What does the metres squared represent in the dipole moment? If I were to do an experiment measuring the magnetic field strength (in tesla) of an iron core solenoid magnet, how would I determine the value for the dipole moment in the formula displayed in https://en.wikipedia.org/wiki/Dipole#Field_of_a_static_magnetic_dipole?
What does the metre squared represent?
 A: First of all, let us consider only a solenoid (without iron core) of length ${l}$ and area of cross-section ${A}$. The magnetic moment of the solenoid is given by:- $${M}_{solenoid}=({n}{l}){I}{A}$$
Where,n=number of turns per unit length of solenoid and
${I}$ = current flowing in the solenoid
Now, the net magnetic field is:-
$${B}={\mu_0}{n}{I}$$
Now if an iron core is introduced, iron being ferromagnetic,all the domains will get arranged in a particular direction due to the magnetising external field of the solenoid. Thus the iron core will develop magnetic moment of its own.
$$M_{iron}= {\chi}{H}{(A×l)}$$
Where, $\chi$ is the magnetic susceptibility and
${H}$ is the magnetic intensity.
I am not going deep into H and $\chi$(which depends upon the material and is much greater than 1 for ferromagnetic materials)
$$M_{total}= M_{iron} + M_{solenoid}$$
And the net magnetic field will be:-
$${B}={\mu}{n}{I}$$
Where ${\mu}$ is the magnetic permeability of iron

Where,
$${k}{\mu_0}={\mu}$$
Now as magnetic moment is the product of current, area and number of turns, it bears an unit A.m²
A: The magnetic dipole moment of a current loop is the current flowing around the loop times the area of the loop. This explains why the units for a magnetic dipole moment are $\text{A}\cdot\text{m}^2$.
