Regarding the $E × B$ drift in the Earth's magnetic field So I have a burning question: The only reason that the E x B drift doesn't generate an electric current is because both the electrons and the positive ions move towards the same direction (towards Earth's ionosphere) therefore a charge separation isn't formed? Are opposite velocities the deal breaker for charge separation or am I missing something else? 
 A: In uniform crossed magnetic and electric fields, the motion of a charge can be separated into a gyromotion and a constant drift. A constant motion implies zero force, so the charge doesn't affect the drift velocity.
The drift velocity is derived in George K. Parks, Physics of Space Plasmas (1991) pages 93-97, and the derivation is a bit too long to reproduce here, so I'll just give the answer:
$$\mathbf{v} = \frac{\mathbf{E} \times \mathbf{B}}{B^2}.$$
Note that charge doesn't appear in this expression. The Lorentz force on a particle with charge $q$ is
$$ \mathbf{F} = q\mathbf{E} + q\mathbf{v} \times \mathbf{B}.$$
If you substitute the drift velocity in this equation and use $\left(\mathbf{E} \times \mathbf{B}\right) \times \mathbf{B} = \left(\mathbf{E}\cdot\mathbf{B}\right)\mathbf{B} - B^2\mathbf{E}$), you can confirm that $\mathbf{F} = 0.$
A: 
The only reason that the E x B drift doesn't generate an electric current is because both the electrons and the positive ions move towards the same direction...

The ExB-drift is independent of charge, so yes, both ions and electrons will undergo the same ExB-drift velocity.

...therefore a charge separation isn't formed?

Counter streaming particles of opposite charge do not a charge separation make.  Rather, such a scenario leads to a current given by:
$$
\mathbf{j} = \sum_{s} \ n_{s} \ q_{s} \ \mathbf{V}_{s}
$$
where $n_{s}$ is the number density, $q_{s}$ is the total charge (including sign), and $\mathbf{V}_{s}$ is the bulk velocity vector of species $s$.
As an example, if electrons and protons flow in the same direction both at $\mathbf{V}_{o}$, there would be zero net current (assuming quasi-neutrality) because the two species have the opposite charge.  If they flow relative to each other, i.e., $\mathbf{V}_{e} \neq \mathbf{V}_{i}$, then there will be a net current.

Are opposite velocities the deal breaker for charge separation or am I missing something else?

Charge separation in a plasma is generally difficult because it results in electric fields.  Since a plasma is an ionized gas of freely moving electrons and ions, any electric field will quickly act to eliminate itself by doing work on the charged particles.  Most plasmas have incredibly high conductivities, so the lifetime of charge separations are very short.
I think you are confusing flow with charge separation, which are two different phenomena.  Particle species in plasmas can flow relative to each other without generating any charge separation.
