# What does the metres squared represent in the dipole moment?

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

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 $$\text{A}\cdot\text{m}^2$$.