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MYP 3 · Science

Electricity and Magnetism

100 questions across 5 sub-topics

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Electric Current and Circuits Series and Parallel Circuits Voltage and Resistance Magnets and Magnetic Fields Electromagnets

Electric Current and Circuits 20 questions

QUESTION 1 4 marks Criterion A
Easy
XZY

The diagram shows a simple circuit.

a. Name component X.
[1]
b. Name component Y.
[1]
c. Name component Z.
[1]
d. Describe what happens when the switch is closed.
[1]
Show complete worked solution
(a)
Component X is a switch — it opens or closes the circuit, controlling whether current can flow.
(b)
Component Y is a cell (battery) — it provides the potential difference (voltage) that pushes charge around the circuit.
(c)
Component Z is a bulb (lamp) — it transfers electrical energy to light and heat energy when current flows through it.
(d)
When the switch is closed, the circuit becomes a complete (closed) loop, so current flows and the bulb lights up.
QUESTION 2 2 marks Criterion A
Easy

State the unit used to measure electric current, and name the instrument used to measure it. State how this instrument is connected in a circuit.

Show complete worked solution

Electric current is measured in amperes (amps, A), using an ammeter. An ammeter is always connected in series in the circuit, so that all of the current being measured passes through it.

QUESTION 3 2 marks Criterion A
Easy

A current of $2\,\text{A}$ flows through a wire for $30\,\text{s}$. Calculate the charge that passes through the wire.

Show complete worked solution

Step 1 — State the formula:

$$ Q = I \times t $$

Step 2 — Substitute the values:

$$ Q = 2 \times 30 $$

Answer: $Q = 60\,\text{C}$ (coulombs)

QUESTION 4 3 marks Criterion A
Medium

A charge of $15\,\text{C}$ passes through a lamp in $5\,\text{s}$.

a. Calculate the current flowing through the lamp.
[2]
b. State this current in milliamps (mA).
[1]
Show complete worked solution
(a)
Rearranging $Q = It$ for current: $$ I = \frac{Q}{t} = \frac{15}{5} $$ Answer: $I = 3\,\text{A}$
(b)
$$ 3\,\text{A} = 3 \times 1000 = 3000\,\text{mA} $$
QUESTION 5 4 marks Criterion A
Medium

State whether each material is a good electrical conductor or an insulator, and explain the general reason metals conduct electricity well.

Materials: copper, rubber, iron, plastic.

a. Classify each of the four materials.
[2]
b. Explain why metals are good conductors of electricity.
[2]
Show complete worked solution
(a)
Conductors: copper and iron (both metals). Insulators: rubber and plastic.
(b)
Metals contain free (delocalised) electrons that are not fixed to one atom and can move easily through the metal's structure. When a voltage is applied, these free electrons flow, forming an electric current. Insulators like rubber and plastic do not have free electrons able to move, so charge cannot flow through them.
QUESTION 6 3 marks Criterion A
Medium
A1A2A3

The diagram shows a single-loop series circuit with one bulb, and three ammeters, $A_1$, $A_2$ and $A_3$, placed at different points around the loop. Ammeter $A_1$ reads $0.4\,\text{A}$.

a. State the readings on $A_2$ and $A_3$.
[2]
b. Explain your answer.
[1]
Show complete worked solution
(a)
$A_2 = 0.4\,\text{A}$ and $A_3 = 0.4\,\text{A}$.
(b)
In a single-loop (series) circuit there is only one path for charge to flow, so exactly the same current passes every point in the loop — current is not used up as it goes around.
QUESTION 7 4 marks Criterion A
Hard

A current of $250\,\text{mA}$ flows through a circuit for $2$ minutes.

a. Calculate the charge that flows, in coulombs.
[2]
b. At the same current, how long would it take for $90\,\text{C}$ of charge to flow? Give your answer in minutes.
[2]
Show complete worked solution
(a)
Convert to base units first: $I = 250\,\text{mA} = 0.25\,\text{A}$, $t = 2\,\text{min} = 120\,\text{s}$.$$ Q = I \times t = 0.25 \times 120 $$Answer: $Q = 30\,\text{C}$
(b)
$$ t = \frac{Q}{I} = \frac{90}{0.25} = 360\,\text{s} $$$$ 360\,\text{s} \div 60 = 6\,\text{min} $$Answer: $6$ minutes
QUESTION 8 7 marks Criterion B
Medium

You want to investigate how the number of cells in a circuit affects the current flowing through a fixed bulb.

a. State the independent and dependent variables.
[2]
b. State two variables you would control, and explain why for one of them.
[2]
c. Describe a method, including equipment, to collect the data.
[3]
Show complete worked solution
(a)
Independent variable: number of cells (e.g. $1, 2, 3$). Dependent variable: current through the bulb, measured with an ammeter.
(b)
Control the bulb used (same bulb every time) and the wires/connections. Why control the bulb: a different bulb has different resistance, which would change the current for reasons unrelated to the number of cells, making the test unfair.
(c)
  1. Set up a series circuit with one cell, the bulb, and an ammeter.
  2. Close the switch and record the ammeter reading.
  3. Add one more identical cell in series, keeping everything else the same, and record the new reading.
  4. Repeat for $3$ and then $4$ cells.
  5. Repeat each reading $3$ times and calculate a mean current for each number of cells, to improve reliability.
QUESTION 9 6 marks Criterion B
Medium

You are given an unknown material and asked to test whether it conducts electricity, using a simple circuit tester.

a. State the equipment you would use and how you would build the tester.
[2]
b. State the independent and dependent variables in this test.
[2]
c. State two variables you should keep the same for a fair test between materials.
[2]
Show complete worked solution
(a)
Equipment: a cell (or battery), a bulb, connecting wires, and two crocodile clips. Connect the cell, bulb and one crocodile clip lead in a series loop, leaving a gap between the two crocodile clips so the material sample can be placed across the gap to complete (or not complete) the circuit.
(b)
Independent variable: the material being tested. Dependent variable: whether the bulb lights (a simple conductor/insulator test).
(c)
Keep the same cell/voltage and the same distance/contact area between the crocodile clips each time — otherwise a difference in the test setup, not the material itself, could explain a different result.
QUESTION 10 7 marks Criterion B
Hard

A student investigates how current changes as more cells are added to a circuit. Their ammeter readings for $2$ cells are inconsistent between repeats: $0.41\,\text{A}$, $0.60\,\text{A}$, $0.40\,\text{A}$.

a. Suggest a likely source of error that would explain this inconsistency.
[2]
b. Suggest an improvement to the method that would reduce this error.
[3]
c. Explain how this improvement would make the results more reliable.
[2]
Show complete worked solution
(a)
A likely cause is a loose or poor connection somewhere in the circuit (e.g. a crocodile clip not gripping a wire firmly) — this would add unpredictable extra resistance in some trials but not others, explaining why one reading ($0.60\,\text{A}$) is very different from the other two.
(b)
Check and firmly re-secure every connection before each reading, and use fixed terminals or soldered/screw connections rather than loose crocodile clips where possible. A digital multimeter with stable probes could also replace a basic ammeter for more consistent contact.
(c)
With secure, consistent connections, the only thing changing between trials would be the number of cells, so repeated readings at the same number of cells should be very close together — giving a much more reliable pattern to base a conclusion on.
QUESTION 11 3 marks Criterion C
Easy
Point in loopPQR
Current (A)0.500.500.50
a. Describe the pattern in this data.
[2]
b. Explain what this shows about a series (single-loop) circuit.
[1]
Show complete worked solution
(a)
The current is exactly the same ($0.50\,\text{A}$) at every point measured around the loop.
(b)
It confirms that in a single-loop circuit, current is not used up as it travels around — the same current flows at every point, since there is only one path for charge to take.
QUESTION 12 4 marks Criterion C
Medium
Time (s)05101520
Charge (C)010203040
Show complete worked solution

Step 1 — Choose a suitable interval:

Between $t=0$ and $t=20\,\text{s}$: $\Delta Q = 40 - 0 = 40\,\text{C}$, $\Delta t = 20 - 0 = 20\,\text{s}$.

Step 2 — Apply $I = \frac{Q}{t}$:

$$ I = \frac{40}{20} $$

Answer: $I = 2\,\text{A}$. (Each $5\,\text{s}$ interval also gives $\frac{10}{5}=2\,\text{A}$, confirming the current is constant.)

QUESTION 13 4 marks Criterion C
Medium
Trial12345
Current (A)1.201.221.191.551.21
a. Identify the anomalous reading.
[1]
b. Suggest why it might have occurred.
[1]
c. Calculate the mean current, excluding the anomalous reading.
[2]
Show complete worked solution
(a)
Trial $4$ ($1.55\,\text{A}$) is anomalous.
(b)
It could be caused by a momentary loose connection, a reading/recording error, or a brief surge — it does not fit the otherwise consistent set of readings.
(c)
$$ \text{mean} = \frac{1.20+1.22+1.19+1.21}{4} = \frac{4.82}{4} $$Answer: mean $\approx 1.21\,\text{A}$ (3 s.f.)
QUESTION 14 4 marks Criterion C
Medium
Bulbs in series1234
Current (A)0.600.300.200.15
a. Describe the pattern shown by this data.
[2]
b. Use the pattern to predict the current if $5$ identical bulbs were connected in series.
[2]
Show complete worked solution
(a)
As more identical bulbs are added in series, the current decreases. The pattern is inversely proportional to the number of bulbs: current $\times$ number of bulbs is constant ($0.60\times1=0.60$, $0.30\times2=0.60$, $0.20\times3=0.60$, $0.15\times4=0.60$).
(b)
Following the pattern (current $= 0.60 \div$ number of bulbs): $$ I = \frac{0.60}{5} $$Predicted answer: $I = 0.12\,\text{A}$
QUESTION 15 5 marks Criterion C
Medium

A charge sensor records the total charge that has passed a point in a circuit at different times.

Time (s)0102030
Total charge (C)051215
a. Calculate the current during each $10\,\text{s}$ interval.
[3]
b. State during which interval the current was greatest, and give its value.
[2]
Show complete worked solution
(a)

$0$–$10\,\text{s}$: $I=\dfrac{5-0}{10}=0.5\,\text{A}$

$10$–$20\,\text{s}$: $I=\dfrac{12-5}{10}=0.7\,\text{A}$

$20$–$30\,\text{s}$: $I=\dfrac{15-12}{10}=0.3\,\text{A}$

(b)
The current was greatest between $t=10\,\text{s}$ and $t=20\,\text{s}$, at $0.7\,\text{A}$ — this is the interval where charge increased the fastest ($7\,\text{C}$ in $10\,\text{s}$).
QUESTION 16 6 marks Criterion C
Hard
Trial12345
Time for 50 C to pass (s)24.825.124.930.225.0
a. Identify the anomalous result and explain why it should not be included in an average.
[2]
b. Calculate the mean time, excluding the anomalous result.
[2]
c. Use the mean time to calculate the current, given the charge was $50\,\text{C}$.
[2]
Show complete worked solution
(a)
Trial $4$ ($30.2\,\text{s}$) is anomalous — it is far higher than the other four consistent readings (all close to $25\,\text{s}$), suggesting an error occurred (e.g. a delay in starting/stopping the timer).
(b)
$$ \text{mean} = \frac{24.8+25.1+24.9+25.0}{4} = \frac{99.8}{4} $$Answer: $24.95\,\text{s}$
(c)
$$ I = \frac{Q}{t} = \frac{50}{24.95} $$Answer: $I \approx 2.00\,\text{A}$ (3 s.f.)
QUESTION 17 7 marks Criterion C
Hard
Time interval (s)0–22–55–8
Current (A)352
a. Calculate the total charge that passes in the $8\,\text{s}$ shown, assuming the current is constant within each interval.
[4]
b. Evaluate this method of calculating total charge. What assumption does it rely on, and how might it affect the accuracy of the answer?
[3]
Show complete worked solution
(a)

Using $Q = I \times t$ for each interval:

$0$–$2\,\text{s}$: $Q_1 = 3 \times 2 = 6\,\text{C}$

$2$–$5\,\text{s}$: $Q_2 = 5 \times 3 = 15\,\text{C}$

$5$–$8\,\text{s}$: $Q_3 = 2 \times 3 = 6\,\text{C}$

$$ Q_{total} = 6+15+6 $$

Answer: $Q_{total} = 27\,\text{C}$
(b)
This method assumes the current is exactly constant within each interval and changes suddenly only at the boundaries ($2\,\text{s}$ and $5\,\text{s}$). In reality the current probably changes more gradually. If the current actually varied smoothly within an interval rather than jumping, the true charge could be slightly different from $27\,\text{C}$ — using shorter time intervals (more frequent measurements) would make the step-based calculation more accurate.
QUESTION 18 5 marks Criterion D
Medium

Homes use fuses and circuit breakers, which automatically break (open) a circuit if the current becomes too high.

Discuss one benefit and one drawback of using fuses and circuit breakers.

Show complete worked solution

Benefit: If a fault causes an unusually high current to flow (for example, a short circuit), a fuse or circuit breaker automatically cuts off the current within a fraction of a second. This prevents wires from overheating, which greatly reduces the risk of electrical fires and protects both the household wiring and the appliances connected to it.

Drawback: A blown fuse must be identified and replaced (or a tripped breaker reset) before the circuit works again, which is inconvenient and can leave part of a home without power until it is fixed. If people don't know how to safely reset it, or repeatedly replace a fuse with one rated too high just to stop it blowing, the safety protection can be lost.

QUESTION 19 5 marks Criterion D
Medium

A student plugs three appliances into a single multi-socket adaptor: a heater drawing $6\,\text{A}$, a kettle drawing $5\,\text{A}$, and a toaster drawing $4\,\text{A}$. The socket the adaptor is plugged into is rated at a maximum of $13\,\text{A}$.

Discuss the benefit of using such an adaptor, and the drawback/risk shown by this data.

Show complete worked solution

Benefit: A multi-socket adaptor is very convenient, allowing several appliances to run from a single wall socket without needing extra wiring installed in the home.

Drawback: The total current drawn is $6+5+4=15\,\text{A}$, which is more than the socket's $13\,\text{A}$ rating. Drawing more current than a circuit is designed for causes the wires to heat up more than intended, which is a genuine fire risk — overloading sockets and adaptors like this is a common cause of household electrical fires, which is why appliances should be spread across multiple circuits rather than overloading one.

QUESTION 20 6 marks Criterion D
Hard

Modern society relies heavily on a continuous, reliable supply of electric current — for lighting, heating, communication, hospitals, and industry. Increasingly, this current is generated using renewable sources such as wind and solar power instead of burning fossil fuels.

Evaluate the impact of this shift, discussing a benefit and a concern it raises.

Show complete worked solution

Benefit: Generating electric current from renewable sources such as wind and solar produces far less carbon dioxide and other pollution than burning coal or gas, helping to reduce the environmental impact of the huge amount of electricity modern society depends on, and reducing reliance on finite fossil fuel resources.

Concern: Renewable sources are less predictable — wind turbines generate no current when there is no wind, and solar panels generate little at night or in cloudy weather. Since hospitals, transport systems, and homes need a continuous, reliable current at all times, this intermittency raises real concerns about power cuts unless it is paired with large-scale energy storage (such as batteries) or backup supplies, which is expensive and still being developed at the scale required.

Series and Parallel Circuits 20 questions

QUESTION 1 3 marks Criterion A
Easy

Compare series and parallel circuits.

a. Describe how components are arranged in a series circuit.
[1]
b. Describe how components are arranged in a parallel circuit.
[1]
c. State one key difference in how current behaves in the two types of circuit.
[1]
Show complete worked solution
(a)
In a series circuit, components are connected one after another in a single loop, so there is only one path for current to flow.
(b)
In a parallel circuit, components are connected across separate branches between the same two points, so there is more than one path for current to flow.
(c)
In series, the same current flows through every component. In parallel, the current splits between branches, so different branches can carry different currents.
QUESTION 2 2 marks Criterion A
Easy
V1V29 V battery

The diagram shows two identical bulbs connected in series to a battery.

Show complete worked solution

If one of the bulbs in this series circuit breaks (its filament fails), the circuit is no longer a complete loop. Since there is only one path for current, breaking it anywhere stops current everywhere — so both bulbs go out, even though only one bulb actually failed.

QUESTION 3 2 marks Criterion A
Easy
Branch 1Branch 2

The diagram shows two identical bulbs connected in parallel across a battery.

Show complete worked solution

If one bulb breaks, its branch is broken — but the other branch is a separate, complete path back to the battery. Current can still flow through the working bulb, so it stays lit, unaffected by the broken one.

QUESTION 4 3 marks Criterion A
Medium
A1A2A3

The diagram shows a series circuit with two bulbs and three ammeters, $A_1$, $A_2$ and $A_3$, at different points around the loop.

a. If $A_1$ reads $0.25\,\text{A}$, state the readings on $A_2$ and $A_3$.
[1]
b. Explain why these readings must be the same.
[2]
Show complete worked solution
(a)
$A_2 = 0.25\,\text{A}$ and $A_3 = 0.25\,\text{A}$.
(b)
This is a series circuit — a single loop with only one possible path for charge. Since charge cannot be created, destroyed, or build up anywhere in the loop, exactly the same current must flow at every point around it.
QUESTION 5 3 marks Criterion A
Medium
AAABranch 1Branch 2

The diagram shows two bulbs connected in parallel, with an ammeter $A$ measuring the total (main) current from the battery, and ammeters $A_1$ and $A_2$ measuring the current in each branch. $A_1$ reads $0.30\,\text{A}$ and $A_2$ reads $0.50\,\text{A}$.

Show complete worked solution

At the junction where the branches meet, the current splitting into the branches must add back up to the current leaving the battery (charge is conserved — none is lost or gained at the junction):

$$ A = A_1 + A_2 = 0.30 + 0.50 $$

Answer: $A = 0.80\,\text{A}$

QUESTION 6 3 marks Criterion A
Medium

Two identical bulbs are connected in series to a $9\,\text{V}$ battery. Voltmeter $V_1$, connected across the first bulb, reads $3.5\,\text{V}$.

