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Homework Help: I = Avnq Where am i going wrong?

  1. Oct 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the drift velocity in a copper fuse wire with a diameter of 0.22mm. The current is at 5A.

    Electronic charge is always 1.6*10-19
    Number of de-localised electrons per cubic meter in copper = 1*10-19

    2. Relevant equations
    Current = cross-sectional area*drift velocity*number of de-localised electrons*Electroic Charge

    3. The attempt at a solution
    First i converted the diameter of 0.22mm into radius 0.11mm. Then using pi*r2 i have 3.8mm2. Then i converted this into SI to make it 3.8*10-3m2.

    So at this point here is the infomation i have.
    v= ?
    n= 1*1029

    I then re-arranged the formula of I=Avnq to make v the subject.
    v= __I__
    Then putting in all the numbers i have...

    v= ______5______
    3.8*10-3m2 * 1*1029 * 1.6*10-19

    Tapping this into my calculator got me the answer of 8.2*10-8ms-1

    However this is crazy slow, i know drift velocity is very slow but not this slow. Looking at the answer in the back of the book, for somereason they put it in mm not m, 8.2mms-1. So this in SI should be 8.2*10-3ms-1. This means at some point in my calculations i am 1*105 wrong.

    Me and and friends doing this work have been struggling to find out why we are not getting the correct answer. I hope i made sence explaining this to you. We think our errors must be around finding the cross-sectional area, or we had the wrong infomation to begin, such as the number of de-localised atoms in a cubic meter of copper. Or just some mess up of the numbers along the way. Thanks to the people who can help us out here.
  2. jcsd
  3. Oct 18, 2008 #2


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    Homework Helper

    Hi SMBeresford,

    This number does not look right (but I see you are using a different number below).

    I don't think this is correct. If you are multiplying pi times (0.11mm)2, you don't get 3.8mm2.

    There is a problem with the conversion here. Remember that:

    1 \mbox{mm}^2 \neq 0.001 \mbox{m}^2

    because the mm are squared. What would 1 mm2 be if you converted to m2?
  4. Oct 18, 2008 #3
    Ah yes okay, so pi(0.11mm)2 is actually 0.038mm2

    Looking back at my original answer I was quite close really, i had the right numbers but i had the wrong powers of 10. I'm expecting to get 8.2*10-3m s-1 but i was actually getting the same but *10-8...

    I see your very good point about how mm2 does not go into m2 by just dividing by 100. Seems like an obvious mistake :P

    Sooo, back to your question of what 1mm2 is in m2. It still should have 1 as the diget, but a different power of ten. I've tried alot of different things but im still not sure... what does 1mm2 convert to in m? my best guess so far has been 0.000001m2 but that still seems wrong and makes things worse when i did the same method in the question...

    oh an also hello to the physics forum
  5. Oct 20, 2008 #4


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    Homework Helper

    Sorry, I did not see your reply until now with the server move.

    That's the correct number. 1 mm2=10-6m2. When I used that it gave me the correct answer to your problem.

    The way to think about it is that to convert mm to m, you divide by 1000. Now 1 mm2 is 1 mm*mm, so you have to convert two factors of mm, so you have to divide by 1000 twice.
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