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sustained current through a complete conducting loop."},{"text":"The circuit must be an open conducting path.","correct":false,"explanation":"An open conducting path can develop an induced potential difference, but charges cannot circulate continuously. The circuit must be a complete conducting loop for continuous current."},{"text":"The circuit must be a complete insulating loop.","correct":false,"explanation":"An insulator lacks mobile charge carriers, so it cannot support the induced current. A complete conducting loop is required to provide a closed path for charge flow."},{"text":"The magnetic field must be a complete uniform field.","correct":false,"explanation":"A changing field can induce an emf without being uniform, so field uniformity is not the key requirement. Continuous current requires a complete conducting loop."}]},{"id":"3ce4e00c-5948-48c9-a87f-8a124ba3622d","cardType":"basic","points":1,"front":"In the equation $\\Phi = BA\\cos\\theta$, what does $\\theta$ represent?","back":"The angle between the magnetic field and the normal to the plane of the coil.","mcqOptions":[{"text":"The angle between the magnetic field and the normal to the plane of the coil.","correct":true,"explanation":"The area vector points normal to the coil, so the scalar product defining flux uses the angle between the magnetic field and this normal."},{"text":"The angle between the magnetic field and the plane of the coil.","correct":false,"explanation":"Using the angle to the plane confuses the plane with its area vector and would require a sine expression. The angle is measured from the field to the normal."},{"text":"The angle between the normal and the plane of the coil.","correct":false,"explanation":"The normal and the plane are always perpendicular, making this angle fixed at $90^\\circ$. In the flux equation, the angle is between the field and the normal."},{"text":"The angle between the magnetic field and the current in the coil.","correct":false,"explanation":"The direction of current is not used to calculate magnetic flux and may not exist before induction occurs. The relevant angle is between the field and the coil's normal."}]},{"id":"58bd182f-8cfe-49d9-a4ad-a61a49da3b76","cardType":"basic","points":1,"front":"Under what orientation is the magnetic flux through a coil equal to zero?","back":"When the magnetic field is parallel to the plane of the coil (the normal is at $90^\\circ$ to the field).","mcqOptions":[{"text":"When the magnetic field is parallel to the plane of the coil (the normal is at $90^\\circ$ to the field).","correct":true,"explanation":"A field parallel to the coil's plane is perpendicular to its normal, so $\\cos 90^\\circ=0$ and therefore the magnetic flux is zero."},{"text":"When the magnetic field is perpendicular to the plane of the coil (the normal is at $0^\\circ$ to the field).","correct":false,"explanation":"A field perpendicular to the plane is parallel to the normal, giving $\\cos 0^\\circ=1$ and maximum flux. Zero flux occurs when the field is parallel to the plane."},{"text":"When the magnetic field is at $45^\\circ$ to the plane of the coil (the normal is at $45^\\circ$ to the field).","correct":false,"explanation":"At $45^\\circ$ to the plane, the field has a non-zero component along the normal, so the flux is not zero. The field must be parallel to the plane."},{"text":"When the magnetic field is parallel to the normal of the coil (the plane is at $90^\\circ$ to the field).","correct":false,"explanation":"A normal parallel to the field gives the greatest possible flux, not zero, because $\\cos 0^\\circ=1$. For zero flux, the normal must be at $90^\\circ$ to the field."}]},{"id":"4b4b3127-d1da-4d02-9505-e14468818be0","cardType":"basic","points":1,"front":"Define magnetic flux linkage.","back":"The product of the magnetic flux through a single turn and the total number of turns ($N\\Phi$).","mcqOptions":[{"text":"The product of the magnetic flux through a single turn and the total number of turns ($N\\Phi$).","correct":true,"explanation":"Each of the $N$ turns links flux $\\Phi$, so the total magnetic flux linkage is their product, $N\\Phi$."},{"text":"The quotient of the magnetic flux through a single turn and the total number of turns ($\\frac{\\Phi}{N}$).","correct":false,"explanation":"Dividing by the number of turns would reduce the value as turns are added, contrary to flux linkage. The correct definition is the product $N\\Phi$."},{"text":"The product of the magnetic flux through a single turn and the cross-sectional area ($A\\Phi$).","correct":false,"explanation":"Multiplying flux by area again gives the wrong dimensions because area is already included in magnetic flux. Flux linkage is the product of flux and number of turns, $N\\Phi$."},{"text":"The product of the magnetic flux density through a single turn and the total number of turns ($NB$).","correct":false,"explanation":"Magnetic flux density is not magnetic flux, as flux also depends on area and orientation. Flux linkage is the product of magnetic flux per turn and the number of turns."}]},{"id":"de768449-4788-4cf9-8d62-2da7af87f276","cardType":"cloze","points":null,"front":"The unit of magnetic flux is the [ ], whereas the unit for flux linkage is the weber turn.","back":"The unit of magnetic flux is the weber (Wb), whereas the unit for flux linkage is the weber turn.","mcqOptions":[{"text":"weber (Wb)","correct":true,"explanation":"Magnetic flux is measured in webers, with symbol Wb. One weber is equivalent to one tesla square metre for a perpendicular uniform field."},{"text":"tesla (T)","correct":false,"explanation":"The tesla measures magnetic flux density $B$, not magnetic flux $\\Phi$. Magnetic flux is measured in webers, symbol Wb."},{"text":"volt (V)","correct":false,"explanation":"The volt measures potential difference or emf rather than magnetic flux. The SI unit of magnetic flux is the weber, symbol Wb."},{"text":"henry (H)","correct":false,"explanation":"The henry measures inductance, which relates flux linkage to current, rather than flux itself. Magnetic flux is measured in webers, symbol Wb."}]},{"id":"d4f296c2-5403-4e6b-88ce-998a68c8285e","cardType":"cloze","points":null,"front":"The unit of magnetic flux is the weber (Wb), whereas the unit for flux linkage is the [ ].","back":"The unit of magnetic flux is the weber (Wb), whereas the unit for flux linkage is the weber turn.","mcqOptions":[{"text":"weber turn","correct":true,"explanation":"Flux linkage equals the flux through each turn multiplied by the number of turns, so its stated unit is the weber turn."},{"text":"weber per turn","correct":false,"explanation":"Dividing by turns describes flux per turn rather than the total linkage of all turns. Flux linkage is measured in weber turns."},{"text":"tesla turn","correct":false,"explanation":"Tesla measures flux density, and multiplying it by turns still omits the required area factor. The unit used for flux linkage is the weber turn."},{"text":"weber metre","correct":false,"explanation":"Multiplying webers by metres introduces an unrelated length factor. Flux linkage multiplies magnetic flux by the number of turns, giving the weber turn."}]},{"id":"63f8b912-1d9d-49e9-ac3c-511dfbab7083","cardType":"basic","points":1,"front":"State Faraday's law of electromagnetic induction.","back":"The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linkage.","mcqOptions":[{"text":"The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linkage.","correct":true,"explanation":"Faraday's law states that changing flux linkage induces an emf whose magnitude increases in direct proportion to the rate at which that linkage changes."},{"text":"The magnitude of the induced emf is directly proportional to the value of the magnetic flux linkage.","correct":false,"explanation":"A large but constant flux linkage does not induce an emf. Faraday's law depends on the rate of change of magnetic flux linkage, not its value alone."},{"text":"The magnitude of the induced emf is inversely proportional to the rate of change of magnetic flux linkage.","correct":false,"explanation":"Faraday's law gives a direct, not inverse, relationship: changing flux linkage more rapidly produces a larger induced emf, rather than a smaller one."