Show complete worked solution

In a series circuit, the potential difference (voltage) of the battery is shared between the components, so the individual voltages must add up to the total: $$ V_1 + V_2 = V_{battery} $$ $$ 3.5 + V_2 = 9 $$ Answer: $V_2 = 5.5\,\text{V}$

QUESTION 7 6 marks Criterion A
Hard

Three bulbs, X, Y and Z, are connected in series to a $12\,\text{V}$ battery. The voltage across X is $4\,\text{V}$ and the voltage across Y is $5\,\text{V}$.

a. Calculate the voltage across Z.
[2]
b. The three bulbs are not identical (X, Y and Z have different voltages across them). What does this tell you about how the battery's voltage is shared between components in series?
[2]
c. State what would happen to the readings on an ammeter placed anywhere in this circuit, compared to another ammeter placed elsewhere in the same loop.
[2]
Show complete worked solution
(a)
In series, voltages share the total: $$ V_X + V_Y + V_Z = V_{total} $$ $$ 4 + 5 + V_Z = 12 $$ Answer: $V_Z = 3\,\text{V}$
(b)
The battery's voltage is not necessarily shared equally between components in series — it is shared according to each component's resistance (bulbs with more resistance take a larger share of the voltage). Even though the shares are unequal here, they must still always add up to the total supply voltage.
(c)
Every ammeter placed anywhere in this single series loop would read exactly the same current, since current (unlike voltage) is the same at every point in a series circuit, regardless of how the voltage happens to be shared.
QUESTION 8 7 marks Criterion B
Medium

You want to investigate whether bulbs are dimmer when connected in series compared to parallel, using the same battery and identical bulbs.

a. State the independent and dependent variables.
[2]
b. State two variables you would keep the same for a fair comparison.
[2]
c. Describe a method to compare the two arrangements.
[3]
Show complete worked solution
(a)
Independent variable: circuit arrangement (series or parallel). Dependent variable: brightness of a bulb, measured objectively using the current through it (ammeter reading) as a proxy for brightness.
(b)
Keep the same battery/cells and identical, matching bulbs in both circuits — different bulbs or a different battery voltage could change brightness for reasons unrelated to the circuit arrangement.
(c)
  1. Build a series circuit with two identical bulbs and an ammeter, and record the ammeter reading (and observe brightness).
  2. Using the same battery and identical bulbs, build a parallel circuit and place an ammeter in one branch; record its reading (and observe brightness).
  3. Repeat each measurement $3$ times and calculate a mean current for each arrangement.
  4. Compare the mean currents: a bulb with a higher current through it will be brighter.
QUESTION 9 6 marks Criterion B
Medium

You want to investigate whether adding more bulbs in parallel changes the brightness of the bulbs that are already in the circuit.

a. State the independent and dependent variables.
[2]
b. State a variable you would control, and explain why.
[2]
c. Describe how you would carry out the test.
[2]
Show complete worked solution
(a)
Independent variable: number of parallel branches (bulbs) connected. Dependent variable: current through one particular, fixed bulb (measured with an ammeter placed permanently in that one branch).
(b)
Control the battery/cells used — if the voltage supplied changed between tests, it could change the current through the fixed bulb for a reason unrelated to the number of branches, making the comparison unfair.
(c)
Set up a parallel circuit with one bulb and an ammeter in its branch; record the current. Add a second identical bulb in a new parallel branch (keeping the first branch and battery unchanged) and record the ammeter reading in the first branch again. Repeat, adding branches one at a time, recording the fixed bulb's current each time.
QUESTION 10 8 marks Criterion B
Hard

A student compares the current in a circuit across several trials, using a single battery that has been in continuous use for a while. Their results drift lower across successive trials even though nothing else in the circuit was changed.

a. Suggest a source of error that could explain this drift.
[2]
b. Suggest an improvement to the method that would remove this source of error.
[3]
c. Explain how this improvement leads to more reliable conclusions.
[3]
Show complete worked solution
(a)
The battery's voltage decreases as it is used (it discharges over time), so later trials are effectively being run at a lower supply voltage than earlier ones — this alone would cause the current to drift downward across trials, independent of anything else being tested.
(b)
Replace the battery with a bench power supply set to a fixed voltage, which does not drain or drop in voltage during the investigation, so every trial is genuinely tested under the same conditions. Alternatively, use a fresh battery for each trial, or complete all trials as quickly as possible to minimise discharge.
(c)
With a constant supply voltage guaranteed for every trial, any differences observed in the current between trials can be confidently attributed to the actual variable being tested (e.g. circuit arrangement), rather than to the battery quietly running down — making the comparison a genuinely fair test and the conclusion far more trustworthy.
QUESTION 11 3 marks Criterion C
Easy
Point in loop1234
Current (A)0.450.450.440.45
Show complete worked solution

The readings are all the same, within normal measurement uncertainty ($0.44$–$0.45\,\text{A}$, a difference easily explained by reading a scale to the nearest small division). This confirms that in a series (single-loop) circuit, the current is the same at every point around the loop.

QUESTION 12 4 marks Criterion C
Medium

A parallel circuit has two branches.

CaseBranch 1 (A)Branch 2 (A)Main current (A)
10.200.35?
20.55?0.90
a. Calculate the missing main current in Case 1.
[2]
b. Calculate the missing branch current in Case 2.
[2]
Show complete worked solution
(a)
At the junction, branch currents add up to the main current: $$ 0.20 + 0.35 = 0.55\,\text{A} $$
(b)
$$ \text{Branch 2} = \text{main} - \text{Branch 1} = 0.90 - 0.55 $$Answer: $0.35\,\text{A}$
QUESTION 13 4 marks Criterion C
Medium
Bulbs in series1234
Brightness (arbitrary units)100503325
a. Describe the trend shown by this data.
[2]
b. Explain this trend in terms of voltage.
[2]
Show complete worked solution
(a)
As more identical bulbs are added in series to a fixed battery, each bulb's brightness decreases.
(b)
Each extra bulb added in series shares a smaller portion of the fixed total battery voltage between more components, so each bulb receives less voltage (and less current) than before — giving it less energy per second to convert to light, so it glows more dimly.
QUESTION 14 5 marks Criterion C
Medium
Bulbs in parallel1234
Brightness of each bulb100100100100
Total current from battery (A)0.51.01.52.0
a. Describe what happens to the brightness of each individual bulb as more bulbs are added in parallel.
[2]
b. Explain this, and describe what happens to the total current drawn from the battery instead.
[3]
Show complete worked solution
(a)
The brightness of each individual bulb stays the same (constant at 100) as more bulbs are added in parallel.
(b)
In parallel, each branch is connected directly across the full battery voltage, regardless of how many other branches exist — so each bulb always receives the same voltage and current as if it were alone, and stays equally bright. However, the total current drawn from the battery increases with every extra branch added (rising $0.5\,\text{A}$ each time), since the battery now has to supply current to more branches at once.
QUESTION 15 4 marks Criterion C
Medium

A student measures the voltage across each of three identical bulbs connected in series to a $6\,\text{V}$ battery.

Bulb123
Voltage (V)2.12.02.2
a. Calculate the total of the three measured voltages, and compare it to the battery voltage.
[2]
b. Explain whether this small difference means the rule "series voltages add up to the total" is wrong.
[2]
Show complete worked solution
(a)
$$ 2.1 + 2.0 + 2.2 = 6.3\,\text{V} $$ This is close to, but not exactly, the battery's $6.0\,\text{V}$ — a difference of $0.3\,\text{V}$.
(b)
No — the rule still holds. A difference of $0.3\,\text{V}$ out of $6.3\,\text{V}$ is small (about $5\%$) and is easily explained by the normal reading uncertainty of a voltmeter (e.g. $\pm0.1\,\text{V}$ on each of three readings can easily combine to a $0.3\,\text{V}$ total discrepancy). The data still strongly supports the rule.
QUESTION 16 7 marks Criterion C
Hard

Two circuits are set up and measured. Circuit X: ammeters at three different points around the loop all read $0.40\,\text{A}$; a voltmeter across each of its two bulbs reads $3.0\,\text{V}$ each. Circuit Y: the main ammeter reads $0.90\,\text{A}$, while two branch ammeters read $0.50\,\text{A}$ and $0.40\,\text{A}$; a voltmeter across each of its two bulbs reads $6.0\,\text{V}$ each (the full battery voltage).

a. State whether Circuit X is series or parallel, and explain using the data.
[2]
b. State whether Circuit Y is series or parallel, and explain using the data.
[2]
c. Summarise the general rule connecting current and voltage patterns to circuit type.
[3]
Show complete worked solution
(a)
Circuit X is series. The current is identical ($0.40\,\text{A}$) at every point measured, which only happens when there is a single loop with one path for charge; the voltage is also shared between the two bulbs ($3.0\,\text{V}$ each, adding to $6.0\,\text{V}$, consistent with a shared supply).
(b)
Circuit Y is parallel. The branch currents are different from each other ($0.50\,\text{A}$ and $0.40\,\text{A}$) and add up to the main current ($0.90\,\text{A}$), which only happens when current splits between separate paths; each bulb also gets the full $6.0\,\text{V}$, not a shared fraction of it.
(c)
In series circuits: current is the same everywhere, but voltage is shared (divided) between components. In parallel circuits: voltage is the same across every branch, but current is shared (divided) between branches, with the branch currents always adding up to the total (main) current.
QUESTION 17 7 marks Criterion C
Hard

A parallel circuit has two branches. The branch currents are measured as $0.30\,\text{A}$ and $0.45\,\text{A}$, but the main ammeter (measuring the total current from the battery) reads $0.80\,\text{A}$.

a. Calculate the total current expected from the two branch readings.
[2]
b. Evaluate whether the main ammeter's reading of $0.80\,\text{A}$ supports the rule that branch currents add up to the total current.
[3]
c. Suggest one thing the student could do to strengthen their conclusion.
[2]
Show complete worked solution
(a)
$$ 0.30 + 0.45 = 0.75\,\text{A} $$
(b)
The expected total ($0.75\,\text{A}$) and the measured total ($0.80\,\text{A}$) differ by only $0.05\,\text{A}$, which is a small difference (about $6\%$) well within the kind of reading uncertainty expected from analogue ammeters (typically $\pm0.05\,\text{A}$ or more). This data still reasonably supports the rule, rather than contradicting it.
(c)
Repeat all three measurements (both branches and the main current) several times and use mean values, or use a more precise digital ammeter, to check whether the small $0.05\,\text{A}$ gap is just measurement uncertainty or a genuine, repeatable discrepancy.
QUESTION 18 5 marks Criterion D
Medium

The wiring in a house is designed as a set of parallel circuits, not one single series loop.

Discuss a benefit and a drawback of wiring a house this way.

Show complete worked solution

Benefit: In a parallel circuit, each appliance or light is on its own independent branch and receives the full mains voltage regardless of what else is switched on. If one appliance breaks or is switched off, the others keep working normally — imagine if a whole house went dark every time one light bulb blew, as would happen in series!

