EMF induced in a straight current-carrying conductor

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Summary:

can emf be produced if the wire moves in another direction apart from downwards through the field.
An emf is induced in straight current carrying conductor as it moves at at right angles to a uniform and constant magnetic field. My textbook used direction 1 in the image shown to demonstrate this. I asked my teacher if direction two would be possible and he didn't understand me. So I want to know if direction 2 will also produce an emf as it follows the movement at right angles rule I was given so i think it should but it would be across the sides of the wire unlike direction 1 which would be along the length of the wire. And if not please explain why.
InkedWire-cutting-a-magnetic-field_LI.jpg

(wire is the green line)
 

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  • #2
sophiecentaur
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Hi and welcome to PF.
The easiest 'explanation' is that lines of force need to be 'cut' as a result of the motion. Your idea of '2' movement doesn't involve any cutting of lines so the will be no induced emf. By Cutting, I mean motion so that the line of the wire is at right angles to the direction of the motion and the direction of the field lines. Slightly different angles will produce a lower value of emf until case '2' gives you zero emf. This is very idealised, of course.
Have you 'done' Vector Multiplictation yet? There is a much better answer that involves describing the motion of the wire and the field lines in terms of vectors but the 'cutting' word is the first stab at a description.
 
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thank you for the welcome. i think direction 2 would still satisfy everything being at right angles to each other and the 'cutting' you mentioned. My diagram might not be the best, i don't mean moving it parallel to the magnetic fields lines like from left to right( let's call this direction 3) .To clarify, in direction 1 your moving up and down, while in direction 2 your moving forwards and backwards. So my thinking was that if your moving forwards and backwards while the field lines run from left to right, then the emf would be induced in the up down direction, everything at right angles to each other like the x-y-z plane. this wouldn't generate a detectable emf if it was connected to a voltmeter at the ends of the wire, but i think it would induce an emf at the sides of the wire.
 
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If the field is uniform, only direction one (of the three orthogonal directions) will produce a current in the wire. Moving parallel to the wire axis will slightly segregate charge as you describe. Moving toward the pole face produces no force.
You need to understand the vector cross product .
 
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  • #5
sophiecentaur
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Gotcha, now. You are right ( well thought out) but the induced emf will be lateral to (across) the wire and won’t cause a current to flow along the wire. It will just displace some charges across the width of the wire. The charge imbalance is limited by that induces emf.
 
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Gotcha, now. You are right ( well thought out) but the induced emf will be lateral to (across) the wire and won’t cause a current to flow along the wire. It will just displace some charges across the width of the wire. The charge imbalance is limited by that induces emf.
thank you, I understand that the emf will be across the width of the wire. Please what does this line mean ' The charge imbalance is limited by that induces emf '
 
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sophiecentaur
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It needed editing to “induced emf”. Internal fields in the wire cancel the emf.
 
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It needed editing to “induced emf”. Internal fields in the wire cancel the emf.
how exactly ?
 
  • #9
sophiecentaur
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The free electrons in the wire move slightly in one direction and that means there is a force (field of the fixed protons)., attracting them back to where they started. The higher the induced emf, the greater the displacement and the greater the resulting restoring force, until equilibrium is reached. Then the movement stops.
 
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The free electrons in the wire move slightly in one direction and that means there is a force (field of the fixed protons)., attracting them back to where they started. The higher the induced emf, the greater the displacement and the greater the resulting restoring force, until equilibrium is reached. Then the movement stops.
thanks
 

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