International GCSE Physics · Pearson Edexcel 4PH1 · Topic 6
Magnetism and electromagnetism
Original exam-style questions with mark schemes. Questions marked “Paper 2 only” use content assessed only on Paper 2. Notes 6(b) →Notes 6(c) →Notes 6(d) →
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Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 13 October 2026 · new questions added regularlyHow every question is checked →
1IGCSE-PHY-6-001Magnetic and non-magnetic materialsPapers 1 & 2Grade 4–51 mark
A student holds a bar magnet close to four different objects. Which object will be attracted to the magnet?
Show mark scheme
B (a steel paper clip);
A is incorrect because aluminium is a metal but it is not a magnetic substance
C is incorrect because copper is a metal but it is not a magnetic substance
D is incorrect because brass (an alloy of copper and zinc) is not a magnetic substance
3IGCSE-PHY-6-003Hard and soft magnetic materials and induced magnetismPapers 1 & 2Grade 4–58 marks
A student hangs a chain of four soft-iron nails from the north pole of a bar magnet. Each nail holds up the one below it. The student then pulls the top nail away from the magnet. The student repeats the experiment with four steel sewing needles.
(a)
Name one magnetic substance other than iron.
(1)
(b)
Describe the difference between a magnetically hard material and a magnetically soft material.
(2)
(c)
Explain why the iron nails attract each other while the top nail touches the magnet.
(2)
(d)
When the top nail is pulled away from the magnet, the iron nails fall apart but the steel needles stay together. Explain this observation.
(2)
(e)
State which material, soft iron or steel, should be used to make the needle of a compass.
(1)
Show mark scheme
(a)
nickel / cobalt / steel;
Allow
named alloys containing these e.g. alnico, neodymium (magnet) alloys
Do not accept
copper
aluminium
any other non-magnetic metal
(b)
(magnetically) hard material keeps/retains its magnetism (once magnetised) / used for permanent magnets;
(magnetically) soft material is easily magnetised and (easily) loses its magnetism (when the field is removed);
Allow
hard material is difficult to demagnetise
soft material is a temporary magnet
Do not accept
answers about physical hardness / being easy to bend
(c)
(magnetism is) induced in the nails by the magnet / field of magnet;
each nail becomes a (temporary) magnet / has poles so attracts the next nail;
Allow
idea that the top of each nail becomes a S pole and the bottom an N pole
Do not accept
nails are charged
static electricity
(d)
(soft) iron loses its (induced) magnetism (when removed from the field);
steel is (magnetically) hard so it keeps / retains its magnetism;
4IGCSE-PHY-6-004Practical: field pattern between two like polesPapers 1 & 2Grade 4–58 marks
A student investigates magnetic field patterns using plotting compasses. The student first plots the field around a single bar magnet. The student then places two bar magnets on a sheet of paper with their north poles facing each other, 8.0 cm apart, as shown in the diagram. Point X is halfway between the magnets.
(a)
Describe how the student can use a plotting compass to draw one magnetic field line around a single bar magnet.
(3)
(b)
State the direction of the magnetic field lines outside a bar magnet.
(1)
(c)
The student places a plotting compass at point X. State what the compass shows at X and explain why.
(2)
(d)
The student repeats the investigation by sprinkling iron filings around the magnets. Give one advantage and one disadvantage of using iron filings instead of plotting compasses.
(2)
Show mark scheme
(a)
any three from:
MP1. place compass next to (a pole of) the magnet;
MP2. mark (a dot at) the position the needle points to;
MP3. move the compass so (the tail of) the needle is at the (previous) dot and repeat;
MP4. join the dots (with a smooth curve);
MP5. add an arrow (in the direction the needle points);
Allow
marks can be awarded from a labelled diagram
draw round magnet on paper as part of MP1
Do not accept
use of iron filings for this part
(b)
(from) north (pole) to south (pole);
Allow
N to S
Do not accept
S to N
(c)
compass does not point in any particular direction / needle not deflected (by the magnets) / no (resultant) field at X;
fields of the two magnets are equal (in size) and opposite (in direction) / cancel;
Allow
neutral point for MP1
compass points north due to Earth's field only for MP1
Do not accept
the fields attract
compass points towards one magnet
(d)
advantage: shows the whole (field) pattern at once / quicker / more detail (than individual lines);
disadvantage: does not show the direction of the field;
Allow
filings show where field is strongest (more concentrated)
5IGCSE-PHY-6-005Inducing a voltage with a moving magnetPapers 1 & 2Grade 4–58 marks
A coil of wire is connected to a sensitive centre-zero ammeter. When the north pole of a bar magnet is pushed into the coil, the ammeter needle moves briefly to the right.
(a)
State what is induced in the coil to make the needle move.
(1)
(b)
Describe what the ammeter needle does when (i) the magnet is held still inside the coil (ii) the north pole is pulled out of the coil (iii) the south pole is pushed into the coil.
(3)
(c)
The student now pushes the north pole into the coil much more slowly. Describe how the movement of the ammeter needle is different from before.
(2)
(d)
The magnet is replaced by a second, smaller coil connected to a battery and a switch. The smaller coil is held still inside the first coil. The needle moves briefly when the switch is closed, but returns to zero while the switch stays closed. Explain these observations.
(2)
Show mark scheme
(a)
voltage / potential difference / e.m.f.;
Allow
current
Do not accept
charge
magnetism
(b)
(i) no deflection / reads zero;
(ii) (moves briefly) to the left / opposite direction;
(iii) (moves briefly) to the left / opposite direction;
Allow
ECF is not applied between (ii) and (iii)
Do not accept
moves to the right for (ii) or (iii)
(c)
(needle still moves) to the right / same direction but the deflection is smaller;
(the deflection) lasts for a longer time / needle moves less far but for longer;
Allow
smaller induced voltage / current for MP1 if direction also stated
idea that rate of change of field is smaller for MP1 if linked to smaller deflection
Do not accept
needle moves to the left
needle does not move at all
needle moves further
(d)
closing switch causes the magnetic field (through the first coil) to change / increase, inducing a voltage;
steady current gives a constant field so no voltage induced / field must be changing to induce a voltage;
6IGCSE-PHY-6-021Changing the current and the field togetherPapers 1 & 2Grade 4–51 mark
A horizontal wire carries a current between the poles of a magnet. The force on the wire has size F and acts vertically upwards. The current in the wire is doubled. At the same time the magnet is turned round so that the direction of the magnetic field is reversed. Which of these describes the new force on the wire?
Show mark scheme
B (2F downwards);
A is incorrect because reversing the field reverses the direction of the force, so it cannot still act upwards
C is incorrect because doubling the current doubles the size of the force; it does not stay at F
D is incorrect because reversing the field reverses the force; it does not cancel the effect of the current
7IGCSE-PHY-6-022Factors affecting a generator's induced voltagePapers 1 & 2Grade 4–51 mark
A simple a.c. generator has a rectangular coil of wire that rotates between the poles of a permanent magnet. Which change will NOT increase the size of the induced voltage?
Show mark scheme
D (rotating the coil in the opposite direction at the same speed);
A is incorrect because more turns on the coil gives a larger induced voltage
B is incorrect because a stronger magnetic field gives a larger induced voltage
C is incorrect because rotating the coil faster changes the field through the coil more quickly, giving a larger induced voltage
8IGCSE-PHY-6-023Errors in a drawing of a bar magnet's fieldPapers 1 & 2Grade 4–58 marks
A student plots the magnetic field around a bar magnet using a plotting compass and then draws the field lines from memory. The diagram shows the student's drawing. It has four field lines, labelled W, X, Y and Z. P and Q are two points in the field. The drawing contains two errors.
(a)
Identify the two errors in the drawing and explain why each is wrong.
(4)
(b)
State and explain at which point, P or Q, the magnetic field is stronger.