},{"text":"The magnitude of the induced emf is directly proportional to the time taken to change magnetic flux linkage.","correct":false,"explanation":"Longer change times produce a smaller emf for the same change in flux linkage. The emf is proportional to the rate of change, which includes division by time."}]},{"id":"2e55c516-ac0e-497b-b645-f17bee186f2e","cardType":"cloze","points":null,"front":"The magnitude of induced emf in a coil is given by the equation: $\\varepsilon = \\text{[...]}$.","back":"The magnitude of induced emf in a coil is given by the equation: $\\varepsilon = \\boxed{-\\frac{\\Delta(N\\Phi)}{\\Delta t}}$.","mcqOptions":[{"text":"$$-\\frac{\\Delta(N\\Phi)}{\\Delta t}$","correct":true,"explanation":"Faraday's law gives the rate of change of flux linkage, $N\\Phi$, while the negative sign expresses Lenz's law: the induced emf opposes the change producing it."},{"text":"$$-\\frac{\\Delta t}{\\Delta(N\\Phi)}$","correct":false,"explanation":"This reverses the rate, giving time divided by changing flux linkage and incorrect dimensions for emf. Faraday's law requires $-\\Delta(N\\Phi)/\\Delta t$."},{"text":"$$-\\frac{N\\Phi}{\\Delta t}$","correct":false,"explanation":"Using $N\\Phi$ rather than its change incorrectly assumes the initial flux linkage is zero. Faraday's law uses $-\\Delta(N\\Phi)/\\Delta t$."},{"text":"$$-\\frac{\\Delta\\Phi}{N\\Delta t}$","correct":false,"explanation":"Dividing the flux change by $N$ makes the predicted emf decrease as turns are added. Flux linkage is $N\\Phi$, so the correct expression is $-\\Delta(N\\Phi)/\\Delta t$."}]},{"id":"67813aa4-f383-4c4f-9bde-72fd45a69741","cardType":"basic","points":1,"front":"What is the physical meaning of the minus sign in Faraday's law equation?","back":"It represents Lenz's Law, indicating that the induced emf opposes the change that created it.","mcqOptions":[{"text":"It represents Lenz's Law, indicating that the induced emf opposes the change that created it.","correct":true,"explanation":"The minus sign expresses Lenz's law: the induced emf drives a current whose magnetic effect opposes the change in flux producing it."},{"text":"It represents Lenz's Law, indicating that the induced emf reinforces the change that created it.","correct":false,"explanation":"The induced effect does not reinforce the original change, because that would violate Lenz's law. The correct interpretation is that it opposes the change creating it."},{"text":"It represents Faraday's Law, indicating that the induced emf opposes the change that created it.","correct":false,"explanation":"Faraday's law relates emf to the rate of change of flux, while the minus sign specifically represents Lenz's law and the opposition to that change."},{"text":"It represents Lenz's Law, indicating that the induced emf opposes the magnetic field that created it.","correct":false,"explanation":"The induced emf opposes the change in magnetic flux, not necessarily the existing magnetic field itself. Confusing these two ideas gives the wrong physical meaning."}]},{"id":"1db45cfe-2d53-4ea8-a380-dc950694aad3","cardType":"basic","points":1,"front":"State Lenz's law.","back":"The direction of an induced emf is such that it opposes the change in magnetic flux that produced it.","mcqOptions":[{"text":"The direction of an induced emf is such that it opposes the change in magnetic flux that produced it.","correct":true,"explanation":"Lenz's law states that the induced emf drives current in a direction producing a magnetic effect that opposes the change in flux responsible for induction."},{"text":"The direction of an induced emf is such that it reinforces the change in magnetic flux that produced it.","correct":false,"explanation":"Reinforcing the change would make the flux change grow rather than oppose itself, contradicting Lenz's law. The induced emf opposes the change in magnetic flux."},{"text":"The direction of an induced emf is such that it opposes the magnetic field that produced it.","correct":false,"explanation":"Lenz's law concerns opposition to a change in magnetic flux, not automatically opposition to the whole magnetic field. A steady field alone produces no induced emf."