If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?

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

The magnetic force acting on a charged particle of charge $-2\, \mu C$ in a magnetic field of $2\, T$ acting in the $y$-direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$,is:

If an electron and a proton having same momenta enter perpendicular to a magnetic field,then

Ionized hydrogen atoms and $\alpha$-particles with same momenta enter perpendicular to a constant magnetic field $B$. The ratio of the radii of their paths $r_{H}: r_{\alpha}$ will be

If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is