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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/aqa/magnetic-flux-and-flux-linkage-a-level-only/flashcards","label":"Magnetic flux and flux linkage (A-level only)","subjectCode":"a_level_physics__aqa","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"ee080a6e-6463-4c7f-9ae4-f05cca43d78a","title":"Magnetic flux and flux linkage (A-level only) Flashcards","topicId":"8f8cded6-7fe8-5aa4-bdb9-9890064ea66e","topicSlug":"magnetic-flux-and-flux-linkage-a-level-only","cards":[{"id":"00292b9b-de4c-4df7-879b-4d8a59630a38","cardType":"basic","points":1,"front":"State the standard SI unit for magnetic flux ($̐$).","back":"Weber ($\\text{Wb}$)","mcqOptions":[{"text":"Weber ($\\text{Wb}$)","correct":true,"explanation":"Magnetic flux is measured in Webers. One Weber represents the flux through an area of $1\\text{ m}^2$ in a perpendicular flux density of $1\\text{ T}$."},{"text":"Tesla ($\\text{T}$)","correct":false,"explanation":"Tesla measures magnetic flux density, not magnetic flux. Magnetic flux accounts for the field passing through an area and is measured in Webers ($\\text{Wb}$)."},{"text":"Henry ($\\text{H}$)","correct":false,"explanation":"Henry is the SI unit of inductance, so it does not measure magnetic flux. The standard unit of magnetic flux is the Weber ($\\text{Wb}$)."},{"text":"Newton ($\\text{N}$)","correct":false,"explanation":"Newton is the SI unit of force, including magnetic force, but not of magnetic flux. Magnetic flux is measured in Webers ($\\text{Wb}$)."}]},{"id":"d70c13ed-27ec-458d-99fb-082172ab38d1","cardType":"basic","points":2,"front":"How do magnetic flux density ($B$) and magnetic flux ($̐$) differ?","back":"

Flux density: field strength or concentration in Tesla.
Magnetic flux: total field passing through a specific area in Webers.

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

Flux density: field strength or concentration in Tesla.
Magnetic flux: total field passing through a specific area in Webers.

","correct":true,"explanation":"Flux density describes the concentration or strength of a magnetic field at a location, whereas magnetic flux measures the total field passing through a specified area."},{"text":"

Flux density: total field passing through a specific area in Webers.
Magnetic flux: field strength or concentration in Tesla.

","correct":false,"explanation":"This reverses the two quantities. Flux density is the field concentration measured in Tesla, while magnetic flux is the total field through an area measured in Webers."},{"text":"

Flux density: field strength or concentration in Webers.
Magnetic flux: total field passing through a specific area in Tesla.

","correct":false,"explanation":"The descriptions are appropriate, but the units have been exchanged. Flux density is measured in Tesla, while total magnetic flux is measured in Webers."},{"text":"

Flux density: total force acting on a conductor in Newtons.
Magnetic flux: field strength at a single point in Tesla.

","correct":false,"explanation":"This confuses magnetic quantities with force and swaps their spatial meanings. Flux density is field concentration in Tesla; flux is the total field through an area in Webers."}]},{"id":"45421423-93e1-49f4-81d3-06c232941f46","cardType":"cloze","points":null,"front":"One Weber ($1 \\text{ Wb}$) is equivalent to one $\\text{[...]}$.","back":"One Weber ($1 \\text{ Wb}$) is equivalent to one $\\boxed{ext{T m}^2}$.","mcqOptions":[{"text":"$$ext{T m}^2$","correct":true,"explanation":"Magnetic flux equals flux density multiplied by perpendicular area. Therefore, the Weber is equivalent to a Tesla multiplied by a square metre."},{"text":"$$ext{T m}$","correct":false,"explanation":"This uses length rather than area, missing the squared metre required when flux density is multiplied by area. One Weber is equivalent to one $\\text{T m}^2$."},{"text":"$$ext{T m}^{-2}$","correct":false,"explanation":"This divides by area instead of multiplying by it. Since magnetic flux is flux density times perpendicular area, one Weber equals one $\\text{T m}^2$."},{"text":"$$ext{N A}^{-1}$","correct":false,"explanation":"This corresponds to force per unit current and lacks the required length factor. One Weber is equivalent to one $\\text{T m}^2$."}]},{"id":"39d5c3d1-83ff-4ed6-8d80-1c1d5fbf3e60","cardType":"basic","points":1,"front":"State the equation for magnetic flux linkage through a coil with $N$ turns.","back":"$$N̐ = BAN$","mcqOptions":[{"text":"$$N̐ = BAN$","correct":true,"explanation":"For a coil whose area is perpendicular to the field, flux per turn is $BA$. Multiplying by the number of turns $N$ gives the total flux linkage."},{"text":"$$N̐ = BA$","correct":false,"explanation":"This gives the magnetic flux through one turn only and omits the number of turns. Flux linkage through the whole coil is $N̐ = BAN$."},{"text":"$$N̐ = rac{BA}{N}$","correct":false,"explanation":"Dividing by the number of turns reverses the required relationship. Each turn contributes flux, so the total linkage is found by multiplying $BA$ by $N$."},{"text":"$$N̐ = rac{BN}{A}$","correct":false,"explanation":"This incorrectly divides by area rather than multiplying by it. Flux through each turn is $BA$, and multiplication by $N$ gives $N̐ = BAN$."}]},{"id":"67a684a0-1dbd-4390-940e-0d22e3b5e30c","cardType":"basic","points":1,"front":"Identify the standard unit used for magnetic flux linkage.","back":"Weber-turns ($\\text{Wb turns}$) or Webers ($\\text{Wb}$)","mcqOptions":[{"text":"Weber-turns ($\\text{Wb turns}$) or Webers ($\\text{Wb}$)","correct":true,"explanation":"Flux linkage is magnetic flux multiplied by the dimensionless number of turns. It may therefore be stated as Weber-turns or simply Webers."},{"text":"Tesla-turns ($\\text{T turns}$) or Teslas ($\\text{T}$)","correct":false,"explanation":"Tesla measures magnetic flux density rather than flux linkage. Flux linkage is flux multiplied by the number of turns, so its unit is Weber-turns or Webers."},{"text":"Weber-metres ($\\text{Wb m}$) or Weber-metres squared ($\\text{Wb m}^2$)","correct":false,"explanation":"Flux linkage does not introduce an additional length or area factor; it multiplies flux by a number of turns. Its unit is Weber-turns or Webers."},{"text":"Ampere-turns ($\\text{A turns}$) or Amperes ($\\text{A}$)","correct":false,"explanation":"Ampere-turns describe magnetomotive force associated with current in a coil, not magnetic flux linkage. Flux linkage is measured in Weber-turns or Webers."}]},{"id":"deac2645-be0e-471d-ace1-a573e0c312a1","cardType":"cloze","points":null,"front":"In the equation $N̐ = BAN \\text{cos}\\theta$, the angle $\\theta$ is measured between the [ ] and the magnetic field lines.","back":"In the equation $N̐ = BAN \\text{cos}\\theta$, the angle $\\theta$ is measured between the normal to the coil and the magnetic field lines.","mcqOptions":[{"text":"normal to the coil","correct":true,"explanation":"The area vector points perpendicular, or normal, to the plane of the coil. The cosine form uses the angle between this normal and the magnetic field lines."},{"text":"plane of the coil","correct":false,"explanation":"Using the plane itself is the common angle convention error. In the cosine equation, the angle is measured between the normal to the coil and the field lines."},{"text":"current in the coil","correct":false,"explanation":"Current direction does not define the angle in the magnetic flux equation. The relevant angle is between the normal to the coil and the magnetic field lines."},{"text":"edge of the coil","correct":false,"explanation":"An edge lies within the coil’s plane and does not uniquely define its orientation. The angle is measured from the normal to the coil to the field lines."}]},{"id":"86c7119f-3cab-4404-aee5-c4276b8c7c49","cardType":"cloze","points":null,"front":"In the equation $N̐ = BAN \\text{cos}\\theta$, the angle $\\theta$ is measured between the normal to the coil and the [ ].","back":"In the equation $N̐ = BAN \\text{cos}\\theta$, the angle $\\theta$ is measured between the normal to the coil and the magnetic field lines.","mcqOptions":[{"text":"magnetic field lines","correct":true,"explanation":"In the cosine form of the flux-linkage equation, the angle is defined between the coil’s normal area vector and the direction of the magnetic field lines."},{"text":"plane of the coil","correct":false,"explanation":"This would compare the normal with the plane, which is always a right angle and does not describe field orientation. The second direction must be the magnetic field lines."},{"text":"current in the coil","correct":false,"explanation":"The current direction is irrelevant to the geometrical angle used to calculate magnetic flux linkage. The angle is between the coil’s normal and the magnetic field lines."},{"text":"axis of rotation","correct":false,"explanation":"An axis of rotation may coincide with the normal in some arrangements, but it is not the second direction in the definition. The angle is measured to the magnetic field lines."}]},{"id":"adc45372-aad5-4fa6-8427-94e7fcfd0ccd","cardType":"basic","points":1,"front":"A coil's plane is parallel to the magnetic field lines. What is the magnetic flux linkage?","back":"Zero","mcqOptions":[{"text":"Zero","correct":true,"explanation":"When the coil’s plane is parallel to the field lines, its normal is perpendicular to them. Therefore $\\text{cos }90^\\circ=0$, giving zero flux linkage."},{"text":"Maximum","correct":false,"explanation":"Maximum linkage occurs when the coil’s normal is parallel to the field, meaning its plane is perpendicular to the field. A parallel plane gives zero linkage."},{"text":"Half the maximum","correct":false,"explanation":"Half the maximum would require the angle between the field and the coil’s normal to satisfy $\\text{cos }\\theta=0.5$. Here the angle is $90^\\circ$, so linkage is zero."},{"text":"Equal to $BAN$","correct":false,"explanation":"The value $BAN$ applies when the coil’s normal is parallel to the field. Here the plane is parallel, so the normal is perpendicular and the linkage is zero."}]},{"id":"bcb8a505-46af-4ec7-9b80-c11a8e5e591c","cardType":"basic","points":2,"front":"Why is flux linkage zero when the plane of the coil is parallel to the magnetic field?","back":"The normal to the coil is at $90^\\text{o}$ to the field, and $\\text{cos}(90^\\text{o}) = 0$.","mcqOptions":[{"text":"The normal to the coil is at $90^\\text{o}$ to the field, and $\\text{cos}(90^\\text{o}) = 0$.","correct":true,"explanation":"Magnetic flux linkage is $BAN\\cos\\theta$, where $\\theta$ is measured between the field and the coil’s normal. Here $\\theta=90^\\text{o}$, so the cosine factor makes the linkage zero."