Lorentz force and charged partice's motion of equation in cylindrical coordinates 
*

*Is the Lorentz force
$$\textbf{F} = q(\textbf{E} + \textbf{v} \times \textbf{B})$$
same in Descartes and in cylindrical coordinates? 

*Moreover do the motion of equation of a charged partice with $\textbf{E} = 0$
\begin{align*}
m\dot{\textbf{v}} &= q(\textbf{v} \times \textbf{B})\\
\dot{\textbf{r}} &= \textbf{v}
\end{align*}
has different a form in the above mentioned two coordinate systems?
I have a feeling that the cross product changes somehow.
 A: When you work in a curvilinear coordinate system, like cylindrical coordinates, the direction of the unit vectors changes with respect to the underlying cartesian background we naturally like to think in. It turns out the force law you wrote down is called "covariant," it does not depend on a choice of coordinate system. And this is indeed a guiding principle in physics that really came into focus with the work of Einstein. That is not to say that the way one goes about actually computing numerical quantities might not be different in different coordinate systems. But part of the reason the law is written in terms of a cross product is that the cross product does not depend on a coordinate system, it is a scheme for mapping vectors into vectors. How you choose to represent those vectors doesn't change the mapping.
Let me show you how you would go about answering this question via brute force to give you some intuition.
First, you should note that the cross product itself is not a quantity that depends on coordinate system, it's a map that assigns a vector to any two other vectors in 3 dimensions. We can choose to represent vectors in any coordinate system, figure out what this law says and transform the answer into another coordinate system and see what the law looks like in the new coordinates. We want to start with Cartesian coordinates: 
$$ \vec{r} = (x,y,z), $$
and move to Cylindrical coordinates: 
$$ \vec{r} = (r,\theta,z). $$
If we think about vectors as arrows in space attached to the origin, then we can work out that the relationship between the coordinate representation $(r, \theta, z)$ and the coordinate representation $(x,y,z)$ is $r=\sqrt{x^2 + y^2}$, $\tan(\theta) = \frac{y}{x}$, and $z=z$. Also, as we move around in space, the direction of increasing $r$ depends on $\theta$. When $\theta = 0$, moving in $\hat{r}$ is the same as moving in $\hat{x}$, at $\theta = \frac{\pi}{2}$, moving in $\hat{r}$ is the same as moving in $\hat{y}$, and in general you can work out by that the direction of increasing $r$ is $\hat{r} = \frac{x\hat{x} + y\hat{y}}{\sqrt{x^2 + y^2}} = \cos(\theta)\hat{x} + \sin(\theta)\hat{y}$. And for the direction of increasing $\theta$ we have  $\hat{\theta} = \frac{-y\hat{x} + x\hat{y}}{\sqrt{x^2 + y^2}} = -\sin(\theta)\hat{x} + \cos(\theta)\hat{y}$
Since you know the relationship between the coordinates, then you can work out what the formula for the components of the cross product between two vectors in a curvilinear coordinate system should be. Starting from two arbitrary vectors in Cartesean coordinates:
$$\vec{a} = (a_1, a_2, a_3) \\
\vec{b} = (b_1,b_2,b_3)$$
and the definition of the cross product 
$$ \vec{a} \times \vec{b} = \hat{x}(a_2b_3 - b_2 a_3) - \hat{y}(a_1b_3 - b_1 a_3) + \hat{z}(a_1b_2 - b_1 a_2)$$
all that is left to do is plug in the substitution of the components of $\vec{a}$ and $\vec{b}$ in cylindrical coordinates and the same for the unit vectors. 
If you go through with this, you will see that it turns out that the same sort of determinant rule works even in cylindrical coordinates. 
$$\vec{a} \times \vec{b} = \hat{r}(a_2b_3 - b_2 a_3) - \hat{\theta}(a_1b_3 - b_1 a_3) + \hat{z}(a_1b_2 - b_1 a_2),$$
where the $a_i$ and $b_i$ are the components of $\vec{a}$ and $\vec{b}$ in cylindrical coordinates this time. This is something you can try to understand as you learn more mathematics, but it's actually a very important feature of the cross product. The determinant rule works in any coordinate system where the unit vectors are orthogonal.