Drawback: Parallel wiring requires much more cable (a separate path back to the supply for each branch/circuit) compared to a single simple loop, which increases the cost and complexity of installing the electrics in a building.

QUESTION 19 5 marks Criterion D
Medium

Older strings of decorative lights connected all the bulbs in series; if one bulb failed, the whole string went dark. Most modern LED light strings connect the bulbs in parallel (or in small parallel groups).

Discuss a benefit and a drawback of the modern parallel design.

Show complete worked solution

Benefit: If one LED fails in a parallel string, only that LED goes out — the rest of the string keeps working, since each LED has its own independent path to the supply. This makes faults far easier to notice and fix, and is much less frustrating for the user than a whole string going dark.

Drawback: A parallel design needs more wiring to give every LED its own connection to both supply wires, which increases the manufacturing cost and complexity of the light string compared to a single simple series loop.

QUESTION 20 6 marks Criterion D
Hard

Because household circuits are wired in parallel, each circuit (e.g. one for upstairs sockets, one for kitchen appliances, one for lighting) can be protected by its own individual fuse or circuit breaker in the consumer unit (fuse box).

Discuss a benefit and a drawback of this design choice for building safety.

Show complete worked solution

Benefit: If a fault occurs on one circuit (for example, a short circuit in a kitchen appliance), only that circuit's breaker trips, cutting power to just that part of the building. The rest of the house — lighting, other rooms — keeps working normally, which is both safer (the fault is isolated) and more convenient than losing power to the whole building.

Drawback: This design is more complex and expensive to install than a single circuit for the whole building, since it requires multiple breakers, more cable runs, and careful planning by an electrician to decide which sockets and lights belong to which circuit — increasing both material and labour costs.

Voltage and Resistance 20 questions

QUESTION 1 2 marks Criterion A
Easy

State Ohm's law, including the equation, and the unit of each quantity in it.

Show complete worked solution

$$ V = IR $$ where $V$ is potential difference (voltage) in volts (V), $I$ is current in amps (A), and $R$ is resistance in ohms ($\Omega$).

QUESTION 2 3 marks Criterion A
Easy
RAV

The diagram shows a resistor $R$ connected to a cell, with an ammeter measuring the current and a voltmeter connected across the resistor. A current of $2\,\text{A}$ flows through a resistor of resistance $5\,\Omega$.

Show complete worked solution

Step 1 — State the formula:

$$ V = IR $$

Step 2 — Substitute the values:

$$ V = 2 \times 5 $$

Answer: $V = 10\,\text{V}$

QUESTION 3 2 marks Criterion A
Easy

A voltmeter reads $12\,\text{V}$ across a component, and an ammeter reads $3\,\text{A}$ through it. Calculate the resistance of the component.

Show complete worked solution

Rearranging $V = IR$ for resistance: $$ R = \frac{V}{I} = \frac{12}{3} $$Answer: $R = 4\,\Omega$

QUESTION 4 3 marks Criterion A
Medium

A resistor of $1500\,\Omega$ is connected to a $6\,\text{V}$ supply.

a. Calculate the current flowing, in amps.
[2]
b. State this current in milliamps.
[1]
Show complete worked solution
(a)
Rearranging $V=IR$ for current: $$ I = \frac{V}{R} = \frac{6}{1500} $$Answer: $I = 0.004\,\text{A}$
(b)
$$ 0.004\,\text{A} = 0.004 \times 1000 = 4\,\text{mA} $$
QUESTION 5 4 marks Criterion A
Medium

Explain how each change below affects the resistance of a metal wire, if all other factors are kept the same.

a. Making the wire longer.
[2]
b. Making the wire thicker (greater cross-sectional area).
[2]
Show complete worked solution
(a)
A longer wire has more resistance. The free electrons must travel further and collide with more of the fixed metal ions along the way, so it is harder for charge to flow — resistance increases roughly in proportion to length.
(b)
A thicker wire has less resistance. A larger cross-section gives the free electrons more space and more possible paths to flow through at once, making it easier for charge to flow.
QUESTION 6 4 marks Criterion A
Medium

Two resistors, $R_1 = 4\,\Omega$ and $R_2 = 6\,\Omega$, are connected in series to a $20\,\text{V}$ battery.

a. Calculate the total resistance of the circuit.
[2]
b. Calculate the current flowing in the circuit.
[2]
Show complete worked solution
(a)
In series, resistances simply add: $$ R_{total} = R_1 + R_2 = 4 + 6 $$Answer: $R_{total} = 10\,\Omega$
(b)
$$ I = \frac{V}{R_{total}} = \frac{20}{10} $$Answer: $I = 2\,\text{A}$
QUESTION 7 7 marks Criterion A
Hard

Three resistors, $R_1 = 2\,\Omega$, $R_2 = 3\,\Omega$ and $R_3 = 5\,\Omega$, are connected in series to a $20\,\text{V}$ battery.

a. Calculate the total resistance of the circuit.
[2]
b. Calculate the current flowing in the circuit.
[2]
c. Calculate the voltage across each resistor, and show that they add up to the battery voltage.
[3]
Show complete worked solution
(a)
$$ R_{total} = 2+3+5 $$Answer: $R_{total} = 10\,\Omega$
(b)
$$ I = \frac{V}{R_{total}} = \frac{20}{10} $$Answer: $I = 2\,\text{A}$
(c)
Since current is the same throughout a series circuit ($I=2\,\text{A}$), use $V=IR$ for each: $$ V_1 = 2\times2 = 4\,\text{V} \qquad V_2 = 2\times3 = 6\,\text{V} \qquad V_3 = 2\times5 = 10\,\text{V} $$ Check: $4+6+10 = 20\,\text{V}$, which matches the battery voltage exactly ?.
QUESTION 8 7 marks Criterion B
Medium

You want to investigate how the length of a wire affects its resistance.

a. State the independent and dependent variables.
[2]
b. State two variables you would control, and explain why for one.
[2]
c. Describe the method and equipment you would use.
[3]
Show complete worked solution
(a)
Independent variable: length of wire (e.g. $10, 20, 30, 40, 50\,\text{cm}$). Dependent variable: resistance, calculated from voltmeter and ammeter readings using $R=\frac{V}{I}$.
(b)
Control the thickness (diameter) and material of the wire, keeping the same reel of wire throughout. Why: a thicker or different-material wire has different resistance for reasons unrelated to length, which would make the comparison unfair.
(c)
  1. Tape the test wire along a metre ruler and connect one end to the circuit with a crocodile clip.
  2. Connect a second crocodile clip at the $10\,\text{cm}$ mark, completing a circuit with a low-voltage power supply and an ammeter in series, and a voltmeter connected across the length of wire being tested.
  3. Record the ammeter and voltmeter readings and calculate resistance using $R = \frac{V}{I}$.
  4. Move the second crocodile clip to $20\,\text{cm}$, $30\,\text{cm}$, etc., repeating the measurement and calculation each time.
  5. Use a low current/short connection times to avoid the wire heating up, which would otherwise change its resistance.
QUESTION 9 6 marks Criterion B
Medium

You want to investigate how the resistance connected in a circuit affects the current that flows, using a fixed power supply.

a. State the independent and dependent variables.
[2]
b. State a variable you would keep the same, and explain why.
[2]
c. Describe how you would carry out the test.
[2]
Show complete worked solution
(a)
Independent variable: resistance (using different labelled resistors, e.g. $2\,\Omega, 4\,\Omega, 6\,\Omega, 8\,\Omega$). Dependent variable: current, measured with an ammeter.
(b)
Keep the supply voltage constant throughout. If the voltage changed between tests, it would affect the current for a reason unrelated to the resistor being tested, making the comparison unfair.
(c)
Set up a series circuit with the power supply, an ammeter, and one resistor; record the current. Replace the resistor with the next value (keeping voltage the same) and record the new current. Repeat for each resistor, and repeat each measurement several times to calculate a mean.
QUESTION 10 7 marks Criterion B
Hard

A student investigates how the resistance of a filament lamp changes as more current flows through it. They take one reading of voltage and current at each of several different currents, but their results seem inconsistent with what they expected.

a. Suggest why a filament lamp's resistance might not stay constant during this test, unlike a normal fixed resistor.
[2]
b. Suggest an improvement to the method to more reliably study how resistance changes with current.
[3]
c. Explain how this improvement leads to a fairer test.
[2]
Show complete worked solution
(a)
As more current flows, the filament heats up. The resistance of the metal filament increases with temperature (hotter metal atoms vibrate more and get in the way of flowing electrons more often), so the resistance itself is changing throughout the test, not staying fixed like an ordinary resistor's.
(b)
Take each voltage/current reading quickly after switching on at each current setting, before the filament has time to heat up further and change its resistance mid-reading, and allow the lamp to cool back down fully between each test. Repeating each specific current setting a few times and averaging would also improve reliability.
(c)
Reading quickly and allowing full cooling between tests ensures each measurement reflects the resistance at a genuinely known, consistent temperature/current combination, rather than a resistance that is still changing as the reading is taken — making each data point comparable to the others.
QUESTION 11 3 marks Criterion C
Easy

A voltmeter reads $4\,\text{V}$ across a resistor, and an ammeter reads $0.5\,\text{A}$ through it.