},{"text":"The direction of an induced emf is such that it maintains the original magnetic flux that produced it.","correct":false,"explanation":"The induced effect acts against the change, but it does not necessarily restore or maintain the original flux exactly. The required statement is opposition to the change in flux."}]},{"id":"68079f34-5101-433d-bc22-b4aa8fcbadd4","cardType":"basic","points":2,"front":"If a North pole is moved away from a coil, what pole is induced on the face of the coil nearest the magnet?","back":"A South pole (to attract the magnet and oppose the decrease in flux).","mcqOptions":[{"text":"A South pole (to attract the magnet and oppose the decrease in flux).","correct":true,"explanation":"As the North pole moves away, the flux through the coil decreases. A South pole is induced on the near face to attract the magnet and oppose that decrease."},{"text":"A North pole (to repel the magnet and oppose the decrease in flux).","correct":false,"explanation":"A North pole would repel the departing North pole and make the separation increase, rather than opposing the change. The near face must become South to attract it."},{"text":"A South pole (to repel the magnet and oppose the increase in flux).","correct":false,"explanation":"The pole type is associated with attraction here, not repulsion, and the flux is decreasing rather than increasing. The correct near face is South to attract the magnet."},{"text":"No pole is induced (because the magnet is moving away from the coil).","correct":false,"explanation":"Motion away still changes the magnetic flux through the coil, so an emf and induced current can occur. The near face becomes South to oppose the decreasing flux."}]},{"id":"b9196b03-dafb-4ff9-b86f-50d4af865115","cardType":"basic","points":2,"front":"If a South pole moves towards a coil, what pole is induced on the face of the coil nearest the magnet?","back":"A South pole (to repel the magnet and oppose the increase in flux).","mcqOptions":[{"text":"A South pole (to repel the magnet and oppose the increase in flux).","correct":true,"explanation":"A South pole approaching the coil increases the magnetic flux. The near face becomes South, causing repulsion and opposing the increase in flux."},{"text":"A North pole (to attract the magnet and oppose the increase in flux).","correct":false,"explanation":"Attraction would help the South pole move closer and increase the flux further, contrary to Lenz's law. The near face must be South so the poles repel."},{"text":"A North pole (to repel the magnet and oppose the decrease in flux).","correct":false,"explanation":"Although unlike poles attract and like poles repel, the approaching magnet causes increasing rather than decreasing flux. The induced near face is South to repel the South pole."},{"text":"No pole is induced (because the magnet is moving towards the coil).","correct":false,"explanation":"Approaching motion changes the magnetic flux and therefore induces an emf. The coil's near face becomes South, producing repulsion against the incoming South pole."}]},{"id":"e7bac54f-620d-453f-9c26-4b5101542e76","cardType":"cloze","points":null,"front":"To determine the direction of induced current from a coil's magnetic poles, use the [ ].","back":"To determine the direction of induced current from a coil's magnetic poles, use the right-hand grip rule.","mcqOptions":[{"text":"right-hand grip rule","correct":true,"explanation":"The right-hand grip rule links conventional current direction around a coil to the magnetic pole at its end, allowing the induced current direction to be determined."},{"text":"Fleming's right-hand rule","correct":false,"explanation":"Fleming's right-hand rule determines the direction of induced current in a moving conductor using motion, field and current. A coil's poles require the right-hand grip rule."},{"text":"left-hand grip rule","correct":false,"explanation":"The left-hand rule is associated with motor effects and force on a current-carrying conductor, not identifying coil poles from current direction. The right-hand grip rule applies here."},{"text":"right-hand corkscrew rule","correct":false,"explanation":"The corkscrew rule can describe the magnetic field around a current, but this course convention identifies a coil's poles using the right-hand grip rule."}]},{"id":"2ed358ba-bc21-45d7-b94b-41e6b7b95585","cardType":"basic","points":1,"front":"What is the formula for the emf induced in a straight conductor moving through a magnetic field?","back":"$$\\varepsilon = B\\ell v$","mcqOptions":[{"text":"$$\\varepsilon = B\\ell v$","correct":true,"explanation":"For a straight conductor moving at right angles through a magnetic field, the motional emf equals magnetic flux density multiplied by length and speed."},{"text":"$$\\varepsilon = \\frac{B\\ell}{v}$","correct":false,"explanation":"Dividing by speed gives the wrong dependence: faster motion produces a larger rate of flux cutting and therefore a larger emf. The correct expression multiplies by $v$."},{"text":"$$\\varepsilon = Bv\\ell^2$","correct":false,"explanation":"The conductor length enters linearly, not as a squared quantity. Squaring it introduces an incorrect extra length factor; the correct formula is $B\\ell v$."},{"text":"$$\\varepsilon = \\frac{B}{\\ell v}$","correct":false,"explanation":"Putting length and speed in the denominator reverses the physical dependence. A longer conductor and faster motion both increase the induced emf, giving $B\\ell v$."}]},{"id":"25dc7722-e595-4f2a-810d-823d43002727","cardType":"basic","points":1,"front":"What spatial condition must be met to use $\\varepsilon = B\\ell v$ directly?","back":"The conductor length, velocity, and magnetic field must be mutually perpendicular.","mcqOptions":[{"text":"The conductor length, velocity, and magnetic field must be mutually perpendicular.","correct":true,"explanation":"The direct expression assumes the conductor, its velocity, and the magnetic field are pairwise perpendicular, so the relevant cross-product factors all have maximum magnitude."},{"text":"The conductor length, velocity, and magnetic field must be mutually parallel.","correct":false,"explanation":"Parallel vectors produce no magnetic force from the relevant cross product, so this arrangement would not give the maximum emf. The three directions must be mutually perpendicular."},{"text":"The conductor length and velocity must be parallel, while the magnetic field is perpendicular.","correct":false,"explanation":"The conductor's length must be perpendicular to its velocity for the stated expression to apply directly. Making them parallel removes the motional emf contribution."},{"text":"The conductor length and magnetic field must be parallel, while the velocity is perpendicular.","correct":false,"explanation":"If the conductor lies parallel to the magnetic field, charges experience no magnetic force along the conductor. Direct use of the formula requires all three directions to be mutually perpendicular."}]},{"id":"c5a8435b-85fc-4b47-8a51-b2564b454b04","cardType":"basic","points":2,"front":"In an AC generator, at what point in the rotation is the induced emf zero?","back":"When the flux linkage is maximum (the rate of change of flux linkage is zero).","mcqOptions":[{"text":"When the flux linkage is maximum (the rate of change of flux linkage is zero).","correct":true,"explanation":"Induced emf depends on the rate of change of flux linkage. At maximum or minimum flux linkage, the graph has zero gradient, so the emf is zero."},{"text":"When the flux linkage is zero (the rate of change of flux linkage is maximum).","correct":false,"explanation":"Zero flux linkage occurs when the rate of change is greatest, so the induced emf is maximum rather than zero. Zero emf occurs at maximum flux linkage."},{"text":"When the rate of change of flux linkage is maximum (the flux linkage is zero).","correct":false,"explanation":"A maximum rate of change produces maximum induced emf, not zero emf. The emf vanishes when the flux linkage reaches an extreme value and its gradient is zero."},{"text":"When the flux linkage is maximum (the rate of change of flux linkage is maximum).","correct":false,"explanation":"At a maximum flux linkage, the instantaneous rate of change is zero, not maximum. Confusing the value with the gradient leads to this incorrect generator position."}]},{"id":"13e274a6-fb9a-4796-aed7-a62335f4770f","cardType":"basic","points":3,"front":"List three factors that increase the maximum emf produced by an AC generator.","back":"

