$A$ particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly into a region of uniform magnetic field $B$ and executes a circular arc of radius $R$. Then,$\frac{q}{m}$ equals:

$A$ $1 \mu\text{C}$ charge is moving with velocity $\vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k}) \text{ m/s}$ in a region of magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \text{ T}$. The magnitude of the force acting on it is $\sqrt{\alpha} \times 10^{-6} \text{ N}$. The value of $\alpha$ is . . . . . . .

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field,then the orbit will be

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

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.