},{"text":"The plane of the coil is at $90^\\text{o}$ to the field, and $\\text{cos}(90^\\text{o}) = 0$.","correct":false,"explanation":"This confuses the plane with the direction used to define the angle. The coil’s normal is at $90^\\text{o}$ to the field, and $\\cos(90^\\text{o})=0$, giving zero flux linkage."},{"text":"The normal to the coil is at $0^\\text{o}$ to the field, and $\\text{cos}(0^\\text{o}) = 0$.","correct":false,"explanation":"The angle in the flux-linkage equation is measured from the coil’s normal, not its plane. The normal is at $90^\\text{o}$ to the field, so the cosine factor is zero."},{"text":"The normal to the coil is at $90^\\text{o}$ to the field, and $\\text{sin}(90^\\text{o}) = 0$.","correct":false,"explanation":"Using sine instead of cosine applies an equation written for the angle to the coil’s plane. Here the normal-field angle is $90^\\text{o}$, and $\\cos(90^\\text{o})=0$."}]},{"id":"4a4335aa-0794-469a-9e3b-0085aed6c45a","cardType":"basic","points":1,"front":"If the angle between the plane of the coil and the magnetic field is $20^\\text{o}$, what value of $\\theta$ is used in $N̐ = BAN \\text{cos}\\theta$?","back":"$$70^\\text{o}$ ($90 - 20$)","mcqOptions":[{"text":"$$70^\\text{o}$ ($90 - 20$)","correct":true,"explanation":"The equation uses the angle between the magnetic field and the normal to the coil. The normal is $90^\\text{o}$ from the plane, so $\\theta=90^\\text{o}-20^\\text{o}=70^\\text{o}$."},{"text":"$$20^\\text{o}$ ($20 - 0$)","correct":false,"explanation":"This uses the stated angle to the plane directly. In the cosine equation, $\\theta$ is measured to the normal, so the required complementary angle is $70^\\text{o}$."},{"text":"$$110^\\text{o}$ ($90 + 20$)","correct":false,"explanation":"Adding the plane-field angle to $90^\\text{o}$ selects the opposite normal rather than the acute normal-field angle normally used here. The required complementary angle is $70^\\text{o}$."},{"text":"$$40^\\text{o}$ ($20 + 20$)","correct":false,"explanation":"Doubling the stated angle has no geometrical basis. Because the normal is perpendicular to the plane, the equation requires $90^\\text{o}-20^\\text{o}=70^\\text{o}$."}]},{"id":"ff2eb48b-07d7-4b90-a843-69d69e452bf4","cardType":"cloze","points":null,"front":"When converting units for area, $1 \\text{ cm}^2 = \\text{[...]} \\text{ m}^2$ and $1 \\text{ mm}^2 = 10^{-6} \\text{ m}^2$.","back":"When converting units for area, $1 \\text{ cm}^2 = \\boxed{10^{-4}} \\text{ m}^2$ and $1 \\text{ mm}^2 = 10^{-6} \\text{ m}^2$.","mcqOptions":[{"text":"$$10^{-4}$","correct":true,"explanation":"Since $1\\text{ cm}=10^{-2}\\text{ m}$, squaring both sides gives $1\\text{ cm}^2=(10^{-2})^2\\text{ m}^2=10^{-4}\\text{ m}^2$."},{"text":"$$10^{-2}$","correct":false,"explanation":"This applies the linear centimetre conversion without squaring it. Area has two dimensions, so $(10^{-2})^2=10^{-4}$ is the required conversion factor."},{"text":"$$10^{4}$","correct":false,"explanation":"This reverses the direction of the unit conversion. A square centimetre is smaller than a square metre, so its value in square metres is $10^{-4}$, not $10^4$."},{"text":"$$10^{-6}$","correct":false,"explanation":"This is the conversion factor for square millimetres rather than square centimetres. Squaring $1\\text{ cm}=10^{-2}\\text{ m}$ gives the correct factor $10^{-4}$."}]},{"id":"90f6f144-992c-479f-994e-fb136ed7da8e","cardType":"cloze","points":null,"front":"When converting units for area, $1 \\text{ cm}^2 = 10^{-4} \\text{ m}^2$ and $1 \\text{ mm}^2 = \\text{[...]} \\text{ m}^2$.","back":"When converting units for area, $1 \\text{ cm}^2 = 10^{-4} \\text{ m}^2$ and $1 \\text{ mm}^2 = \\boxed{10^{-6}} \\text{ m}^2$.","mcqOptions":[{"text":"$$10^{-6}$","correct":true,"explanation":"Since $1\\text{ mm}=10^{-3}\\text{ m}$, the conversion must be squared for area: $1\\text{ mm}^2=(10^{-3})^2\\text{ m}^2=10^{-6}\\text{ m}^2$."},{"text":"$$10^{-3}$","correct":false,"explanation":"This uses the linear millimetre conversion without squaring it. Because area has two dimensions, $(10^{-3})^2=10^{-6}$ is the correct factor."},{"text":"$$10^{-4}$","correct":false,"explanation":"This is the square-centimetre conversion factor, not the square-millimetre factor. A millimetre is $10^{-3}$ metres, so squaring gives $10^{-6}$."},{"text":"$$10^{6}$","correct":false,"explanation":"This reverses the conversion factor. A square millimetre is much smaller than a square metre, so it equals $10^{-6}\\text{ m}^2$, not $10^6\\text{ m}^2$."}]},{"id":"0178de89-d38e-4763-8f74-d4481d6a0afa","cardType":"basic","points":1,"front":"Describe the orientation of a coil relative to the magnetic field to achieve maximum flux linkage.","back":"The plane of the coil must be perpendicular (normal) to the magnetic field lines.","mcqOptions":[{"text":"The plane of the coil must be perpendicular (normal) to the magnetic field lines.","correct":true,"explanation":"Maximum linkage occurs when the coil’s normal is parallel to the field, making $\\theta=0^\\text{o}$ and $\\cos\\theta=1$. Therefore, the plane itself is perpendicular to the field lines."},{"text":"The plane of the coil must be parallel (aligned) to the magnetic field lines.","correct":false,"explanation":"A plane parallel to the field has its normal at $90^\\text{o}$ to the field. This makes $\\cos\\theta=0$, so linkage is zero rather than maximum; the plane must be perpendicular."},{"text":"The normal of the coil must be perpendicular (normal) to the magnetic field lines.","correct":false,"explanation":"This confuses the coil’s plane with its normal. Maximum linkage requires the normal to be parallel to the field, meaning the plane must instead be perpendicular to the field lines."},{"text":"The plane of the coil must be at $45^\\text{o}$ (diagonal) to the magnetic field lines.","correct":false,"explanation":"At $45^\\text{o}$, the cosine factor is only $1/\\sqrt{2}$, so the linkage is below its maximum. The plane must be perpendicular to the field lines."}]},{"id":"0f61a7e7-9b99-41e6-832e-0fae1395f37b","cardType":"basic","points":1,"front":"What does a search coil connected to an oscilloscope produce when rotated at a constant speed in a uniform magnetic field?","back":"A sinusoidal alternating voltage.","mcqOptions":[{"text":"A sinusoidal alternating voltage.","correct":true,"explanation":"Constant-speed rotation makes the flux linkage vary sinusoidally with time. Faraday’s law therefore gives an induced voltage that alternates direction and has a sinusoidal waveform."},{"text":"A constant direct voltage.","correct":false,"explanation":"A steady voltage would require a constant rate of change of flux linkage. During uniform rotation, that rate varies sinusoidally and reverses sign, producing a sinusoidal alternating voltage."},{"text":"A square-wave alternating voltage.","correct":false,"explanation":"A square wave would require abrupt switching between voltage levels. Smooth constant-speed rotation causes a smooth sinusoidal change in flux linkage, so the induced voltage is sinusoidal."},{"text":"A constant zero voltage.","correct":false,"explanation":"The induced voltage is not always zero because the flux linkage changes as the coil rotates. Its rate of change varies and reverses, producing a sinusoidal alternating voltage."}]},{"id":"4ef710cc-4510-42a9-b302-6371a6f4638e","cardType":"basic","points":2,"front":"A $2.0 \\text{ cm}$ by $3.0 \\text{ cm}$ rectangular loop is perpendicular to a $0.50 \\text{ T}$ field. Calculate the magnetic flux.","back":"

$A = 0.02 \\times 0.03 = 6.0 \\times 10^{-4} \\text{ m}^2$
$̐ = BA = 0.50 \\times (6.0 \\times 10^{-4}) = 3.0 \\times 10^{-4} \\text{ Wb}$

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

$A = 0.02 \\times 0.03 = 6.0 \\times 10^{-4} \\text{ m}^2$
$̐ = BA = 0.50 \\times (6.0 \\times 10^{-4}) = 3.0 \\times 10^{-4} \\text{ Wb}$

","correct":true,"explanation":"Converting both lengths to metres gives an area of $6.0\\times10^{-4}\\text{ m}^2$. The field is normal to the loop, so $\\Phi=BA=3.0\\times10^{-4}\\text{ Wb}$."},{"text":"

$A = 0.02 \\times 0.03 = 6.0 \\times 10^{-4} \\text{ m}^2$
$̐ = BA = 0.50 \\times (6.0 \\times 10^{-4}) = 3.0 \\times 10^{-3} \\text{ Wb}$

","correct":false,"explanation":"The area conversion is correct, but the final power of ten is not. Multiplying $6.0\\times10^{-4}$ by $0.50$ gives $3.0\\times10^{-4}\\text{ Wb}$, not $10^{-3}\\text{ Wb}$."},{"text":"

$A = 0.02 \\times 0.03 = 6.0 \\times 10^{-4} \\text{ m}^2$
$̐ = A/B = (6.0 \\times 10^{-4})/0.50 = 1.2 \\times 10^{-3} \\text{ Wb}$

","correct":false,"explanation":"Dividing by the magnetic flux density reverses the relationship $\\Phi=BA$. The area should be multiplied by $0.50\\text{ T}$, giving $3.0\\times10^{-4}\\text{ Wb}$."},{"text":"

$A = 0.2 \\times 0.3 = 6.0 \\times 10^{-2} \\text{ m}^2$
$̐ = BA = 0.50 \\times (6.0 \\times 10^{-2}) = 3.0 \\times 10^{-2} \\text{ Wb}$

