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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/eduqas/electromagnetic-induction/flashcards","label":"Electromagnetic Induction","subjectCode":"a_level_physics__eduqas","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"979ff143-3213-4678-8183-ab670f94124a","title":"Electromagnetic Induction Flashcards","topicId":"eda11e21-ce64-5238-b7c2-535c2f12ca0f","topicSlug":"electromagnetic-induction","cards":[{"id":"a43cc879-abe1-46b1-abe0-ca04b79edca3","cardType":"basic","points":1,"front":"What is the unit of magnetic flux density $B$?","back":"Tesla (T)","mcqOptions":[{"text":"Tesla (T)","correct":true,"explanation":"Magnetic flux density is measured in teslas, where $1\\ \\text{T} = 1\\ \\text{Wb m}^{-2}$. The tesla quantifies the strength of a magnetic field over a given area."},{"text":"Weber (Wb)","correct":false,"explanation":"The weber measures magnetic flux, not magnetic flux density. Magnetic flux density is flux per unit area, so the correct unit is the tesla (T)."},{"text":"Henry (H)","correct":false,"explanation":"The henry is the unit of inductance, linking induced emf with changing current. Magnetic flux density instead has the unit tesla (T)."},{"text":"Newton (N)","correct":false,"explanation":"The newton measures force, which can result from a magnetic field acting on a current. The field strength itself is measured in teslas (T)."}]},{"id":"a862bcc5-1216-469b-a9d1-3c0e16b264f0","cardType":"cloze","points":null,"front":"An induced [ ] can exist in an open circuit, but an induced current requires a complete conducting path.","back":"An induced emf can exist in an open circuit, but an induced current requires a complete conducting path.","mcqOptions":[{"text":"emf","correct":true,"explanation":"An induced emf is an energy supplied per unit charge and can be present across an open circuit. Current only flows when charges have a complete conducting path."},{"text":"current","correct":false,"explanation":"Current requires a closed conducting path, so it cannot be sustained in an open circuit. The induced quantity that can exist across the gap is an emf."},{"text":"magnetic flux","correct":false,"explanation":"Magnetic flux is a property of the magnetic field through an area and does not depend on whether the circuit is open. The correct term is emf."},{"text":"resistance","correct":false,"explanation":"Resistance describes opposition to current and can exist in circuit components whether or not the circuit is complete. The induced quantity present in an open circuit is emf."}]},{"id":"b62f78c9-f9bc-4193-9d29-d7838c11236e","cardType":"cloze","points":null,"front":"An induced emf can exist in an open circuit, but an induced [ ] requires a complete conducting path.","back":"An induced emf can exist in an open circuit, but an induced current requires a complete conducting path.","mcqOptions":[{"text":"current","correct":true,"explanation":"An induced current is the movement of charge around a conducting circuit, so a complete conducting path is necessary. An induced emf can exist even when that path is open."},{"text":"emf","correct":false,"explanation":"An induced emf can exist across an open circuit, so it is not the quantity requiring a complete path. The induced current requires the conducting path."},{"text":"voltage","correct":false,"explanation":"Voltage is another term for potential difference and can exist across an open circuit. The quantity that needs a complete conducting path to flow is current."},{"text":"charge","correct":false,"explanation":"Charge can accumulate at the ends of an open circuit without flowing around it. A complete conducting path is required for an induced current."}]},{"id":"ee68f2aa-bfbe-42d6-97d7-5799dc3e3ae8","cardType":"basic","points":1,"front":"In the equation $\\phi = AB\\cos\\theta$, between what two lines is the angle $\\theta$ measured?","back":"Between the magnetic field and the normal to the area.","mcqOptions":[{"text":"Between the magnetic field and the normal to the area.","correct":true,"explanation":"In $c6 = AB\\cos\\theta$, $c6$ is greatest when the field is parallel to the area normal. Therefore, $\\theta$ is measured between the magnetic field and the normal."},{"text":"Between the magnetic field and the plane of the area.","correct":false,"explanation":"The angle in the cosine formula is measured from the normal, not directly from the plane. Using the plane would require the complementary angle and a sine relationship."