

due to a change of density;

conduction is transfer of thermal energy via intermolecular collisions;

radiation is the transfer of thermal energy via electromagnetic waves

(IR part of the electromagnetic spectrum in this situation)/OWTTE; 6 max

(d) (i) [1]foreachvalidandrelevantpointe.g.

in evaporation the faster moving molecules escape;

this means the average K.E. of the sample left has fallen;

a fall in average K.E. is the same as a fall in temperature; 3 max

(ii) energy lost by evaporation = 50 % 2.2 106

J;

= 1.1 106

J;

correct substitution intoE = ml

to give mass lost = 1.1 106

J / 2.26 106

J kg1

= 0.487 kg

= 487 g; 3 max

(iii) [1]foranyvalidandrelevantfactors[2max]e.g.

area of skin exposed;

presence or absence of wind;

temperature of air;

humidity of air etc.;

[1]for appropriate and matching explanations[2 max]e.g.

increased area means greater total evaporation rate;

presence of wind means greater total evaporation rate;

evaporation rate depends on temperature difference;

increased humidity decreases total evaporation rate etc.; 4 max[25]

24. (a) =2

2

4d

kQor F=

20

2

16 d

Q

; 1 max

(b)4

F; 1 max

(c) the spheres can be assumed to be point charges; 1 max

(d) each time the charge is shared it halves in value on each sphere;

ifFis proportional to the product of the charges then Fwill be reduced by

41 each time /

OWTTE; 2 max

i.e. Look for the halving in value of the charge and the effect is

has on F.

(e)


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8

+

+

Mark diagram and description together looking for these points:

unlabelled diagram;

description / labelling;

force between the charges produces a twist in suspension;

amount of twist is a measure of the force; 4 max

Award good answers[4 max]and weak answers[1 max].

[9]

25. C[1]


= 1200 W (30 60) s;

= 2.2 10 J 2 max

(ii) use of E = mc;

to get = 2.2 106

/ (4200 70) K;

= 7.5 K; 3 max


convection;

conduction;

radiation;

[1]for an appropriate (matching) piece of information / outline

for each process up to[3 max].

e.g. convection is the transfer of thermal energy via bulk movement of a gasdue to a change of density;


radiation is the transfer of thermal energy via electromagnetic waves

(IR part of the electromagnetic spectrum in this situation)/OWTTE; 6 max

(d) (i) [1]foreachvalidandrelevantpointe.g.

in evaporation the faster moving molecules escape;

this means the average K.E. of the sample left has fallen;

a fall in average K.E. is the same as a fall in temperature; 3 max


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(ii) energy lost by evaporation = 50 % 2.2 106

J;

= 1.1 106

J;

correct substitution intoE = ml


J / 2.26 106

J kg1

= 0.487 kg

= 487 g; 3 max

(iii) [1]foranyvalidandrelevantfactors[2max]e.g.area of skin exposed;


temperature of air;


[1]for appropriate and matching explanations[2 max]e.g.





26. D[1]

27. (a) there are two types of charge: positive and negative;

if they are moved apart, this is charge separation /OWTTE; 2 max

(b) electric field strength is the force per unit charge;

felt by a positive test charge placed in the field /OWTTE; 2 maxAccept mathematical definitions, but do not award any marks

for just copying the formula from the data booklet. Answers

need to define the terms to receive credit.

(c) straight lines from Earth to negative;

equally spaced;

edge effect shown;

thundercloud

500 m

ground

3 max


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10

(d) (e) correct substitution to calculate (= 2.86 106 C m2);

E= 20 / (8.85 102

7 106) V m

1;

= 3.2 105

V m1

;

= 3 105

V m1

; 3 max

(ii) [1]foreachappropriateandsensibleassumptione.g.

edge of cloud parallel to ground;

cloud and ground effectively infinite length;

permittivity of air the same as vacuum;

ground and cloud surface both flat, etc.; 2 max

(e) potential difference = 3.2 105

500 V;

= 1.6 108

V; 2 max

Accept 1.5 108 V.

(f) (i) average current = 20 C / 20 ms

= 1000 A; 1

(ii) average p.d. = 1.6 108

V / 2 = 0.8 108

V;

energy released = average p.d charge = 0.8 108 20;

= 1.6 109

J; 3 max

[2] if maximum p.d. used to get 3.2 109 J.

[18]

28. (a) Q is destructive interference;

since the difference in path lengths from the speakers to Q;

must be a21 difference; 3 max

Accept (n +21 ) difference / OWTTE.

(b) frequency = 2 Hz time between loud points = 0.5 s;

wavelength of sound, = 330 / 360 = 0.917 m;

distance between loud points = D / a;

= (0.917 10.0 / 4.0) m

= 2.29 m;

Accept solutions using n= d sin .

since speed = distance / time;

= 2.29 / 0.5 ms1

= 4.58 ms1

5 ms 6 max

N.B. The solution above is, of course, an estimation. If theanswer attempts full solution (i.e. does not resort to smallangle estimations) but still gets confused with maths, full marks

can be awarded if appropriate progress is made or

understanding shown.

(c) realization that the set-up is to do with beats;

two signal generators give different frequencies that are 2 Hz apart;

must be 359 Hz and the other 361 Hz; 3 max[12]


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29. (a) [1]foreachappropriateandvalidpointe.g.

thermal energy is the K.E. of the component particles of an object;

thus measured in joules;

the temperature of an object is a measure how hot something is (it can

be used to work out the direction of the natural flow of thermal energy

between two objects in thermal contact) / measure of the average

K.E. of molecules;

it is measured on a defined scale (Celsius, Kelvin etc.); 4 max


= 1200 W (30 60) s;

= 2.2 106

J 2 max

(ii) use of E= mc;

to get = 2.2 106

/ (4200 70) K;

= 7.5 K; 3 max


convection;

conduction;

radiation;

[1]for an appropriate (matching) piece of information / outline

for each process up to[3 max].

e.g. convection is the transfer of thermal energy via bulk movement

of a gas due to a change of density;


radiation is the transfer of thermal energy via electromagnetic

waves (IR part of the electromagnetic spectrum in this

situation) /OWTTE; 6 max

(d) (i) energy lost by evaporation = 50 % 2.2 106

J;

= 1.1 106 J;correct substitution intoE= ml


J / 2.26 106

J kg1

= 0.487 kg

= 487 g; 3 max

(ii) [1 max]foranyvalidandrelevantfactore.g.

area of skin exposed;


temperature of air;


[1 max]for an appropriate and matching explanation e.g.





30. D[1]


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31. B[1]

32. B[1]

33. B[1]

34. A[1]

35. (a) (i) zero deflection of leaf; negative charge on cap; 2

(award[0]if charge shown elsewhere)

(ii) diagram unchanged from diagram 2; 1(iii) leaf raised; negative charge on leaf and cap; 2

(disregard number of () signs)

(b) (i) energy per unit charge; (ratio idea necessary)

to move positive test charge between points; 2

(ii) leaf undeflected when charge on electroscope /vice versa;

leaf deflects when charge moved towards/away from electroscope;

hence gives a measure of potential; 3

Allow any sensible relevant comments leading to a valid

conclusion.

(c) (i) use of e.m.f. = energy / charge;

=)108.5(

)101.8(3

3

= 1.4 V; 2

Award[0]for formula E =Q

Fseen or implied even if answer is

numerically correct.

(ii) p.d. across internal resistance = 0.2 V; OR current =6

2.1= 0.2A;

resistance r=

2.12.0 6.0; total resistance =

2.04.1 = 7.0 ;

= 1.0 ; internal resistance = 76 = 1.0 ; 3

Accept any other valid route.

(iii) idea of use of ratio of resistances;

energy transfer = 6/7 8.1 103

= 6.9(4) 10 J; 2



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(iv) charge carriers/electrons have kinetic energy / are moving;

these carriers collide with the lattice/lattice ions; (do not allow friction)

causing increased (amplitude of) vibrations;

this increase seen as a temperature rise;

i.e. a transfer to thermal energy; 5

Allow any other relevant and correct statements.

(d) chemical energy in battery electrical energy / kinetic energy of electrons;electrical energy magnetic energy / kinetic energy of iron pieces;

(gravitational) potential energy of iron pieces; 3[25]

36. C[1]

37. C[1]

38. A[1]

39. (a) any particle has wave-like properties / other appropriate statement;

where wavelength =p

hwith h andp identified; 2

Can be back credited from (b).

(b) use ofE=m

p2

2

; OR 21 mv2 = qVor v =

m

qV2 ;

5.0 103

1.6 1019

2 9.1 1031

=p2 v =

31

319

1011.9

100.5106.12

p = 3.8 1023

; = 4.1(9) 107

m s1

;

=)1082.3(

)103.6(23

34

; =

731

34

1019.41011.9

1063.6

;

= 1.7(4) 1011

m; = 1.7(4) 1011

m; 4

Award incorrect calculation of p or v but then clear and correct

evaluation of [2 max].

[6]

40. (a) (i) zero deflection of leaf;

negative charge on cap; 2

(award[0]if charge shown elsewhere)

(ii) diagram unchanged from diagram 2; 1

(iii) leaf raised; negative charge on leaf and cap; 2

(disregard number of () signs)


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(b) (i) energy per unit charge; (ratio idea necessary)

to move positive test charge between points; 2

(ii) leaf undeflected when charge on electroscope /vice versa;

leaf deflects when charge moved towards / away from electroscope;

hence gives a measure of potential; 3

Allow any sensible relevant comments leading to a valid

conclusion.[10]

41. (i) use of e.m.f. = energy / charge;

=)108.5(

)101.8(3

3

= 1.4 V; 2

Award[0]for formula E =Q

Fseen or implied even if answer is

numerically correct.

(ii) p.d. across internal resistance = 0.2 V; OR current =6

2.1= 0.2 A;

resistance r=

2.1

2.0 6.0 ; total resistance =

2

4.1= 7.0 ;

= 1.0 ; internal resistance = 76 = 1.0 ; 3


(iii) idea of use of ratio of resistances;

energy transfer = 6/7 8.1 103

= 6.9(4) 103

J; 2


(iv) charge carriers/electrons have kinetic energy / are moving;

these carriers collide with the lattice/lattice ions; (do not allow friction)

causing increased (amplitude of) vibrations;

this increase seen as a temperature rise;

i.e. a transfer to thermal energy; 5

Allow any other relevant and correct statements.

[12]

42. (i) (induced) e.m.f. proportional to rate of change of magnetic flux (linkage);

(do not allow induced current)

as current increases, magnetic field in coil increases;

thus change in flux linkage and e.m.f. induced; 3

(ii) direction of (induced) e.m.f. such as to tend to oppose;

the change producing it;

induced e.m.f. must oppose e.m.f. of battery / growth of current in circuit; 3

(iii) energy is supplied by the battery;

in making charge move against the induced e.m.f.; 2[8]


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43. A[1]

44. B[1]

45. A[1]

46. D[1]

47. B[1]

48. (a)

V

I00 1

Any reasonable curve in the right direction.

(b) (i) from the value ofV/Iat any point on the curve; 1

Do not accept just from V/I.

(ii) non-ohmic because the resistance (V/Iat each point)

is not constant /OWTTE; 1

(c) (i) 50; 1

(ii) recognize that the voltage must divide in the ratio 3 : 1;

to give R = 150 ; 2]#

Or answer could be solved via the current.

