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    A current of 0.6A flows in a wire with 5 x 10^28 free electrons per unit volume. If the drift velocity of these electrons is 8.0 x 10^-4 m/s, calculate the diameter of the wire.

    So i'm using I = nAVq

    Am i right in saying:

    I = 0.6 Amps

    n = 5 \times 10^28 m^3

    A =

    V = 8 \times 10^-4 m/s

    q =  ?

    what is q?
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    q = charge on an electron

    the equation is sometimes written I = nave (swap e for q) when electrons are involved
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    q = charge in C carried by each charge carrier
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    so in terms of my question q or e whatever.. will be:
    (5 x 10^28) x (1.6 x 10^-19) ??
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    (Original post by neomilan)
    so in terms of my question q or e whatever.. will be:
    (5 x 10^28) x (1.6 x 10^-19) ??
    :yep:
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    (Original post by EierVonSatan)
    :yep:
    are my I n and V correct though?
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    (Original post by neomilan)
    are my I n and V correct though?
    I, n and v look good
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    (Original post by EierVonSatan)
    I, n and v look good
    ah ok good

    i get 1.875 x 10^-36 m^2 for my A

    hopefully thats correct and then i'll just divide by pi and square root it and then times by 2 to get my diameter
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    (Original post by EierVonSatan)
    I, n and v look good
    One more question: if charge passing a pair of headlamps are 6000 each and charge passing a pair of side lamps are 600 each
    will the charge in passing the battery be 6000+6000+600+600 ?
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    (Original post by neomilan)
    One more question: if charge passing a pair of headlamps are 6000 each and charge passing a pair of side lamps are 600 each
    will the charge in passing the battery be 6000+6000+600+600 ?
    Yeah, the total charge in a system is the sum of all the smaller charges
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    thanks! -- i have another question --> http://www.thestudentroom.co.uk/show...php?p=15342283
 
 
 
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