G495 - Field and Particle Pictures: June 11th

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  1. Matt G's Avatar
    • Junior Member
    • Posts: 33
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by silvertear)
    anyone have anything remotely similar to this?
    I said that the rock has an irregular amorphous structure meaning that when a force is applied to it no crystal slip can occur meaning that the stress builds up until it suddenly snaps.

    For the flux question I halved the flux linkage.
  2. silvertear's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by Matt G)
    For the flux question I halved the flux linkage.
    I think I halved the flux, then divided it.

  3. Oromis263's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by In One Ear)
    I said the opposite, potential is zero since if you imagine putting an electron mid way between both, it won't be attracted either way since it feels the same attractive force from both sides. Then i put that the field strength doubled (since i had to use the 58 value somewhere) and i didn't really think that the field strength halving made any sense- it kinda makes sense that overlapping two equal fields doubles the field strength but i'm uncomfortable with the interactions of fields tbh.
    Because protons are of the same charge, there fields will cancel, I'll find the page in the book that shows it. A friend of mine asked me the exact same question last night but in terms of gravity, quite luckily, and his mark scheme did say potential added. I'll have to try find out tomorrow.

    (Original post by chaosdestro0)
    OK that was hard.
    Section A was alright, Section B was a bit hard and section C was solid.
    Worst of all, half way through the exam my left ear had a high pitch noise and I think it's perforated. Had this before in year 10.

    And what was with the second to last question, I know I did it wrong but I will probably loose 1 mark and it will follow through. I know a lot of people who converted cm^2 to m^2 wrongly and I did the same, but even if you converted it correctly it was like 24 *10^-5 and they said it was > 5*10^-5.
    4cm^2 = 4/10000 m^2, then you run the numbers through? So 0.15 times 1/2500 gave 6x10^-5Wb for me.
  4. chaosdestro0's Avatar
    • Adored and Respected Member
    Re: G495 - Field and Particle Pictures: June 11th
    yeah most people I know put 0.6/(5/3) and that turned out fine. That proof of P was quite hard, basically you had to use
    E=FD
    F=ma
    Then C=D/t
    It came out like MADT/D Then that simplified to mv
  5. Rotravis's Avatar
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    • Posts: 31
    Re: G495 - Field and Particle Pictures: June 11th
    What did everyone work out for the flux in that question that then asked you for flux linkage? It said something like "show that it is greater than 10^-5 but my answer was something like 6 x 10^-3 Wb?

    Going on from that, would flux linkage be 6 x 10^-3 x 200 turns? (which is 6/5)

    Then if the flux linkage halved, emf would be (3/5)/(5/3)? That's what I ended up with.
  6. In One Ear's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by chaosdestro0)
    OK that was hard.
    Section A was alright, Section B was a bit hard and section C was solid.
    Worst of all, half way through the exam my left ear had a high pitch noise and I think it's perforated. Had this before in year 10.

    And what was with the second to last question, I know I did it wrong but I will probably loose 1 mark and it will follow through. I know a lot of people who converted cm^2 to m^2 wrongly and I did the same, but even if you converted it correctly it was like 24 *10^-5 and they said it was > 5*10^-5.
    What were 4cm^2 = 4x(10^-2)^2 m^2 x a load of crap?

    Which gave 6x10^-5 (>5x10^-5)?
  7. chaosdestro0's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by Oromis263)
    Because protons are of the same charge, there fields will cancel, I'll find the page in the book that shows it. A friend of mine asked me the exact same question last night but in terms of gravity, quite luckily, and his mark scheme did say potential added. I'll have to try find out tomorrow.



    4cm^2 = 4/10000 m^2, then you run the numbers through? So 0.15 times 1/2500 gave 6x10^-5Wb for me.
    Ahh cool was probably just one mark drop for that, on that proton charge I said it doubled because I looked at the second question and it was comparing. So I knew that it would look odd putting zero.
  8. EricEdwardSelvaraj's Avatar
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    • Posts: 87
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by lukas1051)
    BTW how did everyone show the units of K (one of the very first questions) I got really confused and I don't know why...
    F=kqQ/r^2
    F*r^2=kqQ
    F*r^2/qQ=k

    The units of F=N, r=m, q=C; now substitute

    N*m^2/C^2=k
    Hence, Nm^2C^-2

    Hope it helps.
  9. lukas1051's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    Did anyone else think it was a weird paper in the sense that a lot of things didn't come up?

    Binding energy didn't come up at all with the exception of the graph, nuclear equations didn't really come up, fundamental particles didn't really come up, energy levels didn't come up at all, scattering didn't really come up...
  10. chaosdestro0's Avatar
    • Adored and Respected Member
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by In One Ear)
    What were 4cm^2 = 4x(10^-2)^2 m^2 x a load of crap?

    Which gave 6x10^-5 (>5x10^-5)?
    I know what I did wrong.
    I did (4*10^-2)^2 when it should have been 4*(10-2)^2. So that's probably two marks gone or maybe one.
  11. Amy-Rose's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    Sad to see a lack of binding energy questions!

    And not a single mention of radon in section C.
  12. lukas1051's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by EricEdwardSelvaraj)
    F=kqQ/r^2
    F*r^2=kqQ
    F*r^2/qQ=k

    The units of F=N, r=m, q=C; now substitute

    N*m^2/C^2=k
    Hence, Nm^2C^-2

    Hope it helps.
    Ahh of course, missed out a q, FFFFUUUUUUU
  13. EricEdwardSelvaraj's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by Rotravis)
    What did everyone work out for the flux in that question that then asked you for flux linkage? It said something like "show that it is greater than 10^-5 but my answer was something like 6 x 10^-3 Wb?

    Going on from that, would flux linkage be 6 x 10^-3 x 200 turns? (which is 6/5)

    Then if the flux linkage halved, emf would be (3/5)/(5/3)? That's what I ended up with.
    I cannot remember exactly, but I got similar values. And being an idiot I am, I did 1.8x10^-3/3mm to get 0.6s. This is wrong, it should have been 3mm/1.8^-3.

    I would be gutted if I lost the whole 4 marks because of this; hopefully ecf can be applied.
  14. chaosdestro0's Avatar
    • Adored and Respected Member
    Re: G495 - Field and Particle Pictures: June 11th
    Apart from Section C it was fine, the breaking stress question was odd.
    I started by saying that there is a force between the atoms and that when it reaches the breaking stress, the bonds break leaving them to slide past each other. I had to blag that as I had no clue, and wow that tissue question was rather strange.
    I said
    Energy is proportional to number of Protons
    So rate of change of protons per cm^2 is proportional to Energy per cm^2, then said the maximum rate of change of protons was when the gradient was maximum. And I got like 13cm.
    Last edited by chaosdestro0; 11-06-2012 at 17:04.
  15. In One Ear's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by chaosdestro0)
    I know what I did wrong.
    I did (4*10^-2)^2 when it should have been 4*(10-2)^2. So that's probably two marks gone or maybe one.
    Yeah, seems i have my fair share of mistakes too. Ah well, i thought the paper was tough enough so we should get decent boundaries. Its just so hard to call performance in a physics paper because whilst u may have understood all the questions, wether what you wrote is deemed creditable is a bit hard to predict.
  16. Vit4's Avatar
    • New Member
    • Posts: 2
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by chaosdestro0)
    yeah most people I know put 0.6/(5/3) and that turned out fine. That proof of P was quite hard, basically you had to use
    E=FD
    F=ma
    Then C=D/t
    It came out like MADT/D Then that simplified to mv
    i did p = Ns and then subbed in wd = f*d which is Nm. and so c=m s^-1. so Ns=Nms/m, which makes Ns on both sides
  17. conorordan's Avatar
    • Full Member
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by lukas1051)
    BTW how did everyone show the units of K (one of the very first questions) I got really confused and I don't know why...
    E=\frac{F}{q}\; \textup{and}\; E=\frac{kQ}{r^2}

    \textup{so}\; \frac{F}{q}=\frac{kQ}{r^2}

    k=\frac{Fr^2}{Qq} which has the units \textup{Nm}^2\textup{C}^{-2}
  18. nebelbon's Avatar
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    Re: G495 - Field and Particle Pictures: June 11th
    Didn't do as well as i'd hoped
  19. chaosdestro0's Avatar
    • Adored and Respected Member
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by In One Ear)
    Yeah, seems i have my fair share of mistakes too. Ah well, i thought the paper was tough enough so we should get decent boundaries. Its just so hard to call performance in a physics paper because whilst u may have understood all the questions, wether what you wrote is deemed creditable is a bit hard to predict.
    Yeah I mean Section C was poo poo and I found some of the questions that came up to be quite new concepts. Answerable but unpredictable to revise for.
    Who would have expected that proton question to come, pretty much nobody.
  20. In One Ear's Avatar
    • Exalted and Worshipped Member
    • Posts: 1,056
    Re: G495 - Field and Particle Pictures: June 11th
    (Original post by Oromis263)
    Because protons are of the same charge, there fields will cancel, I'll find the page in the book that shows it. A friend of mine asked me the exact same question last night but in terms of gravity, quite luckily, and his mark scheme did say potential added. I'll have to try find out tomorrow.



    4cm^2 = 4/10000 m^2, then you run the numbers through? So 0.15 times 1/2500 gave 6x10^-5Wb for me.
    Sucks to be me then. Out of interest, could you elaborate on how you thought you "cheated" on the horizontal movement question? I quoted you earlier but you may not have noticed given the volume of people quoting you .
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