Increasing the magnetic flux density ($B$)
Increasing the coil area ($A$) or number of turns ($N$)
Increasing the speed of rotation

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

Increasing the magnetic flux density ($B$)
Increasing the coil area ($A$) or number of turns ($N$)
Increasing the speed of rotation

","correct":true,"explanation":"The maximum emf is proportional to magnetic flux density, coil area, number of turns and angular speed. Increasing any of these increases the rate of change of flux linkage and therefore the maximum emf."},{"text":"

Decreasing the magnetic flux density ($B$)
Decreasing the coil area ($A$) and number of turns ($N$)
Decreasing the speed of rotation

","correct":false,"explanation":"These changes reduce the magnetic flux linkage and its rate of change, so they reduce the maximum emf. Increasing magnetic flux density, coil area or turns, and rotation speed gives the required increase."},{"text":"

Increasing the wire resistance
Increasing the coil area ($A$) but decreasing the number of turns ($N$)
Decreasing the speed of rotation

","correct":false,"explanation":"Resistance affects the current rather than directly increasing the generated maximum emf, while fewer turns and slower rotation reduce emf. The relevant increases are in $B$, $A$ or $N$, and rotation speed."},{"text":"

Decreasing the magnetic flux density ($B$)
Increasing the wire resistance and decreasing the number of turns ($N$)
Keeping the speed of rotation constant

","correct":false,"explanation":"Decreasing $B$ and $N$ lowers flux linkage, while resistance and constant speed do not provide the needed increase. Maximum emf rises when $B$, $A$ or $N$, and rotation speed increase."}]},{"id":"cb0c93fb-7210-44e7-9594-cd3d4050b5fc","cardType":"basic","points":3,"front":"Describe how a transformer induces a voltage in the secondary coil.","back":"

Alternating current in the primary coil creates a changing magnetic flux.
The iron core links this changing flux to the secondary coil.
A changing flux linkage in the secondary induces an emf.

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

Alternating current in the primary coil creates a changing magnetic flux.
The iron core links this changing flux to the secondary coil.
A changing flux linkage in the secondary induces an emf.

","correct":true,"explanation":"Alternating primary current produces changing magnetic flux. The iron core links this flux to the secondary, where the changing flux linkage induces an emf according to Faraday’s law."},{"text":"

Direct current in the primary coil creates a constant magnetic flux.
The iron core blocks this flux from the secondary coil.
A constant flux linkage in the secondary induces an emf.

","correct":false,"explanation":"A transformer requires changing flux, normally produced by alternating current. A constant flux linkage does not induce an emf, and the iron core links rather than blocks the flux."},{"text":"

Alternating current in the primary coil creates a changing magnetic flux.
The iron core converts this flux directly into charge.
A constant flux linkage in the secondary induces an emf.

","correct":false,"explanation":"The core provides a magnetic path and does not convert flux directly into charge. The secondary emf is caused by changing, not constant, flux linkage produced by the alternating primary current."},{"text":"

Alternating current in the secondary coil creates a changing magnetic flux.
The iron core links this changing flux to the primary coil.
A changing flux linkage in the primary induces an emf.

