$A$ particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform electric and magnetic fields, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis, under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

$A$ magnetic field $\overrightarrow{B} = B_0 \hat{j}$ exists in the region $a < x < 2a$ and $\overrightarrow{B} = -B_0 \hat{j}$ in the region $2a < x < 3a$,where $B_0$ is a positive constant. $A$ positive point charge moving with a velocity $\overrightarrow{v} = v_0 \hat{i}$,where $v_0$ is a positive constant,enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like,

$A$ particle with $10^{-11} \, C$ of charge and $10^{-7} \, kg$ mass is moving with a velocity of $10^8 \, m/s$ along the $y$-axis. $A$ uniform static magnetic field $B = 0.5 \, T$ is acting along the $x$-direction. The force on the particle is:

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

The velocity of certain ions that pass undeflected through crossed electric field $E = 7.7 \, kV/m$ and magnetic field $B = 0.14 \, T$ is.....$km/s$