Does a Falling Wire Near a Magnet Experience a Magnetic Force?

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A falling wire near a magnet does not experience a slowing effect due to the absence of a closed loop, preventing current flow and back electromotive force (emf). While magnetic forces act on the charged particles at the ends of the wire, these charges cannot move, resulting in no induced current. The magnetic force is present because the motion of the wire is perpendicular to the magnetic field, but the wire's orientation means it does not generate emf. The relevant equation for emf (emf = vBL) indicates that without mutual perpendicularity of velocity, magnetic field, and length, no emf is induced. Consequently, the wire will not slow down as it falls.
physmath96
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Suppose a single wire (in a horizontal position) falls toward a pole of a magnet. Does the wire slow down?
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IMHO I don't think so, as the straight wire has no loop back connection to each end.
Thus, no current flow, no back emf.
But a copper "ring" would be a different story.

Just my thoughts, could be wrong.
 
pallidin said:
IMHO I don't think so, as the straight wire has no loop back connection to each end.
Thus, no current flow, no back emf.
But a copper "ring" would be a different story.

Just my thoughts, could be wrong.

That's what I initially thought. There is a magnetic force acting on the charged particles at the ends of the wire, but these charges have nowhere to go and thus, there is no current. However, I was wondering if the magnetic forces acting on the ends of the wire could exert torque on the wire...
 
physmath96 said:
There is a magnetic force acting on the charged particles at the ends of the wire...

How do you come to that assumption? Why are there "charged" particles on the ends of the wire in your scenario? The copper wire has many charged particles but is electrically neutral in your description.
 
pallidin said:
Why are there "charged" particles on the ends of the wire in your scenario? The copper wire has many charged particles but is electrically neutral in your description.

The wire has charged particles, right? If the wire falls, then the charged particles (which are confined to the wire) are also falling. If charged particles move in a direction perpendicular to the magnetic field, a magnetic force acts on them. In this case, the magnetic field (at the ends of the wire) is perpendicular to the motion of the charged particles in the wire. Thus, there should be a magnetic force acting on the ends of the wire, right?
 
Well conductors (e.g., metals) have electrons in their atoms. Some of the electrons are quite mobile, i.e., they can readily move if an external force is applied, e.g., through potential difference applied to the conductor, or if the conductor is moved through a magnetic field.

Perhaps this will help.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/conins.html

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir.html

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/hall.html

The answer one seeks lies therein.
 
Astronuc said:
Well conductors (e.g., metals) have electrons in their atoms. Some of the electrons are quite mobile, i.e., they can readily move if an external force is applied, e.g., through potential difference applied to the conductor, or if the conductor is moved through a magnetic field.

In this case, a wire segment is moving through a magnetic field. The equation that applies to this case is: emf = vBL, which holds only when v (velocity), B (magnetic field), and L (length) are mutually perpendicular. Looking at the diagram, one can see that the three values are not mutually perpendicular, and thus, no emf will be induced. The wire will not slow down.
 
physmath96 said:
In this case, a wire segment is moving through a magnetic field. The equation that applies to this case is: emf = vBL, which holds only when v (velocity), B (magnetic field), and L (length) are mutually perpendicular. Looking at the diagram, one can see that the three values are not mutually perpendicular, and thus, no emf will be induced. The wire will not slow down.

I don't think he was wondering about the emf. I think he wanted to know what force would be applied, if any. In this case, a force will only be applied if v and B are not parallel (qv x B).
 
Disinterred said:
I don't think he was wondering about the emf. I think he wanted to know what force would be applied, if any. In this case, a force will only be applied if v and B are not parallel (qv x B).

At the ends of the wire, v is perpendicular to B - thus, there is a magnetic force acting on the charges there. However, the wire itself is parallel to B, and as a result, the charges have no space to move - there is no current.
 
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