[6]


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49. (a)

sufficient arrows to show decreasing radial field;

direction;

no field in the centre; 3

(b) (i) useE= k2

r

qto showE= 4.0 10

4V m

1; 1

(c) (i) along a field line; 1

(ii) F= ma = qE;

a =m

qE;

= 1.8 1011

4.0 104

= 7.2 1015

m s2

; 3

(iii) decreasing;

electric field strength is decreasing so force on electron is decreasing; 2

Do not penalize the candidate if they state field is decreasing.Award the right answer, with the wrong reason[0].

(iv) increase in KE =21 mv2 = 4.5 1031 36 1012 = 1.6 1017 J;

= qV;

to give V= 100 V; 3[13]

50. B[1]

51. D

[1]

52. C[1]

53. A[1]


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54. (a) (i) out of the paper ; 1

(ii) to the left ; 1For (i) and (ii) award[1 max]if labels are missing.

(b) (i) E=Blv = 0.2 0.3 5.5 = 0.33 V; 1

(ii) F= BIL = 0.2 0.8 0.30 = 0.048 N; 1

(c) P= Fv = 0.048 5.5 = 0.26 W;

=EI= 0.33 0.80 = 0.26 W; 2[6]

55. (a) the work done per unit charge;

in bringing a small positive charge;

from infinity to that point; 3A completely accurate definition is necessary for[3 max].

(b) (i)

sufficient arrows to show decreasing radial field;

direction;

no field in the centre; 3

(ii) three concentric circles;

with increasing radii; 2

(c) since field strength is the gradient of potential;Emust be decreasing;

orsome recognition that if the field were constant for example;

the lines would be equally spaced;

OWTTE; 2 max

Allow[1 max]for bald statement spacing betweenequipotentials is increasing / OWTTE.


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18

(d)0

0

V

a r

constant Vinside;

something that resembles a 1/rdependency outside; 2

V at surface must equal V inside.

(e) use V= kr

qto show V=1800 V; 1

(f) (i) along a field line; 1

(ii) use V= krq to find Vat 0.30 m =270 V;

use2

1mv

2= Ve;

to give v =31

19

101.9

106.115302

;

to give v = 2.3 107

m s1; 4

[18]

56. C[1]

57. D[1]

58. C[1]

59. B

[1]

60.D[1]

61. C[1]


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19

62. B[1]

63. (a) (i) when connected to a 3 V supply, the lamp will be at normal brightness;

and energy is produced in the filament at the rate of 0.60 W;Lookfortheideathat3Vistheoperatingvoltageandtheidea

ofenergytransformation.orwhen connected to a 3 V supply, the lamp will be at normal brightness;

and the resistance of the filament is 15 / the current in the filament is

0.20 A; 2 max

(ii) I=V

P;

to giveI= 0.20 A; 2

(b) (i) at maximum value, the supply voltage divides between the

resistance of the variable resistor, internal resistance and the

resistance of the filament;i.e.responsemustshowtheideaofthevoltagedividingbetweenthe

variousresistancesinthecircuit.Donotpenaliseifresponsesdonot

mentioninternalresistancehere.

at zero resistance, the supply voltage is now divided between the filament resistance

and the internal resistance of the supply; 2

(ii) when resistance of variable resistor is zero, e.m.f. =Ir+ Vlamp;

3.0 = 0.2 r+ 2.6;

to give r= 2.0 ; 3

(c) (i) 3.3; 1

(ii) 13

; 1(d) at the higher pd, greater current and therefore hotter; the resistance of a

metal increases with increasing temperature; OWTTE; 2 max

(e)I

V00

correct approximate shape (i.e. showing decreasing gradient with increasing V); 1


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20

(f) parallel resistance of lamp and YZ is calculated from12

1

4

11

R;

to giveR = 3.0 ;

3.0 V therefore divides between 3.0 and 12.0;

to give pd across the lamp = 0.60 V;

Giverelevantcreditifanswersgoviathecurrentsi.e.

calculation of total resistance = 15.0 ;total current = 0.20 A;

current in lamp = 0.15 A; 4[18]

64. (a) the force exerted per unit charge;

on a small positive (test) charge; 2

Accept either small or test or both.

(b) (i) substitute for r= a 2 ;

intoE= 2r

kQto getE= 22a

kQ; 2

(ii) ; 1

(iii) Efor each component =2

a

kQ;

add vectorially;

to getEtot= 22

a

kQ; 3

Award[1]if not added vectorially i.e. Etot= 2 2a

kQ

[8]

65. B[1]

66. D[1]

67. B[1]


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21

68. (a) all particles have a wavelength associated with them /OWTTE;

the de Broglie hypothesis gives the associated wavelength as =p

h;

where h is the Planck constant andp is the momentum of the particle; 3

If answers just quote the formula from the data book then award

[1]for showing at least they recognize which formula relates to

the hypothesis.

(b) (i) KE = Ve = 850 1.6 1019

J = 1.4 1016

J; 1

(ii) useE=m

p

2

2

to getp = mE2 ;

substitutep = 1631 104.1101.92 = 1.6 1023 N s; 2

(iii) =p

h;

substitute =23

34

106.1

106.6

= 4.1 10

11m; 2

[8]

69. (a) move the ring over the end of the magnet /OWTTE; 1i.e.magnetstationary,ringmoved.

(b)

diagram showing wire wrapped around part of the ring;

and appropriate connections to battery and variable resistor;

as the current is changed by altering the value of the resistance;

a current is induced in the ring; 4Mark diagram and description togetherlook for any sensible

description of the production of transformer induced currents.

(c) (i) the emf induced in the ring;

is equal/proportional to the rate of change of magnetic flux

linking the ring; 2

(ii)

clockwise;

Lenzs law: induced current is such as to oppose the change / OWTTE;

current in this direction induces a field in the opposite direction to the

changing field /OWTTE; 3

(iii) area = 3.14 (1.2)2

102

= 4.5 102

m2;

rate of flux change = 4.5 102

m 1.8 103

= emf = 8.1 105

V;


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22

current =2

5

105.1

)101.8(

= 5.4 mA; 3

[13]

70. D[1]

71. B[1]

72. D[1]

73. D[1]

74. C[1]

75. C[1]

76. B[1]

77. (a) (i) ; 1

(ii)

general shape: at least one circle around each wire and one loop

around both wires;

appropriate spacing of lines: increasing separation with distancefrom wires;

correct direction of field; 3

(b) velocity increases;

acceleration increases;

because the force is getting larger the closer the wires get together; 3Watch for ECF if force is drawn in wrong direction in (a) (i) i.e.

velocity increases, acceleration decreases, force gets smaller.

[7]

78. (a) (i) EI; 1

(ii) I2r; 1

(iii) VI; 1


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23

(b) (from the conservation of energy),EI=I2r+ VI;

therefore, V=EIr/E= V+Ir; 2

(c)

V

A

correct position of voltmeter;

correct position of ammeter;

correct position of variable resistor; 3

(d) (i) E= VwhenI= 0;soE= 1.5 V; 2

(ii) recognize this is when V= 0;

intercept on thex-axis = 1.3 (0.1) A; 2

(iii) ris the slope of the graph;

sensible choice of triangle, at least half the line as hypotenuse;

=6.0

7.0;

= 1.2 (0.1)

or

when V= 0E=Ir;

r=I

E

=3.1

5.1;

= 1.2 3

(e) R = 1.2;

I=2.12.1

5.1

= 0.63A;

P =I2R = (0.63)

2 1.2 = 0.48W / 0.47 W; 3

[18]

79. D[1]


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24

80. B[1]

81. D[1]

82. (a) force exerted per unit mass;

on a small / point mass; 2

(b) from the law of gravitation, the field strength2R

MG

m

F ;

= g0 to give GM= g0R2; 2

N.B. To achieve full marks, candidates need to state thatm

F=

g0.

(c) downwards; (accept 90 to B field or down the wire) 1

(d) F=Bevcos; 1

(e) work done in moving an electron the length of the wire is

W= FL =BevLcos;e.m.f. = work done per unit charge;

therefore,E=BLvcos;

or

electric field =e

F=Bvcos;

e.m.f.E= electric field L;

to giveE=BLvcos; 3

Award[2 max]if flux linkage argument is used.