(2)
(c)
The north pole of a second bar magnet is brought towards the end of the magnet labelled N. State what happens.
(1)
(d)
The student then holds a small steel paper clip at P. State what happens to the paper clip.
(1)
Show mark scheme
(a)
arrow on line X is in the wrong direction / X points from S to N;
field lines (outside a magnet) go from N to S / show direction of force on a north pole;
lines Y and Z cross;
field lines cannot cross (because the field has only one direction at a point);
Allow
arrow on X points towards the north pole
a compass at the crossing point would have to point two ways at once
Do not accept
X is too far from the magnet
the lines should touch the magnet at the middle
field lines go from S to N
(b)
P;
field lines closer together (near P) / P is near a pole;
Allow
field is strongest at the poles / ends
second mark only if P is chosen
Do not accept
Q because it is above the middle of the magnet
(c)
they repel / push apart;
Allow
move away from each other
Do not accept
attract
(d)
attracted (towards the magnet / to the pole);
Allow
it sticks to the magnet / becomes magnetised and is attracted
9IGCSE-PHY-6-024Separating steel and aluminium cans at a recycling centrePapers 1 & 2Grade 4–57 marks
At a recycling centre, crushed drinks cans travel along a conveyor belt. Some cans are made of steel and some are made of aluminium. At the end of the belt the cans pass under a rotating drum that contains strong permanent magnets. Steel cans are lifted onto the drum and carried round to a separate container. Aluminium cans fall off the end of the belt into a different container.
(a)
Explain why the steel cans are lifted by the drum but the aluminium cans are not.
(2)
(b)
Sometimes a steel can is held onto the drum by another steel can underneath it, even though it does not touch the drum. Explain why.
(2)
(c)
The magnets in the drum are turned round so that their opposite poles face the belt. State why the steel cans are still lifted by the drum.
(1)
(d)
The magnets in the drum are made from a magnetically hard material. Explain why a magnetically hard material is used.
(2)
Show mark scheme
(a)
steel is a magnetic material / is attracted by magnets;
aluminium is not a magnetic material / not attracted;
Allow
steel contains iron (which is magnetic)
Do not accept
aluminium is too light
aluminium is not a metal
(b)
magnetism is induced in the first (inner) can / first can becomes magnetised by the field;
(so) first can attracts the second can;
Allow
first can becomes a temporary magnet
Do not accept
the second can is magnetised by the conveyor belt
the cans are charged
(c)
magnetism induced in the can always gives an opposite pole nearest the magnet / magnetic materials are attracted by either pole;
Allow
the induced poles in the can also reverse
steel (unmagnetised) is attracted to both N and S poles
Do not accept
the cans are now repelled
steel cans are permanent magnets
(d)
(hard material) keeps its magnetism / difficult to demagnetise;
so the magnets remain strong / permanent magnets do not need re-magnetising;
Allow
hard to magnetise but stays magnetised
Do not accept
hard materials are easily magnetised and easily lose magnetism
10IGCSE-PHY-6-006Left-hand rule: force on a wirePapers 1 & 2Grade 6–71 mark
The diagram shows a wire between the poles of a magnet. The north pole is at the top of the page and the south pole is at the bottom. The current in the wire is directed out of the page. In which direction is the force on the wire?Show mark scheme
C (to the right);
A is incorrect because the force is perpendicular to the field, not along it towards a pole
B is incorrect because this is the direction given by using the right hand or by reversing the current
D is incorrect because the force is perpendicular to the field, not along the field lines
11IGCSE-PHY-6-007Force on a rod rolling on railsPapers 1 & 2Grade 6–78 marks
A copper rod rests across two horizontal copper rails. The rails are connected to a cell. A magnet is arranged so that its north pole is below the rails and its south pole is above them, so the magnetic field acts vertically upwards through the rod. The diagram shows the apparatus viewed from above. When the circuit is complete the rod starts to roll along the rails.
(a)
Explain why there is a force on the rod when there is a current in it.
(2)
(b)
Use the left-hand rule to determine the direction in which the rod rolls.
(1)
(c)
State the direction the rod rolls when (i) the cell is reversed (ii) the cell is reversed and the magnet is turned upside down.
(2)
(d)
Give two changes that would make the force on the rod larger.
(2)
(e)
The teacher replaces the magnet with two flat magnets, one above and one below the rails, to produce a uniform field. State how the poles of the two magnets should face each other.
(1)
Show mark scheme
(a)
current (in rod) produces a magnetic field (around the rod);
(this field) interacts with the field of the magnet (producing a force);
Allow
the two fields combine / add/ cancel
Do not accept
the rod is attracted to the magnet
the rod is magnetised
(b)
to the right / away from the cell;
Do not accept
upwards
towards the magnet
(c)
(i) to the left / towards the cell;
(ii) to the right / same direction as in (b);
Allow
ECF from (b) for opposite and same direction respectively
(d)
any two from:
increase the current / use a cell with a larger voltage;
use a stronger magnet / stronger field;
use a longer rod in the field / rails further apart (with the field covering them);
12IGCSE-PHY-6-008The simple d.c. motor and loudspeakerPapers 1 & 2Grade 6–79 marks
The diagram shows the coil of a simple d.c. motor viewed end-on from the commutator end. The coil is between the poles of a magnet. Side P of the coil carries a current out of the page and side Q carries a current into the page.
(a)
State the direction of the force on (i) side P (ii) side Q.
(2)
(b)
Deduce whether the coil starts to turn clockwise or anticlockwise as seen in the diagram.
(1)
(c)
Explain why the motor needs a split-ring commutator to keep turning in the same direction.
(2)
(d)
The current in the coil is doubled. The magnet is not changed. Explain the effect on the motor.
(2)
(e)
A loudspeaker also uses the force on a current-carrying coil in a magnetic field. The coil is attached to a paper cone. A student connects a loudspeaker to a d.c. supply instead of an a.c. supply and leaves the supply switched on. Describe and explain what happens to the cone.
(2)
Show mark scheme
(a)
(i) upwards / towards top of page;
(ii) downwards / towards bottom of page;
Allow
ECF: (ii) opposite to (i)
Do not accept
towards N or S pole
(b)
clockwise;
Allow
ECF from (a)
(c)
(commutator) reverses the current (in the coil) every half turn;
so forces on the sides (reverse and) keep turning the coil in the same direction;
Allow
idea that otherwise the coil would stop / rock back and forth at the vertical position
Do not accept
commutator reverses the magnetic field
commutator makes the current a.c. from the supply
(d)
(magnetic) force on each side of the coil increases / doubles;
(so) larger turning effect / coil turns faster;
Allow
motor speeds up for MP2
more powerful motor for MP2
Do not accept
motor turns in the opposite direction
force on the coil reverses
(e)
force on the coil is in one direction only / does not reverse (because the current does not change direction);
(so) cone moves (once) and stays displaced / does not vibrate, so no (continuous) sound is produced;
Allow
a single click when the supply is switched on (for MP2)
cone pushed out (or in) and held there for MP2
Do not accept
cone vibrates at a lower frequency
d.c. would damage the loudspeaker (as the only point)
13IGCSE-PHY-6-009Practical: induced voltage and number of turnsPapers 1 & 2Grade 6–710 marks
A student investigates how the peak voltage induced in a coil depends on the number of turns on the coil. The coil is wound around a vertical plastic tube and connected to a data logger. A bar magnet is dropped through the tube and the data logger records the peak induced voltage. The student repeats the experiment with coils of different numbers of turns. Number of turns / –: 50, 100, 150, 200, 250, 300 Peak voltage / V: 0.11, 0.21, 0.38, 0.42, 0.53, 0.63
(a)
State the independent variable and two control variables in this investigation.
