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The existance of Electric field

  1. May 31, 2014 #1
    1. The problem statement, all variables and given/known data
    inductance-E_zps4bccb489.png
    B is increasing with rate: dB/dt
    when C is center of circle 1, Induced Electric Field is 0, but when i take another circle, in C, existing Induced Electric Field

    2. Relevant equations
    FaradaysLaww_zpsef8d3c97.png
    I used this equation to prove

    3. The attempt at a solution
    Do Induced Electric Fields exist in C??
     
  2. jcsd
  3. May 31, 2014 #2

    Simon Bridge

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    Please show your working.

    When the B field changes, there should be an accompanying electric field so that $$\nabla\times\vec E = -\frac{\partial \vec B}{\partial t}$$

    If the loops you have drawn are conducting, then either of them should have an induced current.
     
  4. May 31, 2014 #3
    what i mean is: in C, Existing of Induced Electric Fields
    I have: E = (r/2).dB/dt( r is radius of circle 1)=> C doesn't have electric field
    But when E' = (r'/2).dB/dt(r' is radius of circle 2) => C still have electric field( C belongs to circle2)
    I think when B change, every where in space also have electric field, am i right
     
  5. May 31, 2014 #4

    Simon Bridge

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    It is not always the case. The trouble is that your calculations make no sense.

    You are trying to use Faradays Law:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html#c3

    I do not see how you have done your calculations.
    I do not see how you got your conclusions.

    i.e.
    If r>0, and |dB/dt|>0 then E≠0 so why do you say this means that there is no electric field?

    Also - how did you use the fact that both B and E are vectors?

    I think you need to be more careful in your definitions ... do you intend that the B field is defined throughout all space?
    You also need to go through the calculation one step at a time.
     
    Last edited: May 31, 2014
  6. May 31, 2014 #5
    induced-electric-fields_zpsddb3feb7.png
    C have electric field, isn't it?
     
  7. May 31, 2014 #6

    Simon Bridge

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    Oh I think I get you.
    The calculation depends on where you put the center of the circle ...

    But you've forgotten something - you are only calculating the contribution to the induced electric field due to the flux inside the circle.

    In the first example you discovered that the flux at point C has no contribution to the field there.
    In the second you found a non-zero contribution to the flux.
    To get the total E field at C, you need to add the contributions from everywhere.

    I don't think the calculation is actually valid for the B field you have in mind - but I can't tell for sure.
     
  8. May 31, 2014 #7

    TSny

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    If the uniform magnetic field is confined to a circular region, then the induced E will be zero at the center, C, of the region. You should be able to convince yourself that there is no contradiction with applying the integral form of Faraday's law to any closed path. (Note that for a circular path not centered at C, then you will not be able to pull E out of the integral.)

    If the uniform B field extends to infinity in all directions, then there is no unique answer to the induced E field distribution. So, it's not a well-defined problem.

    A similar situation occurs if you imagine all of infinite space filled with a uniform charge density ρ and try to find the E field produced by ρ using Gauss' law. There will be an infinite number of different E field distributions that satisfy Gauss' law.
     
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