Electromagnetic Induction and Transformers - Worksheets, Questions and Revision

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GCSE · Physics

P7b Electromagnetic Induction and Transformers

AQA 8464 · Calculator allowed · about 92 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key ideas: electromagnetic induction and transformers

Original text written for Revision Library.

When there is relative motion between a conductor and a magnetic field, or when the magnetic field through a coil changes, a potential difference is induced across the ends of the conductor; if the conductor is part of a complete circuit, a current is induced too. This is called electromagnetic induction, or the generator effect. The size of the induced potential difference increases with a stronger magnetic field, a faster speed of movement, and more turns on a coil; its direction reverses if the direction of motion or the polarity of the field is reversed. This effect is used in an alternator, which generates alternating current (a.c.) using slip rings, and in a dynamo, which generates direct current (d.c.) using a split-ring commutator; it is also used in moving-coil microphones, which convert sound into a varying electrical signal. A transformer uses electromagnetic induction to change the size of an alternating potential difference; because it needs a constantly changing magnetic field, it only works with a.c., not d.c. Step-up and step-down transformers are used throughout the National Grid to transmit electrical power efficiently across the country.

Electromagnetic induction (the generator effect) can be demonstrated using a bar magnet and a coil of wire connected to a centre-zero galvanometer.

(a)State what is meant by the term electromagnetic induction.(2)

(Total for Question 1 is 2 marks)

A magnet is moved in and out of a coil of wire connected to a sensitive voltmeter.

(a)State three factors that increase the size of the potential difference induced across the coil.(3)

(Total for Question 2 is 3 marks)

Required practical. Priya investigates the factors that affect the potential difference induced across the ends of a coil of wire. She sets up a coil of 100 turns connected to a voltmeter sensor and a data logger, which records the peak potential difference induced as a bar magnet is pushed into the coil. For each trial she pushes the magnet through a fixed distance of 0.50 m by hand, trying to keep the speed as constant as possible, and calculates the average speed of the push using a stopwatch. She repeats this for a range of speeds. Her results are shown below.

Speed of magnet (m/s): 0.20, 0.40, 0.60, 0.80, 1.00
Peak induced pd (V): 0.05, 0.10, 0.15, 0.22, 0.25

Figure (to be drawn): A bar magnet is held above a hollow tube wound with a coil of 100 turns of insulated copper wire. The ends of the coil are connected to a voltmeter sensor linked to a data logger, which records the peak potential difference each time the magnet is pushed into the coil.

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

(b)State two variables Priya should control to make this a fair test.(2)

(c)Describe the relationship between the speed of the magnet and the peak induced potential difference shown by the data in the table.(2)

(d)One of the readings does not fit the pattern shown by the rest of the data. Identify this anomalous result and suggest one reason it may have occurred.(2)

(e)Suggest one improvement to the method that would increase the accuracy of the peak induced pd readings, and explain how it would improve the results.(2)

(f)Based on the data (ignoring the anomalous result), describe the shape of the graph of peak induced pd (y-axis) against speed of the magnet (x-axis).(1)

(g)Using the data (ignoring the anomalous result at 0.80 m/s), calculate the gradient of the graph of peak induced pd against speed of the magnet. Give the unit of the gradient.(2)

(Total for Question 3 is 13 marks)

Priya then pulls the magnet back out of the coil (instead of pushing it in), moving it at the same speed as before.

(a)State what happens to the direction of the induced current, and explain why.(2)

(Total for Question 4 is 2 marks)

Priya then repeats the original investigation (pushing the magnet into the coil), but uses a bar magnet with the opposite polarity (its north and south poles swapped over), moving it at the same speed as before.

(a)State and explain the effect this has on the induced current, compared with the original magnet.(2)

(Total for Question 5 is 2 marks)

An alternator and a dynamo both use the generator effect to generate electricity, but they produce different types of electrical output. Compare how an alternator and a dynamo work, and explain why they produce different types of output. In your answer, refer to: the components used to connect the rotating coil to the circuit; the shape of the potential difference-time graph produced by each generator.

(Total for Question 6 is 6 marks)

In an alternator, the speed at which the coil rotates can be changed.

(a)State the effect of increasing the speed of rotation of the coil on (i) the frequency of the alternating output, and (ii) the peak potential difference produced.(2)

(Total for Question 7 is 2 marks)

An alternator's rotating coil completes 50 full rotations every second, producing an alternating potential difference at UK mains frequency. Use the equation: frequency = 1 / time period (f = 1 / T).