(3)
(b)
Explain why the student drops the magnet from the same height above the coil each time.
(2)
(c)
Identify the anomalous result and state what the student should do about it.
(2)
(d)
Use the data to describe the relationship between the peak voltage and the number of turns.
(2)
(e)
Predict the peak voltage for a coil of 400 turns.
(1)
Show mark scheme
(a)
independent variable: number of turns (on coil);
any two from:
(same) magnet / strength of magnet;
(same) drop height / speed of magnet;
(same) diameter of coil / tube;
(same) orientation of magnet / pole entering first;
Allow
same tube for diameter
Do not accept
peak voltage as control variable
same coil (since turns change)
(b)
height affects the speed of the magnet (through the coil);
(greater) speed gives a larger (induced) voltage (so keep it constant for fair test);
Allow
rate of change of field for speed
Do not accept
so the magnet hits the coil
to make it accurate (unqualified)
(c)
(0.38 V at) 150 turns;
repeat the measurement / ignore it (when finding pattern);
Do not accept
take a mean including the anomaly
(d)
(peak voltage) directly proportional to number of turns;
evidence from data e.g. doubling turns doubles voltage (100 → 0.21 V, 200 → 0.42 V) / V ÷ N constant (≈0.0021 V);
Allow
'as turns increase voltage increases' for 1 mark only if no proportionality
14IGCSE-PHY-6-010Hand-cranked generator lighting a lampPapers 1 & 2Grade 6–710 marks
A hand-cranked generator has a coil of wire that turns between the poles of a permanent magnet. The generator is connected to a lamp. When the handle is turned steadily, the voltage across the lamp is 6.0 V and the current in it is 0.45 A.
(a)
Explain why a voltage is induced in the coil when the handle is turned.
(2)
(b)
Other than turning the handle faster, give two changes to the generator that would increase the induced voltage.
(2)
(c)
Calculate the power of the lamp. Give the unit.
(2)
(d)
Calculate the energy transferred to the lamp in 3.0 minutes.
(3)
(e)
The generator produces an alternating current. State what is meant by an alternating current.
(1)
Show mark scheme
(a)
coil / wire moves through / cuts the magnetic field / field through coil changes;
(so) a voltage is induced (in the coil);
Allow
e.m.f. / p.d. for voltage
Do not accept
current is induced in the magnet
(b)
any two from:
more turns (on the coil);
stronger magnet / field;
coil with larger area;
Allow
iron core inside the coil
Do not accept
bigger magnet (unqualified)
thicker wire
(c)
e.g. P = 0.45 × 6.0
(P =) 2.7;
W / watt(s);
Allow
J/s
unit mark independent of value
(d)
(t =) 180 (s) seen;
substitution into E = I × V × t;
e.g. E = 0.45 × 6.0 × 180
(E =) 490 (J);
Allow
486 (J)
ECF from (c): E = (c) × 180
8.1 (J) for 2 marks (time not converted, POT/time error)
(e)
current that (repeatedly) changes / reverses direction;
15IGCSE-PHY-6-012Bicycle dynamo: output graph and speedPapers 1 & 2Grade 6–710 marks
A bicycle dynamo contains a permanent magnet with one north pole and one south pole. The magnet rotates inside a fixed coil of wire. The magnet is turned by a small drive wheel that presses against the bicycle tyre, so the drive wheel rolls along the tyre without slipping. The circumference of the drive wheel is 9.0 cm. The drive wheel and magnet make one complete turn for each cycle of the output voltage. The graph shows how the output voltage of the dynamo varies with time when the bicycle moves at a steady speed.
(a)
Explain why the rotating magnet induces an alternating voltage in the coil.
(3)
(b)
Use the graph to determine the frequency of the output voltage.
(2)
(c)
Calculate the speed of the edge of the tyre.
(3)
(d)
The cyclist rides at twice the speed. Describe how the graph of output voltage against time would change.
(2)
Show mark scheme
(a)
(rotating magnet) causes the magnetic field through the coil to change;
(changing field) induces a voltage (in the coil);
direction of field (through coil) reverses each half turn / N and S poles pass alternately, so voltage reverses direction;
Do not accept
the coil cuts the magnet
current flows from the magnet into the coil
(b)
(T =) 40 ms / 0.040 s read from graph;
(f =) 25 (Hz);
Allow
ECF for incorrect period read from graph
Do not accept
0.025 Hz
(c)
(drive wheel makes) 25 turns per second / time for one turn = 0.040 s;
substitution into speed = distance ÷ time e.g. 0.090 ÷ 0.040 or 25 × 0.090;
(speed =) 2.3 (m/s);
Allow
2.25 (m/s)
ECF from (b)
225 (m/s) for 2 marks (cm not converted, POT error)
(d)
peak voltage increases (about doubles to 8 V);
time period halves (to 20 ms) / frequency doubles / twice as many cycles shown;
16IGCSE-PHY-6-025Left-hand rule: foil strip between magnet polesPapers 1 & 2Grade 6–71 mark
A strip of aluminium foil is stretched horizontally between two clamps. The strip passes between the north pole and the south pole of a magnet. The diagram shows the arrangement viewed from above. The direction of the current in the strip is shown. In which direction does the strip move when there is a current in it?Show mark scheme
A (vertically downwards, into the page);
B is incorrect because it is the result of using the right hand, or of taking the field direction from S to N
C is incorrect because the force is not along the field lines; the strip is not attracted to the north pole
D is incorrect because the force is at right angles to the field, not along it towards the south pole
17IGCSE-PHY-6-026Vibrating wire driven by an alternating currentPapers 1 & 2Grade 6–79 marks
A thin copper wire is stretched tightly between two supports. The middle of the wire passes through the gap between two flat magnets that face each other, so the magnetic field in the gap is uniform and directed into the page. The wire is connected to a signal generator, which produces an alternating current. The diagram shows the wire viewed from the front, with the direction of the current at one instant.
(a)
State what is produced around the wire when there is a current in it.
(1)
(b)
Use the left-hand rule to determine the direction of the force on the wire at the instant shown.
(1)
(c)
The frequency of the alternating current is 150 Hz. Explain why the wire vibrates and state the frequency of the vibration.
(3)
(d)
The output of the signal generator is increased so that the current is larger. Explain why the wire vibrates with a larger amplitude.
(2)
(e)
The wire is moved so that it passes through a different part of the gap between the magnets, nearer to the top edge of the gap. The current is unchanged. State and explain whether the amplitude of the vibration changes.
(2)
Show mark scheme
(a)
magnetic field;
Allow
circular magnetic field around the wire
Do not accept
electric field
magnetism (unqualified) in the wire
(b)
down(wards) / towards the bottom of the page;
Do not accept
into the page
upwards
(c)
current reverses direction (repeatedly);
(so) force (on wire) reverses direction / wire pushed up and down;
150 Hz;
Allow
force changes direction 150 times each way per second
Do not accept
300 Hz
75 Hz
(d)
larger current gives larger force (on the wire);
(so) wire moves / is displaced further;
Allow
force proportional to current
Do not accept
larger current makes the wire hotter so it expands
(e)
(amplitude) does not change / stays the same;
field is uniform so (field) strength is the same everywhere in the gap, so the force (on the wire) is the same;
Allow
field lines (in the gap) are parallel and equally spaced, so the force is the same
second mark only if 'no change' is given
Do not accept
amplitude increases because the wire is nearer a magnet
amplitude decreases because the field is weaker near the edge of the gap
18IGCSE-PHY-6-027Practical: force to pull a magnet off a steel platePapers 1 & 2Grade 6–79 marks
A student investigates how the force needed to pull a flat magnet off a steel plate depends on the number of sheets of paper between the magnet and the plate. The student hooks a newtonmeter onto the magnet and pulls it straight away from the plate. The student records the reading at the moment the magnet comes off. Number of sheets / –: 0, 2, 4, 6, 8, 10 Force to pull magnet off / N: 3.6, 2.5, 1.8, 1.2, 0.85, 0.60
(a)
State the independent variable and the dependent variable in this investigation.
(2)
(b)
Give one variable that the student should control.
(1)
(c)
Describe how the force changes as the number of sheets increases. Use data from the table in your answer.
(2)
(d)
Explain, in terms of magnetic field lines, why less force is needed when there are more sheets of paper between the magnet and the plate.
(2)
(e)
The student repeats the investigation with an aluminium plate instead of the steel plate. Predict the force needed to pull the magnet off the aluminium plate with no paper between them. Explain your answer.
(2)
Show mark scheme
(a)
independent: number of sheets (of paper) / distance between magnet and plate;
dependent: force (needed to pull magnet off) / newtonmeter reading;
Do not accept
magnet as independent variable
(b)
any one from:
same magnet;
same (steel) plate;
same type / thickness of paper;
direction of pull (at right angles to plate);
Allow
same newtonmeter
pull at the same (slow) rate
Do not accept
amount of force
number of sheets
(c)
force decreases as number of sheets increases;
by smaller amounts each time / not (inversely) proportional / non-linear, with data quoted;
Allow
force roughly halves for every 4 sheets added
change between 0 and 2 sheets is 1.1 N but between 8 and 10 sheets is 0.25 N
Do not accept
force inversely proportional to number of sheets
force decreases at a constant rate
(d)
field lines further apart (further from the magnet) so field is weaker (at the plate);
less magnetism induced in the steel (plate) / weaker attraction;
Allow
field strength decreases with distance from the magnet
Do not accept
paper blocks the magnetic field
paper is magnetic
(e)
0 (N) / (almost) zero / magnet does not stick;
aluminium is not a magnetic material;
Allow
no magnetism is induced in aluminium
only the weight of the magnet if the plate is horizontal
19IGCSE-PHY-6-028Magnet oscillating on a spring above a coilPapers 1 & 2Grade 6–710 marks
A bar magnet hangs vertically from a spring with its north pole at the bottom. The lower end of the magnet is just above a coil of wire. The coil is connected to a data logger that records the voltage across the coil. The magnet is pulled down and released. It oscillates up and down, so that its lower end moves into and out of the coil.
(a)
Explain why a voltage is induced in the coil while the magnet oscillates.
(2)
(b)
The data logger shows that the induced voltage is alternating. Explain why.
(2)
(c)
The induced voltage is zero at the instant the magnet is at its lowest point. Explain why.
(2)
(d)
As time passes, the oscillations of the magnet get smaller. State and explain how this affects the peak induced voltage.
(2)
(e)
The coil is now connected to a resistor. The oscillations of the magnet die away more quickly than before. Suggest why, in terms of energy.
(2)
Show mark scheme
(a)
magnet moves relative to coil / magnetic field through coil changes;
changing field (through coil) / cutting field lines induces a voltage;
Allow
coil cuts magnetic field lines
Do not accept
the magnet charges the coil
voltage induced because the magnet is near the coil (no movement)
(b)
magnet changes direction (moves in then out);
(so) field through coil increases then decreases / direction of induced voltage reverses;
Do not accept
the north and south poles enter the coil alternately
(c)
magnet (momentarily) stationary;
field through coil not changing / no field lines cut (so no induced voltage);
Allow
magnet speed is zero at the turning point
Do not accept
the magnet is furthest into the coil so the field is zero
(d)
peak voltage decreases;
magnet moves more slowly / field changes more slowly (through coil);
Allow
lower rate of change of field
Do not accept
voltage stays the same because the same magnet is used
(e)
(induced) current in coil/resistor transfers energy (thermally) to the resistor / surroundings;
this energy comes from the (kinetic / elastic) energy of the oscillating magnet;
Allow
energy is conserved so the magnet's energy decreases faster
20IGCSE-PHY-6-029Shake torch: charge, energy and running timePapers 1 & 2Grade 6–711 marks
A shake torch contains a strong magnet that slides back and forth inside a plastic tube when the torch is shaken. A coil of wire is wound around the middle of the tube. The induced current is converted to direct current and used to charge a rechargeable battery. The battery then lights an LED.
(a)
Explain why shaking the torch induces a voltage in the coil.
(2)
(b)
The torch is shaken for 40 s. The average charging current is 0.15 A. Calculate the charge that flows into the battery. Give the unit.
(3)
(c)
The voltage of the battery is 3.6 V. Calculate the energy stored in the battery during the 40 s.
(2)
(d)
The LED has a power of 0.090 W. Calculate how long the LED stays lit using the energy stored in (c).
(3)
(e)
Suggest one change to the design of the torch that would increase the charging current for the same shaking.
(1)
Show mark scheme
(a)
magnet moves through / relative to the coil;
(so) field through coil changes / coil cuts field lines (inducing voltage);
Do not accept
shaking charges the magnet
friction produces the voltage
(b)
e.g. Q = 0.15 × 40
substitution into Q = I × t;
(Q =) 6.0;
C / coulomb(s);
Allow
A s
unit mark independent of value
Do not accept
J
W
(c)
e.g. E = 6.0 × 3.6
substitution into E = Q × V;
(E =) 22 (J);
Allow
21.6 (J)
ECF from (b): E = (b) × 3.6
E = I × V × t = 0.15 × 3.6 × 40 = 21.6 (J) for both marks
(d)
rearrangement: t = E ÷ P;
e.g. t = 21.6 ÷ 0.090
substitution;
(t =) 240 (s);
Allow
ECF from (c)
244 (s) from using 22 J
4.0 minutes
Do not accept
0.0042 (s) (inverted)
(e)
more turns on the coil / stronger magnet;
Allow
move the magnet faster (e.g. longer tube) only if linked to design
21IGCSE-PHY-6-011Practical: measuring the force on a wire with a balancePapers 1 & 2Grade 8–910 marks
A student measures the force on a current-carrying wire using a top-pan balance. A stiff horizontal wire is fixed to an insulating block on the balance pan. A magnet is held in a clamp so that the wire is between its poles; the magnet does not touch the balance. The diagram shows the arrangement seen with the wire end-on. The direction of the current is not shown. The student records the balance reading for different currents in the wire. Current / A: 0.0, 0.5, 1.0, 1.5, 2.0, 2.5 Balance reading / g: 85.24, 85.36, 85.49, 85.60, 85.72, 85.85 g = 10 N/kg
(a)
Explain how the readings show that the magnetic force on the wire is downwards.
(2)
(b)
Use the left-hand rule to deduce the direction of the current in the wire.
(1)
(c)
Calculate the magnetic force on the wire when the current is 2.5 A.
(3)
(d)
Use the data to show that the force on the wire is directly proportional to the current.
(2)
(e)
The student reverses the current and sets it to 2.0 A. Predict the new balance reading.
(2)
Show mark scheme
(a)
(balance) reading increases (as current increases);
so (extra) force on the pan / wire is downwards / wire pushed down onto pan;
Do not accept
the wire becomes heavier
the mass of the wire increases
(b)
into the page;
Allow
away from the viewer / ⊗
Do not accept
out of the page
(c)
change in reading = 0.61 (g) seen;
conversion to kg and use of W = m × g e.g. 0.00061 × 10;
(F =) 0.0061 (N);
Allow
6.1 × 10⁻³ N
6.1 (N) or 0.061 (N) for 2 marks (POT error)
ECF for a misread change in reading
Do not accept
0.8585 N (total reading used)
(d)
calculates F ÷ I (or Δm ÷ I) for at least two currents;
values (approximately) constant / zero force at zero current so proportional;
Allow
e.g. 0.0025 N/A at 1.0 A and 0.0024 N/A at 2.0 A
equivalent ratio method e.g. doubling current 1.0 → 2.0 A approximately doubles change in reading 0.25 → 0.48 g
Do not accept
'reading increases as current increases' (not proportional)
22IGCSE-PHY-6-013Identifying a magnet, iron and aluminium barPapers 1 & 2Grade 8–96 marks
A student is given three bars, X, Y and Z, that look identical because they are painted. One bar is a permanent magnet, one is unmagnetised soft iron and one is aluminium. The student has no other equipment.