},{"text":"Between the magnetic field and the current in the coil.","correct":false,"explanation":"The flux angle concerns the field direction and the area normal, not the current direction. The current may help create the field but does not define $\\theta$ here."},{"text":"Between the normal to the area and the current in the coil.","correct":false,"explanation":"The current direction is not used to define the angle in magnetic flux. The relevant angle is between the magnetic field and the normal to the area."}]},{"id":"6ea96307-e4ac-4c8d-99ab-227c317dbb1c","cardType":"cloze","points":null,"front":"One Weber ($1\\ \\text{Wb}$) is equivalent to $\\text{[...]}$.","back":"One Weber ($1\\ \\text{Wb}$) is equivalent to $\\boxed{1\\ \\text{T m}^2}$.","mcqOptions":[{"text":"$$1\\ \\text{T m}^2$","correct":true,"explanation":"From $\\phi = BA$, one weber of magnetic flux equals one tesla multiplied by one square metre. Thus, $1\\ \\text{Wb} = 1\\ \\text{T m}^2$."},{"text":"$$1\\ \\text{T m}$","correct":false,"explanation":"Magnetic flux density must be multiplied by area, which has units of square metres, not metres. Therefore, one weber is $1\\ \\text{T m}^2$."},{"text":"$$1\\ \\text{T m}^{-2}$","correct":false,"explanation":"Dividing by area gives the unit of magnetic flux density, teslas. Flux is obtained by multiplying by area, so one weber is $1\\ \\text{T m}^2$."},{"text":"$$1\\ \\text{Wb m}^{-2}$","correct":false,"explanation":"$$\\text{Wb m}^{-2}$ is equivalent to a tesla and represents magnetic flux density. Magnetic flux itself is one weber, equivalent to $1\\ \\text{T m}^2$."}]},{"id":"643ad361-e03f-4b5b-88a7-7dff19f6e2f0","cardType":"basic","points":1,"front":"State the formula for the magnetic flux linkage of a coil with $N$ turns.","back":"$$N\\phi = NBA\\cos\\theta$","mcqOptions":[{"text":"$$N\\phi = NBA\\cos\\theta$","correct":true,"explanation":"Flux linkage equals the flux through one turn multiplied by the number of turns. Since $\\phi = BA\\cos\\theta$, the linkage is $N\\phi = NBA\\cos\\theta$."},{"text":"$$N\\phi = NBA\\sin\\theta$","correct":false,"explanation":"The sine form results when the angle is measured from the plane of the coil. Here the standard flux angle is measured from the normal, giving $N\\phi = NBA\\cos\\theta$."},{"text":"$$N\\phi = BA\\cos\\theta$","correct":false,"explanation":"This expression calculates the flux through one turn but omits the multiplication by $N$. Flux linkage for a coil is therefore $N\\phi = NBA\\cos\\theta$."},{"text":"$$N\\phi = \\frac{NB}{A}\\cos\\theta$","correct":false,"explanation":"Dividing by area is a unit error; magnetic flux is found by multiplying field strength by area. Including the number of turns gives $N\\phi = NBA\\cos\\theta$."}]},{"id":"ea9cc2d9-a015-4424-922a-a74637e85677","cardType":"basic","points":1,"front":"State Faraday's law of electromagnetic induction.","back":"The induced emf is equal to the negative rate of change of flux linkage.","mcqOptions":[{"text":"The induced emf is equal to the negative rate of change of flux linkage.","correct":true,"explanation":"Faraday’s law states that induced emf equals the negative rate of change of flux linkage. The negative sign represents Lenz’s law: the induced effect opposes the change."},{"text":"The induced emf is equal to the positive rate of change of flux linkage.","correct":false,"explanation":"This omits the negative sign required by Lenz’s law. The induced emf opposes the change producing it, so it equals the negative rate of change of flux linkage."},{"text":"The induced emf is equal to the negative rate of change of magnetic field.","correct":false,"explanation":"Faraday’s law applies to flux linkage, which can change through field, area, angle, or turns. The correct statement uses the negative rate of change of flux linkage."},{"text":"The induced emf is equal to the rate of change of flux linkage.","correct":false,"explanation":"Although the rate of change determines the emf’s magnitude, its direction requires the negative sign from Lenz’s law. The full law uses the negative rate of change."}]},{"id":"992ad106-b38c-466e-8d4e-cdd9845d7a46","cardType":"cloze","points":null,"front":"The induced emf is equal to the [ ] of a graph of flux linkage against time.","back":"The induced emf is equal to the negative gradient of a graph of flux linkage against time.","mcqOptions":[{"text":"negative gradient","correct":true,"explanation":"The gradient of a flux-linkage-against-time graph is the rate of change of flux linkage. Faraday’s law adds a negative sign, so the induced emf is the negative gradient."