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More closely spaced lines show that the magnetic flux density is greater in that region."},{"text":"A weaker magnetic field.","correct":false,"explanation":"Confusing close spacing with reduced range reverses the convention. Closely spaced field lines indicate a stronger magnetic field, while widely spaced lines indicate a weaker field."},{"text":"A more uniform magnetic field.","correct":false,"explanation":"Uniformity is indicated by field lines remaining equally spaced, not simply being close together. Closer spacing specifically indicates a stronger magnetic field."},{"text":"A changing magnetic field direction.","correct":false,"explanation":"Field direction is shown by the orientation and arrows of the lines, rather than their spacing. Closer spacing indicates a stronger magnetic field."}]},{"id":"675a5e1e-3599-4e19-816e-0f8d4e7e8472","cardType":"cloze","points":null,"front":"Magnetic flux density $B$ is measured in [ ].","back":"Magnetic flux density $B$ is measured in tesla (T).","mcqOptions":[{"text":"tesla (T)","correct":true,"explanation":"The tesla is the SI unit of magnetic flux density. One tesla is equivalent to one weber per square metre."},{"text":"weber (Wb)","correct":false,"explanation":"The weber measures magnetic flux, which is flux density multiplied by perpendicular area. Magnetic flux density itself is measured in tesla (T)."},{"text":"newton (N)","correct":false,"explanation":"The newton measures force, although magnetic force equations can be used to define flux density. Magnetic flux density is measured in tesla (T)."},{"text":"ampere (A)","correct":false,"explanation":"The ampere measures electric current, which can produce a magnetic field but is not its flux-density unit. Magnetic flux density is measured in tesla (T)."}]},{"id":"08ab6698-d0a8-4c52-bd83-0f9210089965","cardType":"basic","points":1,"front":"In $\\phi = BA\\cos\\theta$, what is $\\theta$ measured between?","back":"The magnetic field direction and the normal to the coil.","mcqOptions":[{"text":"The magnetic field direction and the normal to the coil.","correct":true,"explanation":"The area vector points perpendicular to the coil, so the cosine formula uses the angle between the magnetic field and this normal."},{"text":"The magnetic field direction and the plane of the coil.","correct":false,"explanation":"Using the plane angle directly is a common error and would require a sine expression. In the cosine formula, the angle is measured from the coil’s normal."},{"text":"The current direction and the normal to the coil.","correct":false,"explanation":"Current direction is relevant to magnetic forces and moments, not the definition of flux. The angle is between the magnetic field and the coil’s normal."},{"text":"The magnetic field direction and the axis of the field source.","correct":false,"explanation":"The geometry of the field source does not define the flux angle. The angle must be measured between the field direction and the normal to the coil."}]},{"id":"216ffab8-e34e-4521-a949-fac357ae4bd8","cardType":"cloze","points":null,"front":"For a uniform field through a flat area, magnetic flux is [ ].","back":"For a uniform field through a flat area, magnetic flux is $\\phi = BA\\cos\\theta$.","mcqOptions":[{"text":"$$\\phi = BA\\cos\\theta$","correct":true,"explanation":"Magnetic flux equals flux density multiplied by the area perpendicular to the field. The factor $\\cos\\theta$ selects that perpendicular area when $\\theta$ is measured from the normal."},{"text":"$$\\phi = BA\\sin\\theta$","correct":false,"explanation":"This incorrectly treats $\\theta$ as the angle from the coil’s plane. When $\\theta$ is measured between the field and the normal, the correct relation is $\\phi = BA\\cos\\theta$."},{"text":"$$\\phi = \\frac{B}{A}\\cos\\theta$","correct":false,"explanation":"Dividing by area gives incorrect dimensions and predicts less flux through a larger coil. Flux is $\\phi = BA\\cos\\theta$, so it is proportional to area."