0

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?)

Aravindh Vasu
  • 455
  • @ggcg So are electric fields nom conservative? – Aravindh Vasu Feb 28 '20 at 14:53
  • @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. –  Feb 28 '20 at 14:56
  • @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. –  Feb 28 '20 at 14:58
  • @ggcg Do Kirchoff's law hold when there's a changing electric field in our circuit? https://youtu.be/LzT_YZ0xCFY – Aravindh Vasu Feb 28 '20 at 15:01
  • 1
    @ggcg, You should be sure you understand the meaning of the term conservative field? – The Photon Feb 28 '20 at 16:22
  • @ThePhoton, I do know what a conservative vector field is. In the context of the question energy is conserved. –  Feb 28 '20 at 17:05
  • @ggcg, then you already know that the answer to your question, "why would they be non conservative just because they do not have a source?" is that the field is not non-conservative because it doesn't have a source (and in fact there is a source, it's just that the source is the magnetic field, not a charge). It is non-conservative because it forms loops. – The Photon Feb 28 '20 at 18:20

2 Answers2

2

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.
Ajay Mohan
  • 1,411
0

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.

  • So are electric fields nom conservative? – Aravindh Vasu Feb 28 '20 at 14:57
  • Please see my comment to your question. I am having a hard time understanding how you are drawing that conclusion. –  Feb 28 '20 at 14:59
  • @AravindhVasu, correct. Electrostatic fields are conservative. In general, the electric field is not always conservative. – The Photon Feb 28 '20 at 16:13
  • 1
    $\nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$ is not restricted to electrostatics. It is a general rule (one of Maxwell's equations) in classical EM. – The Photon Feb 28 '20 at 16:17
  • @ThePhoton How can electric field both diverge into a sink and also form a loop? –  Feb 28 '20 at 16:20
  • @Knight, there can be a superposition of fields, one forming a loop (sourced from a changing magnetic field) and one terminating in a charge (sourced by the charge). Any time an EM wave passes through a region with charge present, you'd have this scenario. – The Photon Feb 28 '20 at 16:23
  • @ThePhoton And that loop forming field have the same characteristic as our normal electrostatic field? –  Feb 28 '20 at 16:25
  • @Knight No, they're not electrostatic fields because electrostatics is by definition restricted to the study of the fields produced by charges at rest. (And because creating a scenario where $\frac{d{\bf B}}{dt}$ is constant is very difficult) – The Photon Feb 28 '20 at 16:27
  • @ThePhoton I asked about the characteristic, I meant do they have same effect on charged particles as our electrostatic field have ? –  Feb 28 '20 at 16:29
  • @Knight, yes, see Ajay's posted answer. If they didn't your radio receiver antenna wouldn't work very well. They have the same characteristic in terms of forces on charge. They have different characteristics in terms of forming loops or the curl of the vector field. – The Photon Feb 28 '20 at 16:30
  • @ThePhoton Thank you so much, I found this comment of yours very helpful as a supplement to that footnote of Griffiths. –  Feb 28 '20 at 16:31
  • @ThePhoton, you are correct, I should have included curl(E) = 0 in the answer. –  Feb 28 '20 at 17:01