# Find drop-off rate of magnetic interference from a mass of pure iron on a magnetic compass

How can I find the magnetic interference a stationary 35000 kg block of 100% pure iron would have on a magnetic compass and what the drop off rate of the interference would be.

So if said 35000 kg block of iron was 1 meter away from the compass, 100 meters away, or 1000 kilometers way I would like to calculate the rate of drop off of the interference.

This may seem absurd, but it is very important for a conceptual project I am working on.

For the context of this question, assume everything is perfect, and that we are basically operating in a vacuum and there is no interference from anything else and that all instruments are 100% accurate and infinitely precise. And that I have only a very very basic understanding of physics, mathematics and magnetism.

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Specific density of iron is 7.87 times water, so 35,000 kg of iron = 35/7.87 = 4.45 cubic meters. Assuming the block is spherical, this is a sphere with radius 1.02 meters. So I'm assuming you mean 1 meter from the surface of the block.

At these distances you can more or less approximate the iron sphere as a bar magnet. In terms of dependency on distance, it acts like a dipole. From the wikipedia article "Dipole" http://en.wikipedia.org/wiki/Dipole you have

B is the strength of the field, measured in teslas
r is the distance from the center, measured in metres
λ is the magnetic latitude (equal to 90° − θ) where θ
is the magnetic colatitude, measured in radians or degrees from the dipole axis
m is the dipole moment (VADM=virtual axial dipole moment),
measured in ampere square-metres (A·m2), which equals joules per tesla
μ0 is the permeability of free space, measured in henries per metre.


So the strength depends on the angle $\lambda$ from the axis, and drops off as $1/r^3$.

That means that at 100 meters away the strength will be about 1,000,000 times less, and at 1000km = $10^6$ meters, it will be reduced by a factor of $10^{18}$.

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That is fantastic. Thank you!. Would there be a way to estimate the initial strength of the magnetic field or the iron sphere?\? –  Caleb Larsen Feb 28 '11 at 14:31

The magnetic moment of the sphere equals it's volume times it's average magnetization. Now you have to estimate the magnetization, and here is where the problem gets hard: it depends on the magnetic properties of your material.

I don't know how a sphere of 100% pure iron would behave, but I can tell that the behavior of a sphere of almost-100% pure iron can be hard to predict. That's because magnetic properties can be strongly dependent on microscopic details of the material, including impurities and crystallographic defects. The magnetization can also depend on the magnetic history of the sample, e.g. it's orientation relative to the Earth's magnetic field last time it was cooled below it's Curie temperature.

At the microscopic scale, you should have the saturation magnetization of iron, which is 1.72 MA/m (2.16 T$/\mu_0$) at room temperature. But it's unlikely that the macroscopically averaged magnetization would be anywhere near that, unless your iron is instead a very strong iron-based magnet.

If your iron is sufficiently pure, I would expect quite a soft magnetic behavior from it. If we approximate it by a perfectly soft material (linear with infinite susceptibility), then it's magnetization should be 3 $H_0$, where the factor 3 comes from the spherical shape and $H_0$ is the field externally applied to your sphere, for example the Earth's magnetic field ($\mu_0H_0 \approx 50$ µT).

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