(a)
Explain how the student can use only the three bars to identify which bar is the magnet, which is soft iron and which is aluminium.
(6)
Show mark scheme
(a)
Indicative content:
MP1. bring (ends of) pairs of bars together / test each pair;
MP2. aluminium (is not magnetic so it) is not attracted to either bar / one bar shows no force with both others;
MP3. magnet and iron attract each other (whichever ends are used);
MP4. because the magnet induces magnetism in the iron / iron becomes a temporary magnet;
MP5. (so) attraction alone does not show which is the magnet / repulsion is not possible with only one magnet;
MP6. touch the end of one bar to the middle of the other;
MP7. field of a magnet strongest at the poles / ends and weak at the middle;
MP8. if end of first bar attracts middle of second strongly, first bar is the magnet (otherwise the second is);
Allow
award 1 mark per MP to a maximum of 6 using the levels below to check overall quality
MP8 may be expressed as: little / no attraction when end of iron touches middle of magnet
suspension of a bar to show it aligns north–south as an alternative to MP6–MP8 (max 2 marks)
Do not accept
the magnet repels the iron
aluminium is attracted because it is a metal
use of a compass or iron filings (no other equipment)
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated points e.g. testing pairs or stating aluminium is not magnetic; no method to distinguish the magnet from the iron.
Level 2 (3–4)
Identifies the aluminium and recognises that the magnet and iron attract each other, with some explanation (induced magnetism) OR a partial end-to-middle method lacking reasoning.
Level 3 (5–6)
Logically sequenced method identifying all three bars, including the end-to-middle test explained by the field being strongest at the poles, and why attraction alone is not conclusive.
23IGCSE-PHY-6-030Practical: induced voltage and the speed of a magnetPapers 1 & 2Grade 8–911 marks
A student investigates how the peak voltage induced in a coil depends on the speed of a magnet moving past it. A strong magnet is fixed underneath a trolley. The trolley is released from different heights on a ramp and then travels along a horizontal runway. A flat coil is fixed under the runway and connected to a data logger, which records the peak induced voltage as the magnet passes over the coil. A card of length 10.0 cm on the trolley passes through a light gate just before the coil. The light gate records the time taken for the card to pass through it. Time for card to pass light gate / s: 0.50, 0.25, 0.20, 0.16, 0.125 Peak voltage / mV: 8.1, 15.8, 20.3, 24.7, 32.2
(a)
Calculate the speed of the trolley when the time for the card to pass the light gate is 0.16 s.
(2)
(b)
The student claims: 'The peak voltage is directly proportional to the speed of the magnet.' Evaluate this claim using the data.
(4)
(c)
Explain why the peak voltage is larger when the magnet moves faster.
(2)
(d)
Explain why the light gate is placed just before the coil rather than at the bottom of the ramp.
(2)
(e)
Use your answer to (b) to predict the peak voltage when the trolley moves over the coil at 0.30 m/s.
(1)
Show mark scheme
(a)
e.g. v = 0.100 ÷ 0.16 / 10.0 ÷ 0.16 = 62.5 cm/s
substitution into speed = distance ÷ time;
(v =) 0.63 (m/s);
Allow
0.625 (m/s)
62.5 cm/s
62.5 (m/s) scores 1 (POT error, cm not converted)
Do not accept
62.5 (m/s) for full marks
(b)
speeds calculated for at least three readings;
V ÷ v calculated for at least three readings / doubling speed doubles voltage shown with data;
ratio (approximately) constant at about 40 mV per m/s;
(so) claim supported (within experimental uncertainty) / need reading at origin or more readings to confirm;
Allow
graph of V against v drawn: straight line through origin for marks 3 and 4
ratio calculated using time: V × t constant (≈ 4.0 mV s) for marks 2 and 3
claim supported only because voltage increases with speed (no proportionality test): max 1, for the speed calculation
Do not accept
voltage increases as speed increases, so it is proportional
(c)
field (through coil) changes faster / field lines cut faster;
(so) larger (induced) voltage;
Allow
greater rate of change of magnetic field
Do not accept
faster magnet has more energy so more voltage (without rate idea)
(d)
trolley slows down / speed changes along runway (due to friction);
(so) need speed at the coil / speed at the ramp would be (too) high;
Allow
speed measured at the coil is the speed that affects the voltage
24IGCSE-PHY-6-031Electric guitar pickupPapers 1 & 2Grade 8–96 marks
An electric guitar has steel strings. Under each string is a pickup. A pickup is a small permanent magnet with a coil of fine wire wound around it. The coil is connected to an amplifier. When a string is plucked, the part of the string above the pickup vibrates up and down, towards and away from the pickup, at a frequency of a few hundred hertz. The pickup does not work if the steel strings are replaced by nylon strings.
(a)
Explain how the pickup produces an alternating voltage when a steel string vibrates, and how the size and frequency of this voltage depend on the vibration of the string. Your answer should include why the pickup does not work with nylon strings.
(6)
Show mark scheme
(a)
Indicative content:
MP1. steel is a magnetic material / steel string is attracted by the magnet;
MP2. magnetism is induced in the string (above the pickup) / string becomes a (temporary) magnet;
MP3. vibrating string moves towards and away from the coil, so the magnetic field through the coil changes;
MP4. a changing field through the coil induces a voltage;
MP5. field increases then decreases (as string moves towards then away) so the voltage reverses direction / is alternating;
MP6. frequency of voltage is the same as the frequency of vibration of the string;
MP7. larger amplitude / harder pluck means string moves faster, so field changes faster and voltage is larger;
MP8. voltage decreases as the vibration dies away;
MP9. nylon is not magnetic so no magnetism is induced / nylon does not change the field through the coil, so no voltage;
Allow
award 1 mark per MP to a maximum of 6 using the levels below to check overall quality
MP3 may be expressed as the string's field lines cutting the coil
MP9 can be awarded even if MP1/MP2 are not given explicitly
Do not accept
the string carries a current to the coil
the vibration of the string produces sound in the coil that becomes a voltage
nylon does not work because it is an electrical insulator (without magnetic idea)
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated points, e.g. steel is magnetic, or a voltage is induced when something moves near a coil; no linked explanation.
Level 2 (3–4)
Explains that the vibrating string changes the field through the coil and induces a voltage, with some reference to induced magnetism OR to the alternating nature/frequency OR to the nylon string; reasoning partly linked.
Level 3 (5–6)
Coherent linked explanation: induced magnetism in the steel string → changing field through the coil → induced voltage that alternates at the string's frequency, with size linked to the amplitude/speed of vibration and a correct reason why nylon does not work.
25IGCSE-PHY-6-032Motorised roller blind: efficiency and reversingPapers 1 & 2Grade 8–910 marks
A motorised roller blind uses a small d.c. electric motor to wind the blind up. The motor is connected to a 12 V supply. While the blind is being raised, the current in the motor is 0.75 A and the blind takes 8.0 s to rise. The blind has a mass of 2.4 kg. As it is raised, its centre of mass rises by 0.75 m. g = 10 N/kg
(a)
Calculate the efficiency of the motor system while it raises the blind.
(5)
(b)
To lower the blind, the control circuit reverses the connections to the motor. Explain why this makes the motor turn in the opposite direction.
(3)
(c)
The manufacturer replaces the magnet in the motor with a stronger magnet. The current is unchanged. Explain why the motor can now raise the blind more quickly.
(2)
Show mark scheme
(a)
e.g. E = 0.75 × 12 × 8.0
substitution into E = I × V × t;
(input energy =) 72 (J);
e.g. GPE = 2.4 × 10 × 0.75
(useful energy / GPE =) 18 (J);
substitution into efficiency equation, e.g. 18 ÷ 72;
(efficiency =) 0.25 / 25%;
Allow
power route: input power = 9.0 W, useful power = 2.25 W, efficiency 0.25 – full marks
ECF from incorrect energies for marks 4 and 5
25% with no working scores 5
4.0 or 400% (inverted ratio) scores max 3
Do not accept
efficiency greater than 1 / 100%
(b)
current (in coil) is reversed;
force on each side of the coil reverses (with field unchanged);
(so) turning effect / rotation is reversed;
Allow
using the left-hand rule, reversing the current reverses the force
Do not accept
reversing the connections reverses the magnetic field of the magnet
reversing the voltage makes the motor run as a generator
(c)
larger force (on each side of the coil) for the same current;
26IGCSE-PHY-6-033Exercise bike: generator used as the brakePapers 1 & 2Grade 8–910 marks
Some exercise bikes have no friction brake. Instead, the pedals turn a generator: a coil of wire rotates between the poles of a permanent magnet. The generator is connected to a resistor inside the control unit. When the rider selects a higher 'effort level', the control unit replaces the resistor with one of lower resistance.
(a)
Explain how rotating the coil produces an alternating voltage.
(3)
(b)
The rider keeps pedalling at the same rate and selects a higher effort level. Explain why it becomes harder to turn the pedals.
(5)
(c)
The peak voltage is directly proportional to the number of rotations of the coil per second. When the coil makes 15 rotations per second, the peak voltage is 6.0 V. The display in the control unit only works when the peak voltage is at least 9.0 V. Determine the minimum number of rotations per second needed for the display to work.
(2)
Show mark scheme
(a)
coil (sides) cut field lines / field through coil changes;
(so) voltage is induced;
each side moves in the opposite direction (through the field) every half turn so voltage reverses direction;
Allow
field through the coil reverses every half turn so voltage reverses
Do not accept
current reverses because of the commutator
(b)
any five from:
lower resistance gives larger current (in the coil) (for the same induced voltage);
current-carrying coil in a magnetic field experiences a force / motor effect;
larger current gives larger force (on the coil);
this force opposes / acts against the rotation (so rider must push harder);
more electrical energy transferred (to the resistor) per second / larger electrical power;
(by conservation of energy) rider must do more work per second / supply more power;
Allow
voltage is the same because the coil turns at the same rate
more energy is transferred to the surroundings by heating in the resistor per second
Do not accept
lower resistance increases the friction in the generator
the magnet attracts the coil more (unqualified)
lower resistance gives a smaller current
(c)
correct method, e.g. 15 × 9.0 ÷ 6.0 or 9.0 ÷ (6.0 ÷ 15);
27IGCSE-PHY-6-014Increasing the strength of an electromagnetPaper 2 onlyGrade 4–51 mark
A student makes an electromagnet by winding insulated wire around an iron nail and connecting the wire to a cell. Which change will make the electromagnet pick up more paper clips?
Show mark scheme
A (increasing the number of turns of wire on the nail);
B is incorrect because reversing the current only swaps the poles; the strength is unchanged
C is incorrect because aluminium is not magnetic, so the electromagnet would be weaker
D is incorrect because a lower voltage gives a smaller current and a weaker electromagnet
29IGCSE-PHY-6-035Transformer for low-voltage garden lightsPaper 2 onlyGrade 4–58 marks
Low-voltage garden lights are supplied through a transformer that is connected to the 230 V a.c. mains. The diagram shows the structure of the transformer.
(a)
Name the part labelled Y and the material it is made from.
(2)
(b)
Explain how the diagram shows that this is a step-down transformer.
(2)
(c)
The primary coil has 2300 turns and the secondary coil has 120 turns, so the output is 12 V. The manufacturer makes a version for a country where the mains voltage is 115 V a.c. The output must still be 12 V and the primary coil is not changed. Explain how the number of turns on the secondary coil must be changed. You do not need to use a calculator.
(3)
(d)
The coils are made from copper wire with a thin layer of insulation. Suggest why the wire must be insulated.
(1)
Show mark scheme
(a)
core;
(soft) iron;
Do not accept
steel
copper
magnet
(b)
secondary coil has fewer turns than primary coil;
(so) output voltage is less than input voltage;
Allow
turns ratio Ns : Np less than 1
Do not accept
secondary coil is smaller / thinner
(c)
output (secondary) voltage ÷ input (primary) voltage = secondary turns ÷ primary turns / turns ratio must equal the voltage ratio;
input voltage is halved (230 V → 115 V) (so the voltage ratio Vs/Vp doubles);
(so) number of secondary turns must be doubled / 240 (turns);
Allow
Ns = 2300 × 12 ÷ 115 = 240 for marks 1 and 3
secondary needs twice as many turns for the same output
Do not accept
halve the secondary turns / 60 turns
keep 120 turns because the output is the same
(d)
so current flows through every turn / prevents current passing directly between turns (short circuit) / into the core;
30IGCSE-PHY-6-015Field patterns of a straight wire and a flat coilPaper 2 onlyGrade 6–78 marks
A long straight wire passes vertically through the centre of a horizontal card. Four plotting compasses P, Q, R and S are placed on the card around the wire. The diagram shows the card viewed from above. The current in the wire flows upwards, out of the page. The current is large enough for the effect of the Earth's magnetic field to be ignored.
(a)
Describe the shape of the magnetic field pattern around the wire.
(2)
(b)
(i) State whether the field lines are clockwise or anticlockwise as seen in the diagram. (ii) State the direction in which the needle of compass P points.
(2)
(c)
State what happens to the compass needles when the current is reversed.
(1)
(d)
State how the strength of the magnetic field changes with distance from the wire.
(1)
(e)
The wire is then bent into a flat circular coil. Describe the magnetic field pattern near the centre of the coil.
(2)
Show mark scheme
(a)
(concentric) circles (centred on the wire);
(circles) get further apart further from the wire;
Allow
rings for circles
Do not accept
spiral
radial lines
(b)
(i) anticlockwise;
(ii) towards top of page / up the page;
Allow
ECF (ii) from (i): down the page if clockwise given
Do not accept
towards the wire
away from the wire
(c)
(all) needles reverse direction / turn through 180°;
Allow
point the opposite way
Do not accept
needles stop moving
needles point towards the wire
(d)
(field gets) weaker further from the wire;
Allow
field strength decreases with distance
(e)
any two from:
(nearly) straight and parallel field lines at the centre;
(field lines) perpendicular to the plane of the coil / along the axis of the coil;
circular (loops) around each side of the wire;
Allow
similar to the field of a short bar magnet / solenoid
31IGCSE-PHY-6-016Electromagnetic door lockPaper 2 onlyGrade 6–79 marks
An electromagnetic door lock uses an electromagnet fixed to the door frame. When there is a current in the electromagnet it holds a soft-iron plate fixed to the door, keeping the door shut. The diagram shows the electromagnet. The current flows downwards in the front part of each turn of the coil.
(a)
Explain why the core of the electromagnet is made of soft iron rather than steel.
(2)
(b)
Determine the magnetic pole at end R of the electromagnet.
(1)
(c)
Explain why the soft-iron plate is attracted to end R.
(2)
(d)
An electrician connects the supply the other way round. State and explain whether the lock still holds the door shut.
(2)
(e)
The plate is moved a few millimetres further away from end R. State and explain how this affects the force holding the door shut.