","correct":false,"explanation":"The primary current is the source of the changing flux, and the secondary experiences the induced emf. Reversing the roles of the coils describes the process incorrectly for this question."}]},{"id":"b35c42a2-950c-433d-baaa-7e928cba6094","cardType":"cloze","points":null,"front":"For an ideal transformer, the relationship between voltage and turns is $\\text{[...]} = \\frac{N_s}{N_p}$.","back":"For an ideal transformer, the relationship between voltage and turns is $\\boxed{\\frac{V_s}{V_p}} = \\frac{N_s}{N_p}$.","mcqOptions":[{"text":"$$\\frac{V_s}{V_p}$","correct":true,"explanation":"For an ideal transformer, the voltage ratio equals the turns ratio: $\\frac{V_s}{V_p}=\\frac{N_s}{N_p}$. Therefore the missing voltage ratio is secondary voltage divided by primary voltage."},{"text":"$$\\frac{V_p}{V_s}$","correct":false,"explanation":"This reverses the voltage ratio. The ideal transformer equation matches secondary quantities with secondary quantities, so the correct expression is $\\frac{V_s}{V_p}$."},{"text":"$$\\frac{N_p}{N_s}$","correct":false,"explanation":"This is the inverse turns ratio, often selected when the primary and secondary labels are accidentally swapped. The voltage ratio in the stated equation is $\\frac{V_s}{V_p}$."},{"text":"$$\\frac{I_s}{I_p}$","correct":false,"explanation":"Current ratios are inversely related to voltage and turns ratios for an ideal transformer, so this is not the voltage ratio. The required expression is $\\frac{V_s}{V_p}$."}]},{"id":"7cca9455-4f5b-4318-adc5-186a50f06dac","cardType":"cloze","points":null,"front":"For an ideal transformer, the relationship between voltage and turns is $\\frac{V_s}{V_p} = \\text{[...]}$.","back":"For an ideal transformer, the relationship between voltage and turns is $\\frac{V_s}{V_p} = \\boxed{\\frac{N_s}{N_p}}$.","mcqOptions":[{"text":"$$\\frac{N_s}{N_p}$","correct":true,"explanation":"For an ideal transformer, the secondary-to-primary voltage ratio equals the secondary-to-primary turns ratio. Hence $\\frac{V_s}{V_p}=\\frac{N_s}{N_p}$."},{"text":"$$\\frac{N_p}{N_s}$","correct":false,"explanation":"This reverses the turns ratio and would predict the opposite voltage change. Matching corresponding secondary and primary quantities gives the correct ratio $\\frac{N_s}{N_p}$."},{"text":"$$\\frac{V_p}{V_s}$","correct":false,"explanation":"This is another voltage ratio, not the turns ratio required by the equation. The ideal transformer relationship uses secondary turns divided by primary turns: $\\frac{N_s}{N_p}$."},{"text":"$$\\frac{I_p}{I_s}$","correct":false,"explanation":"Current ratios do not replace the turns ratio in this voltage equation. For an ideal transformer, the correct turns relationship is $\\frac{N_s}{N_p}$."}]},{"id":"0c743636-1ca6-419a-963f-4a7c7bdee745","cardType":"basic","points":1,"front":"Why does a transformer not produce a continuous secondary voltage when connected to a steady DC supply?","back":"A steady DC produces a constant magnetic flux, so there is no rate of change of flux linkage to induce an emf.","mcqOptions":[{"text":"A steady DC produces a constant magnetic flux, so there is no rate of change of flux linkage to induce an emf.","correct":true,"explanation":"After the DC supply becomes steady, the magnetic flux and flux linkage are constant. Faraday’s law requires a rate of change of flux linkage, so no continuous secondary emf is produced."},{"text":"A steady DC produces a changing magnetic flux, so there is a continuous rate of change of flux linkage to induce an emf.","correct":false,"explanation":"Steady DC produces constant, not changing, magnetic flux after the initial switch-on transient. With no continuing rate of change of flux linkage, there is no continuous induced emf."},{"text":"A steady DC produces a constant magnetic flux, so there is a continuous rate of change of flux linkage to induce an emf.","correct":false,"explanation":"The first part is correct, but constant flux means zero rate of change once the current is steady. Only the brief change when the supply is switched on can induce an emf."},{"text":"A steady DC produces a zero magnetic flux, so there is no rate of change of flux linkage to induce an emf.","correct":false,"explanation":"A DC current produces a magnetic field and therefore magnetic flux; it is not zero. The absence of continuous emf comes from the flux being constant, giving no rate of change."}]},{"id":"7f99854c-f775-46da-bdd6-2dbf3d48a405","cardType":"basic","points":1,"front":"What is the power relationship for an ideal (100% efficient) transformer?","back":"Input power equals output power: $V_p I_p = V_s I_s$","mcqOptions":[{"text":"Input power equals output power: $V_p I_p = V_s I_s$","correct":true,"explanation":"An ideal transformer is 100% efficient, so no power is lost. Therefore the electrical input power equals the output power, giving $V_p I_p=V_s I_s$."},{"text":"Input power is greater than output power: $V_p I_p > V_s I_s$","correct":false,"explanation":"This describes power loss in a real transformer, not an ideal one. With 100% efficiency, input and output powers are equal: $V_p I_p=V_s I_s$."},{"text":"Input power is less than output power: $V_p I_p < V_s I_s$","correct":false,"explanation":"An ideal transformer cannot provide more output power than input power. Conservation of energy requires equal powers, so $V_p I_p=V_s I_s$."},{"text":"Input power equals output power: $V_p I_s = V_s I_p$","correct":false,"explanation":"Although ideal input and output powers are equal, power uses the voltage and current from the same side. The correct relationship is $V_p I_p=V_s I_s$."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Electromagnetic induction","title":"Electromagnetic induction Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"9a690527-fe90-5233-b556-e5c219095ca9","walled":"$L38"}],"$L39","$L3a"]