Find drop-off rate of magnetic interference from a mass of pure iron on a magnetic compass 
How can I find the magnetic interference of  a stationary $35000 \ \mathrm{kg}$ 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 $35000 \ \mathrm{kg}$ block of iron was $1 \ \mathrm{m}$ away from the compass, $100\ \mathrm{m}$ away, or $1000 \ \mathrm{km}$ 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.
 A: Specific density of iron is $7.87$ times water, so
$35000 \ \mathrm{kg}$ of iron $= 35/7.87 = 4.45  \ \mathrm{m}^3$
Assuming the block is spherical, this is a sphere with radius $1.02 \ \mathrm{m}$. So I'm assuming you mean $1 \ \mathrm{m}$ 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 :
$$B(m,r,\lambda) = \frac {\mu _0 }{4 \pi} \frac {m}{r^3} \sqrt {1+3 \ \mathrm{sin}^2 (\lambda)}$$
Where,

$B$ is the strength of the field  measured in teslas


$r$ is the distance from the centre , measured in metres.


$\lambda$ is the magnetic latitude (equal to $90° - \theta$ ) 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 $(Am^2)$, which equals joules per tesla.
$μ_0$ 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 $100 \ \mathrm{m}$ away the strength will be about $1,000,000$ times less, and at $1000 \ \mathrm{km} = 10^6 \mathrm{m}$ , it will be reduced by a factor of $10^{18}$.
A: 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).
