Analogue of charge distribution in magnetostatics

Short of magnetic monopoles, what is the analogue of charge distribution in magnetostatics? Can we call it magnetic pole distribution?

Yes there is: the current element $d\vec j = \vec j dV$. However $\vec j$ is restricted by $\vec \nabla \cdot \vec j =0$.

There is no true analogue of electric charge distribution in magnetostatics, since there are no "magnetic charges" or magnetic monopoles, which is elaborated by saying that the magnetic field contains no isolated sources or sinks. Mathematically, this is clear from the differential forms of Maxwell's equations by observing that the divergence of the E field yields charge distribution while the divergence of the magnetic field is zero. Hence, you cannot have a density of charges if charge is undefined.

1. A current density is like a charge distribution

Excluding radiation, electromagnetism is basically about one thing (perhaps a crude summary). It's about being able to know the electric and magnetic fields $\mathbf{E}$, $\mathbf{B}$, in different contexts, when you are given some combination of the electric charge distribution $\rho$, the current density $\mathbf{J}$, the electric scalar potential $\Phi$, or the magnetic vector potential $\mathbf{A}$. Consider Maxwell's equations:

$$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}\rho,$$ $$\nabla \cdot \mathbf{B} = 0,$$ $$\nabla \times \mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t},$$ $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}.$$

You see from the first equation (Gauss' law) that a charge distribution $\rho$ gives rise to an electric field $\mathbf{E}$, and you see from the last equation (Faraday-Lenz' law) that a current density $\mathbf{J}$ gives rise to a magnetic field $\mathbf{B}$. This incredible symmetry extends itself through electromagnetism, one simple example being Poisson's equations for the scalar and vector potentials -

$$\nabla^2\Phi = -\frac{1}{\epsilon_0}\rho$$ and $$\nabla^2\mathbf{A} = -\mu_0 \mathbf{J}.$$

In this way, the current density is very much an analogue of charge distributions for magnetic fields. The current density behaves in the same way with regard to magnetic fields as does the charge distribution for $\mathbf{E}$ fields.

Note also Maxwell's equations in the electrostatic case, where both $\nabla \times \mathbf{E} = \mathbf{0}$ and $\nabla \cdot \mathbf{B} = 0$.

2. If magnetic monopoles exist, then there is a magnetic charge distribution

Looking at Maxwell's equations as I wrote them above, you might notice that those four equations are begging us to discover magnetic monopoles. If we did, it would complete the symmetry and we might have something like

$$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}\rho,$$ $$\nabla \cdot \mathbf{B} = \mu_0\rho_m,$$ $$\nabla \times \mathbf{E} = -\mathbf{J_m} -\frac{\partial\mathbf{B}}{\partial t},$$ $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}.$$ Here I have symbolized $\rho_m$ as a charge distribution due to magnetic monopoles, and $\mathbf{J_m}$ as a current density due to flow of magnetic monopoles (i.e., a magnetic current).

You can, under some circumstances, get away with viewing a magnetized object as having a "magnetic charge density" embedded in & on it. There's a good discussion of this in §§13.3–4 of Zangwill's Modern Electrodynamics, which I will summarize briefly below.

If we look at Maxwell's equations in magnetized matter in the absence of free currents, we have $$\vec{\nabla} \times \vec{H} = 0, \qquad \vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}.$$ Here, $\vec{H}$ is the auxiliary magnetic field, and $\vec{M}$ is the magnetization of the matter present. Since $\nabla \times \vec{H} = 0$, we can define a scalar potential* $\psi$ such that $\vec{H} = - \vec{\nabla} \psi$; and if we define the (fictitious) magnetic charge density to be $$\rho^*_M = - \vec{\nabla} \cdot \vec{M},$$ then we have $$\nabla^2 \psi = - \rho^*_M,$$ i.e., $\psi$ is a solution to Poisson's equation with a "source" $\rho^*_M$, just as $V$ is a solution to Poisson's equation with a "source" $\rho$. Moreover, in a region outside the magnetized matter (where $\vec{M} = 0$), we have $\vec{H} = \vec{B}/\mu_0$; and so the magnetic field outside the matter is exactly the same as though there is a concentration of "magnetic charge" inside the matter.

As noted by Zangwill, this method of calculating magnetic fields is somewhat deprecated today; but it does provide some valuable intuition about why (for example) the field lines of a bar magnet look an awful lot like those of an electric dipole.

*Note to pedants: Assume that the underlying space is simply connected.