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GCSE · Physics

P7a Magnetism and the Motor Effect

AQA 8464 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key ideas: magnetism and the motor effect

Original text written for Revision Library.

Magnets produce a magnetic field, a region around the magnet where a force acts on other magnets or on magnetic materials such as iron, steel, cobalt and nickel. Field lines run from the north pole to the south pole outside a magnet; the closer together the lines, the stronger the field. Like poles repel and unlike poles attract. A magnetically soft material, such as iron, is easily magnetised and demagnetised, so it is used for electromagnet cores; a magnetically hard material, such as steel, keeps its magnetism and is used to make permanent magnets. An electromagnet is a solenoid, a coil of wire that produces a magnetic field only when a current flows through it; its strength depends on the current, the number of turns, and whether it has an iron core. When a current-carrying wire is placed in a magnetic field, it can experience a force - this is called the motor effect. The direction of the force is given by Fleming's left-hand rule, and (Higher Tier) its size can be calculated using F = B * I * L. This effect is used in loudspeakers and in the d.c. electric motor, where a split-ring commutator keeps a current-carrying coil turning in one direction.

Iron and steel are both magnetic materials, but they behave differently when used to make magnets.

(a)State what is meant by a magnetically hard material.(1)

(b)State what is meant by a magnetically soft material.(1)

(c)State which of these materials, iron or steel, would be more suitable for making the core of an electromagnet, and give a reason for your answer.(1)

(Total for Question 1 is 3 marks)

Magnets produce a magnetic field around them.

(a)Define a magnetic field.(1)

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

(c)Two bar magnets are placed in a line with a small gap between them, north pole facing north pole. Describe what happens to the magnetic field lines in the gap between the two north poles, and explain what this shows about the force between the poles.(2)

(d)State two examples of materials that are not magnetic.(2)

(Total for Question 2 is 6 marks)

A plotting compass contains a small magnet that is free to rotate.

(a)State why a plotting compass needle points towards magnetic north when it is far away from any other magnets.(1)

(b)When the same plotting compass is placed close to a bar magnet, its needle instead points along the magnetic field lines of the bar magnet. Explain, in terms of induced magnetism, why the compass needle behaves in this way.(3)

(Total for Question 3 is 4 marks)

Required practical. A student is investigating the magnetic field pattern produced by a current-carrying wire. The wire passes vertically through a horizontal sheet of card and is connected in series with a variable resistor, an ammeter and a power supply, so the size of the current can be set and measured. The student uses a small plotting compass placed on the card at different points around the wire.

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

(b)The student reverses the direction of the current in the wire. State what happens to the pattern of magnetic field lines around the wire.(1)

(c)State one variable the student should keep the same to make sure any comparison of field patterns at different currents is a fair test.(1)

(d)Suggest one improvement the student could make to increase the accuracy of the plotted field pattern.(1)

(Total for Question 4 is 6 marks)

An electromagnet is a solenoid (a coil of wire) that produces a magnetic field only when a current flows through it.

(a)State three factors that affect the strength of the magnetic field produced by an electromagnet.(3)

(b)Explain, in terms of magnetic domains, why adding an iron core inside a solenoid increases the strength of the magnetic field it produces.(2)

(Total for Question 5 is 5 marks)

A solenoid carrying a current produces a magnetic field.

(a)Describe the shape of the magnetic field pattern outside a current-carrying solenoid, compared with the field pattern of a bar magnet.(2)

(b)Higher Tier only. State a method the student could use to work out which end of a current-carrying solenoid is its north pole, without using a compass.(2)

(Total for Question 6 is 4 marks)

Fleming's left-hand rule can be used to find the direction of the force on a current-carrying wire in a magnetic field.

(a)State what each of the following represents in Fleming's left-hand rule: (i) the First finger, (ii) the SeCond finger, (iii) the thuMb.(3)

(b)A straight wire is held so that a conventional current flows through it from left to right. The wire lies in a uniform magnetic field that points into the page. Use Fleming's left-hand rule to state the direction of the force on the wire.(2)

(c)State one everyday device that uses the motor effect.(1)

(Total for Question 7 is 6 marks)

Higher Tier only. A straight wire carries a current of 3.0 A. A 0.15 m length of the wire lies at right angles inside a uniform magnetic field of flux density 0.40 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 * current * length (F = B * I * L). Calculate the force on this length of wire.

