EM: induction in a moving conductor

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Homework Statement



An infinitely long wire contains a current I. A metal rod with length L moves with speed v. Determine the emf in the rod.

[PLAIN]http://img227.imageshack.us/img227/5149/52966616.jpg [Broken]

With [tex]\epsilon = -\frac{d\Phi_b}{dt}[/tex], I would have said emf = 0 at first. The current I generates a magnetic field out of the page which varies with the perpendicular distance from it. The magnetic flux through the rod would then be the integral of this field over d to d + L times the area of the rod. I don't get how that can generate an emf, since according to the above equation it needs a changing magnetic flux. But the flux doesn't change when the rod moves parallel to I, since B is the same on lines parellel to I. It only changes with the perpendicular distance from it, so the only way it would generate emf if is the conductor would move vertically in this picture. That's not correct apparently, and the best explanation I could find was that there is a force on the free electrons in the conductor due to the magnetic field (since the charges now have a velocity). That seems reasonable, but how does that fit into Faraday's law? How is there a changing magnetic flux?

Also, if the conductor were rotated by 90 degrees (parallel to I), there wouldn't be an emf. Is this because the force that is being generated is now pushing charges to the bottom or top of the conductor and not along the length?

In the following case they say the emf generated in the right perpendicular rod is opposite to that in the left perpendicular rod.

[PLAIN]http://img828.imageshack.us/img828/534/33791951.jpg [Broken]

I can somewhat imagine that if you were to only regard the first rod, there would be a current directed upwards due to this force, and since all four rods are connected it would go downards in the left rod. But there is also a force on its free electrons, just like on the right rod. How do you explain a downward current with an upward force on the left rod? And again, how is there a changing magnetic flux? B does not change, A does not change and the angle between B and a vector A perpendicular to the surface does not change either.

Having troubles understanding this because my textbook has no mention of this type of induction. It only talks about closed circuits moving out or into magnetic field and then I see ow the magnetic flux changes, but not in this case.

edit: actually, thinking it over again, is the change in flux due to a magnetic field arising from the induced current in the conductor? Then I think I understand the first part - but still not sure about the second, i.e. how there is an upward force on the electrons of the left rod but yet a downwards current. Or does this only occur at the beginning when the system starts moving, and somehow the steady state is an anti-clockwise current?
 
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  • #2
vela
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Homework Statement



An infinitely long wire contains a current I. A metal rod with length L moves with speed v. Determine the emf in the rod.

[PLAIN]http://img227.imageshack.us/img227/5149/52966616.jpg [Broken]

With [tex]\epsilon = -\frac{d\Phi_b}{dt}[/tex], I would have said emf = 0 at first. The current I generates a magnetic field out of the page which varies with the perpendicular distance from it. The magnetic flux through the rod would then be the integral of this field over d to d + L times the area of the rod. I don't get how that can generate an emf, since according to the above equation it needs a changing magnetic flux. But the flux doesn't change when the rod moves parallel to I, since B is the same on lines parellel to I. It only changes with the perpendicular distance from it, so the only way it would generate emf if is the conductor would move vertically in this picture. That's not correct apparently, and the best explanation I could find was that there is a force on the free electrons in the conductor due to the magnetic field (since the charges now have a velocity). That seems reasonable, but how does that fit into Faraday's law? How is there a changing magnetic flux?
To calculate the flux, you need an area. (In this problem, the rod is approximated to be a one-dimensional object. It has no width, so it has no area, just a length.) It turns out the area you need for the calculation is the area swept out by the rod. Since the rod is moving, this area increases with time. It's this increase in area which results in the changing flux.
Also, if the conductor were rotated by 90 degrees (parallel to I), there wouldn't be an emf. Is this because the force that is being generated is now pushing charges to the bottom or top of the conductor and not along the length?
Right.
In the following case they say the emf generated in the right perpendicular rod is opposite to that in the left perpendicular rod.

[PLAIN]http://img828.imageshack.us/img828/534/33791951.jpg [Broken]

I can somewhat imagine that if you were to only regard the first rod, there would be a current directed upwards due to this force, and since all four rods are connected it would go downards in the left rod. But there is also a force on its free electrons, just like on the right rod. How do you explain a downward current with an upward force on the left rod? And again, how is there a changing magnetic flux? B does not change, A does not change and the angle between B and a vector A perpendicular to the surface does not change either.

Having troubles understanding this because my textbook has no mention of this type of induction. It only talks about closed circuits moving out or into magnetic field and then I see ow the magnetic flux changes, but not in this case.
If you look at it from the force point of view, you can see that electrons in both rods are pushed in the same direction by the magnetic force. The separation in charge creates a net electric field which results in an electric force on the charges. Eventually, the electric force and magnetic forces balance, and the charges stop moving, i.e. no current.

From the flux point of view, this is what you'd expect because, as you pointed out, the flux through the closed loop isn't changing, so no emf around the loop, so no current flow.

You can also say there's the same emf induced in each rod, but because they point in opposite directions as you go around the loop, they cancel each other out (like having two batteries pointing in opposite directions in a circuit) and no current flows.
edit: actually, thinking it over again, is the change in flux due to a magnetic field arising from the induced current in the conductor? Then I think I understand the first part - but still not sure about the second, i.e. how there is an upward force on the electrons of the left rod but yet a downwards current. Or does this only occur at the beginning when the system starts moving, and somehow the steady state is an anti-clockwise current?
Steady state is no current. There's a transient current when the system first starts moving, but once the forces balance, there's no current.
 
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  • #3
srn
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Awesome, that clarifies a lot. Thanks vela!
 

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