· Web viewrefers to the force that acts on a current-carrying conductor when it is in a magnetic field and that this force lifts the ring upwards (into an area where the magnetic

Hinchley Wood School

Q1.

Figure 1

A circular coil of diameter 140 mm has 850 turns. It is placed so that its plane is perpendicular to a horizontal magnetic field of uniform flux density 45 mT, as shown in Figure 1.

(a) Calculate the magnetic flux passing through the coil when in this position.

......................................................................................................................

......................................................................................................................(2)

(b) The coil is rotated through 90° about a vertical axis in a time of 120 ms.

Calculate

(i) the change of magnetic flux linkage produced by this rotation,

.............................................................................................................

.............................................................................................................

(ii) the average emf induced in the coil when it is rotated.

.............................................................................................................

.............................................................................................................

.............................................................................................................(4)

(Total 6 marks)

Page 1


Q2. The Large Hadron Collider (LHC) uses magnetic fields to confine fast-moving charged particles travelling repeatedly around a circular path. The LHC is installed in an underground circular tunnel of circumference 27 km.

(a) In the presence of a suitably directed uniform magnetic field, charged particles move at constant speed in a circular path of constant radius. By reference to the force acting on the particles, explain how this is achieved and why it happens.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(4)

(b) (i) The charged particles travelling around the LHC may be protons. Calculate the centripetal force acting on a proton when travelling in a circular path of circumference 27 km at one-tenth of the speed of light. Ignore relativistic effects.

answer = ................................ N(3)

(ii) Calculate the flux density of the uniform magnetic field that would be required to produce this force. State an appropriate unit.

Page 2


answer = ...................................... unit ........................(3)

(c) The speed of the protons gradually increases as their energy is increased by the LHC.State and explain how the magnetic field in the LHC must change as the speed of the protons is increased.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(Total 12 marks)

Q3.(a) State Lenz’s law.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(b) Figure 1 shows two small, solid metal cylinders, P and Q. P is made from aluminium. Q is made from a steel alloy.

Figure 1

Page 3


(i) The dimensions of P and Q are identical but Q has a greater mass than P. Explain what material property is responsible for this difference.

...............................................................................................................

...............................................................................................................

...............................................................................................................(1)

(ii) When P and Q are released from rest and allowed to fall freely through a vertical distance of 1.0 m, they each take 0.45 s to do so. Justify this time value and explain why the times are the same.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(c) 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 centre of the tube, as shown in Figure 2.

Figure 2

Page 4


The time taken for Q to pass through the tube is much longer than that taken by P.

(i) Explain why you would expect an emf to be induced in the tube as Q passes through it.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(ii) State the consequences of this induced emf, and hence explain why Q takes longer than P to pass through the tube.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(3)

Page 5


(d) 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.

P: ...................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................

Q: ...................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(3)

(Total 13 marks)

Q4. (a) A transformer operating on a 230 V mains supply provides a 12 V output. There are 1150 turns on the primary coil.

(i) Calculate the number of turns on the secondary coil.

answer = ........................... turns(1)

(ii) A number of identical lamps rated at 12 V, 24 W are connected in parallel across the secondary coil. The primary circuit of the transformer includes a 630 mA fuse.Calculate the maximum number of lamps that can be supplied by the transformer if its efficiency is 85%.

answer = .......................... lamps(2)

Page 6


(iii) The transformer circuit includes a fuse. Explain why this is necessary.

...............................................................................................................

...............................................................................................................

...............................................................................................................(1)

(iv) Why is the fuse placed in the primary circuit rather than in the secondary circuit?

...............................................................................................................

...............................................................................................................

...............................................................................................................(1)

(b) The figure below shows an experimental arrangement that can be used to demonstrate magnetic levitation. The iron rod is fixed vertically inside a large coil of wire. When the alternating current supply to the coil is switched on, the aluminium ring moves up the rod until it reaches a stable position ‘floating’ above the coil.

(i) By reference to the laws of electromagnetic induction explain

• why a current will be induced in the ring,

• why the ring experiences a force that moves it upwards,

• why the ring reaches a stable position.