(a)State the frequency of the potential difference produced when the coil rotates 50 times per second.(1)

(b)The coil is then slowed so that it rotates 25 times per second. Calculate the new frequency, and state what happens to the period of the output potential difference-time graph.(2)

(Total for Question 8 is 3 marks)

A graph of potential difference (y-axis) against time (x-axis) is drawn for the output of a generator, but the graph is not labelled to say whether the generator is an alternator or a dynamo.

(a)Explain how you could tell, just from the shape of the graph, whether it was produced by an alternator or by a dynamo.(2)

(Total for Question 9 is 2 marks)

A moving-coil microphone changes sound into a varying electric current using electromagnetic induction.

(a)Describe how the microphone does this. Your answer should refer to: the diaphragm; the coil; the permanent magnet.(4)

(Total for Question 10 is 4 marks)

A moving-coil microphone and a loudspeaker both contain a coil and a permanent magnet, but they work in opposite ways.

(a)State one similarity and one difference between how a microphone and a loudspeaker work.(2)

(Total for Question 11 is 2 marks)

A transformer changes the size of an alternating potential difference.

(a)State what a transformer is used for.(1)

(b)Explain why a transformer only works when connected to an alternating current (a.c.) supply, and not to a direct current (d.c.) supply.(2)

(Total for Question 12 is 3 marks)

Many household electrical devices, such as phone and laptop chargers, contain a small transformer.

(a)State why a phone charger contains a transformer, even though it is a small, low-power device.(2)

(Total for Question 13 is 2 marks)

A transformer has more turns on its secondary coil than on its primary coil.

(a)State whether this is a step-up or a step-down transformer.(1)

(b)State the effect this transformer has on the size of the potential difference and the size of the current, between the primary and secondary coils.(2)

(Total for Question 14 is 3 marks)

Higher Tier only. A step-up transformer has 300 turns on its primary coil and 1500 turns on its secondary coil. The potential difference across the primary coil is 12 V. 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 potential difference across the secondary coil.(3)

(Total for Question 15 is 3 marks)

Higher Tier only. A doorbell transformer steps down the UK mains supply of 230 V. The primary coil has 4600 turns, and the transformer must output a secondary potential difference of 12 V. 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)

(Total for Question 16 is 3 marks)

The National Grid transmits electrical power across the country using alternating current (a.c.), stepping the potential difference up and down using transformers at different points in the network.

(a)Explain why the National Grid must use alternating current (a.c.), rather than direct current (d.c.), for this system to work.(3)

(Total for Question 17 is 3 marks)

Electricity generated at a power station is transmitted across the country using the National Grid. At the power station, a step-up transformer increases the potential difference from 25,000 V (25 kV) to 400,000 V (400 kV) before transmission. The current leaving the power station (the primary current) is 800 A. Assume the transformer is 100% efficient. Use the Physics Equations Sheet. Use the equations: potential difference across primary coil * current in primary coil = potential difference across secondary coil * current in secondary coil (Vp * Ip = Vs * Is); power = potential difference * current (P = V * I).

Figure (to be drawn): A simplified National Grid pathway: power station -> step-up transformer -> pylons/transmission cables (high pd, low current) -> step-down transformer -> homes and businesses (lower, safer pd).

(a)Calculate the current in the transmission cables (the secondary current, Is).(3)

(b)Calculate the power transmitted through the cables. Give your answer in megawatts (MW).(3)

(c)Explain why transmitting electricity at a high potential difference and low current (rather than a low potential difference and high current) reduces the energy wasted as heat in the transmission cables.(3)

(Total for Question 18 is 9 marks)

Higher Tier only. A laptop charger contains a transformer. The primary coil is connected to the 230 V mains supply and has 2300 turns. The secondary coil outputs a potential difference of 20 V, with a secondary current of 2.3 A. Assume the transformer is 100% efficient. Use the Physics Equations Sheet. Use the equations: 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); potential difference across primary coil * current in primary coil = potential difference across secondary coil * current in secondary coil (Vp * Ip = Vs * Is).

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

(b)Calculate the current in the primary coil.(3)

(Total for Question 19 is 6 marks)

In real transformers, some energy is wasted as heat rather than being usefully transferred from the primary coil to the secondary coil.

(a)State the energy store to which this wasted energy is transferred.(1)

(b)Give one reason why energy is wasted as heat in a real transformer.(1)

(Total for Question 20 is 2 marks)

A transformer is supplied with 500 W of electrical power at its primary coil. The useful power output at its secondary coil is 460 W. Use the equation: efficiency = useful output power / total input power.