(f) F= Gr

mv

r

Mm 2

2 ;

such that v2

=r

Rg

r

GM2

0 ;

v2

=6

122

107.6

10)4.6(10

to give v = 7.8 10

3m s

1; 3

(g) L =cosBv

E;

=93.0108.7103.6

1036

3

=2.2 10

4m; 2

[14]


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25

83. (a)

general shape: at least one circle around each wire and one loop aroundboth wires;

appropriate spacing of lines: increasing separation with distance from wires;

correct direction of field; 3

(b) constant separation / parallel;

current measured in terms of force per unit length of the wires;

oracceptcompletedefinition:

wires 1 m apart;

force between them is 2 107

N per metre length when current is 1 A; 2

(c) velocity increases;

acceleration increases;because the force is getting larger the closer the wires get together; 3

If answered in terms of a repulsion force then[2 max].

[8]

84. B[1]

85. C[1]

86. A[1]

87.B[1]

88. D[1]

89. (a) component X, battery, ammeter all in series and including means of

varying current; with voltmeter in parallel across component X; 2

(b) (i) 4.0 A; 1

(ii) use ofR =I

V, and not gradient of graph;

resistance = 1.5; 2

(c) (i) straight-line through origin, quadrants 1 or 3 or both;

correct gradient, i.e. passes through V= 4.0 V,I= 2.0 A; 2(ii) p.d.s across X and across R will be 3.7 V (0.1V ) and 6.0 V;

Award[0]if only one p.d. is correct.


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26

total p.d. = 9.7 V; 2

(d) (i) large change in resistance with temperature change;

unique value ofR at any temperature;

not dissipate thermal energy;

small physical size / small thermal capacity; 2 max

(ii) measure resistance at two known temperatures;

divide scale into equally sized units;any further detail

e.g.t/ =)(

)(

0100

0

RR

RRt

100orscale is empirical (for this

thermometer only)or fixed point specified; 3[14]

90. (a) concentric circles;

separation increases (at least three circles required to see this);

correct direction (anticlockwise); 3

(b) (i) current in one turn produces magnetic field in region of the other turn;gives rise to a force on the wire;

Newtons third / idea ofviceversa (gives rise to attraction) /

idea ofviceversa (gives rise to shortening); 3

(ii) use ofB =r2

0I (givesB = 1.0 104I);

use ofF=BIL;

= 1.0 104

I2 2 3.0 10

2;

this force is equal to mg;

henceI2

= 52.04, andI= 7.2 A; 5[11]

91. D[1]

92. D[1]

93. C[1]

94. A[1]

95. A[1]


27/27

96. (a) component X, battery, ammeter all in series and including means of varying

current;

with voltmeter in parallel across component X; 2

(b) (i) 4.0 A; 1

(ii) use ofR =

I

V, and not gradient of graph;

resistance = 1.5 ; 2

(c) (i) straight-line through origin, quadrants 1 or 3 or both;

correct gradient, i.e. passes through V= 4.0 V,I= 2.0 A ; 2

(ii) p.d.s across X and across R will be 3.7 V ( 0.1V) and 6.0 V;

Award[0]if only one p.d. is correct.

total p.d. = 9.7 V; 2[9]

97. (a) (i) e.m.f. (induced) proportional to;

rate of change /cutting of (magnetic) flux (linkage); 2

(ii) magnetic field / flux through coil will change as the current changes; 1

(b) (i) sinusoidal and in phase with current; 1

(ii) sinusoidal and same frequency;

with 90 phase difference to candidates graph for; 2

(iii) e.m.f. is reduced;

because B is smaller; 2

Award[0]for e.m.f. is reduced if argument fallacious.

(c) advantage: no direct contact with cable required;

disadvantage: distance to wire must be fixed; 2[10]

chapter 12 test question answers

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