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    Hi guys, I have a physics question here that I cant seem to work out.

    The current in a tungsten wire is 1.20 A, and the drift velocity of electrons is 2.35mm s-1.
    The number of free electrons per m^3 in tungsten is 6.3 x 10^28 m-3.
    What is the diameter in mm?
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    Using the equation: I=nAvq

    I=1.20A
    n=6.3x10^28
    v=2.35x10^-3 (convert to m, then convert back after the working out)
    q= 1.6x10^-19 (q is the charge carried by each charge carrier - therefore the charge of each electron)
    A=?

    rearrange to give I/nvq = A

    1.20/(6.38x10^28) x (2.35x10^-3) x (1.6x10^-19) = 5.06585613x10^-8

    Area = pi x radius^2, so divide A by pi, find the square root of it and that gives you the radius. Then, multiply by 2 to find the diameter.

    d= 2.5 x 10^-4m >> divide by 10^-3 to convert to mm, so d= 0.25mm
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    (Original post by Gus_C)
    Using the equation: I=nAvq

    I=1.20A
    n=6.3x10^28
    v=2.35x10^-3 (convert to m, then convert back after the working out)
    q= 1.6x10^-19 (q is the charge carried by each charge carrier - therefore the charge of each electron)
    A=?

    rearrange to give I/nvq = A

    1.20/(6.38x10^28) x (2.35x10^-3) x (1.6x10^-19) = 5.06585613x10^-8

    Area = pi x radius^2, so divide A by pi, find the square root of it and that gives you the radius. Then, multiply by 2 to find the diameter.

    d= 2.5 x 10^-4m >> divide by 10^-3 to convert to mm, so d= 0.25mm
    Cheers mate
 
 
 
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