},{"text":"gradient","correct":false,"explanation":"The gradient gives the rate of change of flux linkage but does not include the direction required by Lenz’s law. The induced emf is the negative gradient."},{"text":"positive gradient","correct":false,"explanation":"A positive gradient would predict the same sign as the flux change, ignoring the opposition described by Lenz’s law. The correct term is negative gradient."},{"text":"negative intercept","correct":false,"explanation":"The intercept is the flux linkage at zero time and is unrelated to the induced emf. The emf is determined by the negative gradient of the graph."}]},{"id":"4da9efaa-05f3-4dac-91d2-8ef5cd774d3a","cardType":"basic","points":1,"front":"What is the unit equivalent to one volt (V) in terms of Weber and seconds?","back":"$$\\text{Wb s}^{-1}$","mcqOptions":[{"text":"$$\\text{Wb s}^{-1}$","correct":true,"explanation":"A volt is equivalent to one weber per second because induced emf equals the rate of change of magnetic flux, giving units of flux divided by time."},{"text":"$$\\text{Wb s}$","correct":false,"explanation":"Multiplying webers by seconds gives the wrong dimensions; the volt is the rate of change of flux, so the correct unit is $\\text{Wb s}^{-1}$."},{"text":"$$\\text{Wb s}^{-2}$","correct":false,"explanation":"Using seconds squared represents a second rate of change rather than emf itself. One volt corresponds to flux divided by one second: $\\text{Wb s}^{-1}$."},{"text":"$$\\text{Wb}^{-1}\\text{ s}$","correct":false,"explanation":"This reverses the flux and time dimensions. Faraday’s law shows that emf has units of magnetic flux per time, namely $\\text{Wb s}^{-1}$."}]},{"id":"da0c1c4b-5643-492d-bae9-364a3bbe394d","cardType":"basic","points":1,"front":"State Lenz's law.","back":"The direction of an induced emf/current is such that it opposes the change that produced it.","mcqOptions":[{"text":"The direction of an induced emf/current is such that it opposes the change that produced it.","correct":true,"explanation":"Lenz’s law states that the induced emf or current produces an effect opposing the change in magnetic flux that caused it, conserving energy."},{"text":"The direction of an induced emf/current is such that it supports the change that produced it.","correct":false,"explanation":"Supporting the original change would cause an ever-increasing flux and violate energy conservation. The correct statement is that the induced emf/current opposes the change that produced it."},{"text":"The direction of an induced emf/current is such that it opposes the magnetic field that produced it.","correct":false,"explanation":"The induced effect does not necessarily oppose the whole magnetic field; it opposes the change in flux. Therefore, the correct wording is that it opposes the change that produced it."},{"text":"The direction of an induced emf/current is such that it maintains the change that produced it.","correct":false,"explanation":"Maintaining or reinforcing the change confuses Lenz’s law with positive feedback. The induced emf/current always opposes the change in flux that produced it."}]},{"id":"a50fa06f-4d98-4394-84e1-ea96ca10c945","cardType":"cloze","points":null,"front":"Lenz's law is a consequence of the conservation of [ ].","back":"Lenz's law is a consequence of the conservation of energy.","mcqOptions":[{"text":"energy","correct":true,"explanation":"Lenz’s law follows from conservation of energy: the induced current creates an opposing effect, so work must be done to continue changing the magnetic flux."},{"text":"charge","correct":false,"explanation":"Charge is conserved in circuits, but this is not the principle determining the direction of induced current. Lenz’s law is a consequence of conservation of energy."},{"text":"momentum","correct":false,"explanation":"Momentum conservation concerns motion and interactions, not the direction of electromagnetic induction. The opposing induced effect is required by conservation of energy."},{"text":"magnetic flux","correct":false,"explanation":"Magnetic flux is the quantity whose change produces induction, not the conservation principle behind Lenz’s law. The relevant conserved quantity is energy."}]},{"id":"7acb2ca5-f875-4cae-b67e-15a51650f386","cardType":"basic","points":1,"front":"According to Lenz's law, what pole is created at the end of a coil when a North pole magnet is moved towards it?","back":"A North pole (to repel the approaching magnet).","mcqOptions":[{"text":"A North pole (to repel the approaching magnet).","correct":true,"explanation":"The coil creates a North pole facing the approaching North pole, producing repulsion. This opposes the magnet’s approach, as required by Lenz’s law."