Uses of magnetism — OCR GCSE Physics
Test yourself on Uses of magnetism with OCR GCSE practice questions.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Uses of magnetism explained
This Higher tier equation gives the force on a straight conductor carrying a current when the conductor sits at right angles to a uniform magnetic field.
Read the full explanation
The force in newtons equals magnetic flux density in tesla multiplied by current in amperes multiplied by the length of conductor in metres within the field: F = B × I × L. For example, a 0.5 T field acting on a 2 A current in 0.3 m of wire gives F = 0.5 × 2 × 0.3 = 0.3 N. Learners apply the equation, keep units consistent and recognise that the force is greatest at right angles and zero when the conductor is parallel to the field. Rearranging to find B, I or L is also expected.
apply: potential difference across primary coil (V) / potential difference across secondary coil (V) = number of turns in primary coil / number of turns in secondary coil (content found only in the physics separate science qualification) (Higher tier only)
This transformer equation links the two coil voltages to the two turn counts. Write it as Vp/Vs = Np/Ns, where Vp is the potential difference across the primary coil, Vs the potential difference across the secondary coil, Np the number of turns in the primary coil and Ns the number of turns in the secondary coil. A step-down transformer has fewer secondary turns, so Vs is smaller than Vp. Example: Np = 400, Ns = 20, Vp = 230 V gives Vs = 230 × 20/400 = 11.5 V. Rearrange by cross-multiplying: Vp × Ns = Vs × Np. Keep voltages in volts and turns as whole counts; the ratio is dimensionless. This content appears only in the separate physics qualification and is Higher tier only.
describe how a magnet and a current- carrying conductor exert a force on one another (Higher tier only)
A current-carrying conductor creates its own magnetic field around it. When this field interacts with the field of a nearby magnet, the two fields combine and the conductor experiences a force; by Newton's third law the magnet experiences an equal and opposite force. The force is greatest when the conductor is at right angles to the magnetic field and is zero when it is parallel to the field. Reversing the current or reversing the magnet's poles reverses the direction of the force. This is the motor effect, exploited in loudspeakers and electric motors. This content is Higher tier only.
show that Fleming’s left-hand rule represents the relative orientations of the force, the current and the magnetic field (Higher tier only)
Fleming's left-hand rule is a memory aid for the motor effect: a current-carrying conductor in a magnetic field can experience a force. Hold the left hand with thumb, first finger and second finger mutually at right angles. The first finger points along the magnetic field (from north to south pole), the second finger points along the conventional current, and the thumb then points along the force on the conductor. Because the three directions are perpendicular, rotating the hand rotates all three together, so the rule shows their relative orientations rather than a fixed compass direction. To show the rule works, take a known field direction and a known current direction, set the fingers accordingly, and read off the force direction; then check that reversing the current reverses the force while reversing the field also reverses it.
apply the equation that links the force on a conductor to the magnetic flux density, the current and the length of conductor to calculate the forces involved (Higher tier only)
The force on a straight conductor carrying current at right angles to a uniform magnetic field is given by F = B × I × L, where F is the force in newtons (N), B is the magnetic flux density in tesla (T), I is the current in amperes (A) and L is the length of conductor in the field in metres (m). The equation applies when the conductor is perpendicular to the field; if it lies parallel to the field the force is zero. To calculate, convert length to metres and current to amperes, then multiply the three values, keeping units consistent. For example, B = 0.20 T, I = 3.0 A and L = 0.50 m give F = 0.20 × 3.0 × 0.50 = 0.30 N. Rearranging gives B = F ÷ (I × L), I = F ÷ (B × L) and L = F ÷ (B × I).
explain how the force exerted from a magnet and a current-carrying conductor is used to cause rotation in electric motors (to include: an understanding of how electric motors work but knowledge of the structure of a motor is not expected) (Higher tier only)
A current-carrying conductor placed in a magnetic field experiences a force, because the field of the current interacts with the field of the magnet. The direction of that force is given by Fleming's left-hand rule: first finger points along the field from north to south, second finger along the conventional current, and the thumb shows the direction of the force. In a motor, a coil carrying a current sits between the poles of a magnet. The two sides of the coil carry current in opposite directions, so the force on one side acts upwards while the force on the other acts downwards. This pair of forces on opposite sides of the coil produces a turning effect, so the coil rotates about its axis. A commutator reverses the current every half turn so the forces keep driving rotation in the same direction.
