The EMF induced in straight current-carrying conductor

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Why is an emf induced in straight current carrying conductor as it moves at at right angles to a uniform and constant magnetic filed. By Faraday's law, this e.m.f. is equal to dΦ/dt but I do not understand how a wire cutting a uniform magnetic field experiences a change in magnetic flux . Its area is constant and magnetic flux density is constant so the magnetic flux felt by the wire Φ=BA is constant.

Wire-cutting-a-magnetic-field.png

Wire(green) is moving down at right angles to the field
 
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  • #2
BvU
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Google "Lorentz force"

And indicate the direction of motion of the wire in the drawing
 
  • #3
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Faraday's law can be written as:
emf = d/dt BAN
Assume B is constant and N =1 we can write:
emf = B dA/dt
In this example dA/dt is the rate of change of area in other words the area sliced out by the wire in one second. If the wire has length l and is moving up or down with velocity v the area sliced out per second will be lv. Make a sketch and hopefully you will see it.
 
  • #4
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Faraday's law can be written as:
emf = d/dt BAN
Assume B is constant and N =1 we can write:
emf = B dA/dt
In this example dA/dt is the rate of change of area in other words the area sliced out by the wire in one second. If the wire has length l and is moving up or down with velocity v the area sliced out per second will be lv. Make a sketch and hopefully you will see it.
but this doesn't work for a coil, if i replaced the straight wire with a coil already in the magnetic field and moved it the same way with no rotation no emf would be induced. why is it different for a coil and a straight current carrying conductor
 
  • #5
Merlin3189
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Perhaps because the coil has two sides and the emf in each side opposes when they both move in the same direction.
induction.png
 
  • #6
Dale
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By Faraday's law, this e.m.f. is equal to dΦ/dt but I do not understand how a wire cutting a uniform magnetic field experiences a change in magnetic flux . Its area is constant and magnetic flux density is constant so the magnetic flux felt by the wire Φ=BA is constant.
Well, there isn’t really a loop here, so Faraday’s law doesn’t really make a difference in this case. If you make the “loop” with zero area by going backwards and forwards along the same line then by Faraday’s law you get 0 EMF and any voltage distribution along the wire is consistent with 0 EMF since you will go forwards and backwards across the same distribution. So basically, you need to look for a different law.

The law to look at is the Lorentz force law.


if i replaced the straight wire with a coil already in the magnetic field and moved it the same way with no rotation no emf would be induced.
That is correct. No EMF is induced in such a coil moving through a uniform magnetic field.
 
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For a closed loop moving at steady velocity within a uniform B field there will be no internal current but, as predicted by the Lorentz law, there will be a build up of electrons, most concentrated at one end of the loop, and a shortage of electrons, most concentrated at the other end. Hence there will be an emf. A current can be drawn by adding external circuitry eg a piece of wire connected to the ends of the loop.
 
  • #8
Dale
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Hence there will be an emf.
Not around the whole loop. The EMF on one side cancels out the EMF on the other for a total of 0 EMF around the loop.
 
  • #9
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Well, there isn’t really a loop here, so Faraday’s law doesn’t really make a difference in this case. If you make the “loop” with zero area by going backwards and forwards along the same line then by Faraday’s law you get 0 EMF and any voltage distribution along the wire is consistent with 0 EMF since you will go forwards and backwards across the same distribution. So basically, you need to look for a different law.

The law to look at is the Lorentz force law.


That is correct. No EMF is induced in such a coil moving through a uniform magnetic field.
Thank you I think I understand, just to clarify I can make a "loop" while moving the wire forward and backward( to make the loop) as I move it up and down between the magnets, it will produce no emf because the direction of the current will be the same as wire's movement whether forward or backward.
 
  • #10
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For a closed loop moving at steady velocity within a uniform B field there will be no internal current but, as predicted by the Lorentz law, there will be a build up of electrons, most concentrated at one end of the loop, and a shortage of electrons, most concentrated at the other end. Hence there will be an emf. A current can be drawn by adding external circuitry eg a piece of wire connected to the ends of the loop.
Thanks
 
  • #11
Dale
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I can make a "loop" while moving the wire forward and backward( to make the loop)
Not exactly. In this context a "loop" is a closed path at a given moment in time. So if you have to spend time to make a closed path then it isn't a "loop" in the meaning of this law.

There are other contexts where the sort of loop you are describing is perfectly acceptable, such as describing the work done on a point particle moving in a conservative force field, but in Maxwell's equations the surface and loop integrals are at a given instant of time. Those loops can change over time, either by stretching or moving, but at every moment they are a complete closed path.
 

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