},{"text":"A South pole (to attract the approaching magnet).","correct":false,"explanation":"A South pole would attract the approaching North pole and reinforce its motion, rather than oppose the change. Lenz’s law requires a North pole facing it."},{"text":"A North pole (to attract the approaching magnet).","correct":false,"explanation":"A North pole repels another North pole, so it cannot attract the approaching magnet. The coil creates a North pole specifically to oppose the approach."},{"text":"A South pole (to repel the approaching magnet).","correct":false,"explanation":"Although repulsion is the correct opposing effect, a South pole would attract the approaching North pole. The coil must form a North pole at that end."}]},{"id":"dfce509c-8edf-4f51-a697-47b5ca62e921","cardType":"basic","points":1,"front":"Why is there no induced emf when a magnet is stationary inside a coil?","back":"The flux linkage is constant (the rate of change of flux linkage is zero).","mcqOptions":[{"text":"The flux linkage is constant (the rate of change of flux linkage is zero).","correct":true,"explanation":"Faraday’s law states that induced emf depends on the rate of change of flux linkage. A stationary magnet produces constant flux linkage, so the induced emf is zero."},{"text":"The flux linkage is increasing (the rate of change of flux linkage is maximum).","correct":false,"explanation":"A stationary magnet does not make the flux linkage increase, so this incorrectly predicts an emf. The flux linkage is constant and its rate of change is zero."},{"text":"The magnetic field is zero (there is no magnetic flux through the coil).","correct":false,"explanation":"A stationary magnet can produce substantial magnetic flux through the coil. No emf is induced because that flux linkage is constant, not because the field is zero."},{"text":"The rate of change of flux linkage is constant (the flux linkage is changing uniformly).","correct":false,"explanation":"Stationary positioning means the flux linkage is not changing at all. Its rate of change is zero, which is why no induced emf appears."}]},{"id":"56e6b5ff-8fdd-447f-8349-11ad4ee78538","cardType":"basic","points":1,"front":"What formula gives the motional emf induced in a conductor of length $l$ moving at velocity $v$ perpendicular to field $B$?","back":"$$\\epsilon = Blv$","mcqOptions":[{"text":"$$\\epsilon = Blv$","correct":true,"explanation":"For a conductor of length $l$ moving at velocity $v$ perpendicular to a field $B$, the motional emf is the product $Blv$."},{"text":"$$\\epsilon = \\frac{Bl}{v}$","correct":false,"explanation":"Dividing by velocity reverses the dependence on motion; a faster conductor cuts field lines more quickly and produces a larger emf. The correct formula is $\\epsilon = Blv$."},{"text":"$$\\epsilon = Bvl^2$","correct":false,"explanation":"Squaring the conductor length is an algebraic slip and gives incorrect dimensions. Motional emf depends linearly on $B$, $l$, and $v$: $\\epsilon = Blv$."},{"text":"$$\\epsilon = \\frac{B}{lv}$","correct":false,"explanation":"Putting both length and velocity in the denominator reverses the physical relationships and gives incorrect dimensions. The correct expression is $\\epsilon = Blv$."}]},{"id":"492da27f-af5c-4501-a410-3a5c0a5c767e","cardType":"cloze","points":null,"front":"A rod of length $l$ moving at speed $v$ for time $\\Delta t$ sweeps out an area equal to $\\text{[...]}$.","back":"A rod of length $l$ moving at speed $v$ for time $\\Delta t$ sweeps out an area equal to $\\boxed{lv\\Delta t}$.","mcqOptions":[{"text":"$$lv\\Delta t$","correct":true,"explanation":"The rod travels a distance $v\\Delta t$ in time $\\Delta t$. Multiplying this distance by its length $l$ gives the swept area $lv\\Delta t$."},{"text":"$$\\frac{lv}{\\Delta t}$","correct":false,"explanation":"Dividing by time gives the rate of area sweep rather than the area swept. The distance travelled is $v\\Delta t$, so the area is $lv\\Delta t$."},{"text":"$$l+v\\Delta t$","correct":false,"explanation":"Adding the rod length to its travel distance does not calculate an area. Area requires multiplying the length $l$ by the distance $v\\Delta t$, giving $lv\\Delta t$."},{"text":"$$\\frac{l\\Delta t}{v}$","correct":false,"explanation":"Dividing by speed incorrectly makes the swept area smaller for faster motion. The rod travels $v\\Delta t$, so multiplying by $l$ gives $lv\\Delta t$."}]},{"id":"3e42be78-88b5-4454-bd96-2fcff7cb86aa","cardType":"basic","points":2,"front":"In a rotating coil, what is the value of the induced emf when the coil is face-on to the magnetic field (flux is maximum)?","back":"Zero (rate of change of flux is zero).","mcqOptions":[{"text":"Zero (rate of change of flux is zero).","correct":true,"explanation":"When the coil is face-on, its flux linkage is momentarily at a maximum. At a maximum the rate of change is zero, so the induced emf is zero."},{"text":"Maximum (rate of change of flux is maximum).","correct":false,"explanation":"Maximum flux does not mean maximum rate of change; at the maximum, the flux curve has zero gradient. Therefore the induced emf is zero."},{"text":"Half maximum (rate of change of flux is half maximum).","correct":false,"explanation":"The emf is determined by the gradient of the flux, not by the flux value itself. At maximum flux the gradient is zero, so the induced emf is zero."},{"text":"Constant (rate of change of flux is constant).","correct":false,"explanation":"During rotation the flux is changing, and at the face-on position its instantaneous rate of change is zero rather than constant. The induced emf is therefore zero."}]},{"id":"86c748e9-23fe-481f-95a7-24d02c7f089c","cardType":"basic","points":2,"front":"In a rotating coil, what is the value of the induced emf when the coil is edge-on to the magnetic field (flux is zero)?","back":"Maximum (rate of change of flux is at its peak).","mcqOptions":[{"text":"Maximum (rate of change of flux is at its peak).","correct":true,"explanation":"Although the magnetic flux is zero at this instant, it is changing most rapidly. Faraday’s law therefore gives the maximum magnitude of induced emf."},{"text":"Zero (there is no magnetic flux through the coil).","correct":false,"explanation":"Zero flux does not imply zero emf because emf depends on the rate of change of flux. Edge-on, the flux changes fastest, so the induced emf is maximum."},{"text":"Half-maximum (the flux is changing at half its peak rate).","correct":false,"explanation":"A half-maximum value occurs at an intermediate coil angle, not when the coil is edge-on. At edge-on, the rate of change of flux and the emf are maximum."},{"text":"Constant (the rate of change of flux remains unchanged).","correct":false,"explanation":"The induced emf varies sinusoidally as the coil rotates, so it is not constant. When the coil is edge-on, its magnitude reaches a maximum."}]},{"id":"c6dc1d5a-015d-4b93-9c90-e1d8f961d041","cardType":"basic","points":1,"front":"State the formula for the peak emf ($\\epsilon_{\\text{max}}$) of an alternating current generator.","back":"$$\\epsilon_{\\text{max}} = NBA\\omega$","mcqOptions":[{"text":"$$\\epsilon_{\\text{max}} = NBA\\omega$","correct":true,"explanation":"Faraday’s law applied to a coil rotating uniformly gives a peak emf equal to the product of turns, flux density, coil area and angular velocity."},{"text":"$$\\epsilon_{\\text{max}} = BA\\omega$","correct":false,"explanation":"Omitting $N$ treats a multi-turn coil as a single loop. Each turn contributes emf, so the correct peak value is $\\epsilon_{\\text{max}} = NBA\\omega$."},{"text":"$$\\epsilon_{\\text{max}} = \\frac{NBA}{\\omega}$","correct":false,"explanation":"Angular velocity multiplies the peak emf because faster rotation changes flux more rapidly. The correct relationship is $\\epsilon_{\\text{max}} = NBA\\omega$, not an inverse relationship."},{"text":"$$\\epsilon_{\\text{max}} = NBA\\omega^2$","correct":false,"explanation":"The rate of flux change is directly proportional to angular velocity, not its square. Therefore the peak emf is $\\epsilon_{\\text{max}} = NBA\\omega$."}]},{"id":"a335aaa8-e6d8-450a-a0f7-6ae456baff9a","cardType":"basic","points":1,"front":"State the relationship between angular velocity $\\omega$ and frequency $f$.","back":"$$\\omega = 2\\pi f$","mcqOptions":[{"text":"$$\\omega = 2\\pi f$","correct":true,"explanation":"One complete cycle corresponds to $2\\pi$ radians. Multiplying the number of cycles per second, $f$, by $2\\pi$ gives the angular velocity."},{"text":"$$\\omega = \\frac{2\\pi}{f}$","correct":false,"explanation":"Dividing by frequency reverses the required proportionality: faster oscillations have greater angular velocity. Since each cycle covers $2\\pi$ radians, $\\omega = 2\\pi f$."},{"text":"$$\\omega = \\frac{f}{2\\pi}$","correct":false,"explanation":"This divides the cycle rate by the radians in one cycle instead of multiplying. As one cycle is $2\\pi$ radians, the correct relation is $\\omega = 2\\pi f$."