Page 7


The quality of your written communication will be assessed in your answer.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(6)

(ii) What would happen to the ring if the alternating current in the coil was increased without changing the frequency? Explain your answer.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(2)

(Total 13 marks)

Page 8


Q5.The diagram below shows the orbits of two Earth satellites, a communications satellite in a geosynchronous orbit and a monitoring satellite in a low orbit that passes over the poles.

(a) The time period, T, of any satellite in a circular orbit around a planet is proportional to r3/2, where r is the radius of its orbit measured from the centre of the planet. For a satellite in a low orbit that passes over the poles of the Earth, T is 105 minutes when r is 7370 km.

(i) Calculate the height above the surface of the Earth, in km, of a satellite in a geosynchronous circular orbit.Give your answer to an appropriate number of significant figures.

height above surface .............................. km(4)

(ii) Calculate the centripetal force acting on the polar orbiting satellite if its mass is650 kg.

Page 9


centripetal force ................................ N(2)

(b) These geosynchronous and polar satellites have different applications because of their different orbits in relation to the rotation of the Earth.

Compare the principal features of the geosynchronous and polar orbits and explain the consequences for possible uses of satellites in these orbits.

In your answer you should explain why:

• a low polar orbit is suitable for a satellite used to monitor conditions on the Earth.

• a geosynchronous circular orbit above the Equator is especially suitable for a satellite used in communications.

The quality of your written communication will be assessed in your answer.(6)

(Total 12 marks)

Page 10


M1. (a) (= BA) = 45 × 10–3 × π × (70 × 10–3)2 (1) = 6.9 × 10–4 Wb (1) (6.93 × 10–4 Wb)

2

(b) (i) NΔ ( = NBA – 0) = 850 × 6.93 × 10–4 (1)

= 0.59 (Wb turns) (1) (0.589 (Wb turns))

(if = 6.9 × 10–4, then 0.587 (Wb turns))

(allow C.E. for value of from (a))

(ii) induced emf ( = N ) = (1)

= 4.9 V (1) (4.91 V)

(allow C.E. for value of Wb turns from (ii)4

[6]

M2. (a) (magnetic) field is applied perpendicular to path

or direction or velocity of charged particles

(magnetic) force acts perpendicular to path

or direction or velocity of charged particles

force depends on speed of particle or on B [or F ∞ v or F = BQv explained]

force provides (centripetal) acceleration towards centre of circle

[or (magnetic) force is a centripetal force]

shows that r is constant when B and v are constant 4

Page 11


(b) (i) radius r of path = = 4.30 × 103 (m) (allow 4.3km)

centripetal force = 3.50 × 10–16(N) 3

(ii) magnetic flux density

= 7.29 × 10-5 T 3

(c) magnetic field must be increased

to increase (centripetal) force or in order to keep r constant

[or otherwise protons would attempt to travel in a path of larger radius]

[or, referring to , B must increase when v increases to keep r constant ]2

[12]

M3.(a) direction of induced emf (or current) opposes change (of magnetic flux) that produces it

2

(b) (i) (volumes are equal and mass of Q is greater than that of P) density of steel > density of aluminium

Allow density of Q greater (than density of P).1

Page 12


(ii) use of s =½ g t2 gives t 2 = (from which t = 0.45 s) Backwards working is acceptable for 1st mark

(vertical) acceleration [or acceleration due to gravity] is independent of mass of falling object[or correct reference to F = mg = ma with m cancelling ]

2 nd mark must refer to mass.Do not allow “both in free fall” for 2nd mark.

2

(c) (i) moving magnet [or magnetic field] passes through tube there is a change of flux (linkage)(in the tube)

[or flux lines are cut or appropriate reference to ɛ = N (Δɸ / Δt)] In this part marks can be awarded for answers which mix and match these schemes.

