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GCSE · Physics

P7 Magnetism and Electromagnetism

AQA 8461/8462/8463 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A student holds two bar magnets close to each other and observes what happens.

(a)Which pair of poles will attract each other when brought close together?(1)

A) N and N
B) S and S
C) N and S
D) Poles do not attract, they only repel

(b)State where the magnetic field of a bar magnet is strongest.(1)

(Total for Question 1 is 2 marks)

Magnets can be described as permanent or induced.

(a)Define what is meant by a permanent magnet.(1)

(b)Name one magnetic material, other than iron, that could be used to make a magnet.(1)

(c)A steel paperclip becomes magnetised when it is held close to a strong permanent magnet, but loses most of its magnetism once the magnet is taken away. State the term used to describe this type of magnetism.(1)

(Total for Question 2 is 3 marks)

A student wants to plot the magnetic field pattern around a bar magnet using a plotting compass and a pencil.

(a)Describe how the student could use the plotting compass to plot a magnetic field line around the bar magnet.(3)

(b)State the direction of a magnetic field line at any point.(1)

(c)The field lines drawn are closer together near the poles of the magnet than they are further away. State what this shows about the magnetic field near the poles.(1)

(Total for Question 3 is 5 marks)

A long, straight wire carries an electric current.

(a)State the shape of the magnetic field pattern produced around a long, straight, current-carrying wire.(1)

(b)State two factors that affect the strength of the magnetic field at a point near the wire.(2)

(c)Higher Tier only. State the rule that can be used to work out the direction of the magnetic field around a straight current-carrying wire.(1)

(Total for Question 4 is 4 marks)

A solenoid is a coil of wire that produces a magnetic field when a current flows through it.

(a)Describe the magnetic field pattern produced by a solenoid carrying a current.(2)

(b)State two ways of increasing the strength of the magnetic field produced by a solenoid.(2)

(c)An electromagnet, rather than a permanent magnet, is used in a scrapyard crane to lift and move scrap metal. Explain why an electromagnet is more useful than a permanent magnet for this purpose.(2)

(Total for Question 5 is 6 marks)

The coil of an electromagnet has a resistance of 3.2 Ω. A current of 2.5 A flows through the coil. Use the Physics Equations Sheet. Use the equation: power = current² x resistance (P = I² R)

(a)Calculate the power dissipated as heat in the coil.(2)

(b)Suggest one reason why this heating effect could be a problem for an electromagnet that is used continuously.(1)

(Total for Question 6 is 3 marks)

A straight wire carrying a current is placed between the poles of a strong magnet, at right angles to the magnetic field.

(a)State what happens to the current-carrying wire.(1)

(b)Name the rule that can be used to predict the direction of this force.(1)

(c)For Fleming's left-hand rule, with the thumb and first two fingers of the left hand held at right angles to each other, state what is represented by: (i) the First finger, (ii) the SeCond finger, (iii) the thuMb.(3)

(Total for Question 7 is 5 marks)

A straight wire runs vertically, from the bottom to the top of the page, and carries a current upwards. The wire lies between the poles of a magnet, positioned so that the magnetic field points horizontally across the page, from left to right.

(a)Using Fleming's left-hand rule, state the direction of the force on the wire (choose from: out of the page, into the page).(1)

(b)The current in the wire is reversed, so that it now flows downwards. State and explain the effect this has on the force on the wire.(2)

(Total for Question 8 is 3 marks)

A straight wire of length 0.15 m carries a current of 4.0 A. It is held at right angles to a magnetic field of flux density 0.50 T. Use the Physics Equations Sheet. Use the equation: force on a conductor (at right angles to a magnetic field) carrying a current = magnetic flux density x current x length (F = B I L)

(a)Calculate the force on the wire.(2)

(b)The current in the wire is increased to 8.0 A, with the flux density and length unchanged. State the effect this has on the force, and calculate the new force on the wire.(2)

(Total for Question 9 is 4 marks)

A simple direct current (d.c.) electric motor consists of a rectangular coil of wire that rotates between the poles of a permanent magnet. The ends of the coil are connected to a split-ring commutator.