},{"text":"$$\\omega = \\pi f$","correct":false,"explanation":"Using only $\\pi$ counts half a revolution rather than one complete cycle. A full cycle contains $2\\pi$ radians, so $\\omega = 2\\pi f$."}]},{"id":"ef658e10-5a13-4676-a91f-340b9c7febfd","cardType":"cloze","points":null,"front":"If the rotation frequency of a generator is doubled, the peak emf increases by a factor of [ ] and the output frequency doubles.","back":"If the rotation frequency of a generator is doubled, the peak emf increases by a factor of 2 and the output frequency doubles.","mcqOptions":[{"text":"2","correct":true,"explanation":"Peak emf obeys $\\epsilon_{\\text{max}} = NBA\\omega$, while $\\omega = 2\\pi f$. Doubling frequency therefore doubles angular velocity and peak emf."},{"text":"4","correct":false,"explanation":"A factor of 4 would follow from incorrectly assuming emf depends on frequency squared. Peak emf is directly proportional to frequency, so doubling frequency gives a factor of 2."},{"text":"$$\\frac{1}{2}$","correct":false,"explanation":"This incorrectly treats peak emf as inversely proportional to frequency. Faster rotation increases the rate of flux change, so doubling frequency increases peak emf by a factor of 2."},{"text":"1","correct":false,"explanation":"A factor of 1 would mean peak emf is independent of rotation speed. Because peak emf is proportional to angular velocity, doubling frequency gives a factor of 2."}]},{"id":"98cc83fb-0c8a-4903-bb9c-66a7c5d8ba55","cardType":"cloze","points":null,"front":"If the rotation frequency of a generator is doubled, the peak emf increases by a factor of 2 and the output frequency [ ].","back":"If the rotation frequency of a generator is doubled, the peak emf increases by a factor of 2 and the output frequency doubles.","mcqOptions":[{"text":"doubles","correct":true,"explanation":"Each complete coil rotation produces one complete emf cycle. Doubling the rotation frequency therefore produces twice as many output cycles per second, so the output frequency doubles."},{"text":"quadruples","correct":false,"explanation":"Quadrupling incorrectly applies a squared relationship. Output frequency is directly proportional to the generator’s rotation frequency, so doubling the rotation frequency makes the output frequency double."},{"text":"halves","correct":false,"explanation":"Halving assumes an inverse relationship between rotation and output frequencies. Faster rotation produces more emf cycles each second, so doubling the rotation frequency makes the output frequency double."},{"text":"remains constant","correct":false,"explanation":"A constant output frequency would require the cycle rate to be independent of rotation. Each rotation generates one cycle, so doubling rotation frequency doubles output frequency."}]},{"id":"9bffa814-a16b-412b-adad-43c89f87d2b4","cardType":"basic","points":2,"front":"Describe how to determine the direction of an induced current when a North pole is pulled away from a coil.","back":"

The near end of the coil becomes a South pole
This attracts the retreating North pole to oppose the motion

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

The near end of the coil becomes a South pole
This attracts the retreating North pole to oppose the motion

","correct":true,"explanation":"Lenz’s law requires the induced field to oppose the decreasing flux. The near end therefore becomes South, attracting the retreating North pole and opposing its departure."},{"text":"

The near end of the coil becomes a North pole
This repels the retreating North pole to oppose the motion

","correct":false,"explanation":"Making the near end North produces repulsion, which would assist the North pole’s retreat rather than oppose it. The near end must become South and attract the retreating pole."},{"text":"

The near end of the coil becomes a South pole
This repels the retreating North pole to oppose the motion

","correct":false,"explanation":"A South pole cannot repel a North pole because unlike poles attract. The near end does become South, but it attracts the retreating North pole to oppose the motion."},{"text":"

The near end of the coil becomes a North pole
This attracts the retreating North pole to oppose the motion

","correct":false,"explanation":"Two North poles repel rather than attract. Lenz’s law instead makes the coil’s near end South, allowing it to attract the retreating North pole and oppose the motion."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Electromagnetic Induction","title":"Electromagnetic Induction Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"eda11e21-ce64-5238-b7c2-535c2f12ca0f","walled":"$L39"}],"$L3a","$L3b"]