We know that for a current carrying loop magnetic moment can be calculated by
$$\overrightarrow{M} = I\times \overrightarrow{A}$$
Where I is current in loop and A is the area enclosed by the loop
And magnetic moment for a bar magnet is $$\overrightarrow{M} = m \times \overrightarrow{l}$$
Where m is pole strength and l is the length between the poles of bar magnet
Here my question is that is there some kind of relation between them about why both of them is described about magnetic moment and how these two formulas are actually related, if their is some connection between these two terms/formulas?
Best Answer
The most general formula for the magnetic dipole moment of a volume current distribution is $$ \vec{\mu} = \frac{1}{2} \iiint \vec{r} \times \vec{J} \, d \tau. $$ This has obvious generalizations to a surface current or a line current: $$ \vec{\mu} \sim \frac{1}{2} \iint \vec{r} \times \vec{K} \, d a \sim \frac{1}{2} \int \vec{r} \times \vec{I} \, d \lambda. $$ (I will use $\vec{\mu}$ rather than $\vec{M}$ for the dipole moment because it is more common to use $\vec{M}$ for the magnetization of the medium; see below.)
Connection to current loop formula
We can connect the last of these to the formula you wrote above by noting that if we parametrize a current loop by its arc length $\lambda$, then $d\vec{r}/d\lambda$ is a unit vector pointing along the path of the loop. This means that $$ \vec{I} = I \frac{d\vec{r}}{d\lambda} $$ This then means that we have $$ \vec{\mu} = I \left[\frac{1}{2} \int \vec{r} \times d\vec{r} \right] $$ and it can be shown that the quantity in brackets is equal to the vector area $\vec{a}$ of the loop. Thus, the fundamental equations above reduce to the equation $\vec{\mu} = I \vec{a}$ for a current loop.
Connection to bar magnet formula
The magnetization $\vec{M}$ of a medium is related to the bound current densities in that medium by $$ \vec{J}_b = \vec{\nabla} \times \vec{M}, \qquad \vec{K}_b = \vec{M} \times \hat{n} $$ and so in the absence of free currents we have $$ \vec{\mu} = \frac{1}{2} \iiint \vec{r} \times (\vec{\nabla} \times \vec{m}) \, d \tau + \frac{1}{2} \iint \vec{r} \times (\vec{M} \times \hat{n}) \, d a $$ Via a significant amount of algebra (see Zangwill's Modern Electrodynamics, §13.2.4 for the gory details), it is possible to show that $$\vec{\mu} = \iiint \vec{M} \, d\tau.$$
For a bar magnet, we typically have a cylinder with cross-sectional area $A$ and length $\ell$, with a uniform magnetization $\vec{M}$ along its axis. Call this axis the $z$-axis for concreteness, so that $\vec{M} = M \hat{z}$. Then in this case we have $$ \vec{\mu} = A \ell M \hat{z} = (AM) \vec{\ell}. $$ So if we identify $AM$ (the magnetization times the pole area) as the "pole strength" $m$ of the magnet,1 we end up with $\vec{\mu} = m \vec{\ell}$, as expected.
1 This definition can be further motivated by defining a "fictitious magnetic charge" such that $\rho_B = - \vec{\nabla} \cdot \vec{M}$ and $\sigma_B = \vec{M} \cdot \hat{n}$. Under this definition, the magnetic field of any magnetized object in the region where $\vec{M} = 0$ acts just like an electric field created by these same charge distributions. Moreover, for bar magnet, we have $\rho_B = 0$ and a "magnetic charge" on each pole of $\sigma A = \pm M A = \pm m$; and if $A$ is "small", then the magnetic field looks like the electric field of a dipole with "charges" $\pm m$. See Zangwill §13.4 for details & motivation of this construction.