(a)Describe the role of the split-ring commutator in the motor.(2)

(b)Explain, in terms of forces, why the coil rotates continuously.(3)

(Total for Question 10 is 5 marks)

A moving-coil loudspeaker converts a varying (alternating) electrical current into sound. Explain, using ideas about the motor effect, how the loudspeaker produces sound from an electrical signal.

(Total for Question 11 is 6 marks)

Moving a conductor relative to a magnetic field, or changing the magnetic field around a conductor, can generate electricity. This is called the generator effect.

(a)State what is meant by the 'generator effect'.(2)

(b)State three factors that affect the size of the potential difference induced.(3)

(Total for Question 12 is 5 marks)

A student investigates how the speed of a bar magnet, as it is moved through a coil of wire, affects the potential difference induced across the coil. The coil is connected to a data logger, which measures potential difference.

(a)Identify the independent variable and the dependent variable in this investigation.(2)

(b)State one variable that the student should control (keep the same) to make this a fair test.(1)

(c)Describe the result you would expect to see if the student increases the speed of the magnet.(1)

(Total for Question 13 is 4 marks)

An alternating current (a.c.) generator produces a current that continuously reverses direction as its coil rotates in a magnetic field.

(a)Explain why the direction of the induced current reverses as the coil rotates.(2)

(Total for Question 14 is 2 marks)

A transformer is used to change the size of an alternating potential difference.

(a)Describe the basic structure of a transformer.(2)

(b)Explain why a transformer only works with alternating current (a.c.) in the primary coil, not direct current (d.c.).(3)

(Total for Question 15 is 5 marks)

A step-down transformer converts UK mains electricity (potential difference 230 V) to a potential difference of 12 V, for use with a low-voltage lamp. The primary coil has 4600 turns. Use the Physics Equations Sheet. Use the equation: potential difference across primary coil / potential difference across secondary coil = number of turns on primary coil / number of turns on secondary coil (Vp / Vs = Np / Ns)

(a)Calculate the number of turns needed on the secondary coil.(3)

(b)State whether this is a step-up or a step-down transformer, and explain how you can tell from the number of turns on each coil.(2)

(Total for Question 16 is 5 marks)

For the transformer in Question 16, assume the transformer is 100% efficient, so power input equals power output. Use the Physics Equations Sheet. Use the equation: potential difference across primary coil x current in primary coil = potential difference across secondary coil x current in secondary coil (Vp Ip = Vs Is)

(a)The current in the secondary coil is 2.5 A. Calculate the current in the primary coil.(3)

(b)Explain why the current in the primary coil is smaller than the current in the secondary coil.(2)

(Total for Question 17 is 5 marks)

Electrical power is transmitted around the country through the National Grid using overhead cables. Each cable has a resistance of 5.0 Ω. Use the Physics Equations Sheet. Use the equation: power = current² x resistance (P = I² R)

(a)Calculate the power dissipated as heat in the cable when it carries a current of 10 A.(2)

(b)The same amount of power could instead be delivered to consumers using a lower transmission voltage, which requires a current of 100 A in the same cable. Calculate the power dissipated as heat in the cable in this case.(2)

(c)Using your answers to parts (a) and (b), explain why the National Grid transmits electrical power at a high voltage and a low current, rather than at a low voltage and a high current.(3)

(Total for Question 18 is 7 marks)

In the investigation described in Question 13, the student was concerned that her results might not be very reliable, because it was difficult to move the magnet through the coil at a controlled, repeatable speed by hand, and the potential difference changed very quickly as the magnet passed through.

(a)Suggest one improvement to the method that would make the speed of the magnet more controlled and repeatable.(1)

(b)Suggest why using a data logger, rather than reading a voltmeter by eye, improves the reliability of the potential difference readings.(2)

(c)State one hazard in this investigation and a precaution that could be taken to reduce the risk.(1)

(Total for Question 19 is 4 marks)

Mark scheme · P7 Magnetism and Electromagnetism

Question 1

(a) B1 C (unlike poles, N and S, attract)
(a) Answer: C
(b) B1 at the poles (the ends) of the magnet, oe
(b) Answer: At the poles (the ends) of the magnet.

Question 2

(a) B1 a magnet that produces its own magnetic field and does not need this to be induced by another magnet or an electric current, oe
(a) Answer: A magnet that produces its own magnetic field, which is not induced by another magnet or a current.
(b) B1 steel / cobalt / nickel (any one)
(b) Answer: Steel (or cobalt, or nickel).
(c) B1 induced magnetism / an induced magnet, oe
(c) Answer: Induced magnetism - the paperclip becomes an induced magnet.

