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The Earth's magnetic field is essentially that of a magnetic dipole...

  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data

    The Earth's magnetic field is essentially that of a magnetic dipole. If the field near the North Pole is about 10^-4 T, what will it be (approximately) 13,000 km above the surface at the North Pole?

    2. Relevant equations

    [tex]B = \frac{\mu _0}{2\pi}\frac{\mu}{(R^2+x^2)^{3/2}}[/tex]

    for x >> R

    [tex]B = \frac{\mu _0}{2\pi}\frac{\mu}{x^3}[/tex]

    3. The attempt at a solution

    I know how to find the magnetic field of a dipole on its axis for a circle. My textbook uses the second formula to solve this problem. But I cannot understand what "x" and "R" are for the earth, and how we assumed x>>R.
     
  2. jcsd
  3. May 12, 2015 #2

    haruspex

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    Wherever you got that equation from (link?), it should state what the variables mean and what constraints there are on its validity. If you can't provide a link, please quote them.
    The form of it (##R^2+x^2##), suggests R and x are orthogonal, which doesn't sound right for the given question.
     
  4. May 13, 2015 #3
    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html#c3

    This is the same formula except pi*r^2*I is changed with mu(magnetic dipole moment) in mine.
     
  5. May 13, 2015 #4

    BvU

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    "Dipole field" is the name for the field from a current loop in the limit ##R\rightarrow 0##, with m or ##{\bf \mu} = \pi R^2 \;I## constant. So Haru is right and your book is right too.

    That eliminates the R question. Now you should think about where this dipole could possibly be located, and therefore what to take for the x.
     
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