$A$ uniform electric field and a uniform magnetic field are produced,pointing in the same direction. An electron is projected with its velocity pointing in the same direction.

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?

$A$ charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is

Two ions having masses in the ratio $1: 1$ and charges $1: 2$ are projected into a uniform magnetic field perpendicular to the field with speeds in the ratio $2: 3$. The ratio of the radii of circular paths along which the two particles move is :

At $t = 0$,a charge $q$ is at the origin and moving in the $y$-direction with velocity $\vec{v} = v\hat{j}$. The charge moves in a magnetic field that is for $y > 0$ out of the page and given by $B_1\hat{k}$ and for $y < 0$ into the page and given by $-B_2\hat{k}$. The charge's subsequent trajectory is shown in the sketch. From this information,we can deduce that:

The ratio of periods of $\alpha$-particle and proton moving on circular path in uniform magnetic field is . . . . . . .