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Magnetic field strength and distance

  1. Nov 21, 2009 #1
    Gravitational force is divided by distance squared.

    What is the equivalent divisor for a fixed bar magnet?

    Thank you for your kind attention :-)

    (-_-)
     
  2. jcsd
  3. Nov 21, 2009 #2

    jtbell

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    Staff: Mentor

    A bar magnet is approximately a magnetic dipole. Far from a dipole, the field goes like [itex]1/r^3[/itex].
     
  4. Nov 21, 2009 #3
    Hello jtbell :-)

    Math is not my forte so please allow to confirm:

    the divisor for magnetic field strength is distance cubed?

    Or have I misunderstood?

    (-_-)
     
  5. Nov 21, 2009 #4
    Why the magnetic field is propotional to [tex]1/r^3[/tex] .
    Can you prove it?
     
  6. Nov 21, 2009 #5

    jtbell

    User Avatar

    Staff: Mentor

    Yes.

    See for example the field along the axis of a circular current loop, another example of a magnetic dipole:

    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/magnetic/curloo.html#c3 [Broken]

    and take the limit as z >> R.

    Or you can pretend that the magnet consists of two opposite polarity magnetic monopoles and calculate the field on analogy with an electric dipole (two opposite charges separated by distance d). Find the vector sum of fields from the two monopoles and take the limit as r >> d. It's probably easiest to do this along the axis of the dipole, or along a line perpendicular to the midpoint of the dipole.
     
    Last edited by a moderator: May 4, 2017
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