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Induced Electric Field

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data

    https://www.writelatex.com/628580dnzrxr#/1303752/

    3. The attempt at a solution

    As I said in the link above I tried using the surface orthogonal to the cross section of the wire of radius to integrate.

    For the line integral I assume it just reduces to E*2(pi)R but for the surface integral I'm not sure what dA becomes, is it r^2 sin\phi dr d\phi ?
     
  2. jcsd
  3. Jan 5, 2014 #2

    tiny-tim

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    hi decerto! :smile:

    (why are you using an image of your latex? :confused:)
    |E| is constant for constant r, so yes you use a circle of radius R, and any surface with that circle as boundary …

    obviously, the disc is the easiest! :wink:
    neither (look at the dimensions) …

    rsinφ dr dφ :wink:
     
  4. Jan 5, 2014 #3
    Thanks for the help

    I used latex because I had already typed it up for /r/askphysics and I didn't know if you could used latex on this site so I just copied and pasted it into an online compiler.


    Should the area element not be rdrd[itex]\phi[/itex] as you are integrating from 0-2pi and the angle associated with the sin in spherical coordinates only goes from 0-pi?

    Is this the easiest way to do this question?, I'm fairly new to EM so my conceptual understanding and my experience of these problems is pretty terrible at the moment.
     
    Last edited: Jan 5, 2014
  5. Jan 5, 2014 #4

    tiny-tim

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    hi decerto! :smile:

    sorry, i got confused earlier when i said rsinφ dr dφ

    for some reason i thought we were on a sphere :redface:

    for a disc, it should be r dr dφ
    yes!
     
  6. Jan 5, 2014 #5
    Ye i just edited to ask that questions, thanks again for your help.
     
    Last edited: Jan 5, 2014
  7. Jan 12, 2014 #6

    tiny-tim

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    hmm, that can't be right :confused:

    ohh, i should have drawn a diagram at the start, i've got it completely wrong …

    that disc surface only proves that E has no θ component! :redface:

    a useful surface needs to be perpendicular to B, ie perpendicular to θ, ie a radial slice

    so try a rectangle bounded by radii 0 and r :smile:

    (since B vanishes at r = 0, i think we can safely assume that E does too)
     
  8. Jan 12, 2014 #7
    I think you were right the first time, I was integrating out r by integrating to the edge of the wire instead of integrating to a random variable r<R.

    I don't think you can evaluate E from E.dr when E is radial and dr is along a rectangle, you need a circle that will be every tangential to the E field.
     
    Last edited: Jan 12, 2014
  9. Jan 13, 2014 #8

    tiny-tim

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    i still think my rectangles work (though on second thoughts using r and r+dr would be easier)

    they should give you a differential equation for Ez, and then you can find Er by using ∇·E = 0
     
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