(a)Calculate the efficiency of the transformer, giving your answer as a percentage.(2)

(b)Calculate the power wasted as heat by the transformer.(1)

(Total for Question 21 is 3 marks)

Higher Tier only. A company claims that a new transformer core material will improve efficiency to over 100%, so that the transformer 'creates extra electrical energy from nothing'.

(a)Evaluate this claim, using your knowledge of the conservation of energy.(4)

(Total for Question 22 is 4 marks)

Mark scheme · P7b Electromagnetic Induction and Transformers

Question 1

(a) B1 a potential difference (and current, if the circuit is complete) is induced across the ends of a conductor, oe
(a) B1 when there is relative motion between the conductor and a magnetic field, or when the magnetic field through the conductor changes, oe
(a) Answer: A potential difference (and current, in a complete circuit) is induced across a conductor when there is relative motion between the conductor and a magnetic field, or when the magnetic field through it changes.

Question 2

(a) B1 using a stronger magnet (a stronger magnetic field)
(a) B1 moving the magnet faster (a greater speed of movement)
(a) B1 increasing the number of turns on the coil
(a) Answer: A stronger magnet, moving the magnet faster, and using a coil with more turns.

Question 3

(a) B1 independent variable: the speed at which the magnet is pushed into the coil
(a) B1 dependent variable: the peak potential difference induced across the coil
(a) Answer: Independent variable: speed of the magnet. Dependent variable: peak induced potential difference.
(b) B1 the same magnet is used each time (same strength of magnet)
(b) B1 any other correct control variable, e.g. the same coil (same number of turns) is used each time; the magnet is pushed through the same fixed distance (0.50 m) each time; the magnet starts from the same position relative to the coil each time
(b) Answer: E.g. the same magnet is used each time, and the magnet is pushed through the same fixed distance (0.50 m) each time.
(c) B1 as the speed of the magnet increases, the peak induced potential difference increases, oe
(c) B1 the relationship is (approximately) directly proportional, e.g. doubling the speed from 0.20 m/s to 0.40 m/s approximately doubles the pd from 0.05 V to 0.10 V, oe
(c) Answer: As speed increases, the peak induced pd increases, and the relationship is approximately directly proportional (e.g. doubling the speed roughly doubles the pd).
(d) B1 the reading at 0.80 m/s (peak pd = 0.22 V, when a value of about 0.20 V would be expected from the pattern)
(d) B1 any sensible suggested reason, e.g. Priya did not push the magnet at a truly constant speed during that trial (hand-pushing introduces variation); a timing error with the stopwatch for that trial; the magnet was not aligned centrally with the coil for that trial, oe
(d) Answer: The 0.80 m/s reading (0.22 V, expected about 0.20 V) is anomalous, possibly because the speed was not kept truly constant during that particular push.
(e) B1 any sensible improvement, e.g. use light gates to measure the speed of the magnet directly and precisely; use a mechanical/motorised system to push the magnet at a controlled, repeatable speed; repeat each speed several times and calculate a mean pd, oe
(e) B1 linked explanation of how this reduces error, e.g. removes the variation caused by pushing the magnet by hand, giving more precise/repeatable speed values and more reliable pd readings, oe
(e) Answer: E.g. use a mechanical/motorised system to push the magnet at a controlled, repeatable speed, removing the variation caused by pushing by hand and giving more reliable results.
(f) B1 a straight line passing through the origin, oe (showing direct proportion)
(f) Answer: A straight line through the origin.
(g) M1 correct gradient calculation using two points on the line, e.g. (0.25 - 0.05) / (1.00 - 0.20), oe
(g) A1 gradient = 0.25, with unit V / (m/s), i.e. volts per metre per second (accept range 0.24 to 0.26)
(g) Answer: Gradient = 0.25 V / (m/s)

Question 4

(a) B1 the direction of the induced current reverses
(a) B1 because the direction of the induced potential difference/current depends on the direction of motion of the magnet relative to the coil, and reversing the direction of motion reverses this, oe
(a) Answer: The current reverses direction, because reversing the direction the magnet moves reverses the direction of the induced pd/current.

Question 5

(a) B1 the direction of the induced current reverses (the size of the current stays the same, since the speed and strength of the magnet are unchanged)
(a) B1 because reversing the polarity of the magnetic field reverses the direction of the induced potential difference/current, oe
(a) Answer: The current reverses direction (but stays the same size), because reversing the magnet's polarity reverses the direction of the induced pd/current.