Show complete worked solution

$$ R = \frac{V}{I} = \frac{4}{0.5} $$Answer: $R = 8\,\Omega$

QUESTION 12 5 marks Criterion C
Medium
Voltage (V)2468
Current (A)0.40.81.21.6
a. Calculate the resistance for each pair of readings.
[3]
b. What does this data show about this resistor?
[2]
Show complete worked solution
(a)

$R=\dfrac{2}{0.4}=5\,\Omega$

$R=\dfrac{4}{0.8}=5\,\Omega$

$R=\dfrac{6}{1.2}=5\,\Omega$

$R=\dfrac{8}{1.6}=5\,\Omega$

(b)
The resistance comes out as $5\,\Omega$ every time, regardless of the voltage/current used. This shows the component has a constant resistance — it obeys Ohm's law (it is an "ohmic" conductor), as long as its temperature stays constant.
QUESTION 13 4 marks Criterion C
Medium
0123406121824Current (A)Voltage (V)

The graph shows current against voltage for a resistor.

a. Read two points from the line and use them to calculate the gradient of the graph.
[2]
b. Explain what physical quantity this gradient represents, and state its value with a unit.
[2]
Show complete worked solution
(a)
Using the origin $(0,0)$ and the point $(4\,\text{A}, 20\,\text{V})$: $$ \text{gradient} = \frac{\Delta V}{\Delta I} = \frac{20-0}{4-0} = 5 $$
(b)
Since $V = IR$, the resistance $R = \frac{V}{I}$ is exactly the gradient of a voltage-against-current graph. Answer: $R = 5\,\Omega$
QUESTION 14 5 marks Criterion C
Medium
Length (cm)10203040
Resistance ($\Omega$)2468
a. Describe the relationship between length and resistance shown by this data.
[3]
b. Use the pattern to predict the resistance of $50\,\text{cm}$ of the same wire.
[2]
Show complete worked solution
(a)
Resistance is directly proportional to length — doubling the length ($10\to20\,\text{cm}$) doubles the resistance ($2\to4\,\Omega$), and the ratio of resistance to length ($0.2\,\Omega/\text{cm}$) stays constant throughout the table.
(b)
Following the pattern of $+2\,\Omega$ for every extra $10\,\text{cm}$: Answer: $10\,\Omega$
QUESTION 15 5 marks Criterion C
Medium
Trial12345
Current at 6 V (A)1.191.211.200.851.20
a. Identify the anomalous reading and suggest why it might have occurred.
[2]
b. Calculate the mean current, excluding the anomaly.
[1]
c. Use the mean current to calculate the resistance.
[2]
Show complete worked solution
(a)
Trial $4$ ($0.85\,\text{A}$) is anomalous — it breaks from the otherwise very consistent set of readings, possibly due to a temporary loose connection or misread scale.
(b)
$$ \frac{1.19+1.21+1.20+1.20}{4} = \frac{4.80}{4} = 1.20\,\text{A} $$
(c)
$$ R = \frac{V}{I} = \frac{6}{1.20} $$Answer: $R = 5.0\,\Omega$
QUESTION 16 8 marks Criterion C
Hard
0123406121824Current (A)Voltage (V)

The graph shows the voltage-current relationship for a filament lamp, which curves rather than forming a straight line.

a. Using the point $(1\,\text{A}, 2\,\text{V})$, calculate the resistance at this lower current.
[2]
b. Using the point $(3\,\text{A}, 9\,\text{V})$, calculate the resistance at this higher current.
[2]
c. Explain why the resistance is not the same at both points, and why this graph curves instead of forming a straight line through the origin.
[4]
Show complete worked solution
(a)
$$ R = \frac{V}{I} = \frac{2}{1} $$Answer: $R = 2\,\Omega$
(b)
$$ R = \frac{V}{I} = \frac{9}{3} $$Answer: $R = 3\,\Omega$
(c)
The resistance increased from $2\,\Omega$ to $3\,\Omega$ as the current increased. This is because a higher current makes the filament hotter, and a metal filament's resistance increases with temperature. Since the resistance (the gradient of the $V$–$I$ graph) is not constant but keeps increasing as current increases, the graph curves upward more steeply rather than forming the straight line through the origin that a fixed (ohmic) resistor would give.
QUESTION 17 8 marks Criterion C
Hard

A student tests three wires of different thickness (but the same length and material), measuring the resistance of each three times.

ThicknessTrial 1Trial 2Trial 3
Thin (0.2 mm)12.111.912.0
Medium (0.4 mm)6.06.25.9
Thick (0.6 mm)2.83.02.9

(All resistance values in $\Omega$.)

a. Calculate the mean resistance for each thickness.
[3]
b. Describe the relationship between wire thickness and resistance shown by this data.
[2]
c. Comment on the reliability of this data, using the repeated trials.
[3]
Show complete worked solution
(a)
Thin: $\frac{12.1+11.9+12.0}{3}=12.0\,\Omega$. Medium: $\frac{6.0+6.2+5.9}{3}\approx6.0\,\Omega$. Thick: $\frac{2.8+3.0+2.9}{3}\approx2.9\,\Omega$.
(b)
As the wire gets thicker, its resistance decreases — a thicker wire gives electrons a larger cross-sectional area and more paths to flow through, so it is easier for charge to flow.
(c)
For each thickness, the three repeated trials are all very close together (within about $0.3\,\Omega$ of each other), showing the measurements are precise and reliable. The clear, consistent difference between the three mean values (well beyond this small spread) also gives confidence that the trend — thicker wire, lower resistance — is a genuine effect and not just measurement noise.
QUESTION 18 5 marks Criterion D
Medium

Appliances such as toasters and kettles use a resistive heating element (a wire with deliberately high resistance) to convert electrical energy directly into heat.

Discuss a benefit and a drawback of using resistive heating elements in home appliances.

Show complete worked solution

Benefit: Resistive heating elements convert electrical energy directly and efficiently into heat, with a simple, reliable design and no moving parts — making appliances like kettles and toasters fast, convenient, and relatively cheap to manufacture.

Drawback: A high-resistance element carrying a large current gets very hot, so a fault (such as damaged insulation or a short circuit) creates a genuine fire risk. Using these appliances also uses a significant amount of electrical energy, contributing to household energy costs and, depending on how the electricity is generated, to environmental impact.

QUESTION 19 6 marks Criterion D
Medium

Electricity is often transmitted from power stations to homes through very long cables at extremely high voltage, rather than at the lower voltage actually used in homes. This reduces the current flowing in the transmission cables for the same amount of power delivered.

Discuss a benefit and a drawback of this high-voltage transmission approach.

Show complete worked solution

Benefit: Reducing the current in the long transmission cables reduces the energy wasted as heat in the cables' own resistance. This means significantly more of the generated electrical energy actually reaches homes and businesses usefully, rather than being wasted, making the whole electricity supply system more efficient.

Drawback: Very high voltages are dangerous, so this approach requires expensive infrastructure — tall pylons, thick insulation, and transformers to safely step the voltage up and back down — as well as strict safety measures and exclusion zones around power lines to protect the public, adding significant cost and land use.

QUESTION 20 6 marks Criterion D
Hard

Electric heaters and heated blankets use resistive heating to keep people warm, especially in cold weather or for medical comfort.

Evaluate the impact of this technology, discussing both a benefit and a risk it raises.

Show complete worked solution

Benefit: Electric heating provides convenient, controllable warmth on demand, which can be important for health and comfort — particularly for elderly or vulnerable people in cold weather, or for medical uses such as warming a patient. Unlike some other heating methods, it produces no smoke or fumes at the point of use.

Concern: Because resistive heating relies on a large current flowing through a resistive element, any fault — a damaged cable, a covered heater restricting airflow, or a compressed/damaged heating wire in a blanket — can cause dangerous overheating and a fire hazard. There is also an ongoing energy and environmental cost: heaters can use a large amount of electricity, and if that electricity is generated from fossil fuels, this adds to carbon emissions, so the convenience of resistive heating needs to be balanced against these safety and environmental considerations.

Magnets and Magnetic Fields 20 questions

QUESTION 1 3 marks Criterion A
Easy
NSSNMagnet 1Magnet 2gap

The diagram shows two bar magnets, each with its poles labelled.

a. As shown (N facing S), will the magnets attract or repel?
[1]
b. If Magnet 2 were turned around so its N pole faced Magnet 1's N pole, would they attract or repel?
[1]
c. State the general rule for how magnetic poles interact.
[1]
Show complete worked solution
(a)
Attract. Unlike poles (N and S) attract each other.
(b)
Repel. Like poles (N and N) repel each other.
(c)
Like poles repel; unlike poles attract.
QUESTION 2 4 marks Criterion A
Easy

Classify each material as magnetic (attracted to a magnet) or non-magnetic: iron, copper, steel, aluminium, nickel, plastic, cobalt, wood.

Show complete worked solution

Magnetic: iron, steel, nickel, cobalt (these are all ferromagnetic materials/alloys). Non-magnetic: copper, aluminium, plastic, wood.

QUESTION 3 2 marks Criterion A
Easy

Define a magnetic field, and state the direction convention used for magnetic field lines around a bar magnet.

Show complete worked solution

A magnetic field is the region around a magnet (or magnetic material) in which it can exert a magnetic force. By convention, magnetic field lines are drawn pointing from the north (N) pole to the south (S) pole outside the magnet, with arrows showing this direction.

