Electronic – Why is the magnetic field perpendicular to the magnetic force

One way to think of a inductor from the circuit's point of view is that it gives inertia to current. This gives you some intuitive feel for why current builds up slowly as voltage is applied, and why the voltage goes as high as it needs to on a attempt to shut off the current abruptly. This isn't how the physics works, but this can be a very useful mental model for understanding and designing circuits with inductors.

Gauss Law. Polarization Vector

The Gauss Law brings the local relation between the electric field and the sources.
The main sources of electric field are free charges, but we can also consider the contribution of the field produced by the polarized material.

$$ \nabla\cdot\mathbf{E}(\mathbf{r}) =\dfrac{\rho_l(\mathbf{r}) + \rho_P(\mathbf{r})}{\varepsilon_0} $$

where \$\rho_l\$ is the free charge contribution, and \$\rho_P\$ is the contribution due to polarization. But

$$ \rho_P(\mathbf{r})=-\nabla\cdot\mathbf{P}(\mathbf{r}) $$

where \$\mathbf{P}(\mathbf{r})\$ is the Polarization Vector. Then

$$ \nabla\cdot\mathbf{E}(\mathbf{r})=\dfrac{\rho_l(\mathbf{r})-\nabla\cdot\mathbf{P}(\mathbf{r})}{\varepsilon_0} $$

The free charge density is

$$ \rho_l(\mathbf{r})=\nabla\cdot\left(\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})\right) $$

remember that \$\nabla\cdot\mathbf{D}(\mathbf{r})=\rho_l(\mathbf{r})\$, we can write the general form of the Gauss Law:

$$ \mathbf{D}(\mathbf{r})=\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r}) $$

The displacement vector \$\mathbf{D}(\mathbf{r})\$ is the combination of the applied field \$\mathbf{E}\$ and induced field \$\mathbf{P}\$ in the material by the polarization of its molecules. The polarization of a material depends on the external field, and in turn creates an induced field which overlaps the external field. Then there is a relationship between these fields, in particular between the polarization vector and total field (the field that can be measured).

For linear dielectrics (which are the most technological interest materials) applies: \$\mathbf{P}(\mathbf{r})=\chi_e\,\varepsilon_0\,\mathbf{E}(\mathbf{r})\$ and then

$$ \mathbf{D}(\mathbf{r})=\varepsilon_0(1+\chi)\mathbf{E}(\mathbf{r})=\varepsilon_0\,\varepsilon_r\,\mathbf{E(r)}=\varepsilon\,\mathbf{E(r)} $$

where \$\chi\$ is the dielectric susceptibility of the material. \$\varepsilon=\varepsilon_0\,\varepsilon_r=\varepsilon_0(1+\chi)\$ is the permittivity of material, and \$\varepsilon_r\$ is the relative permittivity.

The greater the permittivity of the material, is polarized more strongly and electrical effects are greater.