},{"text":"$$\\phi = \\frac{BA}{\\cos\\theta}$","correct":false,"explanation":"Dividing by the cosine would make the flux increase without limit near $90^\\circ$. The perpendicular component gives $\\phi = BA\\cos\\theta$, which becomes zero there."}]},{"id":"9f288ed5-a191-4e3b-92d3-aba1e9f169bf","cardType":"cloze","points":null,"front":"The unit relation for magnetic flux is [ ].","back":"The unit relation for magnetic flux is $1\\ \\text{Wb} = 1\\ \\text{T m}^2$.","mcqOptions":[{"text":"$$1\\ \\text{Wb} = 1\\ \\text{T m}^2$","correct":true,"explanation":"Magnetic flux is flux density multiplied by perpendicular area, so its unit is tesla square metres. This combined unit is called the weber."},{"text":"$$1\\ \\text{Wb} = 1\\ \\text{T}/\\text{m}^2$","correct":false,"explanation":"This divides flux density by area instead of multiplying by it. Since $\\phi = BA$ for perpendicular incidence, $1\\ \\text{Wb} = 1\\ \\text{T m}^2$."},{"text":"$$1\\ \\text{Wb} = 1\\ \\text{T m}$","correct":false,"explanation":"This uses length rather than area, omitting one factor of metres. Flux is density multiplied by area, giving $1\\ \\text{Wb} = 1\\ \\text{T m}^2$."},{"text":"$$1\\ \\text{Wb} = 1\\ \\text{T}^2\\text{m}^2$","correct":false,"explanation":"This incorrectly squares the magnetic flux density as well as using an area. Flux contains one factor of $B$, so $1\\ \\text{Wb} = 1\\ \\text{T m}^2$."}]},{"id":"e86a650c-bb14-4eaf-a2b5-9876ba4231fa","cardType":"cloze","points":null,"front":"Magnetic flux is maximum when the coil normal is [ ] to the field and zero when it is [ ] to the field.","back":"Magnetic flux is maximum when the coil normal is parallel to the field and zero when it is perpendicular to the field.","mcqOptions":[{"text":"Magnetic flux is maximum when the coil normal is parallel to the field and zero when it is perpendicular to the field.","correct":true,"explanation":"From $\\phi = BA\\cos\\theta$, parallel directions give $\\theta=0^\\circ$ and maximum flux, while perpendicular directions give $\\theta=90^\\circ$ and zero flux."},{"text":"Magnetic flux is maximum when the coil normal is perpendicular to the field and zero when it is parallel to the field.","correct":false,"explanation":"This reverses the orientations by confusing the coil’s normal with its plane. Flux is maximum when the normal is parallel to the field and zero when perpendicular."},{"text":"Magnetic flux is maximum when the coil normal is parallel to the field and zero when it is parallel to the field.","correct":false,"explanation":"The same orientation cannot produce both maximum and zero flux. A parallel normal gives maximum flux, whereas a perpendicular normal gives zero flux."},{"text":"Magnetic flux is maximum when the coil normal is perpendicular to the field and zero when it is perpendicular to the field.","correct":false,"explanation":"A perpendicular normal gives $\\cos90^\\circ=0$, so it cannot produce maximum flux. Maximum flux occurs when the coil normal is parallel to the field."}]},{"id":"52db790d-9f37-45c9-9b4e-27d45ee5f150","cardType":"basic","points":1,"front":"If given the angle between the field and the plane of the coil, what must you do before using $\\phi = BA\\cos\\theta$?","back":"Convert it to the angle from the normal.","mcqOptions":[{"text":"Convert it to the angle from the normal.","correct":true,"explanation":"The cosine formula defines $\\theta$ from the coil’s normal. The plane and normal are perpendicular, so the required angle is the complement of the given plane angle."},{"text":"Use it directly as the angle from the normal.","correct":false,"explanation":"This confuses the angle from the plane with the angle from the normal. Before using the cosine formula, subtract the given angle from $90^\\circ$."},{"text":"Convert it to the angle from the plane.","correct":false,"explanation":"The angle is already given relative to the plane, but the cosine formula requires the angle relative to the normal. Use the complementary angle instead."},{"text":"Subtract it from $180^\\circ$ to find the normal angle.","correct":false,"explanation":"The plane and its normal differ by $90^\\circ$, not $180^\\circ$. Therefore, convert the plane angle to the normal angle by taking its complement."}]},{"id":"11b18336-7246-4bc4-9167-c74aa6e64235","cardType":"cloze","points":null,"front":"For identical turns in a uniform field, magnetic flux linkage is [ ].","back":"For identical turns in a uniform field, magnetic flux linkage is $N\\phi = NBA\\cos\\theta$.","mcqOptions":[{"text":"$$N\\phi = NBA\\cos\\theta$","correct":true,"explanation":"Flux linkage is the flux through one turn multiplied by the number of identical turns. Substituting $\\phi=BA\\cos\\theta$ gives $N\\phi=NBA\\cos\\theta$."},{"text":"$$N\\phi = BA\\cos\\theta$","correct":false,"explanation":"This gives the flux through only one turn and omits the multiplication by the number of turns. The linkage is $N\\phi=NBA\\cos\\theta$."},{"text":"$$N\\phi = NBA\\sin\\theta$","correct":false,"explanation":"The sine factor incorrectly uses the angle from the coil’s plane. With $\\theta$ measured between the field and the normal, the linkage is $N\\phi=NBA\\cos\\theta$."},{"text":"$$N\\phi = \\frac{BA\\cos\\theta}{N}$","correct":false,"explanation":"Dividing by the number of turns makes linkage decrease as turns are added. Identical turns contribute additively, so the correct expression is $N\\phi=NBA\\cos\\theta$."}]},{"id":"e828108c-b359-4d31-90f5-6463bddb766c","cardType":"basic","points":1,"front":"What change is needed to produce an induced e.m.f.?","back":"A change in magnetic flux linkage.","mcqOptions":[{"text":"A change in magnetic flux linkage.","correct":true,"explanation":"An induced e.m.f. is produced when the magnetic flux linkage through a circuit changes. This can result from changing the field, moving the conductor, or changing the coil’s orientation."},{"text":"A constant magnetic flux linkage.","correct":false,"explanation":"A constant flux linkage gives no rate of change, so Faraday’s law predicts zero induced e.m.f. The required condition is a change in magnetic flux linkage."},{"text":"A steady current in the circuit.","correct":false,"explanation":"A steady current alone does not induce an e.m.f. in the circuit. Induction requires changing magnetic flux linkage, such as from changing current or relative motion."},{"text":"A complete circuit.","correct":false,"explanation":"A complete circuit allows induced current to flow, but it is not required for an induced e.m.f. The essential requirement is changing magnetic flux linkage."}]},{"id":"274a10ed-ac39-4a1d-a56b-b54003d12df2","cardType":"basic","points":1,"front":"What condition allows an induced e.m.f. to produce a current?","back":"The circuit must be complete.","mcqOptions":[{"text":"The circuit must be complete.","correct":true,"explanation":"An induced e.m.f. can exist in an open circuit, but a current flows only when there is a continuous conducting path for charge to move around."},{"text":"The circuit must be open.","correct":false,"explanation":"An open circuit prevents charge from circulating, so no induced current flows. The correct condition is a complete conducting path, although an e.m.f. may still be present."},{"text":"The circuit must have high resistance.","correct":false,"explanation":"Resistance affects the size of the current, not whether a current can flow at all. A complete circuit is required; higher resistance simply reduces the current."},{"text":"The circuit must be insulated.","correct":false,"explanation":"Insulation prevents charge movement and therefore stops current. Induced current requires a complete conducting circuit, with resistance determining its magnitude."}]},{"id":"aeb53dbf-490d-44df-afaf-e489a8fc848d","cardType":"cloze","points":null,"front":"Faraday’s law: induced e.m.f. magnitude depends on the rate of change of [ ].","back":"Faraday’s law: induced e.m.f. magnitude depends on the rate of change of magnetic flux linkage.","mcqOptions":[{"text":"magnetic flux linkage","correct":true,"explanation":"Faraday’s law states that the magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage through the circuit."},{"text":"magnetic field strength","correct":false,"explanation":"A changing field strength can cause induction, but Faraday’s law refers to the rate of change of magnetic flux linkage, which also depends on area, angle, and turns."},{"text":"electric charge","correct":false,"explanation":"Electric charge is not the quantity whose rate of change determines induced e.m.f. Faraday’s law uses magnetic flux linkage, measured through the circuit."},{"text":"resistance","correct":false,"explanation":"Resistance determines the current produced by an e.m.f., not the induced e.m.f. itself. The relevant quantity in Faraday’s law is changing magnetic flux linkage."}]},{"id":"f25e4e5e-190d-4977-aa89-35df27759238","cardType":"basic","points":1,"front":"Which equation gives the average induced e.m.f. over a time interval?","back":"$E = -\\frac{\\Delta(N\\phi)}{\\Delta t}$.","mcqOptions":[{"text":"$E = -\\frac{\\Delta(N\\phi)}{\\Delta t}$.","correct":true,"explanation":"The average induced e.m.f. equals the negative change in flux linkage divided by the time interval. The negative sign gives the direction required by Lenz’s law."},{"text":"$E = \\frac{\\Delta(N\\phi)}{\\Delta t}$.","correct":false,"explanation":"This has the correct magnitude but omits the negative sign, which represents Lenz’s law and the direction opposing the change in flux linkage."},{"text":"$E = -\\frac{\\Delta(N\\phi)}{N\\Delta t}$.","correct":false,"explanation":"Dividing by the number of turns a second time incorrectly removes the turns factor. The flux linkage is already $N\\phi$, so only the time interval belongs in the denominator."},{"text":"$E = -\\frac{\\Delta t}{\\Delta(N\\phi)}$.","correct":false,"explanation":"This reverses the required ratio. Faraday’s law uses change in flux linkage divided by the time interval, not time divided by change in flux linkage."}]},{"id":"2b94b8db-f198-4ea1-bb0e-5d9170fe7a82","cardType":"basic","points":1,"front":"On a flux-linkage-time graph, what determines the induced e.m.f.?","back":"The gradient of the graph.","mcqOptions":[{"text":"The gradient of the graph.","correct":true,"explanation":"Faraday’s law relates induced e.m.f. to the rate of change of flux linkage. On a flux-linkage-time graph, that rate of change is represented by the gradient."},{"text":"The area beneath the graph.","correct":false,"explanation":"The area under a flux-linkage-time graph is not the quantity used for induced e.m.f. The e.m.f. depends on how rapidly the graph changes, which is its gradient."},{"text":"The intercept on the vertical axis.","correct":false,"explanation":"The vertical intercept gives the initial flux linkage, not its rate of change. Induced e.m.f. is determined by the graph’s gradient, with its sign giving direction."},{"text":"The maximum value of the graph.","correct":false,"explanation":"A large flux linkage does not necessarily produce a large e.m.f. A constant large value gives no e.m.f.; the relevant feature is the rate of change, or gradient."}]},{"id":"d0cee94a-ca35-4160-8e08-fb3b2267be6a","cardType":"basic","points":1,"front":"When applying Lenz’s law, does the induced current oppose the original field or the change?","back":"It opposes the change that produced it.","mcqOptions":[{"text":"It opposes the change that produced it.","correct":true,"explanation":"Lenz’s law states that the induced current produces a magnetic effect opposing the change in flux linkage that caused the induction, ensuring conservation of energy."},{"text":"It opposes the original field that produced it.","correct":false,"explanation":"The induced effect does not necessarily oppose the original field itself. It opposes the change in flux linkage, so it may reinforce or oppose the original field."