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Assume a conductor in a rectangle shape for simplicity.

Now, if I only choose one side of this rectangle having length L, and apply external electrical field ∑ only to it(along with the wire), what EMF would I create on the conductor? I would simply say ∑*L, however then I had the following idea, and I started to doubt if I create 2∑*L instead.

Here is what bugs me:

(say ∑L=IR, kirchoffs law: R is the conductors resistance, ∑ is the field we apply) Since nucleus of atoms are almost stable, most current will be due to electron movement, accelerating due to the force of the electrical field. Then electrons will create a current I obviously.

However, there is this topic we covered in semiconductors class in university, that is called hole current. Since electrons move from one atom to other atom, the destination atom is should initially be positively charged to be able to get the atom. When electron completes its movement, destination atom is now neutral, but the source atom is positively charged.

Although only one electron moved physically, there is also a positively charged 'hole' moved in the opposite direction, which doubles the equivalent current, making it 2I. Then it means we had created 2∑*L equivalent voltage on the semiconductor by applying only ∑ electrical field.

Do we have ∑*L or 2∑*L voltage as a result of this experiment? And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

Here is the wiki page for electron hole:

https://en.wikipedia.org/wiki/Electron_hole

In here, its stated that we treat differently to metals and semiconductors, (we ignore holes in metals) :

https://en.wikipedia.org/wiki/Charge_carrier

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It appears to me that you are slightly confused with regards to the concept of current in conductors.

Now, if I only choose one side of this rectangle, and apply external electrical field ∑ only to it, what EMF would I create on the conductor? I would simply say ∑, however then I had the following idea, and I started to doubt if I create 2∑ instead

First, your question is dimensionally invalid as $\Sigma$ cannot have the unit for both electric field and of EMF.

Second, electric field is a vector. When you say you apply a field on one side of the loop, do you mean in a direction along the wire? If so, the EMF generated would initially be ${\Sigma}{l}$ where $l$ Is the length of that edge. (At least for a uniform field). If not you find the EMF using the formula $$\oint \Sigma dl$$ Please note that the electrons in the conductor flow for a very brig period before the potential difference along the wire becomes zero.

And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

This is a far more sensible question. Using a semiconductor instead of a conductor wouldn't change things a lot -The final current will still be zero, but there will be a potential difference across the semiconductor. This depends on what the material of the semiconductor is. In any case, a semiconductor in an external field might act an unbiased p-n junction. Which you can find out about here.

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  • $\begingroup$ Why wouldn't there be a continuous flow in conductors? Since it has a square shape with electric field in only one side (along the wire of course), it should accelerate electrons continuously. If there wasn't a loop, I would also expect conductor to polarize and stop the current flow but this is not the case for loops, or am I mistaken? Sorry for the dimensions, I just updated the question. $\endgroup$ – ozgeneral Sep 18 '15 at 13:08
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EMF is actually the potential difference across the conductor. In your case, since $V = \int \Sigma .dr$, therefore the resulting EMF is $\Sigma$ multiplied by the length of the conductor in direction of applied electric field.

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