QUESTION 4 4 marks Criterion A
Medium
NSfield direction: N ? S (outside magnet)

The diagram shows the magnetic field pattern around a bar magnet.

a. State the direction of the field lines shown, in terms of the poles.
[1]
b. State where the magnetic field is strongest, and explain how the diagram shows this.
[2]
c. State where the field is weakest.
[1]
Show complete worked solution
(a)
The field lines point from the N pole to the S pole, outside the magnet.
(b)
The field is strongest close to the poles. This is shown by the field lines being most closely packed together (densest) near the poles — the closer together the lines are drawn, the stronger the field at that point.
(c)
The field is weakest furthest from the magnet, where the field lines are spread further apart.
QUESTION 5 4 marks Criterion A
Medium

An iron nail is placed touching the end of a strong bar magnet. Several paperclips then stick to the free end of the nail, even though the nail was not a magnet before.

a. Explain why the nail is able to attract the paperclips.
[2]
b. Predict what happens to the paperclips if the bar magnet is now taken away from the nail, and explain why.
[2]
Show complete worked solution
(a)
Being close to the bar magnet's strong field induces magnetism in the iron nail — the magnet temporarily aligns the tiny magnetic regions inside the iron, turning the nail itself into a temporary magnet, which is then able to attract and hold the paperclips.
(b)
The paperclips fall off. Iron is a soft magnetic material — its induced magnetism is only temporary and depends on being near the magnet's field; once the magnet is removed, the nail quickly loses almost all of its induced magnetism and can no longer hold the paperclips.
QUESTION 6 4 marks Criterion A
Medium

Compare permanent magnets made from steel with temporary magnets made from soft iron.

a. Describe the key difference in how well each material retains its magnetism.
[2]
b. Give one everyday use of a permanent magnet, and one use where a temporary magnet is more suitable.
[2]
Show complete worked solution
(a)
Steel retains (keeps) its magnetism well once magnetised, making it a good permanent magnet. Soft iron magnetises easily but loses its magnetism almost as soon as it is no longer near a magnetic field, making it only a temporary magnet.
(b)
Permanent magnet example: a fridge magnet or a compass needle, which need to stay magnetised indefinitely. Temporary magnet example: the core of an electromagnet, which needs to lose its magnetism as soon as it is switched off (e.g. to release an object it was holding).
QUESTION 7 6 marks Criterion A
Hard
Geographic NGeographic ScompasscompassEarth

The diagram shows a simplified model of the Earth's magnetic field, with a compass at two different locations near it.

a. Explain why a compass needle lines up in a north-south direction.
[2]
b. Using the like-poles-repel, unlike-poles-attract rule, explain what this tells you about the magnetic pole located near the Earth's geographic North.
[2]
c. State how the diagram shows the field's strength changing at different points around the Earth.
[2]
Show complete worked solution
(a)
A compass needle is itself a small magnet, free to rotate. It experiences a magnetic force from the Earth's own magnetic field, which acts a little like a giant bar magnet. The needle's north-seeking pole turns to align with the field, causing it to point along the field lines — roughly towards geographic north.
(b)
Since the compass needle's north-seeking pole is attracted towards geographic North, and unlike poles attract, the region near the Earth's geographic North must actually behave like a magnetic south pole — a fact that surprises many people, since it is still called the "north magnetic pole" for navigation purposes.
(c)
As with a bar magnet, the field lines are drawn closer together nearer the magnetic poles and further apart further away (e.g. near the equator, between the poles) — showing the field is stronger nearer the poles and weaker further from them.
QUESTION 8 7 marks Criterion B
Medium

You want to plot the magnetic field pattern around a bar magnet using a small plotting compass.

a. State the independent and dependent variables.
[2]
b. State a variable you would keep the same throughout, and explain why.
[2]
c. Describe a method to plot the field pattern.
[3]
Show complete worked solution
(a)
Independent variable: position around the magnet (where the compass is placed). Dependent variable: the direction the compass needle points at that position.
(b)
Keep the same magnet, in the same fixed position, throughout the whole plot. If the magnet moved between readings, the field pattern being mapped would itself change partway through, making the plotted pattern meaningless.
(c)
  1. Place the bar magnet on a sheet of paper and draw around it.
  2. Place the plotting compass just next to one pole and mark a dot at each end of the needle.
  3. Move the compass so its tail is at the dot just made, and mark a new dot at its head; repeat this several times to trace a full curved line from one pole, around, to the other.
  4. Draw an arrow on the line to show the direction the compass pointed (N to S, outside the magnet).
  5. Repeat starting from different points around the magnet to build up the whole field pattern.
QUESTION 9 6 marks Criterion B
Medium

You are given several unlabelled material samples and asked to test which are magnetic.

a. State the independent and dependent variables.
[2]
b. State a variable you would control, and explain why.
[2]
c. Describe how you would carry out the test.
[2]
Show complete worked solution
(a)
Independent variable: the material of each sample. Dependent variable: whether the sample is attracted to the magnet (yes/no).
(b)
Use the same magnet and bring it to the same distance from each sample. A stronger magnet, or holding it closer, could attract a weakly-magnetic material that a weaker test or greater distance would miss, making the comparison between samples unfair.
(c)
Bring the magnet slowly towards each sample from the same starting distance, and record whether the sample is attracted (moves towards or sticks to the magnet) or not. Repeat each sample test a couple of times to check the result is consistent.
QUESTION 10 8 marks Criterion B
Hard

A student measures the "strength" of different magnets by counting how many paperclips each can hold in a chain. They test the same magnet three times and get quite different results: $6$, $9$, and $12$ paperclips.

a. Suggest a reason this method might give such inconsistent results for the same magnet.
[2]
b. Suggest an improvement to the method that would give more objective, reliable measurements.
[3]
c. Explain why this improvement leads to more reliable data.
[3]
Show complete worked solution
(a)
Counting paperclips is a fairly subjective, imprecise method — it depends on exactly how the clips are added (angle, whether they slip), the size and mass of the individual paperclips used, and a judgement call about when a clip has genuinely "stopped" holding.
(b)
Instead of counting paperclips, use a force meter (newton meter) connected to a small steel/iron plate held against the magnet, and measure the force (in newtons) needed to pull the plate away from the magnet. This gives a precise, repeatable numerical value rather than a subjective clip count.
(c)
A force meter gives a continuous, precise numerical measurement (e.g. to $0.1\,\text{N}$) rather than a whole-number count that depends on clip size and judgement. Repeated measurements of the same magnet using a force meter should therefore be far more consistent with each other, since the measurement itself is not affected by inconsistent paperclip behaviour.
QUESTION 11 3 marks Criterion C
Easy
MaterialIronSteelCopperPlasticNickelWood
Attracted to magnet?YesYesNoNoYesNo
Show complete worked solution

The materials attracted to the magnet (iron, steel, nickel) are all ferromagnetic materials — iron itself, or alloys/elements closely related to it. The materials not attracted (copper, plastic, wood) are all non-magnetic. This shows only a small group of materials are magnetic, not metals in general (copper is a metal but is not attracted).

QUESTION 12 5 marks Criterion C
Medium
MagnetTrial 1Trial 2Trial 3
A565
B898
C343
D10911
a. Calculate the mean number of paperclips held by each magnet.
[3]
b. Rank the magnets from strongest to weakest.
[2]
Show complete worked solution
(a)
A: $\frac{5+6+5}{3}=5.33$. B: $\frac{8+9+8}{3}=8.33$. C: $\frac{3+4+3}{3}=3.33$. D: $\frac{10+9+11}{3}=10.0$.
(b)
D $>$ B $>$ A $>$ C (from the highest to lowest mean number of paperclips held).
QUESTION 13 4 marks Criterion C
Medium
Distance from magnet (cm)123456
Paperclips attracted532100
a. Describe the trend shown by this data.
[2]
b. Explain this trend, and state what the data shows happens beyond a distance of $4\,\text{cm}$.
[2]
Show complete worked solution
(a)
As the distance from the magnet increases, the number of paperclips attracted decreases.
(b)
A magnet's field gets weaker with increasing distance from it, so at greater distances there is not enough force to attract a paperclip at all. Beyond $4\,\text{cm}$, the data shows no paperclips are attracted — the field is too weak at that range to have a noticeable effect.
QUESTION 14 4 marks Criterion C
Medium
Trial12345
Paperclips held89838
a. Identify the anomalous result.
[1]
b. Suggest why this result might have occurred.
[2]
c. State what should be done with this reading.
[1]
Show complete worked solution
(a)
Trial $4$ ($3$ paperclips) is anomalous.
(b)
A clip may have been placed incorrectly and slipped off before being counted, or the chain may have accidentally been disturbed during that trial — it does not fit the otherwise consistent pattern of $8$–$9$ clips.
(c)
It should be excluded (treated as an outlier) when calculating the mean, and the trial should ideally be repeated to check the true result.
QUESTION 15 5 marks Criterion C
Medium
Distance from magnet (cm)246810
Compass deflection (°)805530102
a. Describe the trend shown by this data.
[2]
b. Explain this trend in terms of the magnet's field.
[3]
Show complete worked solution
(a)
As the distance from the magnet increases, the compass needle's deflection angle decreases steadily, approaching close to $0°$ by $10\,\text{cm}$.
(b)
Closer to the magnet, its field is much stronger than the Earth's own weak background magnetic field, so it pulls the compass needle strongly away from north, giving a large deflection. As distance increases, the magnet's field weakens rapidly, so it has less and less influence on the needle compared to the Earth's field — until, far enough away, the needle points almost undisturbed, back towards north (deflection close to $0°$).
QUESTION 16 8 marks Criterion C
Hard
Trial12345
Paperclips held6148713
a. Comment on the reliability of this data.
[3]
b. Suggest reasons for this large variation.
[2]
c. Propose a more objective method that would reduce this variation, and explain why it would be more reliable.
[3]
Show complete worked solution
(a)
This data is not very reliable — the five results for the same magnet range from $6$ to $14$ paperclips, a spread of $8$, which is a very large variation (more than double) for repeated tests of one object.
(b)
Likely causes include: the paperclips used may vary in size or mass, the angle/orientation at which each clip was added may differ, and there is a subjective judgement involved in deciding when a clip has "stopped holding" — none of these are controlled precisely in a simple paperclip-counting test.
(c)
Use a force meter (newton meter) to measure the force (in newtons) needed to pull a fixed steel plate away from the magnet, rather than counting clips. This gives a precise numerical reading that does not depend on inconsistent clip size or subjective judgement, so repeated trials of the same magnet should give much more consistent values.
QUESTION 17 7 marks Criterion C
Hard
Nail materialClips held while touching magnetClips held after magnet removed
Soft iron60
Hardened steel43
a. Describe the pattern in the data while each nail was still touching the magnet.
[2]
b. Describe the pattern once the magnet was removed.
[2]
c. Explain this data using the ideas of temporary and permanent magnetism.
[3]
Show complete worked solution
(a)
While touching the magnet, the soft iron nail held more clips ($6$) than the steel nail ($4$) — iron magnetises more easily/strongly when directly in a strong field.
(b)
Once the magnet was removed, the soft iron nail held $0$ clips (lost all its magnetism), while the steel nail still held $3$ clips (kept most of its magnetism).
(c)
This is a clear example of soft (temporary) versus hard (permanent) magnetic materials. Soft iron gains strong induced magnetism easily while in a magnetic field, but loses it almost completely as soon as the field is removed — consistent with its use in devices that need to switch magnetism on and off. Steel is harder to magnetise fully (hence the lower initial count), but once magnetised it retains much of that magnetism even after the original magnet is taken away — exactly why steel, not iron, is used to make permanent magnets.
QUESTION 18 5 marks Criterion D
Medium

Recycling centres often use large magnets on a conveyor belt to automatically separate steel/iron items from other mixed waste.