[Alternative:(conduction) electrons in copper (or tube) acted on by (moving)magnetic field of Q induced emf (or current) is produced by redistributed electrons ]

2

(ii) emf produces current (in copper) this current [allow emf] produces a magnetic field this field opposes magnetic field (or motion) of Q[or acts to reduce relative motion or produces upward force] no emf is induced by P because it is not magnetised (or not magnet)[or movement of P is not opposed by an induced emf or current]

Alternative to 3 rd mark: current gives heating effect in copper and energy for this comes from ke of Q

max 3

(d) time for P is unaffected because there is still no (induced) emf[or because P is not magnetisedor because there is no repulsive force on P] time for Q is shorter (than in (c)) current induced by Q would be smaller because resistance of brass ∝ resistivity and is therefore higher[or resistance of brass is higher because resistivity is greater] giving weaker (opposing) magnetic field[or less opposition to Q’s movement]

Condone “will pass through faster” for 2nd mark.If emf is stated to be smaller for Q, mark (d) to max 2.

max 3[13]

Page 13


M4. (a) (i) use of = gives NS = = 60 (turns) 1

(ii) max output power = 0.85 × 0.630 × 230 (= 123 W)

max number of lamps = 5 (no mark for non-integer answer)

[or efficiency = gives 0.85 = (and max IS = 10.3 (A))

max number of lamps = 5 ]2

(iii) fuse prevents transformer from overheating [or prevents transformer from supplying excessive currents]

1

(iv) (all of) transformer is disconnected from supply when fuse fails [or fuse in secondary circuit would leave primary circuit live]

1

(b) (i) The candidate’s writing should be legible and the spelling, punctuationand grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will beassigned to one of three levels according to the following criteria.

High level (good to excellent) 5 or 6 marks

The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.

Page 14


The candidate states that the ac in the coil produces a constantly changing magnetic field that passes through the ring, causing an emf to be induced according to Faraday’s law.

The candidate recognises that the induced emf will cause a current to flow in the ring, that the current is likely be large because the coil acts as a single conductor with low resistance, and that this current also produces a magnetic field.

The candidate appreciates that Lenz’s law indicates that the direction of the induced current is such as to produce a magnetic field that will oppose the existing field, and that the two fields will interact.

The candidate refers to the force that acts on a current-carrying conductor when it is in a magnetic field and that this force lifts the ring upwards (into an area where the magnetic field is weaker) until the upwards magnetic force is equal to the downwards weight of the ring.

Intermediate level (modest to adequate) 3 or 4 marks

The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.

The candidate is familiar with either or both Faraday’s and Lenz’s laws but only applies one of them to explain what happens in this demonstration. There are correct references to the two forces that act on the ring, and a reasonable explanation of why the ring reaches a stable position.

Low level (poor to limited) 1 or 2 marks

The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.

The candidate refers much more superficially to either Faraday’s or Lenz’s law (or to both of them) but shows some understanding of why the forces acting on the ring cause it to reach equilibrium.

The explanation expected in a competent answer should include a

Page 15


coherent selection of the following points concerning the physical principles involved and their consequences in this case.

Faraday’s law

• An emf is induced whenever there is a change in the magnetic flux passing through a conductor.

• The magnitude of the emf is proportional to the rate of change of magnetic flux linkage.

• The induced emf will cause a current to flow in any complete circuit, such as a single conducting ring.

• Because the ring is made from aluminium, which is a good conductor, a large initial current will be induced in it.

Lenz’s law

• The induced current flows in such a direction as to oppose the increase in magnetic flux when the current is switched on in the coil.

• The current produces a magnetic field in the opposite direction to that produced by the coil.

• These two (alternating) fields interact like the fields between two facing like magnetic poles, giving repulsion.

Forces

• The ring is a current-carrying conductor in a magnetic field, andconsequently it experiences a force.

• This magnetic force acts upwards, in the opposite direction to theweight of the ring.

• As the ring rises, the magnetic field to which it is exposed becomesweaker as it moves away from the coil.

• This reduces the induced current, reducing also the magnetic forceon the ring.

• The ring reaches a stable height when the magnetic force hasdecreased to the point where it is equal to the weight of the ring.

6

(ii) ring would ‘float’ higher [or be expelled upwards]

Page 16


because (initial) current or emf (induced) in ring is greater

or ring moves into weaker field until magnetic force balances weight [or (initially) magnetic force exceeds weight]

2[13]

M5.(a) (i)

from which = 5.73

and rE (= 5.73 × 7370) = 42 200 (km)

height above surface = 42 200 − 6370 = 35 800 or 35 900 (km) answer to 3SF only

Full solution derived from Newton’s law of gravitation is acceptable for all 4 marks.

