I've been studying Faraday's law.

In one example they gave me a circuit and they connected voltemeter with two points $A$ and $B$ where there is Resistor in the right side $\frac{R}{2}$ between the points .

And another resistor in the left side $R$.

the circuit has Coil that is making constant $\epsilon$ , it is siting in the middle of the circuit.

Now when we connect the Voltemeter (Closed Path) according to Kirchhoff's circuit laws and the COIL IS NOT INSIDE the closed path $ V + I\frac{R}{2} = 0$.

This is simple since the potential Difference between a point to itself is $0$.

BUT when they connected the Voltemeter in different way ( closed path too) and now in the closed path there is the Coil they wrote that $\nabla~ \times~ \vec E \not= 0$ and because of that $ V + I\frac{R}{2} = \epsilon$ .

I didn't understand that at all why puting a coil in the middle change the voltage between the point to itself ?!!

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    $\begingroup$ Welcome here. Formulating the question is not an easy task. I think you should put a diagram with resistor R and coil C, explain what you have, what your letters are ($\epsilon$) and then formulate the question. $\endgroup$ – jaromrax Jan 14 at 9:44

I'm assuming the current in the coil is changing in time.

In the second case with the source of electromotive force (coil) near the measurement circuit, the voltmeter is unable to measure voltage (which is defined as determined by electrostatic field only) properly, because the voltmeter is influenced also by induced electric field of the coil (which does not contribute to voltage). The voltmeter then shows total emf for the path covered by its probes and wires (which is due to both electrostatic and induced electric field).


Sorry, but the question is quite unclear without a picture. Please, edit the question and include it. IT can even be a photo, or a bitmap made with paint, altought the clearest you do it, the better.

That said, if I have understood you... your problem is that you think closed loops imply $\Delta V=0$, and that's not true.

You might be surprised, how comes that the difference between one point and ITSELF is not 0? The answer is simple: you are not measuring the difference between one point and the same point; there is a subtle difference. What we measure is from point $P$, and the same point $P$, but after completing one lap!

That's the difference. You are a charge Q. You go hiking through the mountain and then you go back to the starting point. The "net distance" you ran is 0, yes, but you are so more tired.

That is because of the coil. You said it yourself: $\vec{\nabla}\times\vec{E}$ is no longer $\vec{0}$. The electric field ceases to be conservative. So, the work done from taking a charge from one point to itself now depends on the path, and it is not neccesarily 0 in a closed loop. In fact, it is the work done along the path, which depends on $\epsilon$. You can regard potential as work per unit charge, so that's the thing.

Was that the question? If not, please, add a picture and set it clearer. Thanks.


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