Question 6

Level 0 (0): No relevant content.
Level 1 (1-2): Basic, isolated statements are made about an alternator and/or a dynamo, with little or no comparison, e.g. simple identification of a.c. or d.c. output only.
Level 2 (3-4): Relevant points are made about both an alternator and a dynamo, showing some comparison, but the description may be incomplete or the link between construction and output may not be clear.
Level 3 (5-6): A clear and detailed comparison is given, correctly linking the construction of each generator (slip rings or split-ring commutator) to the shape of its output potential difference-time graph, with correct use of scientific vocabulary throughout.
Indicative content:
- Both an alternator and a dynamo generate a potential difference/current by rotating a coil within a magnetic field, using the generator effect.
- An alternator is connected to the circuit using slip rings (and brushes), which allow the coil to rotate freely while keeping the same connections throughout each rotation.
- Because the connections do not change, the induced pd/current reverses direction every half rotation, producing alternating current (a.c.).
- The output graph of an alternator is a smooth sine wave that alternates between positive and negative values (above and below zero).
- A dynamo is connected to the circuit using a split-ring commutator, which swaps the connections every half rotation.
- This swap keeps the output the same polarity/direction, so a dynamo produces direct current (d.c.).
- The output graph of a dynamo varies in size but stays on one side of zero (it does not go negative), forming a series of humps.
- In both cases, the output pd is greatest when the coil sides are moving at right angles to the magnetic field, and zero when the coil sides are moving parallel to the field.

Question 7

(a) B1 (i) the frequency of the alternating output increases
(a) B1 (ii) the peak potential difference produced increases
(a) Answer: Both the frequency of the output and the peak potential difference produced increase.

Question 8

(a) B1 50 Hz
(a) Answer: 50 Hz
(b) B1 new frequency = 25 Hz
(b) B1 the period doubles (increases from 0.02 s to 0.04 s), oe
(b) Answer: 25 Hz; the period doubles, from 0.02 s to 0.04 s.

Question 9

(a) B1 if the graph shows a smooth wave that goes both above and below the zero line (positive and negative values), it was produced by an alternator, oe
(a) B1 if the graph varies in size but stays entirely on one side of the zero line (never goes negative), it was produced by a dynamo, oe
(a) Answer: If the graph alternates above and below zero, it is from an alternator (a.c.); if it stays on one side of zero, it is from a dynamo (d.c.).

Question 10

(a) B1 sound waves cause the diaphragm to vibrate
(a) B1 the diaphragm is attached to a coil of wire, so the coil vibrates/moves with it
(a) B1 the coil moves within the magnetic field of a permanent magnet, so it moves relative to the field
(a) B1 this induces a varying/alternating potential difference (and current) in the coil, by electromagnetic induction, that varies in the same pattern as the sound wave
(a) Answer: Sound makes the diaphragm vibrate; this moves the attached coil within the magnetic field of a permanent magnet, inducing a varying electrical signal (by electromagnetic induction) that matches the pattern of the sound wave.

Question 11

(a) B1 similarity, e.g. both contain a coil that can move relative to the magnetic field of a permanent magnet, oe
(a) B1 difference, e.g. a microphone converts sound into an electrical signal using the generator effect (electromagnetic induction); a loudspeaker converts an electrical signal into sound using the motor effect (the force on a current-carrying coil in a magnetic field) - they are the reverse of each other, oe
(a) Answer: Similarity: both use a coil and a permanent magnet. Difference: a microphone uses the generator effect to turn sound into an electrical signal, while a loudspeaker uses the motor effect to turn an electrical signal into sound (the reverse process).

Question 12

(a) B1 to change (step up or step down) the size of an alternating potential difference (voltage), oe
(a) Answer: To change the size of an alternating potential difference.
(b) M1 an alternating current in the primary coil produces a constantly changing magnetic field in the iron core, oe
(b) A1 this changing field induces a (changing) potential difference in the secondary coil by electromagnetic induction; a direct current would produce a constant, unchanging field, so no pd would be induced, oe
(b) Answer: An a.c. supply produces a constantly changing magnetic field in the core, which induces a pd in the secondary coil. A d.c. supply produces a constant field, so no pd would be induced.

Question 13

(a) B1 the UK mains supply (230 V) is a much higher potential difference than the phone's battery needs (a few volts), oe
(a) B1 the transformer steps down the potential difference from the mains to a much lower, safe value suitable for charging the phone, oe
(a) Answer: The mains supply (230 V) is much higher than a phone battery needs, so a step-down transformer reduces the potential difference to a safe, usable value.