Question 3

(a) B1 place the plotting compass close to one pole of the magnet and mark the position of each end of the compass needle with a pencil dot, oe
(a) B1 move the compass so that its previous end sits on the last dot, mark the new needle position, and repeat this several times moving away from the magnet, oe
(a) B1 join the dots to draw a smooth field line and add an arrow to show its direction; repeat starting from different points around the magnet to build up the full field pattern, oe
(a) Answer: Place the compass near a pole and mark both ends of the needle; move the compass so it starts where the last dot was and mark the new needle position; repeat this to build a chain of dots, then join them into a smooth line with an arrow showing direction, and repeat from several starting points.
(b) B1 the direction that the north pole of a compass placed at that point would point, oe (from north to south outside the magnet)
(b) Answer: The direction a compass needle's north pole would point if placed at that point (field lines go from north to south outside the magnet).
(c) B1 the magnetic field is stronger near the poles; field strength decreases with distance from the magnet, oe
(c) Answer: The magnetic field is stronger where the field lines are closer together, i.e. near the poles.

Question 4

(a) B1 a pattern of concentric circles, centred on the wire, oe
(a) Answer: A pattern of concentric circles centred on the wire.
(b) B1 increasing the size of the current increases the strength of the field, oe
(b) B1 increasing the distance from the wire decreases the strength of the field, oe
(b) Answer: The size of the current (larger current, stronger field) and the distance from the wire (greater distance, weaker field).
(c) B1 the right-hand grip rule: the thumb points in the direction of the (conventional) current and the fingers curl in the direction of the field, oe
(c) Answer: The right-hand grip rule.

Question 5

(a) B1 the field pattern is similar in shape to that of a bar magnet, with a north pole at one end and a south pole at the other, oe
(a) B1 the field is strong and (approximately) uniform inside the solenoid, oe
(a) Answer: Similar to the field of a bar magnet (a north pole at one end, a south pole at the other), and strong and uniform inside the coil.
(b) B1 increase the current flowing through the solenoid, oe
(b) B1 increase the number of turns of wire on the solenoid, or add an iron core, oe
(b) Answer: Increase the current, and increase the number of turns of wire (or add an iron core).
(c) B1 the magnetic field of an electromagnet can be switched on and off by turning the current on and off, so the crane can pick up metal (field on) and release it (field off) exactly when needed, oe
(c) B1 the strength of an electromagnet can also be changed by changing the current, giving more control than a fixed-strength permanent magnet, oe
(c) Answer: Because the electromagnet's field can be switched on and off (and its strength changed) by controlling the current, allowing scrap metal to be picked up and then released on demand.

Question 6

(a) M1 P = 2.5² x 3.2, oe
(a) A1 20 W, cao
(a) Answer: 20 W
(b) B1 the coil could overheat, which could damage the wire's insulation, or energy is wasted as heat rather than doing useful work, reducing efficiency, oe
(b) Answer: The coil may overheat and its insulation could be damaged, and energy is wasted as heat instead of doing useful work.

Question 7

(a) B1 it experiences a force (this is called the motor effect), oe
(a) Answer: It experiences a force (the motor effect).
(b) B1 Fleming's left-hand rule
(b) Answer: Fleming's left-hand rule.
(c) B1 (i) the (magnetic) Field
(c) B1 (ii) the (conventional) Current
(c) B1 (iii) the direction of Motion / the force (thrust)
(c) Answer: First finger = Field, seCond finger = Current, thuMb = Motion/force.

Question 8

(a) B1 out of the page
(a) Answer: Out of the page.
(b) B1 the force now acts into the page (it reverses direction), ft from (a) if (a) is consistent
(b) B1 because reversing the current reverses one of the two quantities used in Fleming's left-hand rule, which reverses the resultant direction of the force, oe
(b) Answer: The force reverses to act into the page, because reversing the current reverses the direction given by Fleming's left-hand rule.

Question 9

(a) M1 F = 0.50 x 4.0 x 0.15, oe
(a) A1 0.3 N, cao
(a) Answer: 0.3 N
(b) B1 the force doubles, because force is directly proportional to current when B and L are unchanged, oe
(b) A1 0.6 N, ft from (a) if a consistent method is shown
(b) Answer: 0.6 N

Question 10

(a) B1 it reverses the direction of the current in the coil every half turn (every 180 degrees), oe
(a) B1 this reverses the direction of the force on each side of the coil at the right moment, so the coil keeps rotating in the same direction, oe
(a) Answer: It reverses the current in the coil every half turn, which reverses the force on each side at the right moment so the coil keeps turning the same way.
(b) B1 the coil carries a current and is in a magnetic field, so each side of the coil experiences a force (the motor effect), oe
(b) B1 the current flows in opposite directions along the two sides of the coil (up one side, down the other), oe
(b) B1 by Fleming's left-hand rule, the forces on the two sides act in opposite directions, forming a turning effect (a couple) that rotates the coil, oe
(b) Answer: Each side of the current-carrying coil is in a magnetic field and experiences a force; because the current flows in opposite directions along the two sides, the forces are in opposite directions, creating a turning effect that rotates the coil.

Question 11

Level 1 (1-2): Simple, isolated statement(s) are made, for example that a current or a force is involved, or that the cone vibrates. There is little or no explanation of how sound is produced, and no reference to the motor effect.
Level 2 (3-4): The coil, magnet and motor effect are described, and it is stated that the coil (and cone) move as a result of the force produced. There is some explanation of how this leads to sound, but the link between the alternating current and the vibration of the cone is incomplete.
Level 3 (5-6): A complete, logically linked explanation is given: the alternating current in the coil, in the magnetic field of the permanent magnet, produces a force (motor effect) that alternates in direction, moving the coil and attached cone in and out; this makes the cone vibrate, creating sound waves in the air; the amplitude and frequency of the current are linked to the loudness and pitch of the sound produced.
Indicative content:
- The loudspeaker contains a coil of wire, attached to the speaker cone, placed within the magnetic field of a permanent magnet
- An alternating current from the electrical signal (e.g. from an amplifier) flows through the coil
- The current-carrying coil is in a magnetic field, so it experiences a force (the motor effect)
- As the current in the coil continuously changes direction (it is alternating), the direction of the force on the coil also continuously alternates
- This makes the coil, and the cone attached to it, move in and out repeatedly (vibrate)
- The vibrating cone makes the surrounding air particles vibrate, producing sound waves
- A larger current produces a larger force, so a bigger vibration (amplitude) and a louder sound
- A higher frequency alternating current produces faster vibrations of the cone, producing a higher-pitched sound

Question 12

(a) B1 a potential difference (voltage) is induced across a conductor, and a current flows if the circuit is complete, oe
(a) B1 this happens when there is relative movement between the conductor and a magnetic field, or when the magnetic field around the conductor changes, oe
(a) Answer: Moving a conductor relative to a magnetic field (or changing the field around it) induces a potential difference across the conductor, and a current if the circuit is complete.
(b) B1 the speed at which the conductor and magnetic field move relative to each other (or the speed of change of the field), oe
(b) B1 the strength of the magnetic field
(b) B1 the number of turns on the coil (if a coil is used), or the length of wire in the field
(b) Answer: The relative speed of movement, the strength of the magnetic field, and the number of turns on the coil (or length of wire in the field).

Question 13

(a) B1 independent variable: the speed of the magnet as it moves through the coil
(a) B1 dependent variable: the induced potential difference across the coil
(a) Answer: Independent variable: speed of the magnet. Dependent variable: induced potential difference.
(b) B1 e.g. the same magnet (same strength), the same coil (same number of turns), or the same distance/path travelled by the magnet through the coil (any one), oe
(b) Answer: For example, use the same magnet and the same coil (same number of turns) for every repeat.
(c) B1 the induced potential difference (reading on the data logger) increases as the speed of the magnet increases, oe
(c) Answer: The induced potential difference increases as the speed of the magnet increases.

Question 14

(a) B1 as the coil rotates, the direction each side of the coil moves relative to the magnetic field reverses every half turn, oe
(a) B1 by the generator effect, reversing the direction of relative movement reverses the direction of the induced potential difference/current, oe
(a) Answer: As the coil rotates, each side moves in a reversed direction relative to the field every half turn, which by the generator effect reverses the induced current's direction.

Question 15

(a) B1 a transformer has a primary coil and a secondary coil of insulated wire, oe
(a) B1 both coils are wound around a common (soft) iron core, oe
(a) Answer: A primary coil and a secondary coil of wire, both wound around a common soft iron core.
(b) B1 an alternating current in the primary coil produces a continuously changing magnetic field in the iron core, oe
(b) B1 this changing magnetic field induces an alternating potential difference in the secondary coil, by the generator effect, oe
(b) B1 a direct current would produce a constant (unchanging) magnetic field, so no potential difference would be induced in the secondary coil, oe
(b) Answer: Alternating current produces a continuously changing magnetic field in the core, which induces a p.d. in the secondary coil by the generator effect; direct current gives a constant field, so no p.d. would be induced.

Question 16

(a) M1 rearranging: Ns = Np x Vs / Vp, oe
(a) M1 Ns = 4600 x 12 / 230, oe
(a) A1 240 (turns), cao
(a) Answer: 240 turns
(b) B1 it is a step-down transformer
(b) B1 because the secondary coil has fewer turns than the primary coil (240 turns compared with 4600 turns), so the secondary potential difference is lower, oe (ft from (a))
(b) Answer: Step-down, because the secondary coil (240 turns) has fewer turns than the primary coil (4600 turns).

Question 17

(a) M1 rearranging: Ip = Vs Is / Vp, oe
(a) M1 Ip = (12 x 2.5) / 230, oe
(a) A1 0.13 A, awrt (2 s.f.)
(a) Answer: 0.13 A (2 s.f.)
(b) B1 since the transformer is 100% efficient, power in the primary coil equals power in the secondary coil (Vp Ip = Vs Is), oe
(b) B1 because Vp is much greater than Vs (230 V compared with 12 V), Ip must be smaller than Is for the two powers to be equal, oe
(b) Answer: Because power in equals power out (Vp Ip = Vs Is) and Vp is much larger than Vs, so Ip must be smaller than Is to keep the two powers equal.

Question 18

(a) M1 P = 10² x 5.0, oe
(a) A1 500 W, cao
(a) Answer: 500 W
(b) M1 P = 100² x 5.0, oe
(b) A1 50000 W (50 kW), cao
(b) Answer: 50000 W (50 kW)
(c) B1 increasing the current in a cable greatly increases the power wasted as heat, because power loss is proportional to current squared (P = I² R), oe
(c) B1 the calculations show that increasing the current by a factor of 10 (from 10 A to 100 A) increases the power lost as heat by a factor of 100 (from 500 W to 50000 W), ft from (a) and (b)
(c) B1 transmitting power at a high voltage keeps the current low for the same power delivered, which minimises the power wasted as heat in the cables, making transmission more efficient, oe
(c) Answer: Because power loss is proportional to current squared, a small increase in current causes a much larger increase in power wasted as heat (as shown by the calculations); transmitting at high voltage keeps the current, and so the heat loss, low.

Question 19

(a) B1 e.g. release the magnet from the same fixed height each time so it falls under gravity, rather than being pushed by hand, or use a mechanism (such as a ramp or motor) to give the magnet the same speed on each repeat, oe
(a) Answer: Release the magnet from the same fixed height each time (or use a ramp/motorised mechanism) so its speed is controlled and repeatable, rather than pushing it by hand.
(b) B1 the potential difference changes very quickly as the magnet passes through the coil, so a data logger can take and record readings much faster than a person can read a meter by eye, oe
(b) B1 this means the data logger is more likely to capture the true peak potential difference accurately, reducing human reaction-time/reading errors and improving reliability, oe
(b) Answer: A data logger takes readings much faster than a person can read a meter, so it can accurately capture the quickly-changing (peak) potential difference, reducing human error and improving reliability.
(c) B1 matched hazard and precaution, e.g. hazard: the falling magnet could cause injury or damage; precaution: catch the magnet in a soft container/tray, or keep the drop height low, oe
(c) Answer: Hazard: the falling magnet could cause injury or damage equipment. Precaution: catch it in a soft container/tray, or keep the drop height low.

Magnetism and Electromagnetism - Worksheets, Questions and Revision