[or Newton ’s law approach for 1st two marks:

∴rE3 = (=7.54 × 1022)

from which rE = 42 200 (km) ]For 3 rd mark, final answer must be expressed in km.3SF mark is independent.

4

(ii) centripetal force (= m ɷ2r) =

= 4800 (4760) (N) If both T and r values for the geosynchronous satellite are substituted, award 0 marks for (ii).

Page 17


[or centripetal force and v =

gives v = 7350 (m s−1) and centripetal force =

= 4800 (4760) (N) ]

[or centripetal force

= 4800 (4770) (N) ]If only one correct T or r value for the polar satellite is substituted, mark (ii) to max 1.

2

(b) The candidate’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marksThe information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.

Four aspects must be considered in a high level answer:-Features of polar orbit.Features of geosynchronous orbit.Why polar orbit is suitable for monitoring.Why geosynchronous orbit is suitable for communication.

The candidate gives a comprehensive comparison of the principal features of the satellite orbits and explains the consequences for the uses of the two types of satellites. There are clear statements showing good understanding of why the polar satellite is suitable for monitoring, and of why the geosynchronous satellite is useful for communications.

Intermediate Level (Modest to adequate): 3 or 4 marksThe information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.

The candidate’s comparison of the principal features of the orbits is less complete and the consequences for the uses of satellites in them are less well understood. The candidate has an acceptable appreciation of why the polar satellite is suitable for monitoring, and of why the geosynchronous satellite is useful for communications.

Low Level (Poor to limited): 1 or 2 marksThe information conveyed by the answer is poorly organised and may not be

Page 18


relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.

The candidate has a much weaker knowledge of the principal features of the orbits and very limited knowledge of consequences for the uses of satellites in them. Understanding of why the polar satellite is suitable for monitoring, and why the geosynchronous satellite is suitable for communications, is limited or absent.

The explanation expected in a competent answer should include a coherent selection of the following points.

Low polar orbit

• Orbital period is a few hours• Earth rotates relative to the orbit• Many orbits with different radii and periods are possible• Orbit height is less than geosynchronous satellite• Speed is greater than that of geosynchronous satellite• Satellite scans the whole surface of the Earth• Applications: surveillance of conditions / installations on Earth, mapping,

weather observations, environmental monitoring• Gives access to every point on Earth’s surface every day• Can collect data from regions inaccessible to man• Contact with transmitting / receiving aerial is intermittent• Aerial is likely to need a tracking facility• Lower signal strength required than that for geosynchronous satellite

Geosynchronous orbit above Equator

• Orbital period matches Earth’ rotational period exactly• Satellite maintains same position relative to Earth• Only one particular orbit radius is possible• Travels west to east above Equator (in same direction as Earth’s rotation)• Orbit height is greater than polar orbit satellite• Speed is less than that of polar orbiting satellite• Scans a restricted (and fixed) area of the Earth’s surface only• Applications: telecommunications generally, cable and satellite TV, radio,

digital information, etc.• Satellite is in continuous contact with transmitting / receiving aerial• Aerial can be in a fixed position• Higher signal strength required than that for polar satellite

max 6[12]

Page 19


E1. The topic of electromagnetism continues to present greater difficulty than most of the remainder of the Unit 4 content. Candidates who had mastered the distinction between magnetic flux and flux linkage, and who appreciated that induced emf = (rate of change of NΦ), readily gained all six marks. Only a small minority of the candidates came into this category, however. When finding the cross-sectional area presented to the flux, there was evidence of the usual confusion between diameter and radius, leading to the loss of one mark on the question. More worrying were those candidates who wrote the area of a circle as 2πr, or as 2πr2. In part (b), examiners took the view that candidates should know that an emf is measured in V – final answers expressed in Wb turns s–1 were not accepted.

