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B field of a charged ring.

  1. Feb 14, 2012 #1
    Lets say I am standing in the middle of a charged ring. And I am standing on a turn table.
    Now I start to rotate in the center. From my point of view do I perceive a B field.
    I mean I would have a velocity component.
     
  2. jcsd
  3. Feb 14, 2012 #2
    If you mean is there a B field in your reference frame then yes a compass would be messed up sitting in the center, but assuming that the radius of the ring is huge compared to you, you won't feel a magnetic force since your rotation doesn't really produce a velocity vector (again assuming you are vanishingly small compared to the ring radius).
     
  4. Feb 14, 2012 #3
    To compute the B field would I just say that the ring is moving we an angular speed and do it that way.
     
  5. Feb 14, 2012 #4

    Dale

    Staff: Mentor

    Your point of view is non inertial. It is not even clear what is meant by "magnetic field" in non inertial frames.
     
  6. Feb 14, 2012 #5
    so its not easy to calculate if we even can.
     
  7. Feb 14, 2012 #6

    Dale

    Staff: Mentor

    The EM tensor transforms as a tensor, so it is "relatively" easy to calculate the force on a test charge in your non-inertial frame. It is not clear how much of that force to attribute to electric field and how much to attribute to magnetic field because the electric and magnetic fields are components of the tensor in an inertial frame. In non-inertial frames interpreting certain components as this or that becomes suspect.
     
  8. Feb 14, 2012 #7

    Bill_K

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    Science Advisor

    There is no such thing as a "noninertial frame". Because a "frame" covers the entire space, and noninertial coordinates must be treated locally. What a rotating observer sees is determined by a different Lorentz transformation at each point.

    At the exact center of the ring, E = B = 0, and you can Lorentz transform it all you want, you still get zero.

    At a distance r from the center there will be a net radial E field, and a rotating platform will have a tangential velocity v = ωr. An observer moving along with the platform at this point will see a B field, B = -v/c x E, pointing upward.
     
  9. Feb 14, 2012 #8

    Dale

    Staff: Mentor

    That would be news to a very large number of peer reviewed authors.
     
  10. Feb 14, 2012 #9
    So the B field would be B = -v/c x E, pointing upward. as Bill K said.
    So If I had a charged sphere And I was driving around it in a circle I could just use this.
    And this would give me the same B field as if the charged sphere was rotating.
     
  11. Feb 14, 2012 #10

    Bill_K

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    I certainly wouldn't approve a paper that used the term. The word "frame" specifically applies to a set of Minkowski coordinates, and its inappropriate use leads to much of the confusion surrounding rotation and/or acceleration in relativity. The correct description is in terms of a timelike congruence of local observers, one through each point. They will necessarily have different velocities, and therefore do not form a single "frame of reference".
     
  12. Feb 14, 2012 #11

    Dale

    Staff: Mentor

    That is quite different from my understanding of the term. However, I must admit that my understanding comes from Wikipedia and PF:
    http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity
    https://www.physicsforums.com/showthread.php?t=168631

    According to my understanding the word "frame" refers specifically to a "frame field" which is a set of four orthonormal vectors at each point in the manifold. These vectors are not coordinates and integral curves of the timelike vectors may not be geodesics. When they are not then the frame is non-inertial.
     
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