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Calculating magnetic field as a function of magnet size

  1. Sep 4, 2012 #1
    I want to determine how large a magnet I need in order to get a given field a certain distance away from the surface.

    An N48 neodymium magnet has a Bremanence of ~1.47T. How does the physical size of the magnet affect the field a given distance away?

    https://www.physicsforums.com/showthread.php?t=519563 seems to imply it's proportional to the volume. However, I'm unclear on how to apply the equations. I'm also surprised that no mention is given to the shape -- shouldn't a flat, thin magnet have its magnetic field fall off more slowly than a long, deep one?
     
  2. jcsd
  3. Sep 4, 2012 #2
    That formula you refer to is for the field of a magnetic dipole and ONLY on the axis of the dipole. There, d is the distance from the dipole, it has nothing to do with the volume. In other words, the field of a static dipole decays as the reciprocal of the cube of the distance.

    As an estimation to your problem, I am assuming that the (manufacturer?) specified field of 1.47 T is specified at the surface of the end of the magnet. Then, as long as you stay on the axis of the magnet, the field at a distance d away from the surface of the magnet is
    [tex]B(d) = \frac{(\frac{H}{2})^3}{(d+(\frac{H}{2}))^3} \cdot 1.47 [\rm{T}] \; ,[/tex]
    where H is the height of the magnet.
     
  4. Sep 4, 2012 #3
    Thanks! How did you derive that?
     
  5. Sep 4, 2012 #4
    No problem; it's just the only way to enforce the d^-3 relation with the field referenced to the surface. Look up "magnetic dipole" for the derivation of the equation from the other thread.
     
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