Electricity and Magnetism
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Electric Current and Circuits 20 questions
The diagram shows a simple circuit.
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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.
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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.
A current of $2\,\text{A}$ flows through a wire for $30\,\text{s}$. Calculate the charge that passes through the wire.
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Step 1 — State the formula:
$$ Q = I \times t $$
Step 2 — Substitute the values:
$$ Q = 2 \times 30 $$
Answer: $Q = 60\,\text{C}$ (coulombs)
A charge of $15\,\text{C}$ passes through a lamp in $5\,\text{s}$.
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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.
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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 $0.4\,\text{A}$.
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A current of $250\,\text{mA}$ flows through a circuit for $2$ minutes.
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You want to investigate how the number of cells in a circuit affects the current flowing through a fixed bulb.
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- Set up a series circuit with one cell, the bulb, and an ammeter.
- Close the switch and record the ammeter reading.
- Add one more identical cell in series, keeping everything else the same, and record the new reading.
- Repeat for $3$ and then $4$ cells.
- Repeat each reading $3$ times and calculate a mean current for each number of cells, to improve reliability.
You are given an unknown material and asked to test whether it conducts electricity, using a simple circuit tester.
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A student investigates how current changes as more cells are added to a circuit. Their ammeter readings for $2$ cells are inconsistent between repeats: $0.41\,\text{A}$, $0.60\,\text{A}$, $0.40\,\text{A}$.
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| Point in loop | P | Q | R |
|---|---|---|---|
| Current (A) | 0.50 | 0.50 | 0.50 |
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| Time (s) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| Charge (C) | 0 | 10 | 20 | 30 | 40 |
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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.)
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Current (A) | 1.20 | 1.22 | 1.19 | 1.55 | 1.21 |
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| Bulbs in series | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Current (A) | 0.60 | 0.30 | 0.20 | 0.15 |
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A charge sensor records the total charge that has passed a point in a circuit at different times.
| Time (s) | 0 | 10 | 20 | 30 |
|---|---|---|---|---|
| Total charge (C) | 0 | 5 | 12 | 15 |
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$0$–$10\,\text{s}$: $I=\dfrac{5-0}{10}=0.5\,\text{A}$
$10$–$20\,\text{s}$: $I=\dfrac{12-5}{10}=0.7\,\text{A}$
$20$–$30\,\text{s}$: $I=\dfrac{15-12}{10}=0.3\,\text{A}$
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Time for 50 C to pass (s) | 24.8 | 25.1 | 24.9 | 30.2 | 25.0 |
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| Time interval (s) | 0–2 | 2–5 | 5–8 |
|---|---|---|---|
| Current (A) | 3 | 5 | 2 |
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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: $Q_{total} = 27\,\text{C}$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.
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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.
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.
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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.
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.
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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
Compare series and parallel circuits.
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The diagram shows two identical bulbs connected in series to a battery.
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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.
The diagram shows two identical bulbs connected in parallel across a battery.
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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.
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.
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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 $0.50\,\text{A}$.
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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: $A = 0.80\,\text{A}$
Two identical bulbs are connected in series to a $9\,\text{V}$ battery. Voltmeter $V_1$, connected across the first bulb, reads $3.5\,\text{V}$.
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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: $V_2 = 5.5\,\text{V}$
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 $5\,\text{V}$.
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You want to investigate whether bulbs are dimmer when connected in series compared to parallel, using the same battery and identical bulbs.
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- Build a series circuit with two identical bulbs and an ammeter, and record the ammeter reading (and observe brightness).
- Using the same battery and identical bulbs, build a parallel circuit and place an ammeter in one branch; record its reading (and observe brightness).
- Repeat each measurement $3$ times and calculate a mean current for each arrangement.
- Compare the mean currents: a bulb with a higher current through it will be brighter.
You want to investigate whether adding more bulbs in parallel changes the brightness of the bulbs that are already in the circuit.
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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.
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| Point in loop | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Current (A) | 0.45 | 0.45 | 0.44 | 0.45 |
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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.
A parallel circuit has two branches.
| Case | Branch 1 (A) | Branch 2 (A) | Main current (A) |
|---|---|---|---|
| 1 | 0.20 | 0.35 | ? |
| 2 | 0.55 | ? | 0.90 |
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| Bulbs in series | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Brightness (arbitrary units) | 100 | 50 | 33 | 25 |
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| Bulbs in parallel | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Brightness of each bulb | 100 | 100 | 100 | 100 |
| Total current from battery (A) | 0.5 | 1.0 | 1.5 | 2.0 |
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A student measures the voltage across each of three identical bulbs connected in series to a $6\,\text{V}$ battery.
| Bulb | 1 | 2 | 3 |
|---|---|---|---|
| Voltage (V) | 2.1 | 2.0 | 2.2 |
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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).
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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 $0.80\,\text{A}$.
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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.
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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.
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.
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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.
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.
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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
State Ohm's law, including the equation, and the unit of each quantity in it.
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$$ V = IR $$ where $V$ is potential difference (voltage) in volts (V), $I$ is current in amps (A), and $R$ is resistance in ohms ($\Omega$).
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 $5\,\Omega$.
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Step 1 — State the formula:
$$ V = IR $$
Step 2 — Substitute the values:
$$ V = 2 \times 5 $$
Answer: $V = 10\,\text{V}$
A voltmeter reads $12\,\text{V}$ across a component, and an ammeter reads $3\,\text{A}$ through it. Calculate the resistance of the component.
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Rearranging $V = IR$ for resistance: $$ R = \frac{V}{I} = \frac{12}{3} $$Answer: $R = 4\,\Omega$
A resistor of $1500\,\Omega$ is connected to a $6\,\text{V}$ supply.
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Explain how each change below affects the resistance of a metal wire, if all other factors are kept the same.
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Two resistors, $R_1 = 4\,\Omega$ and $R_2 = 6\,\Omega$, are connected in series to a $20\,\text{V}$ battery.
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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.
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You want to investigate how the length of a wire affects its resistance.
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- Tape the test wire along a metre ruler and connect one end to the circuit with a crocodile clip.
- 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.
- Record the ammeter and voltmeter readings and calculate resistance using $R = \frac{V}{I}$.
- Move the second crocodile clip to $20\,\text{cm}$, $30\,\text{cm}$, etc., repeating the measurement and calculation each time.
- Use a low current/short connection times to avoid the wire heating up, which would otherwise change its resistance.
You want to investigate how the resistance connected in a circuit affects the current that flows, using a fixed power supply.
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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.
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A voltmeter reads $4\,\text{V}$ across a resistor, and an ammeter reads $0.5\,\text{A}$ through it.
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$$ R = \frac{V}{I} = \frac{4}{0.5} $$Answer: $R = 8\,\Omega$
| Voltage (V) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Current (A) | 0.4 | 0.8 | 1.2 | 1.6 |
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$R=\dfrac{2}{0.4}=5\,\Omega$
$R=\dfrac{4}{0.8}=5\,\Omega$
$R=\dfrac{6}{1.2}=5\,\Omega$
$R=\dfrac{8}{1.6}=5\,\Omega$
The graph shows current against voltage for a resistor.
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| Length (cm) | 10 | 20 | 30 | 40 |
|---|---|---|---|---|
| Resistance ($\Omega$) | 2 | 4 | 6 | 8 |
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| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Current at 6 V (A) | 1.19 | 1.21 | 1.20 | 0.85 | 1.20 |
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The graph shows the voltage-current relationship for a filament lamp, which curves rather than forming a straight line.
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A student tests three wires of different thickness (but the same length and material), measuring the resistance of each three times.
| Thickness | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|
| Thin (0.2 mm) | 12.1 | 11.9 | 12.0 |
| Medium (0.4 mm) | 6.0 | 6.2 | 5.9 |
| Thick (0.6 mm) | 2.8 | 3.0 | 2.9 |
(All resistance values in $\Omega$.)
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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.
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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.
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.
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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.
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.
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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
The diagram shows two bar magnets, each with its poles labelled.
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Classify each material as magnetic (attracted to a magnet) or non-magnetic: iron, copper, steel, aluminium, nickel, plastic, cobalt, wood.
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Magnetic: iron, steel, nickel, cobalt (these are all ferromagnetic materials/alloys). Non-magnetic: copper, aluminium, plastic, wood.
Define a magnetic field, and state the direction convention used for magnetic field lines around a bar magnet.
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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.
The diagram shows the magnetic field pattern around a bar magnet.
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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.
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Compare permanent magnets made from steel with temporary magnets made from soft iron.
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The diagram shows a simplified model of the Earth's magnetic field, with a compass at two different locations near it.
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You want to plot the magnetic field pattern around a bar magnet using a small plotting compass.
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- Place the bar magnet on a sheet of paper and draw around it.
- Place the plotting compass just next to one pole and mark a dot at each end of the needle.
- 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.
- Draw an arrow on the line to show the direction the compass pointed (N to S, outside the magnet).
- Repeat starting from different points around the magnet to build up the whole field pattern.
You are given several unlabelled material samples and asked to test which are magnetic.
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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.
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| Material | Iron | Steel | Copper | Plastic | Nickel | Wood |
|---|---|---|---|---|---|---|
| Attracted to magnet? | Yes | Yes | No | No | Yes | No |
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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).
| Magnet | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|
| A | 5 | 6 | 5 |
| B | 8 | 9 | 8 |
| C | 3 | 4 | 3 |
| D | 10 | 9 | 11 |
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| Distance from magnet (cm) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Paperclips attracted | 5 | 3 | 2 | 1 | 0 | 0 |
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| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Paperclips held | 8 | 9 | 8 | 3 | 8 |
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| Distance from magnet (cm) | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|
| Compass deflection (°) | 80 | 55 | 30 | 10 | 2 |
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| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Paperclips held | 6 | 14 | 8 | 7 | 13 |
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| Nail material | Clips held while touching magnet | Clips held after magnet removed |
|---|---|---|
| Soft iron | 6 | 0 |
| Hardened steel | 4 | 3 |
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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.
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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.
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.
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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.
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.
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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
An electromagnet can be made by coiling insulated wire around an iron core and connecting the ends of the wire to a battery.
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State three factors that can be changed to increase the strength of an electromagnet, briefly explaining each.
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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.
The diagram shows a simple electromagnet.
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Explain why electromagnets are usually built with a soft iron core rather than a steel core.
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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.
Compare an electromagnet to a permanent bar magnet.
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Explain how reversing the terminals of the battery connected to an electromagnet's coil affects the electromagnet.
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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.
An electromagnet is made from a coil of wire around a soft iron core, connected to a battery.
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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.
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- Wind $10$ turns of insulated wire around the iron nail and connect it to the battery via a switch.
- Close the switch and bring the nail's tip to a pile of paperclips; count how many it picks up.
- Open the switch, remove the wire, and re-wind $20$ turns on the same nail; repeat the test.
- Repeat again for $30$ and $40$ turns, keeping the same battery and nail throughout.
- Repeat each turn-count $2$–$3$ times and calculate a mean, to improve reliability.
You want to investigate how the current flowing through a coil (controlled by the number of cells used) affects the strength of an electromagnet.
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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.
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| Number of turns | 10 | 20 | 30 | 40 |
|---|---|---|---|---|
| Paperclips picked up | 2 | 4 | 6 | 8 |
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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.
| Number of cells | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Paperclips picked up | 3 | 6 | 9 | 12 |
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| Number of turns | 10 | 20 | 30 | 40 |
|---|---|---|---|---|
| Paperclips picked up | 2 | 4 | 9 | 8 |
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| Core material | None (air) | Wood | Soft iron | Steel |
|---|---|---|---|---|
| Paperclips picked up | 1 | 1 | 9 | 7 |
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| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Paperclips picked up | 6 | 7 | 6 | 6 | 7 |
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| Turns | Cells | Paperclips picked up |
|---|---|---|
| 10 | 1 | 2 |
| 20 | 1 | 4 |
| 10 | 2 | 4 |
| 20 | 2 | 8 |
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| Number of turns | Paperclips picked up |
|---|---|
| 10 | 3 |
| 20 | 5 |
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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.
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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.
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.
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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.
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.
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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.