How can we make an order-of-magnitude estimate of the strength of Earth's magnetic field? The source of Earth's magnetic field is a dynamo driven by convection current in the molten core.  Using some basic physics principles (Maxwell's equations, fluid mechanics equations), properties of Earth (mass, radius, composition, temperature gradient, angular velocity), and properties of materials (conductivity and viscosity of molten iron) or other relevant facts, is it possible to estimate the strength of the field to order of magnitude (about one gauss)? 
Descriptions I've seen of the geodynamo all refer to extensive numerical computation on a computer, but can we get a rough idea with simple estimation?
 A: Try looking at the figures in a 2005 simulation by Takahashi et.al. in Science magazine, that at least show recurring reversals at http://www.sciencemag.org/cgi/content/full/309/5733/459?cookietest=yes.
Given that the viscosity, structure, and heat generation of the core are all to some degree unknown, and that the process may depend upon parametric amplification and even parametric resonance, this seems a pretty good beginning.
A: Live on earth is protected from solar wind by the earth's magnetic field. Charged particles from the sun (mostly) penetrate the earth's atmosphere with great velocity. These particles can be trapped by a magnetic field to follow circular path's around the magnetic field lines, thereby losing their energy due to collisions or bremstrahlung.
From first principles we can try to make an estimate of the strength of the magnetic field required to trap charged particles arriving with great velocity.
Starting with a lorentz force and a circular movement we have: $Bqv = m\frac{v^2}{r}$, so
$B= \frac{mv}{qr}$.
$v$ is the velocity of the particle, approx. light velocity, order of magnitude $10^8$ m/s. $q$ is the charge of the particle, elementary unit order of magnitude $10^{-19}$ C. $m$ is the mass of the particle, approx. proton mass, order of magnitude $10^{-27}$ kg. $r$ is the radius wherein the particle has to be trapped, at the most 10 km (height of the atmosphere) $10^4$ m. This gives $B \sim 10^{-4}$ T $= 1$ Gauss.
We have to appreciate the intelligent design ...
A: One answer might be Benford's law which state that the probability of lediding digit being 1 among natural constants is approx 30%
How to interpret Benford's law:
Any constant and measurment of "new" phenomen can be in any range. Ok max magnitude for known universe is approx 10^60, but still. 
Picking random number from this range is not evenly distributed but logarithmic distributed.  
example: We pick random number from 1 to 10^10
There is equal probability to pick number from 1 to 10^5 and from 10^5 to 10^10, this way we can prove that there is most likely to pick number that begins whit 1 and least likely whit 9. We can check this rule just by looking on logarithmic scale or doing some math.
Measuring of earth field, gravity or something else obeys this law and this is the reason that is most likely that you will measure 1*10^n.
Other reason that field is 1 gauss is that 1 is just magnitude/order of field. (by the way wikipedia article about this topic says that it is between 0.3 and 0.6 gauss) 
http://en.wikipedia.org/wiki/Earth's_magnetic_field#Field_characteristics
