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Inductance of a straight conductor

  1. Jan 17, 2016 #1

    cnh1995

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    I have read this chapter about calculating internal and external inductance of a straight wire here.
    http://nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/power-system/chapter_1/1_4.html
    Seeing the flux linked with the conductor itself, emf due to that flux induced in upper half and lower half of the wire are in opposite direction. How does this internal inductance work? How is back emf generated by the flux linked with the conductor itself? Thanks in advance!
     
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  3. Jan 18, 2016 #2

    Simon Bridge

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    Have a look through the related:
    https://www.physicsforums.com/threads/inductance-of-a-straight-wire.620356/

    ... the flux is in the conductor - that's the relation.
    It may make more sense if you think of pulses of charge in the conductor rather than a continuous current going all the way through.


    It may just be that you need less hand-waving...
    http://www.ee.scu.edu/eefac/healy/indwire.html (also in the discussion linked above)
    http://www.g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf (A more detailed look at the calculation)
     
  4. Jan 18, 2016 #3

    tech99

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    When the current is switched on, the magnetic field expands cylindrically. It starts with a maximum at the centre of the wire and expands outwards, cutting "shells" of the wire as it does so. If you use Fleming's Right Hand Rule, for a generator, the direction of the induced emf in the outer shells of the wire can be found, taking the radially expanding direction of the field as the Motion. Once the current reaches maximum, the field is completely built, and energy is stored in it until switch off.
     
  5. Jan 18, 2016 #4

    cnh1995

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    Could you please elaborate? How does the field expand cylindrically?
     
  6. Jan 18, 2016 #5

    tech99

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    So you know that the magnetic field surrounds the conductor rather like lagging around a pipe. It gets stronger towards the centre, and it goes right inside the conductor. So the centre is the strongest point. And when you switch on, the field starts in the centre and expands outwards. As it does so, it is cutting layers of the conductor and so causes a back EMF.
    .
     
  7. Jan 27, 2016 #6

    jim hardy

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    A similar question popped up on PF some years ago.

    As always there's more than one way to imagine taking a measurement.

    To my mind, a piece of wire carrying current is part of a current loop that closes somewhere even if it's only throughthe capacitance of space..
    So there's some flux enclosed by that loop
    and inductance is defined as (flux X turns) per ampere
    so if you set turns and amps both equal 1, you could calculate flux and that'd be inductance.

    This fellow did that and published a short essay based on a 1908 NBS article
    http://www.ee.scu.edu/eefac/healy/indwire.html

    upload_2016-1-27_8-25-50.png

    Perhaps we're stumped by QV cross B in that image from your link?
    What is relative motion between V and B ?
     
  8. Jan 27, 2016 #7

    cnh1995

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    If I visualized correctly, the flux inside the conductor will cause back emf opposite to the current direction inside the wire but since this emf will have a closed path, the return path would be on the surface of the conductor. That means, inside the conductor, there is opposition to the flow of current by the forward path of back emf and on the surface, the back emf assists the current. This leads to skin effect as the current tends to flow on the surface.
     
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