(Total for Question 8 is 3 marks)

Higher Tier only. A wire carrying a current of 4.0 A lies at right angles to a magnetic field. A 0.30 m length of the wire experiences a force of 0.60 N. 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 * current * length (F = B * I * L). Calculate the magnetic flux density of the field, and state its unit.

(Total for Question 9 is 4 marks)

Higher Tier only. A straight current-carrying wire can be placed at different angles to a uniform magnetic field.

(a)State the direction of the current relative to the magnetic field lines for which the force on the wire is at its maximum.(1)

(b)State the direction of the current relative to the magnetic field lines for which the force on the wire is zero.(1)

(c)A wire is placed at an angle between these two orientations (neither parallel nor perpendicular to the field). Explain why the force on the wire in this case is smaller than the maximum possible force, but greater than zero.(2)

(Total for Question 10 is 4 marks)

Higher Tier only. A simple d.c. electric motor contains a rectangular coil of wire that rotates between the poles of a permanent magnet.

(a)Describe the function of the split-ring commutator in a d.c. motor.(3)

(b)At one instant, one side of the coil (side X) is moving upwards through the magnetic field, while the opposite side of the coil (side Y) is moving downwards through the field at the same point. Using Fleming's left-hand rule, explain how the forces on side X and side Y work together to keep the coil rotating in the same direction.(3)

(c)State one advantage of using a coil with many turns of wire, rather than a single loop, in a d.c. motor.(1)

(Total for Question 11 is 7 marks)

Higher Tier only. A loudspeaker converts a varying electrical current into sound. It contains a coil of wire attached to a paper cone; the coil sits inside the magnetic field of a permanent magnet. Explain, using ideas about the motor effect, how a loudspeaker produces sound from a varying current in the coil.

(Total for Question 12 is 6 marks)

Higher Tier only. A group of students investigates how the number of turns on a solenoid affects the magnetic flux density it produces at its centre, using a fixed current of 2.0 A and a Hall probe to measure the magnetic flux density, B. Their results: with 50 turns, B = 0.020 T; with 150 turns, B = 0.060 T.

(a)Use the data to show that B is directly proportional to the number of turns, N.(2)

(b)Predict the magnetic flux density produced when the solenoid has 250 turns, assuming the same pattern continues.(2)

(c)Suggest one way the students could improve the accuracy of their results.(1)

(Total for Question 13 is 5 marks)

Higher Tier only. A student wants to increase the speed of rotation of a simple d.c. motor.

(a)State one change the student could make to the motor to increase its speed of rotation, and explain why this change has this effect.(2)

(b)State a different change the student could make to increase the speed of rotation, and explain why this change has this effect.(2)

(Total for Question 14 is 4 marks)

Higher Tier only. A straight wire of length 0.20 m is connected in a circuit to a 12 V battery. The wire has a resistance of 8.0 Ω and lies at right angles to a magnetic field of flux density 0.50 T. Use the Physics Equations Sheet. Use the equations: potential difference = current * resistance (V = I * R); force on a conductor (at right angles to a magnetic field) carrying a current = magnetic flux density * current * length (F = B * I * L).

(a)Calculate the current in the wire.(2)

(b)Calculate the force on the wire due to the magnetic field. (You may use your answer to part (a).)(3)

(Total for Question 15 is 5 marks)

Higher Tier only. In a simple d.c. motor, the turning effect (torque) on the coil is not constant throughout each full rotation.

(a)State the position of the coil, relative to the magnetic field, at which the turning effect on the coil is at its maximum.(1)

(b)State the position of the coil at which the turning effect on the coil is momentarily zero.(1)

(c)Explain why the turning effect on the coil varies as it rotates through one full turn.(2)

(Total for Question 16 is 4 marks)

Higher Tier only. A straight wire of length 12 cm carries a current of 250 mA. The wire lies at right angles to a magnetic field of flux density 0.080 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 * current * length (F = B * I * L). Calculate the force on the wire, in newtons, giving your answer to 2 significant figures.

(Total for Question 17 is 4 marks)

Electromagnets, unlike permanent magnets, can be switched on and off.

(a)State one everyday device that uses an electromagnet, and explain why an electromagnet (rather than a permanent magnet) is suitable for this use.(3)

(Total for Question 18 is 3 marks)

Higher Tier only. A student investigates the force on a current-carrying wire using the following equipment: a straight horizontal wire passes between the poles of a strong magnet, which sits on a top-pan balance; the wire is connected in series with an ammeter, a variable resistor and a power supply. 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 * current * length (F = B * I * L).

(a)Describe how the student could use readings from the top-pan balance to find the size of the force on the wire.(2)

(b)State one variable the student should control to make this a fair test when investigating the effect of current on the force.(1)

(c)The student repeats the experiment for a range of currents and plots a graph of force (y-axis) against current (x-axis). The graph is a straight line through the origin. Explain how the student could use the gradient of this graph, together with the length of wire in the magnetic field, to calculate the magnetic flux density of the magnet.(3)

(Total for Question 19 is 6 marks)

Mark scheme · P7a Magnetism and the Motor Effect

Question 1

(a) B1 a magnetically hard material keeps (retains) its magnetism once magnetised, and is not easily demagnetised, oe
(a) Answer: A magnetically hard material keeps its magnetism after the magnetising field is removed.
(b) B1 a magnetically soft material is magnetised and demagnetised easily; it does not stay magnetised, oe
(b) Answer: A magnetically soft material is magnetised and demagnetised easily and does not stay magnetised.
(c) B1 iron, because it is magnetically soft, so it loses its magnetism as soon as the current is switched off, allowing the electromagnet to be switched on and off, oe
(c) Answer: Iron - it is magnetically soft, so it loses its magnetism as soon as the current is switched off, allowing the electromagnet to be turned on and off.

Question 2

(a) B1 a region (of space) around a magnet, or a current-carrying conductor, in which a force is exerted on other magnets or on magnetic materials, oe
(a) Answer: The region around a magnet (or current-carrying conductor) in which a force acts on other magnets or on magnetic materials.
(b) B1 from the north (seeking) pole to the south (seeking) pole, oe (accept: the direction a free north pole would move at that point)
(b) Answer: From the north pole to the south pole, outside the magnet.
(c) M1 the field lines curve away from each other and do not cross in the gap; a neutral point (a region with no resultant magnetic field) forms there, oe
(c) A1 this shows that the two poles are repelling each other, since like poles repel, oe
(c) Answer: The field lines bend away from each other and do not meet between the two poles; a neutral point (no resultant field) forms there. This shows the two north poles repel, since like poles repel.
(d) B1 one correct non-magnetic material, e.g. wood, plastic, aluminium, copper, glass
(d) B1 a second, different correct non-magnetic material from the list above
(d) Answer: Any two of: wood, plastic, aluminium, copper, glass (or other non-magnetic materials).

Question 3

(a) B1 the compass needle aligns with (the direction of) the Earth's magnetic field, oe
(a) Answer: The compass needle aligns with the Earth's magnetic field.
(b) M1 the compass needle is made from a magnetic material and becomes an induced magnet in the presence of the bar magnet's field, oe
(b) M1 an induced magnet is always attracted to the magnet that induces it (induced magnetism always causes attraction), oe
(b) A1 because the needle is free to pivot/rotate, it turns until it lines up with the (stronger, local) magnetic field of the bar magnet, i.e. it points along the bar magnet's field lines, oe
(b) Answer: The needle is made of a magnetic material, so it becomes an induced magnet in the bar magnet's field. Because induced magnetism always causes attraction, and the needle can rotate freely, it turns until it aligns with the bar magnet's field lines.

Question 4

(a) B1 place the compass on the card close to the wire and mark the position of the (north-seeking end of the) compass needle with a dot, oe
(a) B1 move the compass so its previous end sits next to the last dot, and mark the new needle position with another dot; repeat this to build up a series of dots, oe
(a) B1 join the dots with a smooth curved line and add an arrow to show the field direction; repeat the whole process starting from different points to plot further field lines, oe
(a) Answer: Mark the compass needle direction as a series of dots moving outward from the wire, joining them into a smooth curved line with an arrow showing the field direction; repeat from different starting points to build up the full field pattern.
(b) B1 the field lines reverse direction (the arrows point the opposite way around the wire), but the pattern of concentric circles stays the same shape, oe
(b) Answer: The field direction reverses, but the pattern is still a set of concentric circles around the wire.
(c) B1 the distance from the wire at which each field line is plotted, or the wire itself (its material and thickness), oe
(c) Answer: The distance from the wire at which readings are taken (or the wire used).
(d) B1 use more compass positions/dots that are closer together when plotting each field line, so the curve drawn through them is more accurate, oe (accept: sprinkle iron filings on the card to show the whole pattern at once)
(d) Answer: Use more, closer-together compass readings when plotting each line (or use iron filings to show the whole pattern at once).

Question 5

(a) B1 the size of the current in the coil
(a) B1 the number of turns on the coil (per unit length)
(a) B1 whether the coil has a core, and the material of the core (e.g. adding an iron core)
(a) Answer: The size of the current, the number of turns on the coil, and whether it has a core (e.g. an iron core).
(b) M1 the iron core becomes magnetised (becomes an induced magnet) because of the magnetic field produced by the current in the solenoid, oe
(b) A1 the induced magnetic field of the core is in the same direction as the solenoid's field and adds to it, increasing the overall (total) field strength, oe
(b) Answer: The iron core becomes an induced magnet in the solenoid's field; its own field adds to the solenoid's field, increasing the total field strength.

Question 6

(a) B1 the field pattern outside the solenoid is the same shape as the field pattern of a bar magnet, oe
(a) B1 field lines run from the north pole, curve around the outside, and return to the south pole (accept: it has a north pole at one end and a south pole at the other), oe
(a) Answer: The same shape as a bar magnet's field: lines run from the north pole, curve round the outside, and return to the south pole.
(b) B1 use the right-hand grip rule, oe
(b) B1 curl the fingers of the right hand in the direction of (conventional) current flow around the coil; the thumb then points towards the north pole, oe
(b) Answer: Use the right-hand grip rule: curl the fingers in the direction of current flow around the coil; the thumb points to the north pole.

Question 7

(a) B1 (i) the (magnetic) Field
(a) B1 (ii) the (conventional) Current
(a) B1 (iii) the direction of the force / Motion (thrust) on the wire
(a) Answer: First finger = Field, seCond finger = Current, thuMb = Motion/force.
(b) M1 correctly applies Fleming's left-hand rule (First finger into the page, SeCond finger left to right), oe
(b) A1 the force acts vertically upwards (up the page), cao
(b) Answer: The force acts vertically upwards (up the page).
(c) B1 any correct example, e.g. an electric motor, a loudspeaker, oe
(c) Answer: Example: an electric motor (or a loudspeaker).

Question 8

M1 correct substitution into F = B * I * L: F = 0.40 * 3.0 * 0.15, oe
A1 F = 0.18, cao
B1 correct unit, N (newtons)
Answer: F = 0.18 N

Question 9

M1 correct rearrangement, B = F / (I * L), oe
M1 correct substitution, B = 0.60 / (4.0 * 0.30), oe
A1 B = 0.50, cao (allow ecf from rearrangement)
B1 correct unit, tesla (T)
Answer: B = 0.50 T

Question 10

(a) B1 the current is at right angles (perpendicular) to the magnetic field lines, oe
(a) Answer: The current is at right angles to the magnetic field lines.
(b) B1 the current is parallel to (in the same or opposite direction as) the magnetic field lines, oe
(b) Answer: The current is parallel to the magnetic field lines.
(c) M1 only the component of the current (or of the wire's length) that is at right angles to the field contributes to the force, oe
(c) A1 at this intermediate angle, this perpendicular component is smaller than the full current/length (as at 90 degrees) but is not zero (as at 0 degrees), so the force lies between zero and the maximum, oe
(c) Answer: Only the part of the current/wire acting at right angles to the field produces a force. At an intermediate angle this component is less than the full value (so less than the maximum force) but still greater than zero.

Question 11

(a) B1 it reverses the direction of the current in the coil, oe
(a) B1 every half turn (each time the coil passes through the plane at right angles to the magnetic field), oe
(a) B1 so that the direction of the force (turning effect) on each side of the coil stays the same on each half turn, keeping the coil rotating continuously in the same direction rather than oscillating back and forth, oe
(a) Answer: It reverses the current in the coil every half turn, so the force on each side of the coil keeps acting in a way that turns the coil continuously in the same direction.
(b) M1 side X and side Y carry current in opposite directions, as they are opposite sides of the same loop of wire, while both sides are in the same direction of magnetic field, oe
(b) M1 by Fleming's left-hand rule, this means the force on side X acts in the opposite direction to the force on side Y, oe
(b) A1 because side X and side Y are on opposite sides of the coil (opposite sides of the axis of rotation), these opposite forces both act to turn the coil in the same rotational direction (they form a turning effect/couple), oe
(b) Answer: Current flows in opposite directions on the two sides of the coil, so by Fleming's left-hand rule the forces on the two sides are in opposite directions; because the sides are on opposite sides of the axis, this pair of forces turns the coil in one consistent rotational direction.
(c) B1 a coil with many turns produces a greater total force / bigger turning effect (for the same current), making the motor more powerful/effective, oe
(c) Answer: A coil with many turns produces a bigger turning effect (for the same current), making the motor more powerful.

Question 12

Level 0 (0): No relevant content.
Level 1 (1-2): Simple, isolated statement(s) about the coil, magnet or motor effect are made, with little or no linking to how sound is produced.
Level 2 (3-4): Some of the chain of reasoning linking the varying current to the force on the coil and the movement of the cone is given, but the explanation is incomplete or not fully logically linked.
Level 3 (5-6): A clear, logically linked chain of reasoning is given, from the varying current in the coil, through the motor effect and Fleming's left-hand rule, to the vibration of the cone and the production of a sound wave.
Indicative content:
- the coil carries a current and lies within the magnetic field of the permanent magnet
- the current-carrying coil experiences a force because of the motor effect
- the direction of this force can be found using Fleming's left-hand rule
- as the current in the coil varies in size and/or direction, the size and/or direction of the force on the coil varies
- the changing force makes the coil, and the cone attached to it, move back and forth (vibrate)
- the vibrating cone pushes and pulls the air next to it
- this creates areas of compression and rarefaction in the air, forming a sound wave
- the frequency at which the cone vibrates matches the frequency of the varying current, so the current's frequency affects the pitch of the sound produced

Question 13

(a) M1 calculates B / N for both readings: 0.020 / 50 = 4.0 x 10^-4 (T per turn) and 0.060 / 150 = 4.0 x 10^-4 (T per turn), oe
(a) A1 since B / N is the same (constant) for both readings, B is directly proportional to N, oe
(a) Answer: B / N = 4.0 x 10^-4 T per turn for both sets of data, so B is directly proportional to N.
(b) M1 uses the constant ratio from part (a): B = 4.0 x 10^-4 * 250, oe (allow ecf from (a))
(b) A1 B = 0.10 T, cao
(b) Answer: B = 0.10 T
(c) B1 repeat each reading (at each value of N) and calculate a mean value of B, to reduce the effect of random error, oe (accept: use a more precise/calibrated Hall probe)
(c) Answer: Repeat each reading and calculate a mean, to reduce the effect of random error.

Question 14

(a) B1 increase the current in the coil (e.g. by increasing the p.d./voltage of the supply), oe
(a) B1 this increases the force on each side of the coil (F = B * I * L, force is directly proportional to current), increasing the turning effect on the coil, so it rotates faster, oe
(a) Answer: Increase the current in the coil; this increases the force on the coil (F = B * I * L), increasing the turning effect, so the motor rotates faster.
(b) B1 increase the strength of the magnetic field (e.g. use a stronger magnet), oe
(b) B1 this increases the force on each side of the coil (F = B * I * L, force is directly proportional to magnetic flux density), increasing the turning effect on the coil, so it rotates faster, oe
(b) Answer: Use a stronger magnet; this increases the force on the coil (F = B * I * L), increasing the turning effect, so the motor rotates faster.

Question 15

(a) M1 correct substitution, I = V / R = 12 / 8.0, oe
(a) A1 I = 1.5 A, cao
(a) Answer: I = 1.5 A
(b) M1 correct substitution into F = B * I * L using their value of I from part (a), oe, allow ecf
(b) M1 correct multiplication, F = 0.50 * 1.5 * 0.20, oe
(b) A1 F = 0.15 N, cao (allow ecf from (a))
(b) Answer: F = 0.15 N

Question 16

(a) B1 when the plane of the coil is parallel to the magnetic field (i.e. the sides of the coil are moving at right angles to the field), oe
(a) Answer: When the plane of the coil is parallel to the magnetic field.
(b) B1 when the plane of the coil is at right angles (perpendicular) to the magnetic field (i.e. the sides of the coil are momentarily moving parallel to the field lines), oe
(b) Answer: When the plane of the coil is at right angles to the magnetic field.
(c) M1 as the coil rotates, the angle between the direction the coil sides are moving (or the current) and the magnetic field lines is continually changing, oe
(c) A1 the force on each side is greatest when the current/motion is at right angles to the field and smaller at other angles (zero when parallel), so the resulting turning effect rises and falls once every half turn (is not constant), oe
(c) Answer: As the coil rotates, the angle between its motion and the field keeps changing; the force is greatest at right angles and zero when parallel, so the turning effect rises and falls each half turn.

Question 17

M1 converts units correctly: L = 0.12 m and I = 0.25 A, oe
M1 correct substitution, F = 0.080 * 0.25 * 0.12, oe
A1 F = 0.0024 (N), cao
B1 answer correctly expressed to 2 significant figures with correct unit, 2.4 x 10^-3 N
Answer: F = 2.4 x 10^-3 N (0.0024 N)

Question 18

(a) B1 a valid device is named, e.g. an electric bell, a scrapyard/recycling crane, a relay or circuit breaker, oe
(a) M1 gives a correct linked reason referring to switching the magnetism on and off, oe
(a) A1 the reason is fully and correctly explained for the named device, oe
(a) Answer: Example: a scrapyard crane uses an electromagnet. It is switched on to attract and lift scrap iron/steel, and switched off to release it, which a permanent magnet (which cannot be switched off) could not do.

Question 19

(a) B1 record the reading on the balance with no current flowing (the zero/before reading), oe
(a) B1 switch on the current, record the new balance reading, and calculate the change in the balance's mass reading; convert this change in mass to a force using force = mass * g (g = 9.8 N/kg), oe
(a) Answer: Record the balance reading with no current, then with current flowing; the change in mass reading, converted using force = mass * g, gives the force on the wire.
(b) B1 the strength of the magnet used, or the length of wire between the poles of the magnet (the length of wire in the field), is kept the same, oe
(b) Answer: The strength of the magnet (or the length of wire in the field) is kept the same.
(c) M1 since F = B * I * L, force is directly proportional to current, so the gradient of the force-current graph is equal to B * L, oe
(c) M1 the student needs to know (or measure) the length of wire, L, that lies within the magnetic field, oe
(c) A1 the magnetic flux density can then be found by dividing the gradient by L (B = gradient / L), oe
(c) Answer: Since F = B * I * L, the gradient of a force-current graph equals B * L. Dividing the gradient by the known length of wire in the field (L) gives the magnetic flux density: B = gradient / L.

Magnetism and the Motor Effect - Worksheets, Questions and Revision