An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

An electron having mass $m$ and kinetic energy $K$ enters a uniform magnetic field $B$ perpendicularly. Its frequency will be:

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

$A$ charged particle moves in a magnetic field $\vec B = 10\,\hat i$ with initial velocity $\vec u = 5\hat i + 4\hat j$. The path of the particle will be:

Two charged particles,having the same kinetic energy,are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of the radii of their circular paths is $6:5$ and their respective masses ratio is $9:4$,then the ratio of their charges will be.