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Calculate electric field applied along a wire

  1. Dec 6, 2012 #1
    Hi, i'm doing some revision and i'm having a problem with the following:

    Conductivity of gold is 4.1*10^-8 , what is the electric field along the wire? The current is 13.2A and the radius is 3.8mm

    I've used 1/conductivity to find resistivity of (2.4*10^-8) and used this along with area to find the resistant for 1m

    (2.4*10^-8) *1m/pi*(0.0038)^2

    This gives 5.3*10^-4 which when multiplied by the current gives 0.007 V/m, however i'm told this is the wrong answer, can anyone see where I might have gone wrong?

    Thanks
     
  2. jcsd
  3. Dec 6, 2012 #2

    gneill

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    Staff: Mentor

    I think you've shown an incorrect value for the exponent for the conductivity of gold; probably a typo, since the value for resistivity looks okay.

    Your steps to find the electric potential across a meter length is okay. What's probably "doing you in" on your result is the significant figures. All the values you've worked with in the problem have at least 2 significant figures, and the result you've presented has only one.
     
  4. Dec 6, 2012 #3

    rude man

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    Homework Helper
    Gold Member

    Use j = σE
    j = current density, amp/sq. m
    σ = conductivity, S/m
    E = electric field, V/m

    Vectors in bold.
     
  5. Dec 6, 2012 #4
    The previous poster was right, I did have conductivity of gold wrong, it is actually 4.1*10^7

    So using j=σE

    I found j to be 2.91*10^5 and dividing this by σ 4.1*10^7 my answer is 7.1*10^-3 V/m to 3sf, does this look correct?

    Thanks
     
  6. Dec 6, 2012 #5

    gneill

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    Staff: Mentor

    Yes, that looks okay (but you've specified two sig figs, not three, which is fine for the given values).

    Your first method would give you the same result (and mathematically amounts to the same thing, but "takes the scenic route").
     
  7. Dec 6, 2012 #6

    rude man

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    j is right, the rest I leave to you & gneill. I would suggest including units for each number. It's a powerful checking tool and really makes no sense without them.
     
  8. Dec 7, 2012 #7
    Thanks for all the help guys :)
     
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