Cambridge IGCSE · thinka-original Practice Paper

2025 Cambridge IGCSE Physics (0625) Practice Paper with Answers

Thinka Nov 2025 (V2) Cambridge IGCSE-Style Mock — Physics (0625)

80 marks75 mins2025
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.

Section Structured Theory Questions

Answer all questions. Write your answers in the spaces provided. Show all your working and use appropriate units.
37 Question · 80 marks
Question 1 · Structured Calculation
2 marks
A cyclist accelerates constantly from a speed of \(3.0\text{ m/s}\) to \(11.0\text{ m/s}\) in a time of \(4.0\text{ s}\).

Calculate the acceleration of the cyclist.
Show answer & marking scheme

Worked solution

1. Use the acceleration formula: \(a = \frac{v - u}{t}\)
2. Substitute the given values: \(a = \frac{11.0 - 3.0}{4.0}\)
3. Calculate the final value: \(a = \frac{8.0}{4.0} = 2.0\text{ m/s}^2\).

Marking scheme

C1: For stating or using \(a = \frac{v - u}{t}\) or substituting values correctly
A1: For correct final answer with unit: \(2.0\text{ m/s}^2\) (accept \(\text{m/s}^2\) or \(\text{m } s^{-2}\))
Question 2 · Structured Calculation
3 marks
A piece of glass has a mass of \(75.0\text{ g}\). When it is fully submerged in water inside a measuring cylinder, the water level rises from \(32.0\text{ cm}^3\) to \(62.0\text{ cm}^3\).

Calculate the density of the glass.
Show answer & marking scheme

Worked solution

1. Determine the volume of the glass: \(V = 62.0 - 32.0 = 30.0\text{ cm}^3\)
2. Use the density formula: \(\rho = \frac{m}{V}\)
3. Substitute the values: \(\rho = \frac{75.0}{30.0} = 2.5\text{ g/cm}^3\).

Marking scheme

C1: For calculating volume: \(V = 30.0\text{ cm}^3\)
C1: For recalling/using density formula \(\rho = \frac{m}{V}\)
A1: For correct final answer with unit: \(2.5\text{ g/cm}^3\) (accept \(\text{g/cm}^3\) or \(\text{g cm}^{-3}\))
Question 3 · Structured Calculation
3 marks
A crane lifts a crate of weight \(1500\text{ N}\) vertically upwards through a height of \(8.0\text{ m}\). This task takes a time of \(6.0\text{ s}\).

Calculate the useful power output of the crane.
Show answer & marking scheme

Worked solution

1. Calculate the work done by the crane: \(W = F \times d = 1500 \times 8.0 = 12\,000\text{ J}\)
2. Use the power formula: \(P = \frac{W}{t}\)
3. Substitute the values: \(P = \frac{12\,000}{6.0} = 2000\text{ W}\) (or \(2.0\text{ kW}\)).

Marking scheme

C1: For calculating work done: \(W = F \times d\) or \(12\,000\text{ J}\)
C1: For using power formula: \(P = \frac{W}{t}\) or substituting \(\frac{12\,000}{6.0}\)
A1: For correct final answer with unit: \(2000\text{ W}\) or \(2.0\text{ kW}\)
Question 4 · Structured Calculation
3 marks
A heavy metal block of mass \(15\text{ kg}\) rests on a flat table. The base of the block has a contact area of \(0.050\text{ m}^2\).

The acceleration of free fall \(g = 9.8\text{ m/s}^2\).

Calculate the pressure exerted by the block on the table.
Show answer & marking scheme

Worked solution

1. Calculate the weight (force) of the block: \(F = W = m \times g = 15 \times 9.8 = 147\text{ N}\)
2. Use the pressure formula: \(p = \frac{F}{A}\)
3. Substitute the values: \(p = \frac{147}{0.050} = 2940\text{ Pa}\).

Marking scheme

C1: For calculating weight: \(W = m \times g = 147\text{ N}\)
C1: For recalling/using pressure formula: \(p = \frac{F}{A}\)
A1: For correct final answer with unit: \(2940\text{ Pa}\) (or \(2900\text{ Pa}\) to 2 s.f., accept \(2.94\text{ kPa}\) or \(\text{N/m}^2\))
Question 5 · Structured Calculation
3 marks
An electric heater is switched on for \(5.0\text{ minutes}\). During this time, a charge of \(1200\text{ C}\) passes through the heating element.

Calculate the electric current in the heater.
Show answer & marking scheme

Worked solution

1. Convert the time to seconds: \(t = 5.0 \times 60 = 300\text{ s}\)
2. Use the charge formula: \(I = \frac{Q}{t}\)
3. Substitute the values: \(I = \frac{1200}{300} = 4.0\text{ A}\).

Marking scheme

C1: For converting time to seconds: \(300\text{ s}\)
C1: For recalling/using current formula: \(I = \frac{Q}{t}\) or substituting \(\frac{1200}{300}\)
A1: For correct final answer with unit: \(4.0\text{ A}\)
Question 6 · Structured Calculation
2 marks
A sound wave traveling through water has a frequency of \(1500\text{ Hz}\) and a wavelength of \(1.2\text{ m}\).

Calculate the speed of the sound wave in water.
Show answer & marking scheme

Worked solution

1. Use the wave speed equation: \(v = f \times \lambda\)
2. Substitute the given values: \(v = 1500 \times 1.2\)
3. Calculate the speed: \(v = 1800\text{ m/s}\).

Marking scheme

C1: For recalling/using wave speed equation: \(v = f \times \lambda\)
A1: For correct final answer with unit: \(1800\text{ m/s}\)
Question 7 · Structured Calculation
3 marks
A satellite orbits a planet in a circular path of radius \(8.0 \times 10^6\text{ m}\). The orbital period of the satellite is \(4.0 \times 10^3\text{ s}\).

Calculate the orbital speed of the satellite.
Show answer & marking scheme

Worked solution

1. Use the circular orbital speed formula: \(v = \frac{2\pi r}{T}\)
2. Substitute the values: \(v = \frac{2 \times \pi \times 8.0 \times 10^6}{4.0 \times 10^3}\)
3. Calculate the speed: \(v = \frac{1.6 \times 10^7 \pi}{4.0 \times 10^3} = 4000\pi \approx 12\,566\text{ m/s}\).
4. Rounding to 2 significant figures gives \(1.3 \times 10^4\text{ m/s}\) (or \(13\,000\text{ m/s}\)).

Marking scheme

C1: For stating/using the orbital speed equation: \(v = \frac{2\pi r}{T}\)
C1: For correct substitution: \(v = \frac{2 \times \pi \times 8.0 \times 10^6}{4.0 \times 10^3}\)
A1: For correct final answer: \(1.3 \times 10^4\text{ m/s}\) or \(13\,000\text{ m/s}\) (accept \(12\,500\) to \(12\,600\))
Question 8 · Structured Calculation
3 marks
A radioactive sample contains an isotope with an initial activity of \(2400\text{ counts/s}\). After a time of \(18\text{ hours}\), the activity has decreased to \(300\text{ counts/s}\).

Calculate the half-life of this isotope.
Show answer & marking scheme

Worked solution

1. Find the number of half-lives that have elapsed:
- \(2400 \rightarrow 1200\) (1 half-life)
- \(1200 \rightarrow 600\) (2 half-lives)
- \(600 \rightarrow 300\) (3 half-lives)
2. The total time elapsed is \(18\text{ hours}\) for 3 half-lives.
3. Calculate the duration of one half-life: \(t_{1/2} = \frac{18\text{ hours}}{3} = 6.0\text{ hours}\).

Marking scheme

C1: For determining that 3 half-lives have elapsed
C1: For setting up the relation: \(3 \times t_{1/2} = 18\text{ hours}\)
A1: For correct final answer with unit: \(6.0\text{ hours}\)
Question 9 · Structured Calculation
2 marks
A toy car of mass 0.80 kg is pulled along a smooth horizontal surface by a force of 3.2 N. Calculate the acceleration of the toy car.
Show answer & marking scheme

Worked solution

Using the formula F = ma, we can rearrange to find acceleration: a = F / m. Substituting the given values: a = 3.2 N / 0.80 kg = 4.0 m/s^2.

Marking scheme

C1: for quoting the formula F = ma or rearranged a = F/m. A1: for the correct final answer with appropriate unit: 4.0 m/s^2 (accept 4 m/s^2).
Question 10 · Structured Calculation
2 marks
A trolley of mass 2.0 kg travels at a speed of 3.5 m/s. Calculate the momentum of the trolley.
Show answer & marking scheme

Worked solution

Using the formula p = mv, we substitute the mass and speed: p = 2.0 kg * 3.5 m/s = 7.0 kg m/s.

Marking scheme

C1: for quoting the formula p = mv or showing substitution 2.0 * 3.5. A1: for the correct final answer with appropriate unit: 7.0 kg m/s (accept 7 kg m/s or 7 N s).
Question 11 · Structured Calculation
2 marks
A box of weight 150 N rests on a table. The area of the bottom of the box in contact with the table is 0.25 m^2. Calculate the pressure exerted by the box on the table.
Show answer & marking scheme

Worked solution

Using the formula p = F / A, we substitute the force (weight) and area: p = 150 N / 0.25 m^2 = 600 Pa.

Marking scheme

C1: for quoting the formula p = F/A or showing substitution 150 / 0.25. A1: for the correct final answer with appropriate unit: 600 Pa (accept 600 N/m^2).
Question 12 · Structured Calculation
2 marks
An electric motor does 1200 J of work in 15 seconds. Calculate the power output of the motor.
Show answer & marking scheme

Worked solution

Using the formula P = W / t, we substitute the work and time: P = 1200 J / 15 s = 80 W.

Marking scheme

C1: for quoting the formula P = W/t or showing substitution 1200 / 15. A1: for the correct final answer with appropriate unit: 80 W (accept 80 J/s).
Question 13 · Structured Calculation
2 marks
A potential difference of 12 V is applied across a resistor, producing a current of 0.75 A. Calculate the resistance of the resistor.
Show answer & marking scheme

Worked solution

Using Ohm's law, R = V / I. Substituting the potential difference and current: R = 12 V / 0.75 A = 16 ohms.

Marking scheme

C1: for quoting the formula R = V/I or V = IR. A1: for the correct final answer with appropriate unit: 16 Δ (or 16 ohms).
Question 14 · Structured Calculation
2 marks
A block of metal has a mass of 450 g and a volume of 50 cm^3. Calculate the density of the metal.
Show answer & marking scheme

Worked solution

Using the formula rho = m / V, we substitute the mass and volume: rho = 450 g / 50 cm^3 = 9.0 g/cm^3.

Marking scheme

C1: for quoting the formula rho = m/V or showing substitution 450 / 50. A1: for the correct final answer with appropriate unit: 9.0 g/cm^3 (accept 9 g/cm^3 or 9000 kg/m^3 if conversions are correctly shown).
Question 15 · Structured Calculation
2 marks
A water wave has a wavelength of 0.20 m and a frequency of 5.0 Hz. Calculate the speed of the water wave.
Show answer & marking scheme

Worked solution

Using the wave equation v = f * lambda, we substitute frequency and wavelength: v = 5.0 Hz * 0.20 m = 1.0 m/s.

Marking scheme

C1: for quoting the formula v = f * lambda or showing substitution 5.0 * 0.20. A1: for the correct final answer with appropriate unit: 1.0 m/s (accept 1 m/s).
Question 16 · structured
2 marks
Explain why a double-glazed window with a vacuum between the two glass panes is more effective at reducing thermal energy transfer than one containing a layer of dry air.
Show answer & marking scheme

Worked solution

A vacuum contains no particles (is empty space), which completely prevents thermal energy transfer by conduction and convection, as both processes require a material medium. Although dry air is a poor conductor, it still contains molecules that allow some conduction and can support the formation of convection currents within the gap.

Marking scheme

- [1 mark] State that a vacuum contains no particles (empty space), so conduction and convection cannot occur.
- [1 mark] State that dry air contains particles, which still allows some conduction and/or the formation of convection currents.
Question 17 · structured
2 marks
A ray of light in water approaches the boundary with air. State and explain what happens to this ray of light when the angle of incidence in the water is greater than the critical angle.
Show answer & marking scheme

Worked solution

When light travels from a more optically dense medium (water) to a less optically dense medium (air), and the angle of incidence exceeds the critical angle, the light cannot refract. Instead, it undergoes total internal reflection and is entirely reflected back into the water.

Marking scheme

- [1 mark] State that total internal reflection occurs (the light is reflected back into the water).
- [1 mark] Explain that this is because the light is travelling in a more dense medium towards a less dense medium and the angle of incidence is greater than the critical angle.
Question 18 · structured
2 marks
Explain, in terms of momentum and forces, how an inflatable airbag in a car reduces the risk of serious injury to a passenger during a collision.
Show answer & marking scheme

Worked solution

The airbag increases the time interval \( t \) over which the passenger's momentum changes to zero. Since force is equal to the rate of change of momentum, \( F = \frac{\Delta p}{t} \), a larger value of \( t \) results in a much smaller average force acting on the passenger, reducing the risk of injury.

Marking scheme

- [1 mark] State that the airbag increases the time duration of the collision / impact.
- [1 mark] Explain that this reduces the rate of change of momentum, resulting in a smaller force acting on the passenger (or mention the relationship \( F = \frac{\Delta p}{t} \)).
Question 19 · structured
2 marks
Explain why a person wearing wide snowshoes can walk on deep, soft snow without sinking, whereas the same person wearing ordinary boots sinks into the snow.
Show answer & marking scheme

Worked solution

Snowshoes have a much larger surface area in contact with the snow compared to ordinary boots. According to the formula for pressure, \( P = \frac{F}{A} \), spreading the person's weight (force \( F \)) over a larger area \( A \) results in a much lower pressure \( P \) on the snow, preventing them from sinking.

Marking scheme

- [1 mark] State that the snowshoes have a larger surface area than ordinary boots.
- [1 mark] Explain that the larger area reduces the pressure exerted on the snow because the force (weight) is distributed over a larger area (or mention \( P = \frac{F}{A} \)).
Question 20 · structured
2 marks
Explain why permanent bar magnets are stored in pairs with opposite poles adjacent and with soft iron keepers across their ends.
Show answer & marking scheme

Worked solution

The soft iron keepers become magnetised by induction from the permanent magnets. This creates a continuous, closed magnetic loop for the magnetic field lines. By keeping the magnetic loops closed, the magnetic domains within the permanent bar magnets remain aligned over time, preventing demagnetisation.

Marking scheme

- [1 mark] State that the iron keepers become magnetised by induction, forming closed loops of magnetic field lines.
- [1 mark] Explain that this keeps the magnetic domains aligned and prevents demagnetisation of the permanent magnets.
Question 21 · structured
2 marks
Explain why alpha (\( \alpha \)) particles have a very short range in air (only a few centimetres) compared to gamma (\( \gamma \)) rays, which can travel many metres.
Show answer & marking scheme

Worked solution

Alpha particles have a relatively large mass and a positive charge of \( +2e \), which makes them highly ionising. They interact strongly and frequently with air molecules, losing their kinetic energy rapidly over a short distance. In contrast, gamma rays are electromagnetic waves with no charge and no mass, making them weakly ionising so they interact rarely and travel much further before losing their energy.

Marking scheme

- [1 mark] State that alpha particles are highly ionising (or have a double positive charge / large mass), so they collide frequently with air molecules and lose energy quickly.
- [1 mark] State that gamma rays are weakly ionising (or have no charge / no mass), so they interact rarely and have a much greater range.
Question 22 · structured
2 marks
State and explain how the orbital speed of a planet changes as it moves further away from the Sun in its elliptical orbit.
Show answer & marking scheme

Worked solution

As the distance between the planet and the Sun increases, the gravitational force of attraction holding the planet in its orbit decreases. This weaker force results in a smaller centripetal acceleration, meaning the planet's orbital speed decreases. Alternatively, as the planet moves further away, its kinetic energy is converted into gravitational potential energy, resulting in a slower speed.

Marking scheme

- [1 mark] State that the orbital speed decreases.
- [1 mark] Explain that the gravitational force of attraction from the Sun becomes weaker at greater distances (or explain in terms of kinetic energy converting to gravitational potential energy).
Question 23 · structured
2 marks
Light from distant galaxies is observed to be redshifted. State what redshift is and explain what this observation indicates about the Universe.
Show answer & marking scheme

Worked solution

Redshift is the increase in the observed wavelength (or decrease in frequency) of light emitted by distant galaxies as they move away from the observer. This observation indicates that distant galaxies are moving away from us, and the fact that almost all distant galaxies show redshift indicates that the Universe is expanding.

Marking scheme

- [1 mark] Define redshift as an increase in the observed wavelength of light (or shift of light towards the red end of the spectrum).
- [1 mark] Explain that this indicates galaxies are moving away from us, which shows that the Universe is expanding.
Question 24 · Structured Explanation
2 marks
A small heater is placed at the bottom of a beaker filled with cold water. Explain, in terms of the density of water, how the water throughout the beaker becomes hot.
Show answer & marking scheme

Worked solution

When the water at the bottom of the beaker is heated, it expands and its volume increases. This expansion causes its density to decrease. The less dense warm water rises towards the top. At the same time, the cooler, denser water at the top sinks to the bottom to take its place. This continuous cycle forms a convection current that heats all the water in the beaker.

Marking scheme

1. [1] Heating causes the water to expand and decrease in density.
2. [1] The less dense warm water rises and the cooler, denser water sinks, forming a convection current.
Question 25 · Structured Explanation
2 marks
A ray of light travels inside a semicircular glass block and strikes the straight edge at an angle of incidence greater than the critical angle. Describe and explain the path of the light ray after it meets the straight edge.
Show answer & marking scheme

Worked solution

Since the light ray is travelling in a more dense medium (glass) towards a less dense medium (air) and the angle of incidence is greater than the critical angle, total internal reflection occurs. No light is refracted out of the glass block. Instead, all of the light is reflected back into the glass block, obeying the law of reflection where the angle of reflection equals the angle of incidence.

Marking scheme

1. [1] Total internal reflection occurs / no light refracts out of the block.
2. [1] All the light reflects back into the glass with the angle of reflection equal to the angle of incidence.
Question 26 · Structured Explanation
2 marks
An astronaut on a spacewalk outside the International Space Station cannot hear the sound of a nearby thruster firing, but can see the flash of light from it. Explain this observation by comparing the propagation of light and sound.
Show answer & marking scheme

Worked solution

Sound waves are longitudinal mechanical waves that propagate through the vibration of particles. Because space is a vacuum, there are no particles, so sound cannot travel to the astronaut's ears. On the other hand, light is an electromagnetic wave that consists of oscillating electric and magnetic fields, which does not require any material medium to travel and can easily propagate through a vacuum.

Marking scheme

1. [1] Sound is a mechanical wave / requires a material medium to travel (which is absent in the vacuum of space).
2. [1] Light is an electromagnetic wave / does not require a medium to travel (and can propagate through a vacuum).
Question 27 · Structured Explanation
2 marks
A beam containing both alpha (\(\alpha\)) particles and beta (\(\beta\)) particles is directed into the region between two oppositely charged horizontal plates. State and explain the difference in the deflection of these two types of radiation.
Show answer & marking scheme

Worked solution

Alpha particles are positively charged helium nuclei and are attracted towards the negatively charged plate. Beta particles are negatively charged electrons and are attracted towards the positively charged plate, so they deflect in opposite directions. Additionally, beta particles are deflected much more than alpha particles because their mass is extremely small compared to the mass of alpha particles, resulting in a much larger acceleration for a given force.

Marking scheme

1. [1] Deflection is in opposite directions because alpha is positively charged (deflected towards negative plate) and beta is negatively charged (deflected towards positive plate).
2. [1] Beta particles deflect much more because they have a much smaller mass than alpha particles.
Question 28 · Structured Recall & Diagram
2 marks
State what is meant by the electromotive force (e.m.f.) of an electrical source.
Show answer & marking scheme

Worked solution

The electromotive force (e.m.f.) is defined as the work done by the source (or energy transferred from other forms to electrical energy) per unit charge in driving charge around a complete circuit.

Marking scheme

1 mark for: work done per unit charge (or energy transferred per unit charge). 1 mark for: around a complete circuit.
Question 29 · Structured Recall & Diagram
2 marks
A bar magnet is placed on a sheet of paper. State two rules that must be followed when drawing magnetic field lines to represent the magnetic field of the magnet.
Show answer & marking scheme

Worked solution

When drawing magnetic field lines, the lines must never cross each other, and they must show the direction of the magnetic field from the North pole to the South pole.

Marking scheme

1 mark for: field lines do not cross/intersect. 1 mark for: direction is from North to South (indicated by arrows).
Question 30 · Structured Recall & Diagram
2 marks
Explain, in terms of density changes, why hot air rises above a heater in a cold room.
Show answer & marking scheme

Worked solution

When air is heated, it expands and its volume increases, causing its density to decrease. The surrounding cooler, denser air sinks and pushes the less dense warm air upwards, causing it to rise.

Marking scheme

1 mark for: heated air expands and its density decreases. 1 mark for: less dense warm air is pushed up by the cooler denser air.
Question 31 · Structured Recall & Diagram
2 marks
A student stands at a distance of 85 m from a large flat wall. She claps her hands once and hears an echo. The speed of sound in air is 340 m/s. Calculate the time interval between the clap and hearing the echo.
Show answer & marking scheme

Worked solution

The sound travels to the wall and back, so the total distance is 2 * 85 m = 170 m. Using the formula time = distance / speed, we get time = 170 / 340 = 0.50 s.

Marking scheme

1 mark for: using total distance 170 m or formula t = 2d/v. 1 mark for: 0.50 s (with unit).
Question 32 · Structured Recall & Diagram
2 marks
Describe how the acceleration of an object can be determined from its speed–time graph, and state how the distance travelled can be found from the same graph.
Show answer & marking scheme

Worked solution

On a speed-time graph, the acceleration is given by the gradient (or slope) of the line. The distance travelled by the object is represented by the area under the line.

Marking scheme

1 mark for: acceleration is the gradient/slope. 1 mark for: distance is the area under the graph.
Question 33 · Structured Recall & Diagram
2 marks
An atom of radium-226 ({}^{226}_{88}Ra) decays by emitting an alpha particle to form an isotope of radon (Rn). State the proton number and the nucleon number of this isotope of radon.
Show answer & marking scheme

Worked solution

An alpha particle has a proton number of 2 and a nucleon number of 4. Emitting an alpha particle decreases the proton number of the nucleus by 2 (88 - 2 = 86) and the nucleon number by 4 (226 - 4 = 222).

Marking scheme

1 mark for: proton number is 86. 1 mark for: nucleon number is 222.
Question 34 · Structured Recall & Diagram
2 marks
A ray of light strikes a plane mirror with an angle of incidence of 40 degrees. State the angle of reflection, and state what happens to the speed of the light as it reflects from the mirror.
Show answer & marking scheme

Worked solution

According to the law of reflection, the angle of reflection is equal to the angle of incidence, so it is 40 degrees. Reflection is a boundary phenomenon where light stays in the same medium, so the speed of light remains unchanged.

Marking scheme

1 mark for: 40 degrees. 1 mark for: speed remains unchanged/constant.
Question 35 · Structured Recall & Diagram
2 marks
State the two main physical effects that cause the Earth to experience a cycle of day and night, and a cycle of seasons throughout the year.
Show answer & marking scheme

Worked solution

The daily cycle of day and night is caused by the Earth rotating on its own axis once every 24 hours. The yearly cycle of seasons is caused by the tilt of the Earth's axis of rotation relative to its orbital plane as it travels around the Sun.

Marking scheme

1 mark for: rotation of the Earth (on its axis) causing day and night. 1 mark for: tilt of the Earth's axis / orbit around the Sun causing seasons.
Question 36 · Structured
2 marks
Fig. 1.1 shows a ray of light inside a semi-circular glass block striking the flat boundary with air at an angle of incidence of \(48^\circ\). The critical angle for this glass-air boundary is \(42^\circ\). State what happens to the ray of light at the boundary and name this phenomenon.
Show answer & marking scheme

Worked solution

Since the angle of incidence (\(48^\circ\)) is greater than the critical angle (\(42^\circ\)) and the light is travelling from a more optically dense medium (glass) to a less optically dense medium (air), the ray of light cannot refract out into the air. Instead, it undergoes total internal reflection, reflecting back into the glass block with an angle of reflection equal to the angle of incidence (\(48^\circ\)).

Marking scheme

[1] State that the ray is reflected back into the glass / undergoes reflection (at an angle of \(48^\circ\)). [1] Identify the phenomenon as total internal reflection.
Question 37 · Structured
2 marks
Fig. 2.1 shows a standard bar magnet with its North (N) and South (S) poles. State the direction of the magnetic field lines outside the magnet and explain how the spacing of these lines represents the strength of the magnetic field.
Show answer & marking scheme

Worked solution

The magnetic field lines always point from the North pole to the South pole outside a magnet. The density or spacing of the magnetic field lines represents the strength of the field: where the lines are closer together (such as near the poles), the magnetic field is strongest.

Marking scheme

[1] Direction of field lines: from North (N) to South (S). [1] Spacing: closer/denser lines represent a stronger magnetic field (or vice versa).

Ready to test yourself?

Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.

Practice This Topic

Wondering how well you actually know this?

thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Want more questions like this? Practice unlimited on thinka, instant answers included.

Start Practicing Free