Finding net magnetic and electric force on charged particle 
This is from my textbook, it is not an assigned problem, but I want to understand. 
It says: 

Consider the situation in the figure, in which there is a uniform electric field in the x direction and a uniform magnetic field in the y direction. For each example of a proton at rest or moving in the x, y, or z direction, what is the direction of the net electric and magnetic force on the proton at this instant?

I believe I need to use the equation 
$$ F_{net}=(q\overrightarrow{E})+q(\overrightarrow{v} \times \overrightarrow{B}) $$
But I'm not sure exactly how. We've just started to learn about this, and I want to get a head start. Could anyone put me on the right track?
 A: First of all, I think you might have written the equation for the net force incorrectly: $$\vec{F}_{net} = q\vec{E} + q(\vec{v} \times \vec{B})$$ The second term is $q(\vec{v} \times \vec{B})$ and not $q(\vec{E} \times \vec{B})$.
From the first term of the force equation ($q\vec{E}$), we can see that the electric field will try push the proton parallel to it (so the proton will be pushed a bit in the $\hat{x}$-direction). 
Note that the second term of the equation ($q(\vec{v} \times \vec{B})$) is perpendicular to the velocity (direction of motion) of the proton. 
You might remember this from earlier: when a force acts perpendicular to the motion of a body, the force acts centripetally - that is, the body starts revolving in a circle. Therefore, the magnetic force is a centripetal force.
So we have two ways the proton can be pushed: the electric field pushes it in the $\hat{x}$-direction and the and magnetic field (when the proton is moving) tries to get the proton to revolve around the axis perpendicular to the direction of motion ($\hat{v}$) and the magnetic field ($\hat{B}$) (remember, the second term is a cross product). 
The proton gets pushed in both ways at the same time (in the first example, it doesn't move at first, but then starts to move as the electric field pushes it, so the magnetic force appears too). It might be a bit hard to visualize, but I'll talk about the first example: the proton is both first pushed in the $\hat{x}$-direction (so its velocity is non-zero) and then undergoes the centripetal force (from the magnetic field) which causes it to start revolving around the x-axis. However, just before it dips under the z-axis, the proton stops moving (the electric field caused it to decelerate), causing the electric field to accelerate it again, restarting the process while causing the proton to move a net displacement along the z-axis (since it never rotated back to the origin).

(source: physics-animations.com) 
(Note that the directions in the animation are not the same as in the second example). 
Can you start to visualize how it will move in the other examples?
A: The question asks for direction of force "at this instant" so we need to look at the force on the particle when it has the velocity as given (no need to figure out what happens after that).
Now the electrical force is
$$\vec{F_e} = q\vec{E}$$
so the electrical force is always in the X direction (along $\vec{E}$).
The magnetic force is perpendicular to both velocity and magnetic field, given by the cross product:
$$\vec{F_m}=q\vec{v}\times\vec{B}$$
When the proton is at rest, it does not initially experience a force from the B field (since the magnetic force depends on both B and v) and there will be just the electrical force along x.
For the other examples, the electrical force will be along x again each time (it is independent of velocity) while the magnetic force is always perpendicular to B and v.
This means that when the particle is traveling in the x direction, magnetic force will be in the z direction ($\vec{x}\times\vec{y}$); when particle is traveling in the y direction, there will be no force ($\vec{y}\times\vec{y}=0$), and when it's traveling in the z direction, magnetic force will be towards -x.