Question 14

(a) B1 step-up
(a) Answer: Step-up
(b) B1 the potential difference increases (the secondary pd is greater than the primary pd)
(b) B1 the current decreases (the secondary current is smaller than the primary current)
(b) Answer: The potential difference increases, and the current decreases, from primary to secondary.

Question 15

(a) M1 correct rearrangement: Vs = Vp * (Ns / Np), oe
(a) M1 correct substitution: Vs = 12 * (1500 / 300), oe
(a) A1 Vs = 60 V, cao
(a) Answer: Vs = 60 V

Question 16

(a) M1 correct rearrangement: Ns = Np * (Vs / Vp), oe
(a) M1 correct substitution: Ns = 4600 * (12 / 230), oe
(a) A1 Ns = 240 (turns), cao
(a) Answer: Ns = 240 turns

Question 17

(a) B1 the National Grid relies on transformers to step the potential difference up (for efficient transmission) and step it back down (for safe use in homes), oe
(a) B1 transformers only work with an alternating supply, because they rely on a constantly changing magnetic field to induce a potential difference in the secondary coil, oe
(a) B1 a direct current supply would produce a constant magnetic field, so no potential difference would be induced, and the transformers (and so the whole system) would not work, oe
(a) Answer: The Grid relies on transformers to step the potential difference up and down; transformers only work with a.c., since they need a constantly changing magnetic field to induce a pd. A d.c. supply gives a constant field, so no pd would be induced and the transformers would not work.

Question 18

(a) M1 correct rearrangement: Is = (Vp * Ip) / Vs, oe
(a) M1 correct substitution: Is = (25000 * 800) / 400000, oe
(a) A1 Is = 50 A, cao
(a) Answer: Is = 50 A
(b) M1 uses P = V * I with a matching pair of primary or secondary values, oe
(b) M1 correct substitution, e.g. P = 25000 * 800 (or 400000 * 50), oe
(b) A1 P = 20,000,000 W = 20 MW, cao (allow awrt 20 MW)
(b) Answer: P = 20 MW (20,000,000 W)
(c) B1 the power wasted as heat in the cables is given by P = I² * R, for a fixed cable resistance R, oe
(c) B1 transmitting at a high pd means a much smaller current is needed to transmit the same power, since P = V * I, oe
(c) B1 because power loss depends on current squared, a smaller current greatly reduces the power/energy wasted as heat, making transmission more efficient, oe
(c) Answer: Power wasted as heat is P = I² * R for a fixed cable resistance. A high pd needs a much smaller current to transmit the same power (P = V * I); since power loss depends on the current squared, a smaller current greatly reduces the energy wasted.

Question 19

(a) M1 correct rearrangement: Ns = Np * (Vs / Vp), oe
(a) M1 correct substitution: Ns = 2300 * (20 / 230), oe
(a) A1 Ns = 200 (turns), cao
(a) Answer: Ns = 200 turns
(b) M1 finds the secondary (output) power: Ps = Vs * Is = 20 * 2.3, oe
(b) M1 rearranges Vp * Ip = Vs * Is to give Ip = (Vs * Is) / Vp, and substitutes, oe
(b) A1 Ip = 0.2 A, cao
(b) Answer: Ip = 0.2 A

Question 20

(a) B1 the thermal energy store of the surroundings (the transformer and the air around it heat up), oe
(a) Answer: The thermal energy store of the surroundings.
(b) B1 any one from: resistance in the coils causes heating as current flows through them; eddy currents are induced in the iron core; not all of the magnetic field lines from the primary coil pass through the secondary coil (flux leakage), oe
(b) Answer: E.g. resistance in the coils causes heating as current flows through them.

Question 21

(a) M1 efficiency = (460 / 500) * 100, oe
(a) A1 92%, cao (allow awrt 92%)
(a) Answer: 92%
(b) B1 40 W (= 500 - 460), cao
(b) Answer: 40 W

Question 22

(a) B1 the claim is not possible / is incorrect
(a) B1 the law of conservation of energy states that energy cannot be created or destroyed, only transferred from one store to another, oe
(a) B1 a transformer can only transfer the electrical energy supplied to its primary coil; the useful output energy at the secondary coil can never be greater than the input energy, oe
(a) B1 in reality, some energy is always wasted (e.g. as heat, due to resistance in the coils or eddy currents in the core), so a real transformer is always less than 100% efficient, never more, oe
(a) Answer: The claim is not possible. By the conservation of energy, energy cannot be created; a transformer can only transfer the input energy, so useful output can never exceed input. In reality some energy is always wasted as heat, so efficiency is always below 100%.

Question 22

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