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    I'm not sure of the method in answering this question:

    Q. A piece of element x is 2.0mm wide and 0.25mm thick. The number of conduction electrons per (m^3) is (6.0 x 10^20). What is the average drift speed of the electrons when a current of 1.5 mA flows ?

    Here is what I have so far:

    Rearranging I=nAve gives v = I/nae and the cross section area is 3.14x10^-6

    This gives the drift velocity at 4.98 ms-1. Is this correct or not ?

    Thanks
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    The cross section is rectangular not circular.
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    No. If the element is 2.0mm wide and 0.25mm thick, then A = 5 x 10-7m2.

    I got 31.2ms-1.
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    (Original post by teachercol)
    The cross section is rectangular not circular.
    ^^i doubt it

    youre rearranging is correct, but your answer is wrong

    If you meant that the element is 0.25mm thick and 2.0mm long then the area in m^2 is pi x (0.00025)^2

    e= 1.6 x 10^-19 (i think)

    sub in these values and v will be in m/s (and then you can convert it into whatever you want)

    [i got an answer of 79.6m/s (3sf)]
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    The question clearly states wide and not long. I did the question again and got 31.25 ms-1 as Morbo got. So I guess it depends on whether it is circular or rectangular.

    Anyway, thanks for all your help
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    (Original post by teachercol)
    The cross section is rectangular not circular.

    (Original post by Adhavan)
    ^^i doubt it
    Hall probe elements are normally rectangular slices of silicon - which looks like the case here.

    Area = width x thickness.

    Trust me ...
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    (Original post by teachercol)
    Hall probe elements are normally rectangular slices of silicon - which looks like the case here.

    Area = width x thickness.

    Trust me ...
    Fair enough mate

    pretty much every average drift velocity question ive ever done has involved a cylindrical wire of some kind (which is why i was sceptical to the idea of a rectangular element)
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    I've put my final answer as 31.25 ms-1. I'll ask my teacher to see if this is right tomorrow. Thanks once again for all your help.
 
 
 
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