Discuss a benefit and a drawback of using magnets this way.

Show complete worked solution

Benefit: A magnetic separator can quickly and automatically pull out steel and iron objects from a large, fast-moving stream of mixed waste, without needing anyone to sort it by hand. This makes recycling ferrous (iron-containing) metal far more efficient, reduces landfill waste, and saves the energy and raw materials needed to mine and produce brand-new steel.

Drawback: A simple magnet only attracts magnetic metals like steel and iron — non-magnetic metals such as aluminium and copper are left behind and pass straight through, so additional separation methods (and additional cost/equipment) are still needed to fully sort all recyclable materials.

QUESTION 19 5 marks Criterion D
Medium

For centuries, sailors and travellers relied on magnetic compasses for navigation. Today, most people use GPS on smartphones instead.

Discuss a benefit and a drawback of relying on a magnetic compass compared to GPS.

Show complete worked solution

Benefit: A magnetic compass needs no batteries, satellites, or signal — it works purely using the Earth's own magnetic field, making it a simple, reliable backup that keeps working in remote areas or during power/technology failures where GPS might not be available.

Drawback: A compass can be thrown off by nearby magnetic materials or other magnetic fields (for example, being near a car, an electronic device, or a deposit of magnetic rock), giving an inaccurate reading without the user realising. It is also far less precise than GPS, which can pinpoint an exact location rather than just a direction.

QUESTION 20 6 marks Criterion D
Hard

Older hard disk drives (HDDs) store computer data by magnetising microscopic regions on a spinning disk.

Evaluate the impact of magnetic data storage, discussing a benefit and a concern it raises.

Show complete worked solution

Benefit: Magnetic storage allows huge amounts of data to be stored very cheaply per gigabyte, and — unlike some memory types — it is non-volatile, meaning the stored data is retained even with no power connected, which is essential for reliably keeping files, photos, and programs long-term.

Concern: Because the data relies on tiny magnetised regions, exposing a hard drive to a sufficiently strong external magnetic field can corrupt or erase the stored data — an important risk for anyone storing important files this way. There is also an environmental concern: as hard drives are replaced (increasingly by non-magnetic solid-state drives), the discarded devices contribute to growing electronic waste unless properly recycled.

Electromagnets 20 questions

QUESTION 1 3 marks Criterion A
Easy

An electromagnet can be made by coiling insulated wire around an iron core and connecting the ends of the wire to a battery.

a. State what happens when the switch in the circuit is closed.
[1]
b. State what happens when the switch is opened again.
[1]
c. State one key difference between an electromagnet and a normal permanent bar magnet.
[1]
Show complete worked solution
(a)
Current flows through the coil, and the coil produces a magnetic field — the iron core becomes magnetised, turning the whole device into a magnet.
(b)
The current stops flowing, so the coil no longer produces a magnetic field, and the (soft iron) core loses almost all of its magnetism — the electromagnet switches off.
(c)
An electromagnet's magnetism can be switched on and off (by turning the current on or off), whereas a permanent magnet's magnetism is always present.
QUESTION 2 3 marks Criterion A
Easy

State three factors that can be changed to increase the strength of an electromagnet, briefly explaining each.

Show complete worked solution

1. Number of turns of wire: more turns in the coil produces a stronger magnetic field. 2. Size of the current: a larger current flowing through the coil produces a stronger magnetic field. 3. Core material: using a magnetic core (such as soft iron) inside the coil, instead of no core (air), greatly concentrates and strengthens the field.

QUESTION 3 4 marks Criterion A
Easy
coil leads to cellpaperclipsinsulated wire coil (turns)switch

The diagram shows a simple electromagnet.

a. Name the part labelled as the coil around the nail.
[1]
b. Name the material the core (nail) is usually made from, and why.
[1]
c. Describe what happens to the paperclips when the switch is closed, and then when it is opened again.
[2]
Show complete worked solution
(a)
This is the coil of insulated wire, wound in turns around the core.
(b)
Soft iron — it magnetises strongly when current flows but loses its magnetism almost completely when the current stops, allowing the electromagnet to be switched off.
(c)
When the switch is closed, current flows through the coil, magnetising the iron core, so the paperclips are attracted and picked up. When the switch is opened, the current stops, the core loses its magnetism, and the paperclips fall off.
QUESTION 4 3 marks Criterion A
Medium

Explain why electromagnets are usually built with a soft iron core rather than a steel core.

Show complete worked solution

Soft iron magnetises quickly and strongly when current flows, but loses its magnetism almost immediately once the current is switched off. Steel would stay magnetised even after the current stopped, meaning the electromagnet could not be switched off — this would make it useless for applications like a scrapyard crane, which needs to release its magnetised load by switching the current off.

QUESTION 5 4 marks Criterion A
Medium

Compare an electromagnet to a permanent bar magnet.

a. State two similarities between them.
[2]
b. State two differences between them.
[2]
Show complete worked solution
(a)
Both have a north (N) and south (S) pole, both produce a magnetic field around them, and both follow the same rule that like poles repel and unlike poles attract.
(b)
An electromagnet can be switched on and off (a permanent magnet cannot), and an electromagnet's strength can be varied (by changing the current or number of turns) and its poles can be reversed (by reversing the current) — none of which is possible with a fixed permanent magnet.
QUESTION 6 3 marks Criterion A
Medium

Explain how reversing the terminals of the battery connected to an electromagnet's coil affects the electromagnet.

Show complete worked solution

Reversing the battery terminals reverses the direction the current flows through the coil. This reverses the direction of the magnetic field the coil produces, which swaps the electromagnet's poles — whichever end was the north pole becomes the south pole, and vice versa.

QUESTION 7 6 marks Criterion A
Hard
NSfield direction: N ? S (outside magnet)

An electromagnet is made from a coil of wire around a soft iron core, connected to a battery.

a. Describe the effect on the electromagnet's strength of doubling only the number of turns in the coil, keeping the current the same.
[2]
b. Describe the effect of also doubling the current at the same time (so both the turns and the current are doubled).
[2]
c. Describe the magnetic field pattern that a current-carrying coil like this produces, compared to a bar magnet.
[2]
Show complete worked solution
(a)
Doubling the number of turns (while keeping the current the same) roughly doubles the strength of the magnetic field produced, since each additional turn of current-carrying wire adds to the total field.
(b)
With both the turns and the current doubled, the electromagnet becomes much stronger still than doubling either factor alone — the two effects combine, since both a higher current and more turns each independently increase the field strength.
(c)
A current-carrying coil (solenoid) produces a magnetic field pattern that looks just like a bar magnet's — field lines emerge from one end (acting as the N pole), loop around the outside, and re-enter at the other end (the S pole), with the field densest and strongest near the two ends of the coil.
QUESTION 8 7 marks Criterion B
Medium

You want to investigate how the number of turns in a coil affects the strength of an electromagnet, using the number of paperclips it can pick up as a measure of strength.

a. State the independent and dependent variables.
[2]
b. State two variables you would control, and explain why for one.
[2]
c. Describe the method you would use.
[3]
Show complete worked solution
(a)
Independent variable: number of turns in the coil (e.g. $10, 20, 30, 40$). Dependent variable: strength of the electromagnet, measured by the number of paperclips it can pick up.
(b)
Control the current (same battery/cells) and the core material (same iron nail). Why control the current: a bigger current also increases strength, so if current changed along with the number of turns, it would be unclear which factor actually caused any change observed.
(c)
  1. Wind $10$ turns of insulated wire around the iron nail and connect it to the battery via a switch.
  2. Close the switch and bring the nail's tip to a pile of paperclips; count how many it picks up.
  3. Open the switch, remove the wire, and re-wind $20$ turns on the same nail; repeat the test.
  4. Repeat again for $30$ and $40$ turns, keeping the same battery and nail throughout.
  5. Repeat each turn-count $2$–$3$ times and calculate a mean, to improve reliability.
QUESTION 9 6 marks Criterion B
Medium

You want to investigate how the current flowing through a coil (controlled by the number of cells used) affects the strength of an electromagnet.

a. State the independent and dependent variables.
[2]
b. State a variable you would control, and explain why.
[2]
c. Describe how you would carry out the test.
[2]
Show complete worked solution
(a)
Independent variable: current, controlled via the number of cells used (e.g. $1, 2, 3$). Dependent variable: strength of the electromagnet, measured by the number of paperclips it picks up.
(b)
Control the number of turns in the coil (keep it fixed). More turns also increases strength, so changing it alongside the number of cells would make it unclear whether any change in strength was due to the current or the turns.
(c)
Set up the coil with $1$ cell, close the switch, and count the paperclips picked up. Add a second identical cell in series (keeping the coil the same) and repeat the test. Repeat again with $3$ cells, and repeat each test a couple of times to check for consistency.
QUESTION 10 8 marks Criterion B
Hard