recall that a change in the magnetic field around a conductor can give rise to an induced potential difference across its ends, which could drive a current, generating a magnetic field that would oppose the original change (content found only in the physics separate science qualification) (Higher tier only)
When the magnetic field around a conductor changes, a potential difference is induced across its ends; this is electromagnetic induction. The change can come from moving a magnet near a coil, moving the coil in a field, or changing the current in a nearby circuit. If the conductor forms part of a complete circuit, the induced potential difference drives a current. That induced current generates its own magnetic field, and the direction of this field opposes the original change, in line with Lenz's law. So moving a magnet towards a coil produces a current whose field repels the approaching magnet, resisting the motion. A faster change or a stronger magnet induces a larger potential difference. This content appears only in the physics separate science qualification.
explain how this effect is used in an alternator to generate a.c., and in a dynamo to generate d.c. (content found only in the physics separate science qualification) (Higher tier only)
Both machines rely on electromagnetic induction: a conductor experiences an induced potential difference when the magnetic flux through it changes. In an alternator, a coil rotates between the poles of a magnet, so the flux linkage rises, falls, reverses and rises again each half-turn. The induced p.d. therefore reverses direction every half rotation, giving an alternating current. Slip rings and brushes connect the spinning coil to the external circuit without swapping connections, so the output polarity follows the coil. In a dynamo, the same rotating coil is joined to a split-ring commutator; the commutator swaps the coil-to-circuit connections every half-turn, exactly when the induced p.d. would reverse, so the external current keeps one direction and is direct. Learners should link each structural feature to the resulting output, and may sketch or describe the p.d.
explain how the effect of an alternating current in one circuit, in inducing a current in another, is used in transformers (content found only in the physics separate science qualification) (Higher tier only)
A transformer has two coils wound on a shared iron core. An alternating current in the primary coil creates a magnetic field that continually grows, collapses and reverses, so the magnetic flux linking the secondary coil changes continuously. This changing flux induces an alternating potential difference across the secondary, and if the secondary circuit is closed an alternating current flows. The iron core channels the flux to link the coils efficiently. Because induction needs change, a steady direct current in the primary gives no changing flux and no output, which is why transformers work only with a.c. The ratio of turns sets the voltage: a step-up transformer has more secondary turns, a step-down transformer fewer. For an ideal transformer the power is conserved, so Vp × Ip = Vs × Is, and the turns ratio satisfies Vp/Vs = Np/Ns.
explain how the ratio of the potential differences across the two coils in a transformer depends on the ratio of the numbers of turns in each (content found only in the physics separate science qualification) (Higher tier only)
A transformer has a primary coil of Np turns and a secondary coil of Ns turns. The alternating potential difference across the primary, Vp, produces a changing magnetic flux in the iron core, and this induces an alternating potential difference Vs across the secondary. For an ideal transformer the ratio of the potential differences equals the ratio of the turns: Vp/Vs = Np/Ns. So a step-up transformer has more turns on the secondary (Ns > Np) and gives a larger Vs; a step-down transformer has fewer secondary turns and gives a smaller Vs. Doubling Ns while keeping Np and Vp fixed doubles Vs. Rearranged, Vs = Vp × Ns/Np.
apply the equations linking the potential differences and numbers of turns in the two coils of a transformer (content found only in the physics separate science qualification) (Higher tier only)
The linking equation for an ideal transformer is Vp/Vs = Np/Ns, where Vp and Vs are the potential differences across the primary and secondary coils and Np and Ns are their numbers of turns. Rearranged, Vs = Vp × Ns/Np and Ns = Np × Vs/Vp. To apply it, list the three known quantities, rearrange for the unknown, substitute values with consistent units (potential differences in volts, turns as plain counts) and calculate. For example, with Vp = 12 V, Np = 100 and Ns = 400, Vs = 12 × 400/100 = 48 V. Check that a step-up result is larger than Vp and a step-down result is smaller.
explain the action of the microphone in converting the pressure variations in sound waves into variations in current in electrical circuits, and the reverse effect as used in loudspeakers and headphones (to include: an understanding of how dynamic microphones work using electromagnetic induction) (content found only in the physics separate science qualification) (Higher tier only)
A dynamic microphone turns sound into a current by electromagnetic induction. Sound waves are pressure variations that make a diaphragm vibrate. The diaphragm is attached to a coil sitting in the gap of a permanent magnet. As the coil moves, the magnetic field lines it cuts change, so an e.m.f. is induced across the coil and a varying current flows in the circuit. Louder sounds give larger vibrations and bigger induced currents; higher-pitched sounds give faster vibrations and a higher frequency of current. A loudspeaker or headphone reverses this: an alternating current from the circuit flows through a coil in a magnetic field, the coil experiences a force (the motor effect) that changes direction as the current alternates, and the attached cone vibrates, producing pressure variations heard as sound.
Your focus
- Apply the equation force = magnetic flux density × current × length to calculate force.
- Rearrange the equation to find flux density, current or length.
- Explain how the orientation of the conductor affects the force.
Show all 36 objectives
- Apply the equation Vp/Vs = Np/Ns to find an unknown voltage or turn count.
- Distinguish primary from secondary quantities when substituting values.
- Predict whether a transformer steps voltage up or down from its turn counts.
- Describe how the magnetic field of a current interacts with a magnet's field to produce a force.
- Explain how the direction and magnitude of the force depend on current direction and orientation.
- Apply Newton's third law to the force between conductor and magnet.
- Assign the field, current and force to the correct fingers of Fleming's left-hand rule.
- Explain why the rule represents relative orientations of three perpendicular quantities.
- Apply the rule to predict how the force direction changes when current or field is reversed.
- Use F = B × I × L to calculate the force on a current-carrying conductor.
- Rearrange the equation to determine flux density, current or length.
- Convert quantities to consistent SI units and interpret the result in newtons.
- Explain why a current-carrying conductor in a magnetic field experiences a force.
- Apply Fleming's left-hand rule to determine the direction of the force.
- Explain how opposing forces on a coil produce rotation in an electric motor.
- Recall that a changing magnetic field induces a potential difference across a conductor.
- Explain that the induced potential difference can drive a current that generates an opposing magnetic field.
- Apply the idea of opposing the original change to predict the direction of an induced current.
- Explain how a rotating coil in a magnetic field produces an alternating output in an alternator.
- Explain how a split-ring commutator converts the alternating coil current into a direct external current in a dynamo.
- Describe or sketch the output p.d.–time graphs for an alternator and a dynamo.
- Explain how an alternating current in the primary coil induces an alternating p.d. in the secondary coil of a transformer.
- Explain why a transformer requires alternating rather than direct current.
- Apply the turns ratio and power conservation to describe step-up and step-down transformers.
- State the relationship Vp/Vs = Np/Ns for an ideal transformer.
- Distinguish step-up from step-down transformers using the numbers of turns.
- Predict how changing the number of secondary turns changes the secondary potential difference.
- Rearrange Vp/Vs = Np/Ns to make any one quantity the subject.
- Substitute values with consistent units to calculate an unknown potential difference or number of turns.
- Evaluate whether a calculated result is consistent with a step-up or step-down transformer.
- Explain how a dynamic microphone converts pressure variations in sound waves into variations in current using electromagnetic induction
- Explain how a loudspeaker or headphone converts variations in current into pressure variations in sound using the motor effect
- Relate the amplitude and frequency of the sound to the size and frequency of the induced current
Uses of magnetism exam tips
Marking Points
- Selects the relationship force = magnetic flux density × current × length and substitutes the given values correctly.
- Calculates the force in newtons using consistent units of tesla, ampere and metre.
- Rearranges the equation to determine magnetic flux density, current or length when the force is known.
- Recognises that the force is greatest when the conductor is at right angles to the field and zero when parallel.
- Selects the ratio Vp/Vs = Np/Ns and substitutes the two known voltages and one known turn count correctly.
- Rearranges the equation by cross-multiplication to isolate the unknown quantity before substituting numbers.
- Calculates the secondary voltage as Vp × Ns/Np and states the unit in volts.
- Identifies a step-down transformer from Ns being smaller than Np and predicts a smaller secondary voltage.
- Checks that the ratio of voltages equals the ratio of turns and that the answer is physically sensible.
- States that a current-carrying conductor produces a magnetic field around itself.
- Explains that the conductor's field interacts with the magnet's field to produce a force on the conductor.
- Applies Newton's third law to state that the magnet experiences an equal and opposite force.
- Describes the force as greatest when the conductor is perpendicular to the field and zero when parallel.
- Explains that reversing the current or the magnetic poles reverses the direction of the force.
- States that the first finger represents the magnetic field direction, from north pole to south pole.
- States that the second finger represents the conventional current direction in the conductor.
- States that the thumb represents the direction of the force on the conductor.
- Explains that the three fingers are held mutually at right angles so the rule shows relative orientations of force, current and field.
- Applies the rule to a stated field and current direction to deduce the force direction, and notes that reversing the current reverses the force.
- Selects the equation F = B × I × L and identifies each symbol with its quantity.
- States the units: force in newtons, flux density in tesla, current in amperes and length in metres.
- Substitutes consistent SI values, converting length to metres, and multiplies to obtain the force.
- Rearranges the equation to find flux density, current or length when the force is known.
- Recognises that the force is zero when the conductor is parallel to the field, since the equation applies at right angles.
- States that a current-carrying conductor in a magnetic field experiences a force.
- Applies Fleming's left-hand rule to give the direction of the force on the conductor.
- Explains that the two sides of the coil carry current in opposite directions so the forces on them act in opposite directions.
- Explains that the opposing forces on opposite sides of the coil produce a turning effect that rotates the coil.
- Describes how the current is reversed each half turn so rotation continues in the same direction.
- Keeps the explanation focused on the force and rotation rather than on the physical structure of a motor.
- States that a change in the magnetic field around a conductor induces a potential difference across its ends.
- Explains that the induced potential difference can drive a current when the conductor is part of a complete circuit.
- States that the induced current generates a magnetic field that opposes the original change.
- Identifies that a faster change in magnetic field or a stronger magnet increases the induced potential difference.
- Links a changing magnetic flux through a rotating coil to an induced potential difference, citing electromagnetic induction.
- Explains that the induced p.d. in the coil reverses each half-turn because the coil sides swap which magnetic pole they pass.
- Identifies slip rings and brushes as the connection that lets the output polarity alternate, producing a.c.
- Identifies the split-ring commutator as the feature that reverses the coil connections every half-turn so the external current stays one way, producing d.c.
- Describes or sketches the alternator output as a wave that crosses zero and reverses, and the dynamo output as humps that stay on one side of zero.
- Relates the size of the induced p.d. to the rate of change of flux, for example by faster rotation or a stronger magnet.
- States that an alternating current in the primary coil produces a continuously changing magnetic flux in the core.
- Explains that the changing flux links the secondary coil and induces an alternating potential difference across it.
- Identifies the laminated iron core as the path that links the two coils and improves the transfer of flux.
- Explains that a direct current in the primary would give no changing flux and therefore no induced output.
- Relates the number of turns on each coil to whether the transformer steps the voltage up or down.
- Applies power conservation, Vp × Ip = Vs × Is, or the turns ratio Vp/Vs = Np/Ns, to a numerical case.
- States that the ratio of the potential differences across the two coils equals the ratio of the numbers of turns in each, so Vp/Vs = Np/Ns.
- Identifies a step-up transformer as one with more turns on the secondary coil than on the primary coil, giving a larger secondary potential difference.
- Identifies a step-down transformer as one with fewer turns on the secondary coil than on the primary coil, giving a smaller secondary potential difference.
- Explains that increasing the number of secondary turns while keeping the primary turns and primary potential difference fixed increases the secondary potential difference in proportion.
- Selects the equation Vp/Vs = Np/Ns and rearranges it correctly for the unknown quantity before substituting values.
- Substitutes the known potential differences and numbers of turns into the rearranged equation with consistent units.
- Calculates the unknown potential difference or number of turns and gives the answer with an appropriate unit or as a whole number of turns.
- Checks the answer against the transformer type, confirming a step-up transformer gives a larger secondary potential difference and a step-down transformer a smaller one.
- States that sound waves are pressure variations that make the diaphragm vibrate
- Explains that the coil moves in the magnetic field of the permanent magnet as the diaphragm vibrates
- States that a changing magnetic field through the coil induces an e.m.f. across it, so a varying current flows in the circuit
- Relates larger-amplitude sound to larger coil movement and larger induced current, and higher frequency to faster vibration and higher-frequency current
- Describes the reverse effect in a loudspeaker: an alternating current in a coil in a magnetic field produces a force that changes direction, vibrating the cone to create pressure variations
Examiner Tips
- 💡Write the equation, substitute values with units, then state the unit of the answer.
- 💡Convert any length given in centimetres to metres before multiplying.
- 💡Check whether the question asks for force, flux density, current or length, and rearrange before calculating.
- 💡Write the equation first, then substitute, so the method is visible even if arithmetic slips.
- 💡Label every substituted value with its coil to avoid swapping primary and secondary.
- 💡Sanity-check whether the answer should be larger or smaller than the input voltage for a step-down or step-up transformer.
- 💡Name both fields (the conductor's field and the magnet's field) when explaining the interaction.
- 💡Use Fleming's left-hand rule or the perpendicular/parallel rule to justify the direction and size of the force.
- 💡Link the effect to a real device such as a motor or loudspeaker to show understanding.
- 💡Label a hand diagram with field, current and force before answering, so each finger's role is unambiguous.
- 💡When a question reverses the current or the field, state explicitly that the force direction reverses.
- 💡Check that the three directions you give are mutually perpendicular, since the rule depends on right angles.
- 💡Write the equation, then substitute values with units before calculating to catch conversion slips.
- 💡When asked for a rearranged quantity, divide the force by the product of the other two values.
- 💡Sanity-check the magnitude: a small flux density and modest current give a small force in newtons.
- 💡Name Fleming's left-hand rule and state what each finger or thumb represents before applying it.
- 💡Explain the turning effect by referring to forces on opposite sides of the coil acting in opposite directions.
- 💡Mention the reversal of current each half turn so the coil keeps rotating the same way.
- 💡Keep to the force and rotation; do not spend time describing motor components, which are not required.
- 💡Link the induced potential difference to a change in the magnetic field, not to a steady field.
- 💡State that a complete circuit is needed for the induced potential difference to drive a current.
- 💡Use the idea that the induced field opposes the original change to predict the direction of the induced current.
- 💡Remember that a faster change or stronger magnet gives a larger induced potential difference.
- 💡Structure the answer as one paragraph per machine, naming the component and the output it gives.
- 💡Use the phrase 'reverses every half-turn' explicitly for both the coil p.d. and the commutator action.
- 💡Add a labelled sketch of p.d. against time for each machine to secure the graph mark.
- 💡Tie any statement about magnitude back to rate of change of flux rather than to the mere presence of a magnet.
- 💡Begin by naming the changing flux as the cause and the induced p.d. as the effect, then add the core's role.
- 💡Include the reason transformers need a.c. rather than d.c. to show full understanding.
- 💡When using the turns ratio, state which coil has more turns before quoting step-up or step-down.
- 💡Check power calculations by confirming Vp × Ip equals Vs × Is for an ideal transformer.
- 💡Write the relationship as Vp/Vs = Np/Ns before substituting any numbers, so the ratio is set up correctly.
- 💡Check the direction of the change: more secondary turns than primary turns means a step-up transformer and a larger secondary potential difference.
- 💡Keep the primary potential difference and primary turns fixed when reasoning about how changing the secondary turns alters the secondary potential difference.
- 💡Write down Vp, Vs, Np and Ns and mark which one is unknown before rearranging the equation.
- 💡Convert all potential differences to the same unit, usually volts, before substituting.
- 💡Sanity-check the result: a step-up transformer must give Vs greater than Vp, and a step-down transformer Vs less than Vp.
- 💡Name the key physics idea explicitly: electromagnetic induction for the microphone and the motor effect for the loudspeaker
- 💡Link each stage in order: sound pressure variation, diaphragm vibration, coil movement, changing magnetic field, induced e.m.f. and current
- 💡Use the reverse chain for the loudspeaker so the examiner sees the symmetry between the two devices
- 💡Quantify the link between sound and signal by stating how amplitude affects current size and how frequency affects current frequency
Common Mistakes
- Multiplies the values without checking units, giving an answer in the wrong unit; convert lengths to metres and keep tesla and ampere before calculating.
- Assumes the force is largest when the conductor is parallel to the field; state that the force is greatest at right angles and zero when parallel.
- Forgets to rearrange before substituting when finding current or flux density; isolate the required quantity first, then substitute.
- Inverts the turn ratio and divides by the wrong coil count; keep primary quantities on the same side as each other.
- Mixes the two voltages when substituting; label each value as primary or secondary before calculating.
- Omits the unit or writes the voltage without stating volts; always attach the correct unit to the final answer.
- Claims the force acts only on the conductor; state that the magnet also feels an equal and opposite force.
- Says the force is largest when the conductor is parallel to the field; the force is greatest at right angles.
- Treats the force as independent of current direction; reversing the current reverses the force.
- Uses the right hand instead of the left hand; keep to the left hand for the motor effect.
- Points the first finger along the current and the second along the field; assign the first finger to the field and the second to the current.
- Treats the thumb as the current direction; reserve the thumb for the force on the conductor.
- Assumes the rule gives an absolute compass bearing; recognise that it gives relative orientations that rotate with the hand.
- Leaves the length in centimetres before multiplying; convert to metres so the force comes out in newtons.
- Adds the three quantities instead of multiplying them; apply F = B × I × L as a product.
- Applies the full equation when the conductor is parallel to the field; recognise the force is then zero.
- Mixes units such as milliamperes with tesla and metres; convert all values to SI units first.
- Claims that a stationary current-carrying conductor in a magnetic field always rotates without any opposing force; correct this by explaining that rotation needs forces acting in opposite directions on opposite sides of the coil.
- Confuses the direction of the force with the direction of the current; correct this by applying Fleming's left-hand rule with field, current and force each on their own axis.
- Describes the detailed structure of a motor instead of explaining the force; correct this by concentrating on the interaction between the magnetic field and the current and how it produces rotation.
- Claims that a steady magnetic field induces a potential difference; correct this by stating that the field around the conductor must change.
- Says the induced current's field reinforces the original change; correct this by stating that the induced field opposes the original change.
- Assumes an induced potential difference always drives a current; correct this by noting that a complete circuit is needed for a current to flow.
- States that the alternator produces a.c. because the current changes size only; correct this by explaining that the direction reverses each half-turn.
- Claims the dynamo has no reversal inside the coil; correct this by noting the coil current still alternates and the commutator reverses the external connections.
- Confuses slip rings with the split-ring commutator; correct this by matching each component to the output it produces.
- Says a current is induced without any change in flux; correct this by requiring relative motion or a changing field.
- Says the primary current flows through the core into the secondary; correct this by describing induction through a changing flux, not electrical contact.
- Claims a transformer works with direct current; correct this by requiring a changing flux, which a steady d.c. does not provide.
- Treats the iron core as generating the current; correct this by giving the core the role of linking flux between coils.
- Assumes a step-up transformer increases power; correct this by applying conservation of energy so a higher voltage comes with a lower current.
- Inverts the ratio and writes Vs/Vp = Np/Ns; correct it to Vp/Vs = Np/Ns so the primary quantities sit on the same side.
- Assumes a step-up transformer increases power as well as potential difference; correct this by noting an ideal transformer keeps power the same, so a larger Vs comes with a smaller Is.
- Treats the turns ratio as an additive difference rather than a ratio; correct this by comparing Ns with Np as a fraction, for example Ns/Np = 2 doubles Vs.
- Substitutes values into the ratio the wrong way round and gets a reciprocal answer; correct this by writing Vp/Vs = Np/Ns and matching each value to its coil.
- Mixes units, for example entering a potential difference in millivolts beside one in volts; correct this by converting all potential differences to the same unit first.
- Reports a fractional number of turns; correct this by rounding to a whole number of turns and checking the value is sensible.
- Claims the microphone needs a battery to create the current; explain that the induced current comes from electromagnetic induction as the coil moves in the magnetic field
- Says the magnet moves to push current into the coil; describe the diaphragm and coil moving relative to a fixed magnet so the field through the coil changes
- Treats the loudspeaker as inducing a current from sound; explain that the loudspeaker uses the motor effect, where a current in a magnetic field experiences a force
- Ignores the link between loudness or pitch and the size or frequency of the current; state that amplitude of vibration sets current size and vibration frequency sets current frequency