(2)
Show mark scheme
(a)
soft iron easily magnetised / becomes a strong (temporary) magnet;
soft iron loses its magnetism when current is switched off (so door can be opened) but steel stays magnetised;
Allow
steel is magnetically hard for MP2
Do not accept
soft iron is easier to shape
steel is too heavy
(b)
north / N;
Do not accept
S
(c)
magnetism induced in the plate (by field of the electromagnet);
near side of plate becomes a south pole / opposite pole to R (so attracted);
Allow
ECF from (b) for the opposite pole
Do not accept
the plate is a permanent magnet
the plate is charged
(d)
yes it still holds;
(poles reverse but) the induced pole in the plate also reverses so it is still opposite / still attracts;
Allow
soft iron is attracted by either pole
Do not accept
no, because R becomes a south pole (as the only answer)
(e)
force (on plate) is smaller;
(magnetic) field is weaker further from the pole / field lines further apart so less magnetism induced in the plate;
32IGCSE-PHY-6-017Mobile phone charger transformerPaper 2 onlyGrade 6–710 marks
A mobile phone charger contains a transformer that changes the 230 V a.c. mains supply to 5.0 V a.c. This is then converted to d.c. for the phone. The primary coil has 4600 turns.
(a)
Calculate the number of turns on the secondary coil.
(3)
(b)
The current in the secondary coil is 2.0 A. Calculate the current in the primary coil. Assume the transformer is 100% efficient.
(3)
(c)
Explain why the transformer only works with an alternating input voltage.
(3)
(d)
State why the two coils are wound on an iron core.
(1)
Show mark scheme
(a)
substitution e.g. 230 / 5.0 = 4600 / Ns;
rearrangement e.g. Ns = 4600 × 5.0 / 230;
(Ns =) 100;
Allow
full marks for correct answer with no working
Do not accept
4600 × 230 / 5.0 = 211 600
(b)
substitution into Vp × Ip = Vs × Is e.g. 230 × Ip = 5.0 × 2.0;
rearrangement e.g. Ip = 10 / 230;
(Ip =) 0.043 (A);
Allow
0.0435 (A)
0.04 (A)
92 (A) for 1 mark (inverted ratio)
(c)
a.c. (in primary) produces a changing magnetic field (in the core);
changing field passes through / links the secondary coil;
(changing field) induces a (alternating) voltage in the secondary / d.c. gives constant field so no induced voltage;
Do not accept
current flows through the core from one coil to the other
electrons jump between the coils
(d)
(iron is easily magnetised so it) carries / concentrates / links the magnetic field (from primary to secondary);
Allow
increases the strength of the field (through the secondary)
33IGCSE-PHY-6-019Electron beam deflected by a magnetic fieldPaper 2 onlyGrade 6–79 marks
In an evacuated glass tube, an electron gun produces a narrow beam of electrons travelling to the right. The beam enters a region of uniform magnetic field directed into the page, as shown in the diagram.
(a)
The beam of electrons is an electric current. State the direction of the conventional current in the beam and explain your answer.
(2)
(b)
Use the left-hand rule to determine the direction of the force on the electrons as they enter the field.
(1)
(c)
Explain why the beam follows a curved path in the field.
(2)
(d)
The tube is turned so that the electrons travel parallel to the magnetic field lines. State and explain what happens to the beam.
(2)
(e)
The electron beam is replaced by a beam of positive ions travelling to the right through the same field. State the direction of the force on the ions as they enter the field.
(1)
(f)
State how the path of the electron beam changes if the magnetic field is made stronger.
(1)
Show mark scheme
(a)
to the left;
electrons are negative / conventional current is opposite to electron flow;
Do not accept
to the right
(b)
downwards / towards bottom of page;
Allow
ECF from (a): if current to the right given, upwards
Do not accept
into the page
to the left
(c)
force (always) perpendicular / at right angles to direction of motion;
(so force) changes direction of motion (continuously) / beam curves (downwards);
Do not accept
force slows the electrons down
force pulls electrons towards a pole
(d)
(beam) not deflected / travels in a straight line;
no force when (charged particle) moves parallel to field;
Do not accept
beam deflected more
(e)
upwards / towards top of page;
Allow
opposite direction to (b) for ECF
(f)
(beam) curves more / is deflected more / sharper curve;
34IGCSE-PHY-6-036Practical: comparing electromagnet core materialsPaper 2 onlyGrade 6–710 marks
A student investigates how the material of the core affects an electromagnet. The student uses cores of the same size and shape made of different materials. For each core, the student measures the force needed to pull a small soft-iron disc away from the end of the electromagnet while the current is on. The student then switches the current off and measures the force again. Core material / –: none (air), aluminium, soft iron, steel Force with current on / N: 0.3, 0.3, 5.8, 4.1 Force after current switched off / N: 0.0, 0.0, 0.1, 2.6
(a)
Describe how the student should construct the electromagnet and set the current to the same value for each core.
(3)
(b)
State two other variables that the student must keep the same.
(2)
(c)
Explain why the aluminium core gives the same force as no core.
(2)
(d)
An electromagnet is used on a crane in a scrapyard to lift scrap iron and then drop it. Use the data to explain which core material should be used.
(3)
Show mark scheme
(a)
(insulated) wire coiled / wound around the core;
connected to a (d.c.) supply / cell with a switch;
ammeter (and variable resistor) in series to set the same current;
Allow
diagram showing these features
Do not accept
uninsulated wire
voltmeter in series
(b)
any two from:
number of turns (on the coil);
same disc;
position of disc / distance from core;
size / shape of core;
direction / method of pulling;
Do not accept
current (already given)
core material
(c)
aluminium is not a magnetic material / not magnetised;
(so) it does not add to / strengthen the field of the coil;
Allow
no magnetism induced in aluminium
Do not accept
aluminium blocks the field
aluminium is a poor conductor
(d)
soft iron;
largest force when current is on (5.8 N) / strongest electromagnet;
(almost) no force after switching off (0.1 N) so it releases the scrap, whereas steel keeps 2.6 N;
Allow
soft iron is magnetically soft – loses its magnetism
35IGCSE-PHY-6-038Charging an electric toothbrush without contactsPaper 2 onlyGrade 6–78 marks
An electric toothbrush is charged by standing it on a charging base. Inside the base is a coil connected to an alternating supply. Inside the handle of the toothbrush is a second coil wound around an iron rod. The rod sits inside the base coil when the toothbrush is on the base. The two coils are inside separate plastic cases. There is no electrical connection between the base and the toothbrush.
(a)
Explain how a voltage is induced in the coil inside the handle.
(3)
(b)
Explain why the plastic cases do not stop the charger from working.
(1)
(c)
The toothbrush is held 3 cm above the base, so the iron rod is no longer inside the base coil. The supply to the base is still switched on. Explain why the voltage induced in the handle coil is much smaller.
(2)
(d)
Explain why the iron rod inside the handle coil increases the voltage induced in it.
(2)
Show mark scheme
(a)
alternating current in base coil produces a (changing) magnetic field;
field passes through the (iron rod and) handle coil;
changing field (through handle coil) induces a voltage;
Allow
alternating field / field that changes direction (for mark 1)
Do not accept
current flows from the base coil to the handle coil
electrons jump across the gap
(b)
magnetic field passes through plastic / plastic does not block the magnetic field;
Allow
plastic is not a magnetic material so it has no effect on the field
Do not accept
plastic is an insulator so it carries the current
(c)
(magnetic) field (from base coil) is weaker further from the base coil / less of the field passes through the handle coil;
(so) smaller change of field through the handle coil (so smaller induced voltage);
Allow
field lines are further apart away from the coil
rate of change of field through the handle coil is smaller
Do not accept
no voltage is induced because the coils are not touching
the air blocks the magnetic field
(d)
magnetism is induced in the iron rod / iron rod becomes magnetised (by the field);
(so) stronger (changing) field through the handle coil / rod concentrates the field (so larger voltage);
Allow
iron is a magnetic material / magnetically soft so it is easily magnetised (for mark 1)
36IGCSE-PHY-6-039Rapid charging station for electric carsPaper 2 onlyGrade 6–711 marks
A rapid charging station for electric cars is connected to the local distribution network, which operates at 11 kV. A transformer at the charging station changes 11 kV to 400 V for the charger. The charger delivers a power of 50 kW to a car battery.
