$A$ particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$. The radius of the circular path described by it will be:

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.

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

Two particles $A$ and $B$ having equal charges $+6\,C$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $2\,cm$ and $3\,cm$ respectively. The ratio of mass of $A$ to that of $B$ is

$A$ proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of the radii of their circular paths is

An electric charge moving with uniform velocity has