$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

An $\alpha$-particle and a proton travel with the same velocity in a magnetic field perpendicular to the direction of their velocities. Find the ratio of the radii of their circular paths.

An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is

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

$A$ particle having a charge of $10\,\mu C$ and $1\,\mu g$ mass moves along a circular path of $10\, cm$ radius in the presence of a uniform magnetic field of $0.1\, T$. When the charge is at point $P$,a uniform electric field is applied in the region such that the charge moves tangentially with a constant speed. The value of the electric field is......$V/m$