},{"text":"It reinforces the change that produced it.","correct":false,"explanation":"Reinforcing the change would create positive feedback and allow energy to appear without an input. Lenz’s law requires the induced effect to oppose the change."},{"text":"It opposes the current that produced it.","correct":false,"explanation":"The induced current is determined by the change in magnetic flux, not simply by opposing an existing current. Its magnetic effect opposes the change that caused induction."}]},{"id":"350a7ec0-aea6-40ba-9eb9-eb57c6813658","cardType":"basic","points":1,"front":"What does the minus sign in Faraday’s law represent?","back":"Lenz’s law - the direction of the induced e.m.f.","mcqOptions":[{"text":"Lenz’s law - the direction of the induced e.m.f.","correct":true,"explanation":"The negative sign indicates that the induced e.m.f. acts in the direction predicted by Lenz’s law, opposing the change in magnetic flux linkage that produced it."},{"text":"Faraday’s law - the magnitude of the induced e.m.f.","correct":false,"explanation":"Faraday’s law determines the magnitude from the rate of change of flux linkage, but the minus sign specifically indicates direction according to Lenz’s law."},{"text":"Conservation of charge - the direction of conventional current.","correct":false,"explanation":"Conservation of charge explains charge balance, not the minus sign in Faraday’s law. The sign represents Lenz’s law and the induced e.m.f.’s opposing direction."},{"text":"Fleming’s left-hand rule - the direction of the induced e.m.f.","correct":false,"explanation":"Fleming’s left-hand rule concerns motor forces on current-carrying conductors. The minus sign in Faraday’s law represents Lenz’s law for electromagnetic induction."}]},{"id":"4dd60a3e-be37-403e-b3c2-b04b7e1bf2fb","cardType":"basic","points":2,"front":"Why must the induced current oppose the change that produced it?","back":"Otherwise energy would appear from nowhere.","mcqOptions":[{"text":"Otherwise energy would appear from nowhere.","correct":true,"explanation":"If the induced current reinforced the change causing it, the process would continually increase without an energy input. Opposing the change ensures energy is conserved."},{"text":"Otherwise energy would disappear into nowhere.","correct":false,"explanation":"The issue is not unexplained energy loss but the possibility of energy being created without work. The induced current must oppose the change to conserve energy."},{"text":"Otherwise charge would appear from nowhere.","correct":false,"explanation":"Lenz’s law concerns conservation of energy, not the creation of electric charge. The induced effect opposes the change so that external work is required."},{"text":"Otherwise the magnetic field would become zero.","correct":false,"explanation":"Opposition does not necessarily cancel the entire magnetic field. It limits the change in flux linkage and ensures that energy is not created without an input."}]},{"id":"c52169cd-baf8-4a1b-8dc0-8dc591094831","cardType":"basic","points":1,"front":"For a search coil, what does the area under an e.m.f.-time graph give?","back":"The change in magnetic flux linkage.","mcqOptions":[{"text":"The change in magnetic flux linkage.","correct":true,"explanation":"Faraday’s law gives induced e.m.f. as the rate of change of flux linkage. Integrating e.m.f. over time therefore gives the total change in magnetic flux linkage."},{"text":"The maximum magnetic flux linkage.","correct":false,"explanation":"Confusing the graph’s area with a maximum value ignores the time integration. The area under the graph gives the change in magnetic flux linkage, not its maximum value."},{"text":"The average magnetic flux linkage.","correct":false,"explanation":"An average would require dividing an appropriate total by the time interval. The area under an e.m.f.-time graph directly gives the change in magnetic flux linkage."},{"text":"The rate of magnetic flux linkage.","correct":false,"explanation":"The instantaneous e.m.f., rather than the area, represents the rate of change of flux linkage. The area over a time interval gives the change in magnetic flux linkage."}]},{"id":"edf3d31a-2c0e-4e14-b6e0-557dfd5d0135","cardType":"basic","points":1,"front":"A search coil starts normal-parallel to a field and is pulled completely out. What is $|\\Delta(N\\phi)|$?","back":"$|\\Delta(N\\phi)| = NBA$.","mcqOptions":[{"text":"$|\\Delta(N\\phi)| = NBA$.","correct":true,"explanation":"Initially the coil’s flux linkage has magnitude $NBA$ because its normal is parallel to the field. Outside the field it is zero, so the magnitude of the change is $NBA$."},{"text":"$|\\Delta(N\\phi)| = 2NBA$.","correct":false,"explanation":"The factor of two applies when a coil rotates through $180^\\circ$, changing from $NBA$ to $-NBA$. Pulling it out changes the linkage from $NBA$ to zero, giving $NBA$."},{"text":"$|\\Delta(N\\phi)| = \\frac{NBA}{2}$.","correct":false,"explanation":"Halving the initial linkage incorrectly assumes an intermediate angle or partial removal. Complete removal changes the flux linkage from $NBA$ to zero, so its magnitude changes by $NBA$."},{"text":"$|\\Delta(N\\phi)| = 0$.","correct":false,"explanation":"Zero would mean that the initial and final flux linkages were equal. The linkage falls from $NBA$ inside the field to zero outside it, so the change has magnitude $NBA$."}]},{"id":"3edc6292-6070-47df-8263-ce4a185d3f12","cardType":"basic","points":1,"front":"A search coil is rotated through $180^\\circ$ in a uniform field. What is $|\\Delta(N\\phi)|$?","back":"$2NBA$.","mcqOptions":[{"text":"$2NBA$.","correct":true,"explanation":"A $180^\\circ$ rotation changes the flux linkage from $NBA$ to $-NBA$. The change is $-2NBA$, whose magnitude is therefore $2NBA$."},{"text":"$NBA$.","correct":false,"explanation":"This counts only the initial flux linkage and misses the equally large linkage in the opposite direction. Changing from $NBA$ to $-NBA$ gives a magnitude of $2NBA$."},{"text":"$4NBA$.","correct":false,"explanation":"This incorrectly doubles the change twice. A half-turn changes the linkage from $NBA$ to $-NBA$, a total change with magnitude $2NBA$, not $4NBA$."},{"text":"$0$.","correct":false,"explanation":"Although the flux linkage has the same magnitude after a half-turn, its sign reverses. It changes from $NBA$ to $-NBA$, so the magnitude of the change is $2NBA$."}]},{"id":"b69ed803-9e68-4b0e-80db-4d904163b22c","cardType":"basic","points":2,"front":"Why is the output of a simple a.c. generator alternating?","back":"The induced e.m.f. reverses every half-turn.","mcqOptions":[{"text":"The induced e.m.f. reverses every half-turn.","correct":true,"explanation":"Each half-turn reverses the direction in which the coil sides cut the magnetic field. The induced e.m.f. therefore reverses polarity, producing an alternating output."},{"text":"The induced e.m.f. reverses every full turn.","correct":false,"explanation":"Waiting for a full turn overlooks the polarity reversal after $180^\\circ$. The coil sides reverse their motion through the field every half-turn, so the induced e.m.f. reverses then."},{"text":"The induced e.m.f. increases every half-turn.","correct":false,"explanation":"Rotation does not produce a repeated increase in e.m.f.; its magnitude rises and falls during each half-turn. Its direction reverses every half-turn, which makes the output alternating."},{"text":"The induced e.m.f. vanishes every half-turn.","correct":false,"explanation":"The e.m.f. does pass through zero at particular orientations, but this alone does not define an alternating output. Alternation occurs because the induced e.m.f. reverses every half-turn."}]},{"id":"8503e321-783e-4742-b639-87dde163d61b","cardType":"basic","points":2,"front":"Why is induced e.m.f. zero when a generator’s flux linkage is maximum?","back":"

Gradient of flux linkage is zero.
So rate of change is zero.

","mcqOptions":[{"text":"

Gradient of flux linkage is zero.
So rate of change is zero.

","correct":true,"explanation":"Induced e.m.f. equals the negative rate of change of flux linkage. At a maximum, the flux-linkage graph has zero gradient, so the induced e.m.f. is zero."},{"text":"

Value of flux linkage is zero.
So magnetic field strength is zero.

","correct":false,"explanation":"At a maximum, flux linkage is greatest rather than zero, and the magnetic field can remain non-zero. The e.m.f. vanishes because the flux-linkage gradient is zero."},{"text":"

Gradient of flux linkage is maximum.
So rate of change is maximum.

","correct":false,"explanation":"This confuses the maximum value of flux linkage with its maximum gradient. At the turning point, the gradient and rate of change are zero, so the induced e.m.f. is zero."},{"text":"

Current in the coil is zero.
So magnetic flux density is zero.

","correct":false,"explanation":"Generator e.m.f. is determined by the rate of change of flux linkage, not by assuming zero current or field. At maximum linkage, its gradient is zero."}]},{"id":"b10e53f7-ed31-4741-a6dd-85288450afb1","cardType":"basic","points":2,"front":"Why must a transformer use a.c. rather than steady d.c.?","back":"A.c. produces a changing magnetic flux.","mcqOptions":[{"text":"A.c. produces a changing magnetic flux.","correct":true,"explanation":"The alternating primary current continually changes the magnetic flux in the core. This changing flux links the secondary coil and induces an e.m.f. by electromagnetic induction."},{"text":"A.c. produces a constant magnetic flux.","correct":false,"explanation":"A constant flux would not induce a continuous secondary e.m.f. A.c. is required because its changing current produces changing magnetic flux through the transformer’s secondary coil."},{"text":"A.c. produces a larger primary current.","correct":false,"explanation":"The essential requirement is not simply a large primary current. A transformer needs a changing magnetic flux, produced by a.c., to induce an e.m.f. in the secondary."},{"text":"A.c. produces a lower primary resistance.","correct":false,"explanation":"Primary resistance does not explain why transformers require alternating current. A.c. creates a changing magnetic flux, whereas steady d.c. produces no continuous induced secondary e.m.f."}]},{"id":"22645927-2d55-4329-9f91-dca2b288113c","cardType":"basic","points":2,"front":"Why is a transformer core laminated?","back":"To reduce eddy currents and heating losses.","mcqOptions":[{"text":"To reduce eddy currents and heating losses.","correct":true,"explanation":"Insulated laminations interrupt large circulating current paths within the conducting core. This increases resistance to eddy currents and reduces the thermal energy they dissipate."},{"text":"To reduce flux leakage and heating losses.","correct":false,"explanation":"Laminating the core is not primarily a method of preventing flux leakage. It breaks up conducting paths to reduce eddy currents and the heating losses they cause."},{"text":"To increase core conductivity and heating losses.","correct":false,"explanation":"Increasing conductivity would allow larger eddy currents and therefore greater heating. Laminations electrically separate core layers, increasing resistance to eddy currents and reducing heating losses."},{"text":"To prevent magnetic saturation and heating losses.","correct":false,"explanation":"Magnetic saturation is controlled mainly by core material, flux density and operating conditions. Laminations specifically restrict eddy-current loops, reducing the associated heating losses."}]},{"id":"0d81b420-bf0d-4ca1-9cae-82e1e448284a","cardType":"cloze","points":null,"front":"For an ideal transformer, [ ].","back":"For an ideal transformer, $\\frac{N_s}{N_p} = \\frac{V_s}{V_p} = \\frac{I_p}{I_s}$.","mcqOptions":[{"text":"$$\\frac{N_s}{N_p} = \\frac{V_s}{V_p} = \\frac{I_p}{I_s}$","correct":true,"explanation":"For an ideal transformer, voltage is proportional to turns, while equal input and output power makes current inversely proportional to voltage. This gives the stated three equal ratios."},{"text":"$$\\frac{N_s}{N_p} = \\frac{V_p}{V_s} = \\frac{I_p}{I_s}$","correct":false,"explanation":"This reverses the voltage ratio while leaving the turns ratio unchanged. Secondary-to-primary voltage equals secondary-to-primary turns, so the correct voltage ratio is $V_s/V_p$."},{"text":"$$\\frac{N_s}{N_p} = \\frac{V_s}{V_p} = \\frac{I_s}{I_p}$","correct":false,"explanation":"This treats current as directly proportional to turns and voltage. Conservation of power makes current inversely proportional, so $N_s/N_p$ and $V_s/V_p$ equal $I_p/I_s$."},{"text":"$$\\frac{N_p}{N_s} = \\frac{V_s}{V_p} = \\frac{I_p}{I_s}$","correct":false,"explanation":"This reverses only the turns ratio. Voltage follows the turns ratio in the same direction, so $V_s/V_p$ must equal $N_s/N_p$, as well as $I_p/I_s$."}]},{"id":"a204d9f2-ddf8-4ebb-a574-3ff83e484926","cardType":"basic","points":2,"front":"For an ideal step-up transformer, what happens to voltage and current?","back":"

Voltage increases.
Current decreases.

","mcqOptions":[{"text":"

Voltage increases.
Current decreases.

","correct":true,"explanation":"An ideal transformer conserves power, so increasing the secondary voltage requires the secondary current to decrease by the corresponding factor."},{"text":"

Voltage decreases.
Current increases.

","correct":false,"explanation":"This describes a step-down transformer. For a step-up transformer, the secondary voltage increases while the current decreases because an ideal transformer conserves power."},{"text":"

Voltage increases.
Current increases.

","correct":false,"explanation":"Increasing both quantities would increase output power, violating power conservation in an ideal transformer. Voltage increases, but current decreases in the same ratio."},{"text":"

Voltage increases.
Current stays the same.

","correct":false,"explanation":"The current is not unchanged: an ideal transformer conserves power. When a step-up transformer increases voltage, it must decrease current correspondingly."}]},{"id":"f115e0de-abcd-411e-9613-5a72a859da4b","cardType":"basic","points":1,"front":"With $N_p$ and $V_p$ fixed, how should $V_s$ vary with $N_s$ for an ideal transformer?","back":"$$V_s$ is directly proportional to $N_s$.","mcqOptions":[{"text":"$$V_s$ is directly proportional to $N_s$.","correct":true,"explanation":"The transformer equation gives $V_s/V_p=N_s/N_p$. With $V_p$ and $N_p$ fixed, increasing $N_s$ by any factor increases $V_s$ by the same factor."},{"text":"$$V_s$ is inversely proportional to $N_s$.","correct":false,"explanation":"This reverses the transformer relationship. Since $V_s/V_p=N_s/N_p$, fixed primary quantities mean that increasing $N_s$ increases, rather than decreases, $V_s$ proportionally."},{"text":"$$V_s$ is independent of $N_s$.","correct":false,"explanation":"The secondary voltage depends on the secondary turns through $V_s/V_p=N_s/N_p$. With the primary quantities fixed, $V_s$ is directly proportional to $N_s$."},{"text":"$$V_s$ is proportional to the square of $N_s$.","correct":false,"explanation":"The transformer equation contains the turns ratio to the first power, not its square. Therefore, doubling $N_s$ doubles $V_s$ rather than quadrupling it."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Electromagnetism","title":"Electromagnetism Flashcards","topbar":{"backHref":"/a-level/physics/ocr/a/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"83169c62-2054-5aaa-9ff7-b8d7f76c9005","walled":"$L38"}],"$L39","$L3a"]