Possible duplicate of Why is the electric field created by a battery non-conservative?

Note that for any physical voltmeter, there must be a small current through (a typical hand-held voltmeter has in input impedance of roughly 10 mega-ohms) to read a voltage across. Do you see a closed path for a current through when you connect one probe to the battery and the other probe to your hand?

@Armadillomon: The field generated by a battery isn't conservative to begin with; that's why it is known as "emf" and not "potential difference".

@user7777777, why do you say the field generated by a battery isn't conservative? Are you referring to the electric field due to the battery? Consider an isolated battery - is the electric field due to this battery non-conservative? If so, according to $\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}$, there should be an associated time dependent magnetic field shouldn't there?

@user7777777, I asked the question after I checked the link.

@user7777777, you're avoiding my question. When you write "the field generated by a battery isn't conservative", do you mean the electric field and if so, where is the associated timed dependent magnetic field?

@Alfred Centauri: Yes, I am referring to the electric field. Your reference to Maxwell's equations again shows that you have not read the post that I linked (and its comments). Literally, it says "Why is the electric field created by a battery non-conservative?". Let me quote: *aren't [Maxwell's equations] universally valid, compatible with both special relativity and quantum mechanics?* No, they're not. Maxwell's equations are classical. For example, you're not going to be able to explain the photoelectric effect using Maxwell's equations. straight from the comments.

@user7777777, does this quote answer my question?

@Alfred Centauri: At this point, I have completely no idea what you are trying to ask/doubt. The entire answer is in the post I linked, and you still haven't read it properly. In one sentence, the answer is "Maxwell's equations don't apply to batteries.".

@user7777777, I'm reading the post again and Ben Crowell's answer clearly doesn't imply the electric field of the battery is non-conservative: "So the short answer is that the F inside the battery contains a term from an effective chemical force, and this force is not the same as the electrical force. In fact, it's in the opposite direction." - You can also see this at the Wikipedia article here which shows that emf has terms from effective chemical and thermal forces.

@Alfred Centauri: You are correct about the force being non-conservative; that's why it is called "emf". However, the charges are still affected by this non-conservative force, and by definition, $\mathbf{F} = q \mathbf{E}$, so this force is still part of the electric field. The net electric field is not conservative as it is the sum of a conservative component and the emf, $IR = V + \mathcal{E}$. By your argument, if the field were conservative, charge will not be able to flow along a closed circuit at all.

@user7777777, I haven't made an argument. I've simply asked you where the time dependent magnetic field is that is associated with a non-conservative electric field. If a battery in isolation (no current through) produces a non-conservative electric field, then $\nabla\times\mathbf{E}_{BAT} \ne 0$ and thus $-\frac{\partial\mathbf{B}}{\partial t} \ne 0$. As far as I can tell, you're argument is that this isn't true.