Understanding EMF and Charge Separation in a Changing Magnetic Field

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Icy98
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Homework Statement


When there is a changing magnetic flux, emf is induced in the solenoid. The solenoid is made up of circular loops of wire. My first question is, since emd is induced in the solenoid, is there a site of higher voltage and another site of lower voltage? My second question is, (as provided in the picture attached), it is said that at (c), flux does not change hence there is no emf, but there is separation of charges. How is this possible?

Homework Equations


EMF= -N ( delta phi/ delta time)

The Attempt at a Solution


For my second question, when there is separation of charges, isn't there an emf?[/B]
 

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Icy98 said:
My first question is, since emd is induced in the solenoid, is there a site of higher voltage and another site of lower voltage? My second question is, (as provided in the picture attached), it is said that at (c), flux does not change hence there is no emf, but there is separation of charges. How is this possible?
Have a look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html

So when a winding is moved across the constant magnetic field, an emf will be induced in the right half and the left part of the winding, but the directions of the emf's will be opposite, hence the resulting emf ( the sum of emf's ) in the winding as a whole, will be zero, though there is a voltage difference between the top and bottom of the winding.

And to your attached: Determine whether a current is induced . . . .

Well, we don't know, because we don't know if the coil is loaded by e.g. a resistor, thereby creating a closed current loop.

A magnetic flux induces voltage, Ohm's law must take care of the resulting current ( I = V / R ).
 
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Icy98 said:
For my second question, when there is separation of charges, isn't there an emf?
Think of the left and right-hand sides of the coil as two batteries of identical voltage hooked up + to + and - to -. So there is no current since the two voltages cancel each other.
But the + side is still at a different potential than the - side, by the amount of the battery voltage, right?