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Emf induced in a conducting ring.

  1. Nov 13, 2013 #1
    Consider a magnetic field perpendicular to a conducting ring moving with a velocity, v.When the ring is moving on the ground in translational motion alone, will emf be induced? I am slightly confused because if you consider the two halves of the ring as two rods, emf will be induced in both of them individually, but what about the emf induced in the ring as a whole?
     
    Last edited: Nov 13, 2013
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  3. Nov 13, 2013 #2

    Meir Achuz

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    You have not given a clear picture of the relative orientations and motion, but if the magnetic flux through the ring changes with time, there will be an EMF.
     
  4. Nov 13, 2013 #3
    The ring is performing translational motion and the magnetic field is orthogonal to the motion of the ring. Does it matter if the magnetic flux is changing because I considered the two halves of the ring to be separate rods. That way emf induced in each 'rod' is E= Bl(2R) where R is the radius of the sphere.
     
  5. Nov 13, 2013 #4

    Nugatory

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    Consider the direction of the EMF induced in the two separate rods.

    Suppose we were to connect the left ends of the two rods, and the right ends of the rods, with conducting wires. That's equivalent to the ring. What's the induced EMF across the two connections?
     
  6. Nov 13, 2013 #5

    Meir Achuz

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    Two rods are not a ring. However you try, the EMF is given by the rate of change of the magnetic flux through the ring.
     
  7. Nov 14, 2013 #6
    If you apply Kirchoff's loop rule over here, the total EMF will be zero.
     
  8. Nov 14, 2013 #7

    Philip Wood

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    MA: I agree that looking at the change (if there is any!) in flux through the ring is the best approach. Nugatory is (I think) trying to come at it from the OP's point of view of regarding the ring as joined rods. The emf in each will be in the same direction in space, but looking at the rods as parts of the ring, in opposite senses around the ring.
     
  9. Nov 14, 2013 #8
    Nothing is specified about the magnetic field. So it might also be a constant field which means there won't be any EMF induced.
     
  10. Nov 14, 2013 #9

    Meir Achuz

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    If it's a constant B field, there's no point asking the question.
     
  11. Nov 14, 2013 #10

    Philip Wood

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    With respect, MA, there might be a point, if the OP is trying to reconcile emf's induced in conductors cutting flux (the 2 rods approach) with there being no emf in the ring.
     
  12. Nov 14, 2013 #11
    For the emf to be induced in a moving conductor, you don't require varying magnetic field. So, that's the reason i considered them to be two rods. In that case, emf will be produced in both as in the case of two cells connected in the same circuit. I need help with that part of the problem.
     
  13. Nov 14, 2013 #12

    WannabeNewton

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    A ring is not the same thing as two rods, no matter how you try to construe the system.

    OP, why not just start with two conducting rods connected by wires like Nugatory suggested?

    Now, why does EMF occur in the first place for a conducting system moving non-trivially through an external magnetic field? What does the magnetic field do to the charges in the conducting system that causes an EMF to be induced? And how does this relate to the "two conducting rods connected by wires" system?
     
    Last edited: Nov 14, 2013
  14. Nov 14, 2013 #13

    Philip Wood

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    I did say "joined rods", imagining their ends to be connected together to form a ring.
     
  15. Nov 15, 2013 #14
    When a conducting body moves through the field, the charges in the system experience force 'qvb' and hence, opposite charges move to extreme ends and potential difference is established. this continues until qvb= qe(i.e; force due to electric field in the direction opposite to that of qvb.)
     
  16. Nov 15, 2013 #15

    Philip Wood

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    srijag. Do you now have a satisfactory answer to your original question?
     
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