E2. It was rare for all four marks to be awarded in part (a). The essence of this question was well understood, but poor use of English and an inability to write logically limited the mark that could be given. An alarming proportion of answers made no reference at all to the magnetic field; these students appeared to be answering a more general question about circular motion. Many of the students evidently thought that the purpose of the magnetic force (presumably acting outwards) was to balance the centripetal force, rather than to provide it. Relatively few correct solutions were seen that used r = mv / BQ to show that r is constant when B and v are constant.

The common error in part (b)(i) was failure to deduce the radius of the path of the protons from the 27 km circumference of the LHC. This only meant the loss of one of the three marks, however, provided the principles of the rest of the calculation were correct. Careless arithmetic such as failure to square v, and/or forgetting to convert km to m, was also a frequent source of loss of marks. F = BQv was usually applied successfully in part (b)(ii), where the unit of magnetic flux density was quite well known. Almost inevitably, there was some confusion between flux density and magnetic flux.

The fact that had to be appreciated in part (c) was that in the LHC the radius of the path of the charged particles must remain constant as they are accelerated. A large proportion of students thought that it was necessary to maintain a constant centripetal force for this to happen, whereas it ought to have been clear to them that F must increase as v increases if r is to be constant.

E3.Acceptable statements really needed to refer to both the direction of the induced emf (or current) and to the change (in magnetic flux) that produces the effect. In part (b)(i) an explanation of the greater mass of Q was required, so a simple statement that density was involved was inadequate; candidates had to state that steel (or Q) has a greater density

Page 20


than aluminium (or P). In part (b)(ii) the time of 0.45s was usually justified through the application of s = ut + ½ at2, although some candidates made no attempt to justify this value. Backwards working, such as showing that the distance fallen is approximately 1.0m when the time of fall is 0.45s, was accepted. Explanations of why the two times are equal were expected to refer to acceleration due to gravity being independent of the mass of a falling body.

There was widespread misunderstanding in candidates’ attempts to answer part (c). In part (i), clearly Q is a moving magnet passing through a conducting tube and so the magnet’s flux lines are cut by the tube – hence an emf is induced. A significant number of responses stated that Q would be cutting through the flux lines of the tube. The tube was regularly referred to as a magnet. A very common misapprehension was that when a current is induced in the tube, it is the current that causes the emf. In part (c)(ii) many answers were too trivial, such as ones which referred to the repulsion of poles, or were simply wrong, such as attributing the effect to induced charges. Some responses even suggested that the induced electromotive force acts as a mechanical force to oppose the falling magnet. Examiners were pleased to encounter logical answers stating that the induced emf caused a current to flow in the copper, which then produced a magnetic field to oppose the movement of the falling magnet Q by opposing the magnet’s own field. Relatively few answers made any reference as to why cylinder P would fall without opposition.

Full marks were regularly awarded in part (d), where it was usually seen that the time for P would be unaffected (an explanation was needed for the mark) but that for Q would be shorter. Some candidates thought that the increased resistance of the tube would cause a reduced emf; these answers were subjected to a two mark maximum.

E4. The transformer turns ratio equation was familiar territory for most in part (a) (i), but correct application of the efficiency formula proved to be a greater challenge in part (a) (ii). Many correct answers were seen, and almost all students knew that the number of lamps has to be an integer. Most difficulties arose from mixing up data for the secondary coil with that for the primary (for example, multiplying the primary current by the secondary voltage).

Parts (a) (iii) and (iv) proved to be an exacting test of whether students could think through to the real reasons or had enough practical experience of transformers to know these reasons.Many attempts at part (iii) were general answers about the reason for fitting a fuse in any circuit rather than specifically in a transformer’s circuit. Very few students stated that transformer coils can overheat and become damaged when they handle excessive currents and that they therefore need to be protected.

Similarly, it was only a small minority of answers to part (iv) that were properly valid; that stopping the primary current would isolate the whole transformer from the mains or that a failed fuse in the secondary circuit would leave the primary live.

