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How do we know what the dipole moment of "The Earth Magnet" is?

Physics Asked on November 8, 2021

It is given in my textbook and one other book(that I got from the library) that:-

…The earth has a magnetic dipole of dipole moment about $8.0times 10^{22} mathbf{ J.T^{-1}}$

The book doesn’t state how we know it and it uses the word ‘assume’ which confuses me a little bit. Internet doesn’t seem to help. Sources are welcomed.

One Answer

The short answer is that it's a sort of least squares fit to measurements at the surface. But first, let's suppose that the source of Earth's field is a dipole that is at the Earth's center and pointing towards the geographic North Pole. Given that, it only takes one measurement at the surface to determine what the dipole moment $m$ is. Let's say $R$ is the distance to the center in meters, $B$ is the magnitude of the surface field in teslas, and $lambda$ is the latitude. Then the dipole moment is given by (see Dipole Moment Variation): $$m = frac{4 pi R^3}{mu_0} frac{B}{sqrt{1 + 3sin^2lambda}}.$$

In reality, Earth's magnetic field has a much more complex source and it is only partly dipolar. To find out what that dipolar part is, we need a model for the magnetic field at the surface. This involves fitting large numbers of surface measurements to spherical harmonics, analogous to a polynomial fit but for the surface of a sphere. The three lowest-degree terms give a vector for Earth's dipole moment. It turns out that the best-fitting dipole moment is at an angle of about $11^circ$ off the rotational axis (see the Wikipedia article on Earth's magnetic field).

I know this is rather technical, and I'm sorry for that. I couldn't find a source that explained it in any simpler terms.

Answered by A. Newell on November 8, 2021

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