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

$A$ 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:

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).

$A$ charge $q$ is released in the presence of an electric field $(E)$ and a magnetic field $(B)$. After some time,its velocity is $v$. Then:

Give the definition of $1\, T$ (Tesla) magnetic field.

$A$ particle having charge $10^{-9} \text{ C}$ moving in the $x-y$ plane in fields of $0.4 \hat{i} \text{ N/C}$ and $4 \times 10^{-3} \hat{k} \text{ T}$ experiences a force of $(4 \hat{i} + 2 \hat{j}) \times 10^{-10} \text{ N}$. The velocity of the particle at that instant is . . . . . . $\text{m/s}$.