How does the magnetic field strength (in Teslas) change when two cylindrical magnets are pulled appart? I have two cylindrical magnets aligned such that the opposite poles are facing each other (N-S N-S).
I am trying to find a mathematical relationship that models the change in the magnetic field strength (B - measured in Tesla) at the midpoint of the two magnets when they are pulled apart (with a distance between them denoted r).
I have found many seemingly conflicting resources that say the relation is one of the following:

*

*B=1/r.

*B=1/r^2.

*B=1/r^3.

I am very unsure of which is applicable to my scenario. I should note that I have a high school level understanding of magnetism so I struggle to understand some of the more complex explanations. I would appreciate it very much if someone could provide some insight into this.
 A: A simple model of your scenario would be to replace both your magnets with ideal magnetic dipoles, which is a decent approximation and quite useful because the fields of finitely sized permanent magnets are difficult to write down.  Electromagnetism has an important and useful property known as superposition.  The total electric field in a volume of space is the sum of all electric fields, and likewise for magnetic fields.
Therefore, ignoring any slight effects that each magnet's individual field might have on the domains and field of the other, you can apply superposition and know that the total field in space is the sum of the fields of each magnet.  This gives us a good starting point.  Now we just approximate each magnet as an ideal dipole, which has a neat closed form expression for magnetic field, and we can write down an exact expression for the total field everywhere in space.
The field of a dipole scales as $1/r^3$, so we can assume that the sum of their fields will also scale as $1/r^3$ far away from the magnets. The field between the two is a bit trickier, but you should be able to compute it given your expression for the total field. If the magnets are separated by a distance $a$, then the field at the midpoint should scale as $1/(a/2)^3$.
A: The B field changes as the magnets pull apart. All the answers below are approximate'

*

*When they are still close together they behave like infinite charged planes, so B does not change with distance.

*As they move further apart, the near end of each magnet looks like a point charge, with the far ends of the magnets being negligible.  Then the field falls like 1/r^2.

*As they move still further apart, they begin to look like two magnetic dipoles with B falling like 1/r^3.

A: The answer depends on how $r$ compares to the length $L$ and radius $R$ of each of the magnets. As @klippo's answer notes, for large separations ($r>>R$) the magnetic field strength should be the dipole field that scales as $1/r^3$.  If the magnets are very close together ($r << R$), then just as for the electric field of a parallel plate capacitor, the magnetic field will be constant independent of $r$. In between, when ($R<< r << L$), each pole tip acts like a magnetic monopole and the midpoint magnetic field strength should scale as $1/r^2$.
The field at the midpoint can actually be calculated exactly for identical ideal magnets.
As in the answer to How strong is the magnetic field from a neodymium magnet?, the magnetic field a distance $z$ from a pole face along the symmetry axis of an ideal cylindrical magnet is
$$B=\frac{B_r}{2}\left (\frac{L+z}{\sqrt{{R}^2+{(L+z)}^2}} -\frac{z}{\sqrt{R^2+z^2}}\right )$$
where $B_r$ is its remanence of the magnet.
For two magnets oriented as you describe, separated by a distance $r=2z$, the magnetic field strength from each magnet at the midpoint will be the same, so the combined magnetic field will just be twice that of a single magnet, i.e.
$$B(z)=B_r\left (\frac{L+z}{\sqrt{{R}^2+{(L+z)}^2}} -\frac{z}{\sqrt{R^2+z^2}}\right )$$
This has the expected constant, $1/r^2$, and $1/r^3$ dependances in the ($r<<R$), ($R<< r << L$), and ($r>>L$) limits, respectively:
$$\begin{aligned}
\lim_{z \to 0} B(z)&= B_r\frac{L}{\sqrt{{R}^2+L^2}}\\
\lim_{L \to \infty, R \to 0} B(z)&= B_r \frac{R^2}{2z^2}= B_r \frac{8R^2 }{r^2}\\
\lim_{z \to \infty} B(z)&= B_r \frac{R^2 L}{z^3}= B_r \frac{8R^2 L}{r^3}
\end{aligned}$$
See here and here for the series expansions giving the last two limits.
A: Thank you for the explanations. They are exactly what I am after. However, do you have a non-mathematical explanation to understand the relationship?
