Chapter 15 - Worksheet Answers.pdf - AS-A2-Physics

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A magnet is a piece of magnetic material that has been magnetised. ... Magnets can be made by being placed in a solenoid, by being stroked by a magnet, or by  ...
Magnetism reminders Question 10W: Warm-up Exercise 1. A magnet is a piece of magnetic material that has been magnetised. Unmagnetised material can be attracted by a magnet. Only a magnet will be repelled by a magnet. 2. Magnets can be made by being placed in a solenoid, by being stroked by a magnet, or by hitting when aligned with Earth’s field. The point is that the random arrangement of magnetic moments and domains needs to be aligned. The magnetised steel can be demagnetised by repeated hitting, by heating or by slowly pulling it out of a coil carrying an alternating current. 3. Either sprinkle iron filings in a plane around the magnet and tap the surface gently, or use one or more plotting compasses to investigate the field. 4.

5.

6. The like poles repel, the unlike poles attract. There will be a distance over which the force acts. 7. The field lines are more closely spaced. 8. The meter needle deflects in one direction then returns to zero. 9. The meter needle deflects more. 10. The meter needle remains at zero. 11. The meter needle deflects in the opposite direction.

Magnetic flux Question 20W: Warm-up Exercise 1. Point X near to one of the poles. Point Y well away from poles where lines are well spaced out. 2. P and Q identified as two points where arrowed lines are in opposite directions. 3. Increase the number of turns per unit length, increase the current. 4. C – this has the greatest number of flux lines.

Drawing magnetic circuits 1

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Question 30S: Short Answer Solutions 1.

N

S

B

A

D

C

2.

N

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S

Sketching flux patterns Question 40S: Short Answer 1. The flux from a long magnet looks like this:

N

S

2. The flux from a flat thin magnet looks like this:

S

N

3. The flux from a horseshoe magnet looks like this:

4. The flux pattern is symmetrical about the line dividing the poles and about the line joining their centres:

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S

N

5. The flux pattern is symmetrical about the line dividing the coils and about the line through their centres:

6. The poles are as shown: –

7. The poles are as shown:

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+

S

N

8. Because the coil is long, the current turns at places along its length surround each place in much the same way. A fly at each point would just see coils around it in all directions. So the field is the same along all the central part of the length (not near the ends). Thus it must be straight and uniform. 9. The flux just runs in circles round inside the coil:

10. The flux crosses the air gaps, goes through the rotor, and joins up by going round the casing of the stator:

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flux

flux

Magnet down a tube Question 50S: Short Answer 1. As the magnet falls, the flux of the magnet cuts the copper tube, inducing current in the tube. The current direction is such as to produce a force on the magnet to minimise the change taking place. Since the change inducing the current is the motion of the magnet down the tube, the effect of the force is to reduce the relative motion between the tube and the magnet, so the magnet moves more slowly than it would do in free air. 2. Light gates, datalogger, computer and software. Useful discussion could include some estimate of the likely values of parameters and discussion of appropriate sensitivity of the apparatus used. 3. Same time as in free air, no induced currents in glass. 4. The magnet will fall in the same time as in free air, because the slot prevents eddy currents around the tube. 5. There will be eddy currents around each of the short tubes so there will be a similar delay to that when using a single copper tube. The time delay is slightly shorter due to the presence of plastic rings, which introduce no delay at all.

Changes in flux linkage 6

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Question 60S: Short Answer 1. A, C and D all produce a change in flux and can therefore all produce a deflection on the meter. Motion in direction B produces no change in flux, so there is no induced emf. 2. Action

Outcome

Switch on current in coil 1

Needle flicks (say) left, flux linkage increases

Switch off current in coil 1

Needle flicks right, flux linkage decreases

With current on in coil 1, move towards coil 2

Needle flicks left, flux linkage increases

With current on in coil 1, move away from coil 2

Needle flicks right, flux linkage decreases

With current on in coil 1, increase current in rheostat

Needle flicks (say) left, flux linkage increases

With current on in coil 1, decrease current in rheostat

Needle flicks right, flux linkage decreases

Electromagnetism Question 70S: Short Answer 1. A diagram will be essential and a complete answer needs to include: a circuit to provide high current through the wire and the means of controlling it; means of detecting magnetic field, e.g. filings and card or compass needle etc; evidence of an effect, e.g. filings line up, compass alignment; must use directional detection device for the final part – reverse current and observe effect, i.e. the compass reverses direction. 2. A complete answer will include: a diagram showing lines through the solenoid and uniform in the centre; no uniform region for the flat coil and lines approach the circular round wire; both diagrams to show the correct direction of lines in relation to the current. 3. A complete answer will include: current causes a magnetic field in the solenoid; magnetisation of iron plunger; opposite polarity to solenoid causes attraction. 4. Movement of iron causes the plate to close the starter motor circuit. 5. To pull the plate off contact when the solenoid current stops. 6. Factors affecting field strength are current I and spacing of coils, N coils in length L: N B  I, B  L 7. Calculation using I = 30 A, N = 100, L = 0.50 m:

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B

 o NI 4  10 7 N A –2  100  10 A   2.5 mT 0.50 m L

B

 o NI 4  10 7 N A –2  750  1000 A   0.38 T 2.5 m L

8.

9.



P  I 2 R  10 3 A



2

 (3  10 2 )  30 kW.

10. To remove heat generated by the current due to the resistance of coils. Resistance must be kept low.

Rates of change Question 80S: Short Answer Solutions 1. Flux cuts the coil as the magnet approaches. Flux linkage is constantly changing (due to the non-uniform field of the magnet) so an emf is induced according to Faraday’s law. 2. Increasing speed and stronger field both contribute to the increased rate of change of magnetic flux. 3. The magnet is travelling faster when it leaves the coil so that rate of change of flux is greater. 4. The flux change is in the opposite direction when the magnet leaves the coil compared with when it is entering. 5. The area under the curve represents the total flux change, which is the same leaving as entering. 6. A= r 2, B = 0.40 T: dB V  A dt



0.40 T    1.0  10 2 m  0.3 s



2

 0.42 mV.

7. The ring must be moved parallel to the flux lines. 8. The trace height will be doubled because the flux linkage will be doubled and therefore so will the rate of change of flux. 9. The trace height is smaller – the flux is weaker towards the ends. 10. The trace height will be doubled because the rate of change of flux will be doubled and also the horizontal spacing will be halved because the frequency is doubled, i.e. there are twice as many cycles on screen for the same timebase sweep.

Bugging 8

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Question 90S: Short Answer Solutions 1. When current flows in the phone line, there are magnetic field loops around the current. If the bugging line is placed nearby, this magnetic field also loops around the bugging line. As the current in the phone line modulates with the frequency of the verbal orders, the magnetic field extends to the closed loop of the bugging line and induces a similarly modulated current in it. This induced current is transduced into sound by the headphones. One can think of the telephone wire as the primary coil of a transformer, and the bugging line as the secondary coil of the same transformer. The bug would work even better if, like a transformer, the wires could be wrapped around an iron core.

phone line

Transformers Question 100S: Short Answer 1. The needle kicks in one direction (say to the right) because as the current rises in circuit 1, there is a change in magnetic flux. Some of this flux is linked with circuit 2, inducing a current in this complete circuit. 2. No deflection as the flux is not changing when the current is constant. 3. The needle kicks in the opposite direction to question 1 as there is a flux change in the opposite direction, i.e. flux is collapsing instead of growing. 9

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4. The meter shows a steady deflection in a.c. mode. 5. The flux linkage between the coils is constantly changing due to the constantly changing current. 6. Circuit B is more economical. In circuit B, current flows through the transformer only when the bell push is pressed. In circuit A, current flows through the primary circuit at all times, which wastes some energy. 7. A large flux is linked due to the presence of the iron core and a very large number of turns on the secondary winding. The contact is broken quickly, causing a large change of flux in a short time. According to Faraday’s law, this produces a very high emf for a short time.

The circuit breaker Question 110S: Short Answer Solutions 1.

2.

A supply B

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3. There will be zero flux in the ring as the two coils produce flux in opposite directions. 4. Meter will record zero volts. As there is no net flux in the ring at any time, there can be no induced emf. 5. Meter will record zero volts. As there is no net flux in the ring at any time, there can be no induced emf. 6. If there is a difference in the current between the two coils, there will be a net flux in the iron ring. Because the current is alternating, this net flux will be constantly changing and so there will be an emf induced in coil C. This will cause a current to flow in the iron-cored coil which will attract the pivoted iron rod, thus breaking the circuit and cutting off the current.

Eddy currents and Lenz’s law Question 120S: Short Answer 1. The current is switched on and produces a magnetic flux in the solenoid. Flux lines near the end of the solenoid cut the aluminium ring as the field grows. Change in flux linkage induces an emf in the ring and a current flows. The direction of the induced current is such as to oppose the change inducing it (Lenz’s law) and the force on the ring due to current in the field acts to drive the ring out of the field, i.e. upwards. When the current, and therefore the flux, reaches a steady value, the induced current falls to zero and the ring falls. 2. The current is switched off and the flux collapses; induced current and the force are in the opposite direction. 3. Presence of iron increases the permeance and more flux is generated for the same current. Assuming that the flux changes at the same rate as without the iron, a greater emf will be produced. (Note: there is a potential problem here with self inductance – if it is very large, the current will collapse slowly.) 4. Alternating current generates an alternating force. Although the current direction changes, the force is always such as to push the ring out of the field – an effect which is counteracted by the weight of the ring. The ring hovers when these forces are (more or less) balanced. 5. Points for discussion include the effect of dimensions on resistance and therefore the size of the induced current; also different materials will have different resistivities. Changing dimensions will also affect the weight of the ring and therefore the balance between gravitational and electromagnetic forces. 6. The solid vane swings in the field, and the conductor cutting the flux lines induces eddy currents in the plane of the vane. Currents flow in such a direction as to minimise change – forces act so as to slow the motion of the vane, i.e. always act in the opposite direction to the motion. In the case of the slotted vane the presence of slots limits eddy currents and therefore the magnetic braking forces. 7. The bar which heats up is completely solid – the heating effect is due to induced currents in the core produced by the flux changing at 50 Hz. The bar remaining cool has been laminated, i.e. made up of thin strips of iron insulated from each other by a non-conducting coating.

Flux or flux linkage? 11

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Question 150S: Short Answer 1. B – most lines. 2. Flux linkage depends on both coil and current. Coil A has largest product of turns and flux lines.

Graphs of changing flux and emf Question 170S: Short Answer 1. The graph looks like this:

time

time

2. The graph looks like this: :

time

3. The graph looks like this:

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time

time

time

4. The graph looks like this:

time

time

5. The graph looks like this:

time

6. The graph looks like this:

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time

time

time

7. The graph looks like this:

time

time

8. The graph looks like this:

time

time

Alternating current generators Question 180S: Short Answer 1. The flux is greatest when the magnet is lined up with the pole pieces. The flux is least when the magnet is at right angles to the pole pieces. 2. Approximately 1/4 rotation, so 1/4  1/10 s = 1/40 s. 14

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3. Emf = flux change  turns / time. Flux change = emf  time / turns: 2 V  1 / 40 s flux change =  10 4 Wb. 500 turns 4. Area of coils = 20 mm  20 mm = 400 mm2 = 4  10–4 m2. B-field = flux / area:

B

104 Wb 4  104 m2

 0.25 T.

5. Flux = flux density  area. Area = 20 mm  50 mm = 1000 mm2 = 10–3 m2:   0.5 T  10 –3 m 2  5  10 –4 Wb 6. Centre the coil between the pole pieces, with its plane parallel to the pole pieces: area of coil = r 2  300 mm 2 = 3  10 4 m 2

flux   0.5 T  (3  10 4 m 2 )  1.5  10 4 Wb. 7. emf  time = change of flux. The flux reverses so the change of flux is 3  10–4 Wb: 3  10 4 Wb emf   3 mV / turn. 0.1 s 8. With 500 turns emf = 3 mV per turn  500 turns = 1.5 V. 9. The emf alternates sinusoidally. 10. emf = rate of change of flux linkage = 104 Wb turns per second. 11. Maximum flux = maximum flux linkage / 2 f: 10 4 Wb turns s 1 N max   32 Wb turns approx. 2  50 s 1 12. Flux = flux linkage / turns: 30 Wb turns   0.1 Wb. 300 turns 13. Area =

0.1 Wb  1 m2 . 0.1 T

Sketching field patterns and predicting forces Question 230S: Short Answer 1. The magnets will be pulled together (attract). This result depends on the poles being opposite, so that their fields are in the same direction in between them, going from one to the other. 2. The fields between the offset poles and poles at right angles are as shown below. Forces on the offset poles pull them together and into alignment. Forces on the poles at right angles pull them together and also tend to align them.

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S

N

N

3. The field from the two coils together looks like this:

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4. The complete field is got by copying reflected versions of the top right quadrant:

5. The magnetic field forms circles around the wire, as shown below. This also follows from symmetry.

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current

flux

6. The field encircles the two wires, as shown below. If flux lines shorten, the wires will be pulled together. So like currents attract, not repel.

7. As shown below, at A the two fields are in the same direction and add. At B and C the fields are at right angles and the resultant is tilted.

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A N

S

B C

8. The field shape comes from joining up the resultant fields at adjacent places, as shown below. At X the resultant is zero because the field directions are opposite. At some point they will be equal and opposite.

S

N

=0

9. The force goes downwards in the diagram, at right angles to the direction of the current and to the field of the magnet. wire (current down into screen (paper)) field

S

N force on current

10. As shown below, the forces on the two sides of the coil are opposite. The coil is given a twist (anticlockwise). The field is as shown. The idea that lines of flux shorten and straighten predicts the same twisting forces. current down force S

N force

current up

11. The poles of the coil faces are as shown. They predict the same direction of twisting. All these explanations – force on a wire, shortening and straightening flux, and forces on poles – are equivalent. S S

N N

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Forces and currents Question 240S: Short Answer 1. The wires attract. 2. Reverse the direction of the current in one of the wires. 3. Current flows in downwards and upwards in the two halves of the wire so the wires repel each other and move apart. 4. The loop would take up the shape of a circle. 5. For diametrically opposite segments of foil tape, current flows in opposite directions and so each of these segments repel each other. All these add up to an outwards force on the loop. 6. Flux lines are perpendicular to the plane containing the foil loop, and form closed paths round the foil. 7. The magnetic field should be drawn as circles round the wire with clockwise arrows; lines from N to S, i.e. downwards on diagram, straight between poles tending to curve further away. 8. With the wire between the poles of the magnet the combined field is stronger on the right of the wire and weaker on the left. The diagram should show lines closer together on the right and more spaced on the left, showing the catapult field exerting a force to left on the wire (confirmed by the left-hand rule or vector product rule). 9. (It is now Illegal to use mercury in such an arrangement open to the atmosphere because of dangers from mercury vapour.) Current flows down the point of the star in contact with the mercury trough and is perpendicular to the magnetic field. A force perpendicular to the field and current causes the star to rotate. As one point leaves the mercury, the next one enters and receives an impulse – so rotation continues.

Thinking about the design of a simple d.c. motor Question 250S: Short Answer 1. The answer should give some simple arrangement of a conductor carrying a current arranged to be flowing perpendicular to the field of a bar magnet, e.g. a loop of wire free to swing in a field. When the current is switched on, the wire kicks out of the field. 2. For 20 cm length 20 cm  0.025 N  0.63 N 8 cm 3. If each wire carries 3.0 A this is the same as effective current of 9.0 A, so force F = 3  0.63 N = 1.9 N. 4. There are no forces on the short sides of the coil. Forces on the other two sides are in opposite directions because the current flow in each side is in the opposite direction. More turns of wire on the coil give more force for the same current, i.e. it is the ampere-turns which is important. 20

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Emf in an airliner Question 260S: Short Answer 1. The wings cut the flux lines of the Earth’s magnetic field, inducing an emf between the wing tips. An alternative answer is that the charges in the wings are moving through the Earth’s magnetic field and experience a force which redistributes the charge in the wings. 2. Vertically downward arrow to show the vertical component. 3. Vertical component = (1.7  10–4 T) cos 20 = 1.6  10–4 T. 4. F  qvB  (1.6  10 19 C)  270 m s 1  (1.6  10 21 T)  6.9  10 21 N.

5. emf  vLB  270 m s 1  60 m  (1.6  10 4 T)  2.6 V.

6. At the equator the aircraft is flying parallel to the flux and cuts no (or very little) flux.

The Birmingham maglev Question 270C: Comprehension Solutions 1. Next pole to right is N, then S… etc.

2. Show flux lines on diagram looping from N to S etc. steel suspension rail

S

N

suspension electromagnet

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S

3. The train would drop on to the tracks. 4. Mechanical strength. 5. Aluminium has a lower resistivity than steel. 6. It must be a magnetic material to provide a path for the return flux, i.e. complete the magnetic circuit, and must also be magnetically ‘soft’. 7. The rails have angled joins so that they can slide over one another and remain more or less in contact to preserve the unbroken magnetic circuit and paths for eddy currents. 8. To maintain the magnetic force constant and to maintain stability. 9. Too small an air gap will increase the magnetic force and may result in the magnets sticking to the rail. 10. size of air gap

attractive force

increasing either reduces the other decreasing either increases the other

11. size of air gap

attractive force

current in coils

control system: if air gap changes, current and force are changed in the same sense.

Explaining with induction Question 130X: Explanation–Exposition A good answer will put the different stages in an argument in a helpful order, combining them to provide an argument intelligible to the student's peers. 22

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A bicycle speedometer Question 140X: Explanation–Exposition 1. As the coil rotates, the vertical sides of the coil cut the flux of the magnetic field, inducing an emf; or as the coil rotates, the flux linked with the area of the coil changes due to the change in angle; or forces on the charges in the coil moving through the magnetic field cause a redistribution of the charge resulting in a p.d. between X and Y. 2. In the parallel position, the rate of change of flux linkage is greatest as the rate of change of angle is greatest – or, in this position, the charges in the coil are moving perpendicular to the field and experience maximum force. 3. Each half cycle, any one side of the coil is moving first ‘up’ the field and then ‘down’ the field, so the direction cutting flux correspondingly changes, giving an emf in the opposite direction; or each half cycle, the direction of the force on each charge carrier changes sign because the motion through the field changes direction.

The geophone Question 160X: Explanation–Exposition 1. A good answer would include the relative motion between coil and magnet, and how this comes about. From this start discuss changes in flux linked, then emf induced; diagrams should make the links explicit. The planes of movement need to be made explicit. 2. The sensitivity of the geophone is the ratio of the emf to the velocity of the coil. A good answer would relate sensitivity to particular changes in the design, explaining how these increase the sensitivity. You might choose to alter the number of turns for example, so increasing the emf per mm s–1 of movement. 3. The resolution is the ratio of the change in output to the change in input. A good answer would relate resolution to particular changes in the design, explaining how these increase the resolution. You might choose to increase magnetic flux for example, so that small changes in movement, perhaps only over a small range, produced a large change in emf. 4. Increasing the spring constant or decreasing the mass of the coil and former will increase its natural (or resonant) frequency and allow the device to achieve a higher amplitude at higher driven frequencies.

Electronic ignition Question 190X: Explanation–Exposition 1. As the reluctor spins, so the permeance of the magnetic circuit alters. The magnet is therefore able to drive different amounts of flux through the coil. As the flux increases so a pulse of emf is generated in one sense; as the flux decreases, so a corresponding pulse is generated in the 23

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reverse sense. 2. Pulses of emf are produced in both senses. Only pulses in one direction are required – those that allow the transistor to switch on and permit current through the primary coil of the ignition coil. 3. A large current pulse passes through the primary coil of the pair that make up the ignition coil. This generates a pulse of magnetic flux, linked to the larger number of turns on the secondary coil. Here a large emf is induced, as in the standard transformer action.

The induction motor Question 200X: Explanation–Exposition Solutions 1. The flux has the same shape, but rotated through an angle.

flux rotated

2. If the rotor turns, the relative rotation of flux and rotor is reduced. So by Lenz’s law the eddy currents are in such a direction as to give forces which turn the rotor in the same direction as the rotating flux. The relative rotation becomes less. The conductor is ‘dragged’ along with the moving flux. 3. There are no eddy currents. There is no relative rotation, no eddy currents and no force to keep the rotor turning against friction or a load. 4. This is hard. At the instant shown there is no flux at right angles to the rotating flux. But as the flux rotates this is the direction in which it will next grow. The increasing component of the rotating flux is in this direction. 5. Induced emfs, and so induced eddy currents, circulate around the rate of change of flux producing them. In the same way, a transformer secondary is wound around the changing flux. 6. The poles are at the faces of the circulating eddy currents. The rotor will turn anticlockwise, as the N and S pole pairs are pulled more into alignment. 7. If the eddy currents were in the other direction, the N and S poles on the rotor would interchange, and forces between the poles would turn the rotor clockwise. This would increase the relative rate of rotation of flux and rotor. Lenz’s law says that the forces must decrease the relative rate of 24

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rotation, which is causing the eddy currents. So we choose the poles which give forces in agreement with Lenz’s law. 8. A steel rotor would improve the magnetic circuit and increase the flux. But a steel rotor conducts less well than a copper rotor, so this would reduce the eddy currents. 9. The squirrel cage rotor has laminated iron to increase the flux by improving the magnetic circuit. The eddy currents are carried by copper or aluminium bars set into the laminated iron. Their high conductance ensures large eddy currents.

A variable-speed linkage Question 220X: Explanation–Exposition 1.

rotor

input

output

to constant speed a.c. motor

magnetic field coil

2. There is relative motion between the armature and the rotor. Flux from the armature is linked with the rotor and induces eddy currents in it. 3. More current produces a greater flux. In turn this produces a greater change in flux, so a larger emf, so larger eddy currents. 4. Larger eddy currents lead to stronger induced poles, so larger alignment forces, due to the interaction between this induced flux and the flux produced by the field coils. 5. The eddy currents in the armature produce a magnetic field in a direction to be attracted to the magnetic field in the rotor, so the rotor is dragged round by the armature.

ICT driven by precision motors Question 280X: Explanation–Exposition A good answer will show how the fundamental principles apply, whatever the motor chosen. All of the 25

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major arguments that you might be expected to use are presented diagrammatically on the CD-ROM and in the Advancing Physics A2 student’s book.

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