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I "Biot-Savart equivalent" of Faraday's Law?

  1. Apr 29, 2016 #1
    This is a repeat of this thread https://www.physicsforums.com/threads/biot-savart-version-of-maxwell-faraday-equation.855423/

    That thread was dominated by one verbose poster. I am repeating this thread so as to hear the opinions of others.

    Ampere's Law can be derived from the Biot-Savart Law.

    Faraday's Law is similar to Ampere's Law.

    Is there a "Biot-Savart equivalent" of Faraday's Law?

    I imagine it might look something like this: (not taking into account Coulomb's Law)

    [tex]\frac{d\vec{E}}{dV}=\frac{-\left(\frac{∂\vec{B}}{∂t}\right)×\vec{1_r}}{4\pi r^2}[/tex]
    Last edited: Apr 29, 2016
  2. jcsd
  3. Apr 29, 2016 #2
    Without an equation like the one above, how can we construct the entire E-vector field due to a changing magnetic field?
  4. Apr 30, 2016 #3
    Maxwell's equations include ##∇×B=\mu_0 J## and ##∇×E=-\frac{∂B}{∂t}##

    ##∇×B=\mu_0 J## cannot adequately describe the Biot-Savart Law, the same with ##∇×E=-\frac{∂B}{∂t}##, which is why I've conceived the above.
  5. May 1, 2016 #4


    Staff: Mentor

  6. May 1, 2016 #5
    Yeah, but I don't think that this is a boring question.

    Prof is asleep. No response.
    The wiki doesnt have anything on this.

    I thought of this question because of the parallel between E and M and I also want to construct the entire E-vector field due to a changing magnetic field
  7. May 1, 2016 #6


    Staff: Mentor

    Try finding a prof who isn't asleep maybe an applied mathematics profs or one in EE.
  8. May 2, 2016 #7


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    Well, if the magnetic field is time dependent then also very likely the electric field is, and then you need (generally) the full Maxwell equations. So the Ampere law should be used as the full Ampere-Maxwell Law,
    $$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}.$$
    The Maxwell equations for given charge-current distributions are solved by the retarded propagator (aka Jefimenko equations):

  9. May 2, 2016 #8
    Ooh, thanks for bringing that to my attention. I had conceived Jefimenko's idea independently, now I know of the equations.

    is the formula below (after taking into account Coulomb's Law) consistent with the Jefimenko equations?

    [tex]\frac{d\vec{E}}{dV}=\frac{-\left(\frac{∂\vec{B}}{∂t}\right)×\vec{1_r}}{4\pi r^2}[/tex]
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