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Electric fields originate from charges. But according to faraday's law $$\nabla \times E = - \dfrac{\partial B}{\partial t}$$ changing magnetic fields creates electric fields. How do these electric fields form a loop, doesn't the first law state that electric fields do not form closed loops? Where does this induced electric field originate and end? (and don't eddy currents flow in circles?)

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  • $\begingroup$ @ggcg So are electric fields nom conservative? $\endgroup$ Commented Feb 28, 2020 at 14:53
  • $\begingroup$ @NickD. This statement is not true. If there are no charges around there doesn't need to be an Electric field around. This is one possibility. $\endgroup$
    – user196418
    Commented Feb 28, 2020 at 14:56
  • $\begingroup$ @AravindhVasu, why would they be non conservative just because they do not have a source? Magnetic fields never terminate and motion in their presence is conservative, and well as the field energy. Can you connect your questions to some reasoning? In the case of induced E and B the energy is transferred between the 2 fields. In the case of a given B(t) the energy is provided by whatever generator is controlling the B(t) field. $\endgroup$
    – user196418
    Commented Feb 28, 2020 at 14:58
  • $\begingroup$ @ggcg Do Kirchoff's law hold when there's a changing electric field in our circuit? youtu.be/LzT_YZ0xCFY $\endgroup$ Commented Feb 28, 2020 at 15:01
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    $\begingroup$ @ggcg, You should be sure you understand the meaning of the term conservative field? $\endgroup$
    – The Photon
    Commented Feb 28, 2020 at 16:22

2 Answers 2

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In a footnote in page 305 of Introduction to Electrodynamics (3rd ed.) by Griffiths, the author writes the following:

You could, I suppose, introduce an entirely new word to denote the field generated by changing $\mathbf{B}$. Electrodynamics would then involve three fields:

  1. $\mathbf{E}$ fields, produced by electric charges, satisfying $\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$ and $\nabla \times \mathbf{E} = 0$ .
  2. $\mathbf{B}$ fields, produced by electric currents, satisfying $\nabla \cdot \mathbf{B} = 0$ and $\nabla \times \mathbf{B} = \mu_0\mathbf{J}$.
  3. $\mathbf{G}$ fields, produced by changing magnetic fields, satisfying $\nabla \cdot \mathbf{G} = 0$ and $\nabla \times \mathbf{G} = -\frac{\partial \mathbf{B}}{\partial t}$.

Because $\mathbf{E}$ and $\mathbf{G}$ exert forces in the same way [$\mathbf{F} = q(\mathbf{E} + \mathbf{G})$], it is tidier to regard their sum as a single entity and call the whole thing "the electric field."

  • By "first law", I assume you mean Gauss' law which states the following.
    $$\nabla \cdot \mathbf{E}= \frac{\rho}{\epsilon_0}$$ This equation doesn't hold for the $\mathbf{G}$ fields (borrowing the author's usage of the word) as is evident from the third point of the footnote: $\nabla \cdot \mathbf{G}=0$. It is the $\mathbf{E}$ fields that don't form closed loops, since $\nabla \times \mathbf{E}=0$.
  • Yes, $\mathbf{G}$ is not a conservative vector field in the presence of time varying magnetic fields. This can be seen from the following.
    $$\oint_C \mathbf{G}\cdot d\mathbf{l}=-\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$ Furthermore, you may want to look at this answer that discusses when the Lorentz force field is conservative.
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What is the so called "first law" you are referring to?

As the equation suggests the "source" or origin of the induced field is a change in magnetic field, or magnetic flux. These E fields do not terminate on point sources, they circulate like B fields closing in on themselves.

If you are thinking about Coulomb's law, or the differential equation that states

div(E) is proportional to charge density,

this is an electrostatics law (statics meaning all charges are at rest, held at fixed positions, and there is no time dependence). The electrostatic field line terminate at source points, charges. The case you are asking about involves time dependent changes so the behavior of static field lines does not apply here.

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  • $\begingroup$ So are electric fields nom conservative? $\endgroup$ Commented Feb 28, 2020 at 14:57
  • $\begingroup$ Please see my comment to your question. I am having a hard time understanding how you are drawing that conclusion. $\endgroup$
    – user196418
    Commented Feb 28, 2020 at 14:59
  • $\begingroup$ @AravindhVasu, correct. Electrostatic fields are conservative. In general, the electric field is not always conservative. $\endgroup$
    – The Photon
    Commented Feb 28, 2020 at 16:13
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    $\begingroup$ $\nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$ is not restricted to electrostatics. It is a general rule (one of Maxwell's equations) in classical EM. $\endgroup$
    – The Photon
    Commented Feb 28, 2020 at 16:17
  • $\begingroup$ @ThePhoton How can electric field both diverge into a sink and also form a loop? $\endgroup$
    – user240696
    Commented Feb 28, 2020 at 16:20

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