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    I wrote my answers to Section B ( written paper) in fountain pen :0 Will it not get scanned ?
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    (Original post by -Gifted-)
    I wrote my answers to Section B ( written paper) in fountain pen :0 Will it not get scanned ?
    Lol I did all my work in fountain pen too. It's fine, I did everything in fountain pen last year, no problem whatsoever
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    unofficial mark scheme for section B?
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    (Original post by DesignPredator)
    Here are my answers to Written, the question parts might not be correct.

    Written answers
    1ai) 8.26 (will be a range)
    ii) 2.6x10^-4

    bi)28
    ii)5.9x10^4
    iii) bottom box ticked

    2ai) Derivation
    ii) 6.06x10^6 (3SF only)

    b) Downward Curve staring at 9.81
    Minimum at g=0 at a point slightly offset from halfway closer to Venus
    Rises up to 8.87

    3a) No external forces

    bi) V=mv/N

    bii) Derivation


    ci) 216
    ii) 1.15x10^-19 kgms-1 OR Ns
    iii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

    4a) Small amplitude oscillation as only valid for small angular amplitudes.

    b) -Set up light string with Bob on the end,
    attach top of string to clamp stand boss.
    -Measure length from top to centre of Bob.
    -Release the Bob so it oscillates with small amplitude.
    -Use stop clock to measure time taken for 20 oscillations.
    -Divide this by 20 for mean T
    -Plot T^2 against L
    -Calc Gradient
    g=(4pi^2)/gradient
    -Repeat procedure

    c) -Student value 4x true value
    -Time period half so all values for T^2 will be 1/4 the true
    -Gradient is 4x lower
    -State equation or show that g is inversely proportional to gradient.

    5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
    Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

    b) -Change in flux linkage when current flows.
    -EMF induced in wheel (or whatever it was called)
    -Current induced in the wheel as good conductor.
    -Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

    c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

    More energy used or less effective or heating.
    Either your 5a is wrong or the AQA text book is wrong. Because the text book says induced emf is EQUAL to rate of change of flux linkage. Not proportional. Plus if it was proportional then the equation would need a constant. But all the letters in the equation can be varied
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    (Original post by lucabrasi98)
    Either your 5a is wrong or the AQA text book is wrong. Because the text book says induced emf is EQUAL to rate of change of flux linkage. Not proportional. Plus if it was proportional then the equation would need a constant. But all the letters in the equation can be varied
    Both will be accepted as seen from previous mark schemes and there is a constant, that is -1.
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    (Original post by DesignPredator)
    Here are my answers to Written, the question parts might not be correct.

    Written answers
    1ai) 8.26 (will be a range)
    ii) 2.6x10^-4

    bi)28
    ii)5.9x10^4
    iii) bottom box ticked

    2ai) Derivation
    ii) 6.06x10^6 (3SF only)

    b) Downward Curve staring at 9.81
    Minimum at g=0 at a point slightly offset from halfway closer to Venus
    Rises up to 8.87

    3a) No external forces

    bi) V=mv/N

    bii) Derivation


    ci) 216
    ii) 1.15x10^-19 kgms-1 OR Ns
    iii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

    4a) Small amplitude oscillation as only valid for small angular amplitudes.

    b) -Set up light string with Bob on the end,
    attach top of string to clamp stand boss.
    -Measure length from top to centre of Bob.
    -Release the Bob so it oscillates with small amplitude.
    -Use stop clock to measure time taken for 20 oscillations.
    -Divide this by 20 for mean T
    -Plot T^2 against L
    -Calc Gradient
    g=(4pi^2)/gradient
    -Repeat procedure

    c) -Student value 4x true value
    -Time period half so all values for T^2 will be 1/4 the true
    -Gradient is 4x lower
    -State equation or show that g is inversely proportional to gradient.

    5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
    Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

    b) -Change in flux linkage when current flows.
    -EMF induced in wheel (or whatever it was called)
    -Current induced in the wheel as good conductor.
    -Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

    c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

    More energy used or less effective or heating.
    For 5b, I put emf induced in coil by accident , I think the question was out of 3, would I get at least 1 for saying change in disk rotates, so change in flux ?
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    (Original post by DesignPredator)
    Both will be accepted as seen from previous mark schemes and there is a constant, that is -1.
    B,A,N and T can all be varied. A constant is something like G in the equation for gravitational firled strenth. The -1 isn't really a constant, it's just essentially demonstrating Lenz's law. It implies they're in opposite directions. And IIRC you had to mention lenz' law later in the question.

    I don't recall any mark schemes where they accept both but maybe you're right
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    Im pretty sure one of the mark schemes said it was proportional
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    What if I said Faraday's law was emf is proportional to rate of change of flux, and not flux linkage? (constant of proportionality would be N, number of turns)

    Will I lose any marks?
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    (Original post by mcride98)
    What if I said Faraday's law was emf is proportional to rate of change of flux, and not flux linkage? (constant of proportionality would be N, number of turns)

    Will I lose any marks?
    Not sure, I may have said that too . To make things even worse, I said the induced emf is proportional to the rate of change of flux with time
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    (Original post by DesignPredator)
    Here are my answers to Written, the question parts might not be correct.

    Written answers
    1ai) 8.26 (will be a range)
    ii) 2.6x10^-4

    bi)28
    ii)5.9x10^4
    iii) bottom box ticked

    2ai) Derivation
    ii) 6.06x10^6 (3SF only)

    b) Downward Curve staring at 9.81
    Minimum at g=0 at a point slightly offset from halfway closer to Venus
    Rises up to 8.87

    3a) No external forces

    bi) V=mv/N

    bii) Derivation


    ci) 216
    ii) 1.15x10^-19 kgms-1 OR Ns
    iii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

    4a) Small amplitude oscillation as only valid for small angular amplitudes.

    b) -Set up light string with Bob on the end,
    attach top of string to clamp stand boss.
    -Measure length from top to centre of Bob.
    -Release the Bob so it oscillates with small amplitude.
    -Use stop clock to measure time taken for 20 oscillations.
    -Divide this by 20 for mean T
    -Plot T^2 against L
    -Calc Gradient
    g=(4pi^2)/gradient
    -Repeat procedure

    c) -Student value 4x true value
    -Time period half so all values for T^2 will be 1/4 the true
    -Gradient is 4x lower
    -State equation or show that g is inversely proportional to gradient.

    5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
    Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

    b) -Change in flux linkage when current flows.
    -EMF induced in wheel (or whatever it was called)
    -Current induced in the wheel as good conductor.
    -Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

    c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

    More energy used or less effective or heating.
    Cheers! I can't remember all my answers now but I recall getting a lot of similar answers. Do you remember what question 4(a) asked?
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    Hey cobalt, noticed you have an offer for natsci at cambridge , which college????
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    Was the question which proceeded the energy proof asking to calculate the momentum of the alpha particle. If so, does anyone remember their working out for it?
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    (Original post by lfcrules)
    Hey cobalt, noticed you have an offer for natsci at cambridge , which college????
    Christ's
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    (Original post by C0balt)
    Im pretty sure one of the mark schemes said it was proportional


    All I know is the book said equal to in bold writing. So if it's not then I blame AQA and Nelson Thrones for ****ing me over and over again.
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    thanks a lot
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    (Original post by C0balt)
    Christ's
    Congrats C0balt, I've no doubt you'll make your offer!
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    (Original post by lucabrasi98)
    All I know is the book said equal to in bold writing. So if it's not then I blame AQA and Nelson Thrones for ****ing me over and over again.
    They are both accepted. Stop fretting.
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    (Original post by C0balt)
    I initially got 8.22 but then changed to 8.0 because it gave me an exact value lmao
    Yeah we always get a mark for stating it's the gradient
    I got 8.0 too. Δy/Δx across the whole range returned it.
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    (Original post by Cadherin)
    Congrats C0balt, I've no doubt you'll make your offer!
    Thanks i hope so, all depends on the next 3 exams!
    (Original post by micycle)
    I got 8.0 too. Δy/Δx across the whole range returned it.
    :five:
 
 
 
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