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Earth's Magnetic Field

Discussion in 'Electronic Design' started by Paul Hovnanian P.E., Feb 22, 2006.

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  1. There's a program on NOVA (on our local PBS station) about the earth's
    magnetic field and how its strength is decaying quite rapidly.

    I blame it on all the people using 50 and 60 Hz power systems. We're
    degaussing the earth!

    ;-)
     
  2. Bob

    Bob Guest


    Yep, our magnetic field periodically reverses (every 100K years?). I have
    heard that the sun's field reverses every 11 years. That's why compass sales
    are so slow, there.

    Maybe Edison was right. He was a proponent of DC.

    Bob
     
  3. James Beck

    James Beck Guest

    I don't know which program you watched, but the one I saw where they
    used the old deviation figures from the British Navy to show how the
    magnetic field has been changing over the last 300 years or so was cool.
    That anomaly down by Oz looks ominous.

    Jim
     
  4. Nah, Tesla was right. Remember he didn't want to piddle around in the
    several dozens of Hz for power distribution, he wanted to get up into
    the frequency range where the Earth's core is insensitive to rapid
    imposed field reversals. You can't degauss the ionosphere.


    Mark L. Fergerson
     
  5. Ian Stirling

    Ian Stirling Guest

    The amount of power, and materials you need for a globespanning solenoid
    producing the current field are not actually that high, even lacking
    superconductors.
     

  6. Its all Phil's fault!


    --
    Service to my country? Been there, Done that, and I've got my DD214 to
    prove it.
    Member of DAV #85.

    Michael A. Terrell
    Central Florida
     
  7. Guest

    I blame Genome and his big Geostationary Orbital Magnet Thing:
    http://groups.google.com/group/sci....db17c9cfae9/c27242245fb3f43e#c27242245fb3f43e

    Cheers,
    James Arthur
     
  8. Perhaps some people don't get the funny part... You can't degauss the
    world this way--- all our power circuits have wires going both ways,
    which makes the magetic field drop off at a rather high rate. Also
    you'd need to have the opposite poles of the degausser span a good
    fraction of the Earth's magnet, not easily done. Nerd humor is a very
    special and very dry variety. No Millie Henry jokes, please.
     
  9. Ken Taylor

    Ken Taylor Guest

    I didn't have the capacitor to resist 'er.

    Ken
     
  10. colin

    colin Guest

    If someones worked it out exactly it would be interesting to know.

    Colin =^.^=
     
  11. Guest

    According to Wikipedia:

    "The field is similar to that of a bar magnet, but this similarity is
    superficial. The magnetic field of a bar magnet, or any other type of
    permanent magnet, is created by the coordinated motions of electrons
    (negatively charged particles) within iron atoms. The Earth's core,
    however, is hotter than 1043 K, the Curie point temperature at which
    the orientations of electron orbits within iron become randomized. Such
    randomization tends to cause the substance to lose its magnetic field.
    Therefore the Earth's magnetic field is caused not by magnetised iron
    deposits, but mostly by electric currents in the liquid outer core."

    OK. Here's an approximation based on Ampere's law:

    Integral around a closed loop of (B*dl) = Total current through the
    loop / (e0 x c^2)

    If we choose a path from N magnetic pole to S, along the earth's axis
    and then along a "meridian" just above the surface back around we get a
    loop that encloses the entire current inside the earth. B*dl is a dot
    product, but we assume B is roughly parallel to our path at all points.

    The strength of the earth's magnetic field at the poles is roughly B=60
    x 10^-6 T and at the equator is roughly half that. Let's assume that B
    is a constant 60 x 10^-6 T along the magnetic axis and half that on the
    surface.

    The length of the path along the axis is roughly 6.4x10^6 m and around
    the surface 2.0x10^7 m. Therefore the integral is (6.4x10^6 m) x (60 x
    10^-6 T) + (2.0x10^7 m) x (30 x 10^-6 T) = 984. Multiplying by ((e0 x
    c^2) gives us:

    I = 984 x (10^7/4pi) = 780,000,000 Amps.
     
  12. Ian Stirling

    Ian Stirling Guest

    I did the numbers a while back.
    IIRC.
    The current needed is about a million amp-turns.
    Assuming aluminium, and a meter square cable, that's 1v/35m, or about a
    megavolt, for a total power requirement of a terawatt or so.

    This is only about a decades primary production of aluminium, and 1/10
    current electricity production.

    Insulation is another problem - at 30Kv/cm, that's 30cm of insulation.

    If it's split into 10 'wires' of 30cm diameter, thats maybe another 6m^2
    of plastic, or 240 million cubic meters, or maybe 500 megatons, or maybe
    5 years current production.

    Oil insulation may be better, as it'll allow easier cooling, though I
    think that for a 1m outside diameter 'wire', 3.5Kw/m will mean it's only
    warm on the outside - if it doesn't melt the middle.

    You will want a highly interconnected band of moderately spaced cables,
    so that you can isolate failed segments.
    A cable break would be just _bad_.
     
  13. Ken Smith

    Ken Smith Guest

    There is a spot in the south atlantic that looks like the source may have
    inverted already.

    In the last 20 years, the average field in the San Fransisco area has
    decrease by something like 0.1%
     
  14. Ken Smith

    Ken Smith Guest

    This isn't true for the 3rd harmonic.

    (sticks finger into air)

    Around here there is more 180Hz than 60Hz magnetic field. At Moffet field
    (california) there is about 150nT of 180Hz at the bay end of the runways.

    There is also about 1nT at low frequencies from the BART train system.
    BART runs on the east side of the bay.
     
  15. Guest

    Oops, I used the radius of the earth instead of the diameter, The
    correct result is (12.8x10^6 m) x (60 x 10^-6 T) + (2.0x10^7 m) x (30 x
    10^-6 T) = 1368. Multiplying by ((e0 x c^2) gives:

    I = 1368 x (10^7/4pi) = 1,088,000,000 Amps.

    As a reality check, consider that the field in a solenoid the length of
    the earth's diameter is is:

    B= u0 NI/L = 1.26 x 10^-6 x 1.088 x 10^9 / 12.8 x 10^6 = 106 x 10^-6 T.


    My assumption was that inside the earth it was just 60 micro Tesla, not
    106, but it's in the ballpark.
     
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