$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

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.

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon: