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Suppose I have a steel bar sitting on a desk, inside the Earth's magnetic field. Suppose the bar is solid and it a cylinder, 1 meter tall and 25cm in radius. The Earth's field points uniformly through the long axis of the cylinder.

Is there some way to determine the magnetic contribution coming from the steel bar? Is there even a magnetic signature since the time derivative of the magnetic field is 0? My thinking is yes, because steel is made of iron which is ferromagnetic. Therefore there should be some induced magnetization from the main earth field, which causes the steel to produce it's own field.

In short my question is: is there some way to estimate the field strength $B = B(z) $ along the longitudinal axis for some location $z$?

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  • $\begingroup$ What do you mean by "the magnetic contribution"? $\endgroup$
    – probably_someone
    Commented Jan 13, 2020 at 19:06
  • $\begingroup$ As in, suppose the earth field without anything else is $B = B_0 \hat{z}$. Is it possible to calculate what the magnetic field might be at z = 2m; 1m above the steel bar? Does the steel bar produce an induced magnetic field of its own since it is within the Earth's? $\endgroup$
    – john morrison
    Commented Jan 13, 2020 at 19:07
  • $\begingroup$ Depends on what kind of steel this is, and its magnetic history. $\endgroup$
    – user137289
    Commented Jan 13, 2020 at 22:32

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The math is the same as for a dielectric object in a uniform electric field, described in Polarization density:

enter image description here

There is an explicit solution for a sphere. For a cylinder I think you would need a numerical model.

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A long thin cylinder with a large permeability would acquire an approximately constant magnetization given by $4\pi{\bf M}=\mu{\bf B}_{earth}$ (in Gaussian units). The B field along the axis would be the same as that of a solenoid with B=$\mu B_{earth}$ at its center. For your, relatively fat cylinder, it is harder to calculate.

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