(a)
The charging station is 3 km from the nearest substation. Explain why it is better to step down the voltage at the charging station rather than at the substation.
(3)
(b)
The transformer is 98% efficient. The power output from the transformer is 50 kW. Calculate the current in the primary coil of the transformer.
(4)
(c)
Calculate the ratio of the number of turns on the primary coil to the number of turns on the secondary coil.
(2)
(d)
The battery of a car needs 1.6 × 10⁸ J of energy to be fully charged. Calculate the time, in minutes, needed to fully charge the battery at 50 kW.
(2)
Show mark scheme
(a)
400 V cable would carry a larger current (for the same power) / 11 kV gives a smaller current;
larger current causes more heating / energy loss in the cable;
so keep high voltage over the 3 km and step down close to where it is used;
Allow
P = IV used to show current is 27.5 times bigger at 400 V
thinner/cheaper cables possible at 11 kV
Do not accept
high voltage is safer
high voltage travels faster
less resistance at high voltage
(b)
input power = 50 000 ÷ 0.98;
(input power =) 51 000 (W);
e.g. Ip = 51 020 ÷ 11 000
use of Vp × Ip = input power;
(Ip =) 4.6 (A);
Allow
4.64 (A)
4.5 (A) / 4.55 (A) (efficiency ignored) scores 2
ECF from incorrect input power for marks 3 and 4
Do not accept
4.5 (A) for full marks
125 (A) (secondary current)
(c)
substitution with consistent units, e.g. 11 000 ÷ 400;
37IGCSE-PHY-6-018High-voltage transmission and power lossPaper 2 onlyGrade 8–911 marks
A power station transmits 50 MW of electrical power to a city. The transmission cables have a total resistance of 2.0 Ω. The generators produce a voltage of 25 kV. A step-up transformer increases this to 400 kV before transmission.
(a)
Explain why a step-up transformer is used before the electricity is transmitted.
(3)
(b)
Calculate the power wasted in the cables when the power is transmitted at 400 kV.
(4)
(c)
Show that the power wasted in the cables at 400 kV is about 250 times smaller than it would be if the power were transmitted at 25 kV.
(3)
(d)
Explain why step-down transformers are used before the electricity reaches homes.
(1)
Show mark scheme
(a)
(step-up) increases voltage so current is smaller (for same power);
(smaller current) means less heating in the cables;
(so) less energy / power wasted (in transmission);
Allow
reference to P = IV for MP1
allow thinner cables can be used as alternative to MP3
Do not accept
step-up increases the power
high voltage travels faster
(b)
(I =) 125 (A);
(voltage across cables =) 125 × 2.0 = 250 (V);
use of P = I × V e.g. 125 × 250;
(P =) 31 000 (W);
Allow
31 250 (W)
31 kW
ECF from incorrect current
alternative P = I²R route (125² × 2.0) leading to 31 000 W: full marks
38IGCSE-PHY-6-020Practical: testing the transformer equationPaper 2 onlyGrade 8–911 marks
A student investigates a transformer made from two coils on an iron core. The primary coil has 400 turns and is connected to a 2.0 V a.c. supply. The student changes the number of turns on the secondary coil and measures the output voltage. Secondary turns / –: 100, 200, 300, 400, 500, 600 Output voltage / V: 0.46, 0.93, 1.39, 1.85, 2.31, 2.78
(a)
Calculate the output voltage predicted by the transformer equation for a secondary coil of 500 turns.
(2)
(b)
The student claims: 'My transformer obeys the transformer equation exactly.' Evaluate this claim using the data.
(3)
(c)
Suggest one reason why the measured output voltages are lower than predicted.
(1)
(d)
With 600 secondary turns, the current in the secondary coil is 0.12 A and the current in the primary coil is 0.20 A. Calculate the efficiency of the transformer.
(3)
(e)
The student keeps the same primary coil and supply and wants an output voltage of 6.0 V. Use the data to estimate the number of turns needed on the secondary coil.
(2)
Show mark scheme
(a)
substitution e.g. 2.0 / Vs = 400 / 500;
(Vs =) 2.5 (V);
(b)
(all) measured values smaller than predicted (so claim not correct);
calculation to support e.g. 2.31 / 2.5 = 0.92 / predicted values 0.50, 1.00 … 3.00 V compared;
but Vs (still) proportional to Ns / Vs ÷ Ns constant ≈ 0.0046 (V per turn) / consistent percentage difference;
Allow
difference of about 7–8% for MP2
ECF from (a)
Do not accept
claim correct because voltage increases with turns
(c)
not all the magnetic field (from primary) passes through secondary / energy wasted as heat (in coils or core) / transformer not 100% efficient;
Allow
resistance of coils
Do not accept
voltmeter not zeroed (unless explained as systematic error)
(d)
(output power =) 2.78 × 0.12 = 0.33 (W);
(input power =) 2.0 × 0.20 = 0.40 (W);
(efficiency =) 0.83 / 83%;
Allow
0.834 / 83.4%
ECF for one incorrect power
Do not accept
efficiency greater than 100%
(e)
use of measured output voltage per turn e.g. 6.0 ÷ (2.78 ÷ 600) / 6.0 ÷ 0.0046;
(Ns =) 1300 (turns);
Allow
1280–1310 (turns)
ECF from a value of Vs ÷ Ns calculated in (b)
1200 (turns) from the ideal transformer equation scores 1 only
39IGCSE-PHY-6-040Protons from the Sun in the Earth's magnetic fieldPaper 2 onlyGrade 8–98 marks
The solar wind is a stream of fast-moving charged particles, such as protons and electrons, from the Sun. The diagram shows some of the Earth's magnetic field lines, in the plane of the page. Outside the Earth, the field lines go from the geographic south of the Earth to the geographic north. A proton at X, above the equator, is moving directly towards the Earth. A proton at Y, close to the north polar region, is moving along a field line towards the Earth.
(a)
Use the left-hand rule to determine the direction of the force on the proton at X. Explain how you used the direction of motion of the proton.
(2)
(b)
State the direction of the force on an electron at X moving with the same velocity as the proton.
(1)
(c)
Aurorae are lights produced when charged particles from the Sun hit the upper atmosphere. Aurorae are seen mostly near the poles, and very rarely near the equator. Explain why, using the motion of the protons at X and Y.
(4)
(d)
The solar wind also contains some neutral atoms. Explain why the Earth's magnetic field does not deflect them.
(1)
Show mark scheme
(a)
current is in the direction the proton moves (because it is positive);
into the page;
Allow
conventional current to the left, towards the Earth
Do not accept
out of the page
downwards
(b)
out of the page;
Allow
ECF: opposite to answer to (a)
Do not accept
into the page
(c)
at X motion is at right angles to the field so there is a force;
force (at right angles to motion) changes direction / deflects the protons (so they do not reach the atmosphere);
at Y motion is parallel to the field (lines);
so no force / not deflected so protons reach the atmosphere near the poles;
Allow
near the poles field lines enter the Earth (almost vertically)
particles are guided along field lines towards the poles
Do not accept
the field attracts the protons to the poles because the poles are magnetic
protons are attracted to the north pole because it is negative
(d)
no (overall) charge / neutral so no force from a magnetic field;
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