Earth's Magnetic Field

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!

;-)

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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This isn\'t right.  This isn\'t even wrong. -- Wolfgang Pauli
Reply to
Paul Hovnanian P.E.
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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

Reply to
Bob

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

Reply to
James Beck

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

Reply to
Mark Fergerson

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

Reply to
Ian Stirling

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

Reply to
Michael A. Terrell

I blame Genome and his big Geostationary Orbital Magnet Thing:

formatting link

Cheers, James Arthur

Reply to
dagmargoodboat

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.

Reply to
Ancient_Hacker

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

Ken

Reply to
Ken Taylor

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

Colin =^.^=

Reply to
colin

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.

-- Joe Legris

Reply to
jalegris

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_.

Reply to
Ian Stirling

In article , James Beck wrote: [...]

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%

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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.

-- Joe Legris

Reply to
jalegris

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