","correct":false,"explanation":"Dividing centimetres by 10 converts them to decimetres, not metres, making the area 100 times too large. The correct area is $6.0\\times10^{-4}\\text{ m}^2$ and the flux is $3.0\\times10^{-4}\\text{ Wb}$."}]},{"id":"d32ce412-e690-42d3-b8b3-a1dfda70ecd4","cardType":"cloze","points":null,"front":"Magnetic flux linkage is the product of the [ ] and the number of turns in a coil.","back":"Magnetic flux linkage is the product of the magnetic flux and the number of turns in a coil.","mcqOptions":[{"text":"magnetic flux","correct":true,"explanation":"Flux linkage is defined as $N\\Phi$, where $N$ is the number of turns and $\\Phi$ is the magnetic flux through one turn of the coil."},{"text":"magnetic flux density","correct":false,"explanation":"Magnetic flux density describes the field strength, but it must first be combined with area and orientation to obtain flux. Flux linkage is magnetic flux multiplied by the number of turns."},{"text":"cross-sectional area","correct":false,"explanation":"The coil’s cross-sectional area contributes to flux through $\\Phi=BA\\cos\\theta$, but area alone is not multiplied by the number of turns to define flux linkage. The required quantity is magnetic flux."},{"text":"induced voltage","correct":false,"explanation":"Induced electromotive force depends on the rate of change of flux linkage under Faraday’s law; it is not a factor defining flux linkage. The defining product uses magnetic flux."}]},{"id":"1647f926-2358-4182-afc2-cb61d7f0a39b","cardType":"cloze","points":null,"front":"Magnetic flux linkage is the product of the magnetic flux and the [ ] in a coil.","back":"Magnetic flux linkage is the product of the magnetic flux and the number of turns in a coil.","mcqOptions":[{"text":"number of turns","correct":true,"explanation":"Flux linkage is $N\\Phi$, so the magnetic flux through each turn is multiplied by the number of turns, $N$, in the coil."},{"text":"cross-sectional area","correct":false,"explanation":"Cross-sectional area is used when calculating flux with $\\Phi = BA\\cos\\theta$; flux linkage instead equals flux multiplied by the number of turns."},{"text":"magnetic field strength","correct":false,"explanation":"Magnetic field strength helps determine the flux through one turn, but flux linkage is the magnetic flux multiplied by the number of turns in the coil."},{"text":"angle of rotation","correct":false,"explanation":"The coil’s angle affects the flux through each turn via $\\cos\\theta$, but flux linkage is the product of flux and the number of turns."}]},{"id":"0b955518-f5e1-407d-95b9-7b65e25c57a2","cardType":"basic","points":1,"front":"Define the normal to a coil in the context of flux linkage.","back":"An imaginary line perpendicular ($90^\\text{o}$) to the surface of the coil.","mcqOptions":[{"text":"An imaginary line perpendicular ($90^\\text{o}$) to the surface of the coil.","correct":true,"explanation":"The normal is defined as an imaginary line at right angles to the coil’s surface. The angle in $\\Phi = BA\\cos\\theta$ is measured from this line."},{"text":"An imaginary line parallel ($0^\\text{o}$) to the surface of the coil.","correct":false,"explanation":"This describes a line lying along the coil’s surface, not its normal. The normal is perpendicular, or at $90^\\text{o}$, to the surface."},{"text":"An imaginary line perpendicular ($90^\\text{o}$) to the magnetic field.","correct":false,"explanation":"The normal is defined relative to the coil’s surface, not relative to the magnetic field. It is perpendicular to the surface of the coil."},{"text":"An imaginary line parallel ($0^\\text{o}$) to the magnetic field.","correct":false,"explanation":"The normal can be parallel to the field in a particular orientation, but this does not define it. It is always perpendicular to the coil’s surface."}]},{"id":"f633a767-bf52-4ae6-80bc-0393558e6490","cardType":"basic","points":2,"front":"A search coil has $200$ turns and an area of $5.0 \\text{ cm}^2$. If the flux through one turn is $2.0 \\text{ ̑Wb}$, calculate the flux linkage.","back":"$$N̐ = 200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-4} \\text{ Wb turns}$","mcqOptions":[{"text":"$$N̐ = 200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-4} \\text{ Wb turns}$","correct":true,"explanation":"Flux linkage equals the number of turns multiplied by the flux through one turn. Therefore, $200 \\times 2.0 \\times 10^{-6}$ gives $4.0 \\times 10^{-4}\\text{ Wb turns}$."},{"text":"$$N̐ = 200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-6} \\text{ Wb turns}$","correct":false,"explanation":"This fails to multiply the flux by all $200$ turns. Flux linkage is $200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-4}\\text{ Wb turns}$."},{"text":"$$N̐ = 200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-2} \\text{ Wb turns}$","correct":false,"explanation":"This contains a power-of-ten conversion error, effectively treating the micro prefix incorrectly. Multiplying $200$ by $2.0 \\times 10^{-6}$ gives $4.0 \\times 10^{-4}\\text{ Wb turns}$."},{"text":"$$N̐ = 200 \\times 2.0 \\times 10^{-6} = 2.0 \\times 10^{-4} \\text{ Wb turns}$","correct":false,"explanation":"This result effectively uses only $100$ turns rather than $200$. The stated number of turns gives $200 \\times 2.0 \\times 10^{-6} = 4.0 \\times 10^{-4}\\text{ Wb turns}$."}]},{"id":"3f8b4831-b150-4d2d-94a4-a44e0a675493","cardType":"basic","points":2,"front":"What happens to the magnetic flux through a bucket if you tilt it in the 'rain' analogy?","back":"The flux decreases because the perpendicular area catching the 'rain' is reduced.","mcqOptions":[{"text":"The flux decreases because the perpendicular area catching the 'rain' is reduced.","correct":true,"explanation":"Tilting reduces the area presented perpendicular to the falling 'rain'. Likewise, magnetic flux decreases as the coil’s effective perpendicular area becomes smaller."},{"text":"The flux increases because the perpendicular area catching the 'rain' is increased.","correct":false,"explanation":"Tilting the bucket does not increase the area presented perpendicular to the 'rain'; it reduces it. The magnetic flux therefore decreases rather than increases."},{"text":"The flux stays constant because the surface area catching the 'rain' is unchanged.","correct":false,"explanation":"This confuses the bucket’s actual area with its effective perpendicular area. Tilting leaves the actual area unchanged but reduces the perpendicular area, so flux decreases."},{"text":"The flux decreases because the magnetic field strength of the 'rain' is reduced.","correct":false,"explanation":"Tilting changes the effective perpendicular area, not the strength of the field represented by the 'rain'. Flux decreases because less perpendicular area intercepts the field."}]},{"id":"90effb34-5492-43d4-a945-0cff70011e2c","cardType":"cloze","points":null,"front":"In AQA exams, you must differentiate between [ ] (for a single loop) and flux linkage (for a multi-turn coil).","back":"In AQA exams, you must differentiate between flux (for a single loop) and flux linkage (for a multi-turn coil).","mcqOptions":[{"text":"flux","correct":true,"explanation":"Magnetic flux, $\\Phi$, refers to the field passing through a single loop. For a coil with multiple turns, AQA distinguishes this from flux linkage, $N\\Phi$."},{"text":"flux density","correct":false,"explanation":"Flux density, $B$, measures magnetic flux per unit perpendicular area and is not restricted to a single loop. The required single-loop quantity is flux."},{"text":"flux linkage","correct":false,"explanation":"Flux linkage accounts for all the turns in a multi-turn coil, so it is not the single-loop quantity. A single loop has magnetic flux, $\\Phi$."},{"text":"field strength","correct":false,"explanation":"Magnetic field strength describes the field rather than the total passing through one loop. The quantity for a single loop is magnetic flux."}]},{"id":"88530753-1033-42af-b38e-948bb7d20d25","cardType":"cloze","points":null,"front":"In AQA exams, you must differentiate between flux (for a single loop) and [ ] (for a multi-turn coil).","back":"In AQA exams, you must differentiate between flux (for a single loop) and flux linkage (for a multi-turn coil).","mcqOptions":[{"text":"flux linkage","correct":true,"explanation":"Flux linkage, $N\\Phi$, is the total flux linked by all turns of a coil. It equals the flux through one turn multiplied by the number of turns."},{"text":"flux density","correct":false,"explanation":"Flux density, $B$, describes the magnetic field and is measured in tesla; it does not total the flux across multiple turns. That quantity is flux linkage."},{"text":"magnetic flux","correct":false,"explanation":"Magnetic flux refers to the flux through a single loop, so using it here misses the multiplication by the number of turns. A multi-turn coil has flux linkage."},{"text":"field strength","correct":false,"explanation":"Field strength characterises the magnetic field itself, not the combined flux linked by several turns. The required multi-turn quantity is flux linkage, $N\\Phi$."}]},{"id":"920e0c02-b837-46e9-86b4-df96e6af7747","cardType":"basic","points":1,"front":"State the value of $\\text{cos}\\theta$ when the magnetic field is parallel to the normal of the coil.","back":"$$1$ (as $\\theta = 0^\\text{o}$)","mcqOptions":[{"text":"$$1$ (as $\\theta = 0^\\text{o}$)","correct":true,"explanation":"When the field is parallel to the normal, the angle between them is $0^\\text{o}$. Since $\\cos 0^\\text{o}=1$, the flux is at its maximum."},{"text":"$$0$ (as $\\theta = 90^\\text{o}$)","correct":false,"explanation":"This treats the angle as measured from the coil’s surface rather than its normal. A field parallel to the normal gives $\\theta=0^\\text{o}$ and $\\cos\\theta=1$."},{"text":"$$-1$ (as $\\theta = 180^\\text{o}$)","correct":false,"explanation":"A cosine of $-1$ applies when the field is antiparallel to the chosen normal. Parallel vectors have $\\theta=0^\\text{o}$, so $\\cos\\theta=1$."},{"text":"$$rac{1}{2}$ (as $\\theta = 60^\\text{o}$)","correct":false,"explanation":"This assumes an angle of $60^\\text{o}$ without justification. Parallel field and normal directions have zero angle between them, giving $\\cos 0^\\text{o}=1$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Magnetic flux and flux linkage (A-level only)","title":"Magnetic flux and flux linkage (A-level only) Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"8f8cded6-7fe8-5aa4-bdb9-9890064ea66e","walled":"$L39"}],"$L3a","$L3b"]