Most students find that electromagnetic induction is one of the most demanding topics in the specification. In these circumstances perhaps it should not be surprising that many of the attempts to answer part (b) (i) were very disappointing. Even when pointed at a logical and sequential structured answer by three bullet points, many students could not

Page 21


construct a coherent, ordered response. In assessing the Quality of Written Communication, one aspect that must be taken into consideration is the appropriate use of technical terminology. This was often absent from the responses seen. The term induction has a very special meaning in physics; magnetic induction involves magnetising a material by applying a magnetic field, electrostatic induction involves charge separation by applying an electric field, electromagnetic induction involves producing an emf by applying a changing magnetic field. In all cases, direct contact is unnecessary. Many answers contained the word ‘induced’ used much more carelessly than its technical meaning in physics; a current was stated to be induced in the coil because it was connected to the ac supply, for example. This current was then said to induce a magnetic field. Many students seemed obsessed by effects in the iron rod, rather than in the aluminium ring. The aluminium ring was confused with the coil for example ‘the coil is pushed upwards by the magnetic field’. It was evident that a large proportion of students were familiar with statements of the laws of electromagnetic induction but could not apply them meaningfully to explain what happens in this demonstration. The field produced was regularly referred to as an electric field. The repulsion of the ring was sometimes attributed to Coulomb’s law and the repulsion between charges. Fleming’s left hand rule was confused with his right hand rule, or with the right hand grasp rule.

Broadly, an outline plan of an appropriate answer to this question would be along the following lines. The ac current in the coil produces an alternating magnetic field, which is concentrated in the iron rod and passes through the ring. This changing magnetic field induces an emf in the ring. Because the ring is aluminium it is a good conductor and the emf causes a large current in it. A current-carrying conductor in a magnetic field experiences a force so this current produces a magnetic field whose direction opposes the applied field. Interaction between these fields gives a net upwards repulsion of the ring. As the ring moves upwards the magnetic field becomes weaker and the force on the ring decreases. The ring’s position becomes stable when the upwards magnetic force balances its weight. Answers written in this fashion were rare but not difficult to identify and they were rewarded well.

Answers to part (b) (ii) suffered from the same lack of general understanding as the previous part. It was often realised that the ring would move to a higher position, or be expelled upwards from the rod. Reasons were less well presented. Reference to a larger induced current (or emf) in the ring was considered a prerequisite for an acceptable explanation.

E5.Use of the relation T ε r3/2 provided a swift method to the required answer in part (a)(i), but the mathematics involved appeared to be beyond the skills of many of the candidates. Answers which derived an expression for r 3 from first principles, or which quoted r = (GMT2/4π2)1/3 directly, were equally acceptable. The major error in most answers was a failure to subtract the radius of the Earth from the value of r when finding the height of the satellite. On this occasion the number of significant figures required in the final answer was three, because all of the data had been provided to at least 3SF. Following the expectation of recent papers, a large proportion of answers only included 2SF.

Candidates had been concentrating on the geosynchronous satellite in part (a)(i) and a proportion of them failed to read the question sufficiently carefully in part (a)(ii), thereby

Page 22


failing to spot that it was about the polar orbiting satellite. Their responses therefore quoted T and r values for the geosynchronous orbit, not answering the question set. For those who used the T and r values given in the question, this part usually provided two straightforward marks using F = mω2r, or F = mv2 /r, or F = GMm/r2.

The features of the orbits and the applications of the two types of satellite were quite well known. This gave most candidates a better opportunity to score a rather better mark on a communications question in part (b) than has often been the case in previous Unit 4 examinations. Nevertheless, only a few answers giving a comprehensive and coherent treatment, expected for 5 or 6 marks, were produced. The majority gave some relevant (and sometimes unrelated) facts and many were written sufficiently well to merit an intermediate level mark. In the case of the polar orbit, only a minority of answers made any reference to the fact that the Earth rotates under the orbiting satellite, and that it is this feature which allows the satellite to provide complete coverage of the Earth’s surface. Other features of the polar satellite that were often overlooked were that orbits with different radii are possible, that data can be collected from inaccessible regions, and that contact is intermittent. Features of the geosynchronous orbit not often mentioned included the fact that the radius of the orbit is unique, that the direction of travel is west to east and that the signal strength required is higher than that for a low orbit. A few candidates stated that the orbital period of the geosynchronous satellite is one year.

Page 23

Documents

· Web viewrefers to the force that acts on a current-carrying conductor when it is in a magnetic field and that this force lifts the ring upwards (into an area where the magnetic