$E \times B$ drift plasma physics Do you need to have an electric field in order to have an $\mathbf E \times \mathbf B$ drift? I'm confused because it's seems that there is an induced E-field from the charge separation due to forces acting oppositely on ions and electrons.
 A: Yes, you do need to have electric field in order to get an ExB drift. As the name suggests, this will happen in the direction of the cross product of the electric and magnetic field. 
$\vec{v}_{E \times B} = \frac{\vec{E} \times \vec{B} }{B^2}$
This drift does not depend on the charge of particle, meaning that it will move both ions and electrons in the same direction, and there will be no charge separation in this case. But there are other drifts that will cause such charge separation. For instance, any external force $\vec{F}$ acting upon those charges (for example gravity), will create charge separation, since:
$\vec{v}_{F} =\frac{1}{q} \frac{\vec{F} \times \vec{B} }{B^2}$ .  (note the dependance on $q$, which bares the sign of the charge)
Under the influence of force $\vec{F}$ positive and negative charges will drift in opposite directions, creating charge separation, which in turn will create electric field that will cause the ExB drift. 
The thing with drifts is that you will always have all of them at the same time (as long as you have external forces or fields to cause them), but the question one should ask is about their magnitude and hence their relevance in given situation. For example if you calculate drifts in Earth's ionosphere, you can see that the force drift caused by the gravitation is orders of magnitude smaller than drifts caused by the magnetic field (which cause charge separation) and related ExB drift. 
