$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

Two electrons,$e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the kinetic energy of $e_1$ is double than that of $e_2$. The relation of their frequencies of rotation,$f_1$ and $f_2$ is

An electron,a proton,and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e, r_p$,and $r_{\alpha}$ respectively in a uniform magnetic field $B$. The relation between $r_e, r_p$,and $r_{\alpha}$ is: