# 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
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).