Hi,
I am stuck in unerstanding "motional EMF". Please see the image.

I want to know what will happen after a specific time(large enough). I think after the time all charges are pulled upward the bar and there will be no charge in the region between the bar?
And what will happen if after Vab = constant I suddenly stop pushing the bar?
Will all charge will move downward as initial and finally Vab is reduced to 0V?

 

I want to know what will happen after a specific time(large enough). I think after the time all charges are pulled upward the bar and there will be no charge in the region between the bar?

Why do you think that?

The article is trying to say that the motion of the bar through the magnetic field causes the some separation of charge.

However as soon as you separate charge there exists a force trying to bring them back together.

Charge separation will occur until the (electrostatic) restoring force is equal to the separating force (due to the magnetic field).
Then equilibrium occurs.

At this time no more charge will be separated. ie only enough will be separated to generate a balancing equilibrium restoring force. So the interior of the bar will not be devoid of charge as only so much wuill have been separated. The process does not continue indefinitely, unless charge is removed from, the ends.

The article offers an equilibrium calculation balancing the force separating the charge against the force bringing them back together.

And what will happen if after Vab = constant I suddenly stop pushing the bar?
Will all charge will move downward as initial and finally Vab is reduced to 0V?

If the bar stops moving the force due to the magnetic field will cease, but the electrostatic restoring force will remain. So the E force will cause the charges to return to their original positions. As the charges return the E force diminishes until everything is back to equilibrium and the e force is zero.

 

Thanks!
I want to understand this part before going to the next. In this case the bar is not connected to anything and it doesn't create a closed path. Therefore, when charges are seperated to ends they can't move to anywhere and ultimately a EMF is formed but how about the case bellow:
Now suppose the conducting bar moves through a region of uniform magnetic field (pointing into the page) by sliding along two frictionless conducting rails that
are at a distance w apart and connected together by a resistor with resistance R, as
shown in Figure 10.2.2.
\(\vec{B} = -B\vec{k}\\)

In this case a posetive charge q will be pulled from lower end to upper end but because now there is a closed path, these charges will not be piled up at upper end but move in the path through resistor R and finally go back to lower end.
I think the EMF in this case is formed because the speed/rate of separating charges of separating force is larger than the speed of charges go from upper end through resistor R and go back to lower end?
And as the result there will be some charges stocked at the upper end and EMF is formed?
But how about the case resistor R = 0Ω? In this case how EMF is formed?
Is every charge pulled to upper end will go back to lower end immediately and no EMF is formed?
Finally, could you explain why in the image above, in electromotive force calculation why there is only moving bar contribute to the intergal?

 

Quick reply as it's breakfast time here.

Is the instantaneous Area, A not ydx, that is the product of the length of the bar and a differential (infinitesimal) length δx?
And δx = vδt

 

Quick reply as it's breakfast time here.

Is the instantaneous Area, A not ydx, that is the product of the length of the bar and a differential (infinitesimal) length δx?
And δx = vδt

Yes! But how it relates to my questions?

 

Sorry I've been away all day.

There are three different approaches to explaining your sliding wire experiment. It depends upon what your course has treated as fundamental and what as derived.

If you PM me an email address capable of receiving scans I will send a copy, detailing all three, that should answer all your questions. It is too much to post here.