A student investigates how the strength of an electromagnet is affected by the core material (no core, wood, soft iron, steel), reusing the same steel core straight after testing it without demagnetising it, then testing soft iron afterwards.

a. Suggest a source of error in this method.
[2]
b. Suggest an improvement to the method.
[3]
c. Explain how this improvement makes the comparison between core materials fairer.
[3]
Show complete worked solution
(a)
Steel retains magnetism after the current is switched off. If the steel core is not properly demagnetised before moving on, leftover residual magnetism could linger nearby or affect readings, and — more importantly — reusing the same coil winding without care could mean the number of turns isn't kept perfectly identical between core swaps, affecting the fairness of the comparison.
(b)
Use a separate, identically wound coil for each core to guarantee the same number of turns every time, and demagnetise the steel core (e.g. by tapping it while removed from any field, or briefly reversing the current several times) before and after testing it, so no residual magnetism carries over between tests.
(c)
With identical coils and no residual magnetism carried over, the only thing that differs between tests is the core material itself — so any difference in the number of paperclips picked up can be confidently attributed to the core material, rather than to inconsistent winding or leftover magnetism from a previous test.
QUESTION 11 3 marks Criterion C
Easy
Number of turns10203040
Paperclips picked up2468
Show complete worked solution

The data shows a clear pattern: more turns in the coil means more paperclips picked up. The relationship is directly proportional — paperclips picked up $=\ \text{turns}\div5$ in every case ($10\div5=2$, $20\div5=4$, $30\div5=6$, $40\div5=8$), confirming that increasing the number of turns increases the electromagnet's strength.

QUESTION 12 5 marks Criterion C
Medium
Number of cells1234
Paperclips picked up36912
a. Describe the pattern shown by this data.
[2]
b. Use the pattern to predict the result for $5$ cells, and explain a limitation of extending this pattern too far.
[3]
Show complete worked solution
(a)
As the number of cells (and therefore the current) increases, the number of paperclips picked up increases, and does so directly proportionally — each extra cell adds exactly $3$ more paperclips.
(b)
Following the pattern ($+3$ per cell): $5$ cells would be predicted to pick up $\mathbf{15}$ paperclips. However, this pattern is unlikely to continue indefinitely — with a very large current, the wire could overheat, the core could reach a limit where adding more current barely increases its magnetism further ("saturation"), or the battery/cells might not be able to safely supply that much current at all.
QUESTION 13 4 marks Criterion C
Medium
Number of turns10203040
Paperclips picked up2498
a. Identify the anomalous result.
[1]
b. Explain why this result is anomalous.
[2]
c. State what should be done about this result.
[1]
Show complete worked solution
(a)
The result for $30$ turns ($9$ paperclips) is anomalous.
(b)
It breaks the otherwise steadily increasing pattern ($2,4,\ldots$) by rising more than expected and then being higher than the very next reading ($8$ at $40$ turns), which should be greater still if the pattern continued sensibly.
(c)
It should be treated as an outlier and this trial should be repeated to check the true value at $30$ turns.
QUESTION 14 5 marks Criterion C
Medium
Core materialNone (air)WoodSoft ironSteel
Paperclips picked up1197
a. Describe the difference between the non-magnetic cores (air, wood) and the magnetic cores (iron, steel).
[2]
b. Explain why the magnetic cores make such a large difference, and suggest why soft iron gave a slightly higher reading than steel here.
[3]
Show complete worked solution
(a)
The non-magnetic cores (air/no core, and wood) give a very low strength (only $1$ paperclip) — similar to the coil alone with no core at all. The magnetic cores (soft iron and steel) give a much greater strength ($9$ and $7$ paperclips).
(b)
A magnetic core (iron or steel) concentrates and strengthens the coil's magnetic field far more than air or a non-magnetic material like wood can, which is why they perform so much better. Soft iron gave a slightly higher reading than steel because iron magnetises more easily and completely while current is actively flowing through the coil, whereas steel is somewhat more resistant to becoming magnetised in the first place (though it would retain more of its magnetism afterwards).
QUESTION 15 4 marks Criterion C
Medium
Trial12345
Paperclips picked up67667
a. Calculate the mean number of paperclips picked up.
[2]
b. Comment on the reliability of this data.
[2]
Show complete worked solution
(a)
$$ \text{mean} = \frac{6+7+6+6+7}{5} = \frac{32}{5} $$Answer: $6.4$
(b)
The results only range from $6$ to $7$ (a range of just $1$), which is a small spread for $5$ repeated trials — this suggests the measurement is fairly reliable and consistent.
QUESTION 16 7 marks Criterion C
Hard
TurnsCellsPaperclips picked up
1012
2014
1024
2028
a. Describe the pattern shown when both the number of turns and the number of cells are varied together.
[3]
b. Using this pattern, predict the number of paperclips for $30$ turns and $2$ cells.
[3]
c. State one assumption this prediction relies on.
[1]
Show complete worked solution
(a)
The number of paperclips picked up appears to depend on both turns and cells together: doubling either the turns or the cells alone doubles the strength, and the strength roughly follows (turns $\div 10$) $\times$ (cells) $\times 2$ in every row shown.
(b)
Going from $20$ turns to $30$ turns is an increase of $10$ turns; at $1$ cell, each extra $10$ turns adds $2$ paperclips, and at $2$ cells each extra $10$ turns should add roughly double that, about $4$ paperclips. So: $$ 8 + 4 = 12 $$Predicted answer: approximately $12$ paperclips.
(c)
It assumes the same proportional pattern continues smoothly beyond the tested combinations, with no limiting effects (like core saturation or wire heating) starting to occur at $30$ turns and $2$ cells.
QUESTION 17 8 marks Criterion C
Hard
Number of turnsPaperclips picked up
103
205
a. A student claims: "Doubling the number of turns doubles the electromagnet's strength." Calculate the ratio between the two results and compare it to what the claim predicts.
[2]
b. Discuss possible reasons the data does not show an exact doubling.
[4]
c. State an overall conclusion about the claim, based on this data.
[2]
Show complete worked solution
(a)
Doubling the turns claim predicts $3\times2=6$ paperclips at $20$ turns. The actual result was $5$ paperclips — a ratio of $\frac{5}{3}\approx1.67$, not exactly $2$.
(b)
Counting whole paperclips is an imprecise measurement — the true strength might correspond to something between $5$ and $6$ clips, but a fraction of a clip can't be recorded, which alone could explain part of the gap. It is also possible that using $20$ turns of the same wire (rather than a separate longer wire) slightly increases the coil's own resistance, which could very slightly reduce the current compared to the $10$-turn coil, meaning the field doesn't scale up by exactly the full factor expected from turns alone.
(c)
The data broadly supports the idea that more turns produce a stronger electromagnet, but does not precisely confirm an exact doubling relationship — the increase is close to, but measurably less than, double, most likely due to the limitations of this measuring method.
QUESTION 18 5 marks Criterion D
Medium

Scrapyards use powerful electromagnets on cranes to lift and sort large loads of scrap metal, releasing the load by switching the current off.

Discuss a benefit and a drawback of using electromagnets this way.

Show complete worked solution

Benefit: An electromagnetic crane can quickly and efficiently lift heavy loads of scrap steel/iron and precisely control when to release them (by switching the current off), which is far faster and safer for workers than using mechanical grabs or hooks, and allows magnetic scrap to be automatically separated from non-magnetic material.

Drawback: Powerful electromagnets require a large, continuous supply of electrical current to stay switched on, using significant amounts of energy. There is also a safety risk if power to the electromagnet fails or is interrupted unexpectedly while a heavy load is suspended — the load would suddenly drop.

QUESTION 19 6 marks Criterion D
Medium

MRI (Magnetic Resonance Imaging) scanners use extremely strong electromagnets to produce detailed images of the inside of the human body, without using harmful radiation.

Discuss a benefit and a concern raised by this technology.

Show complete worked solution

Benefit: MRI scanners allow doctors to see detailed, high-quality images of soft tissue, organs, and injuries inside the body without exposing the patient to ionising radiation (unlike X-rays or CT scans), making them a safer imaging option that can be used more freely, including repeated scans, to diagnose and monitor a wide range of medical conditions.

Concern: The magnetic fields used are extremely strong, which raises real safety risks: loose metal objects can become dangerous flying projectiles near the scanner, and the strong field can be dangerous for patients with certain metal implants or devices such as pacemakers. MRI machines are also very expensive to build, install, and run (the powerful electromagnets often require special cooling), limiting how widely available this technology can be, especially in areas with fewer healthcare resources.

QUESTION 20 6 marks Criterion D
Hard

Electromagnets are essential components inside loudspeakers, electric motors, and many other modern devices, including those in electric vehicles.

Evaluate the impact of electromagnet technology in these applications, discussing a benefit and a drawback.

Show complete worked solution

Benefit: Electromagnets, through the motor effect, are at the heart of technologies that have transformed daily life and communication — from loudspeakers that let us hear music and calls, to electric motors that power everything from washing machines to electric vehicles. Electric vehicles in particular can reduce local air pollution and emissions compared to petrol/diesel vehicles, improving air quality where they are used.

Drawback: The overall environmental benefit depends heavily on how the electricity used is generated — if it still comes largely from burning fossil fuels, much of the emissions problem is simply moved elsewhere rather than solved. Building these devices, including the powerful magnets and batteries involved, also requires mining and processing raw materials, which has its own environmental and social costs, and contributes to electronic waste at the end of a device's life.