What is the behavior of a charged particle moving in a region where the electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other?

If the magnetic field is parallel to the positive $y$-axis and the charged particle is moving along the positive $x$-axis (Figure),which way would the Lorentz force be for
$(a)$ an electron (negative charge),
$(b)$ a proton (positive charge).

In the product $\overrightarrow{F} = q(\vec{v} \times \overrightarrow{B})$ where $\overrightarrow{B} = B \hat{i} + B \hat{j} + B_{0} \hat{k}$. Given $q = 1$,$\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$,and $\overrightarrow{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$,what is the complete expression for $\overrightarrow{B}$?

$A$ charge $q$ moves with velocity $\vec{V}$ through an electric field $\vec{E}$ as well as a magnetic field $\vec{B}$. Then the force acting on it is:

Give the definition of magnetic field and provide its unit.

$A$ particle of mass $1 \times 10^{-26} \,kg$ and charge $1.6 \times 10^{-19} \,C$ travelling with a velocity $1.28 \times 10^6 \,ms^{-1}$ along the positive $X$-axis enters a region in which a uniform electric field $E$ and a uniform magnetic field of induction $B$ are present. If $E = -102.4 \times 10^3 \hat{k} \,NC^{-1}$ and $B = 8 \times 10^{-2} \hat{j} \,Wbm^{-2}$, the direction of motion of the particle is: