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I have learned that the $E$-field induced by changing magnetic flux, such as in 'motional emf', is non-conservative in nature. I am also aware that static $E$-fields are conservative in nature.

What is the reason for this difference in the nature of the $E$-fields?

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  • $\begingroup$ because the induced field lines go around in circles. $\endgroup$
    – R. Emery
    Commented Dec 23, 2020 at 12:20

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A force feild is called conservative when it can be expressed in terms of a potential energy (or potential), as $$ \mathbf{F}(\mathbf{x}) = -\nabla U(\mathbf{x}). $$

Helmholtz decomposition means that any field can be expressed as a sum of a potential and a solenoidal components, i.e., as $$ \mathbf{E}(\mathbf{x}) = -\nabla \phi(\mathbf{x}) + \nabla\times\mathbf{A}(\mathbf{x}). $$

Note that $\nabla\cdot(\nabla\times\mathbf{A})\equiv 0$, that is $\nabla\cdot\mathbf{E}=\nabla^2\phi$. Since the Maxwell equation for the electrostatic field is
$$ \nabla\cdot \mathbf{E} = 0, $$ it is fully described by the scalar potential - adding a solenoidal component would not change anything!

On the other hand, since $\nabla\times\nabla\phi\equiv 0$, we have $\nabla\times\mathbf{E}=\nabla\times(\nabla\times\mathbf{A})$. For the magnetic field this means that this field is fully described by a vector potential, since we have a Maxwell equation $$\nabla\times\mathbf{B}=0.$$ For the electric field it means that its solenoidal component is fully determined by the derivative of the magnetic field, since $$ \nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}. $$ In other words the solenoidal (i.e., non-conservative) component of the electric field is solely determined by the magnetic field changing in time.

To summarize, the distinction between the conservative and non-conservative components of the electric field is due to the form of the Maxwell equations, which are the experimentally determined laws of nature.

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  • $\begingroup$ I think I asked a question that's a little out of my pay grade of understanding, as I don't understand parts of your answer. Is there any simpler way to explain it? any analogy that you can think of to simplify this? $\endgroup$ Commented Dec 5, 2020 at 7:32
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The reason static field is conservative is this: static field means, by definition, total Coulomb field of any static configuration of charges. We know Coulomb field is conservative (because of Coulomb's law).

If tommorow there was a non-conservative field discovered that is due to and associated with static charges, that would be a big change, as we could extract energy from it without moving any of the static charges. But nobody has discovered yet such non-conservative field of static charges, and we believe it doesn't exist. It would be a so-called "free lunch" - free energy extraction with no change in configuration of the charged system. In reality, extraction of energy leaves systems changed.

If charges move with acceleration, non-conservative field (circulation integral is not zero) may be present and this can be verified, for example, if magnetic field inside a solenoid changes. Then the induced electric field is observed to be present and acting on current in the coil, it is non-conservative, it goes in circles. Now we can extract energy from that as well, but there is no free lunch this time - the charges are moving, hence work can be done on/extracted from them. Extracting energy from non-conservative electric field is associated with forces on moving charges and thus configuration of the charges in space is affected adequately to the amount of energy extracted. This prevents the free lunch because eventually, the source of the energy is drained.

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I don't know if I'm correct or not but what I understand is that the name 'electric field' dose not really matter. It is just observed that there is some non conservative force field that gets produced when there is varying magnetic field. As for why it is called so is because it has a lot of similarities with the usual electric field.

Do tell what you think about my opinion.

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