Bound and free current or charges I'm pretty confused with the concepts of bound and free charges as well as current.
1.Bound charge density (pb) arises due to polarization of a  dielectric material. Free charge density (pf) have no relationship with the polarisation of material, it solely depends on the current flowing through the material.
The net sum of bound charge density and free charge density is zero as the conductor on the whole has to be electrically neutral.
2.Bound current density (Jb) arises due to magnetisation of a material. Free current density (Jf) has no relationship with the magnetisation of material, it solely depends on the magnetic flux density through the material.
The net sum of bound current density and free curreny density is zero as the conductor on the whole has to be electrically neutral.
Am i right in my approach or I'm wrong?
 A: In electrodynamics of continuous media, the macroscopic charge density $\langle \rho \rangle$ is obtained from the microscopic charge density $\rho$ by averaging over macroscopically small, but microscopically large spatial regions. (For details see standard textbooks like Jackson or Landau-Lifshitz vol. 8). The macroscopic charge density can be decomposed into a free charge density $\rho_{\rm free}$ (corresponding to free charges) and and a bound charge density $\rho_{\rm {\rm pol}}$ (related to the electric polarizability of neutral matter = dielectric materials), i.e. $\langle \rho \rangle = \rho_{\rm free} + \rho_{\rm pol}$. In the dipole approximation, the bound charge density can be written as $\rho_{\rm pol}= - \vec{\nabla} \cdot \vec{P}$, where $\vec{P}$ is the dipole density.
Analogously, the macroscopic current density can be decomposed as $\langle \vec{j} \rangle = \vec{j}_{\rm free}+ \vec{j}_{\rm pol} + \vec{j}_{\rm mag}$. The continuity equation (holding separately for the three parts!) requires $\vec{j}_{\rm pol}= \partial \vec{P} / \partial t$. Again in dipole approximation, $\vec{j}_{\rm mag}$ is given by $\vec{j}_{\rm mag} = c \vec{\nabla} \times \vec{M}$, where the magnetisation $\vec{M}$ is the magnetic dipole density.
Your claim that "the net sum of bound charge density and free charge density is zero as the conductor on the whole has to be electrically neutral" is unfounded, as free charges can, of course, be present on a conductor (or also an isolating material). However, the total sum of the bound charges vanishes.
Also your second claim that "the net sum of bound charge current density and free current density is zero" is obviously wrong.
A: Dealing with polarized materials one can separate all the charges in the world in two groups, a group that are bound charges, which are the polarized materials. They have $\rho_b$ charge density. Which is:
$$
\rho_b = -\nabla\cdot P
$$
where:
$$
P \equiv \text{dipole moment per unit volume}
$$
and any other charge distribution, $\rho_f$ that has nothing to do with the polarization (which is not a current since we are dealing with electrostatics here. there is no current)
This approach to the charges shows useful because:
$$
\epsilon_0\nabla\cdot E = \rho_f + \rho_b \\
\epsilon_0\nabla\cdot E = \rho_f - \nabla\cdot P\\
\nabla\cdot*(\epsilon_0 E + P) = \rho_f\\
\nabla\cdot D = \rho_f
$$
$D$ is called the electrical displacement and solving this equation is easier than solving a polarized space in some cases.
The same analogy is in magnetostatics, we have currents densities that is caused because of magnetic properties of the material, $J_b$, and other currents in the world! Thus having:
$$
J_b =  \nabla\times M
$$
we can write:
$$
\nabla\times \frac1{\mu_0}B = J_b + J_f\\
\nabla\times(\frac1{\mu_0}B - M) = J_f\\
\nabla\times H = J_f
$$
