$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{i}$ with an initial velocity $\vec{v} = v_0 \hat{j}$. The speed of the particle will become $2v_0$ after a time:

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

An electron gun with its collector at a potential of $100 \; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \; mm$ of $Hg)$. $A$ magnetic field of $2.83 \times 10^{-4} \; T$ curves the path of the electrons in a circular orbit of radius $12.0 \; cm$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons,and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e/m$ from the data.

$A$ charged particle moves with velocity $\overrightarrow{v}$ in a uniform magnetic field $\overrightarrow{B}$. The magnetic force experienced by the particle is

An electron is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. The angle $\theta$ between $\vec{v}$ and $\vec{B}$ lies between $0^\circ$ and $\frac{\pi}{2}$. Its velocity vector $\vec{v}$ returns to its initial value in a time interval of: