Force acting on an electron moving with velocity $v$ in a magnetic field $B$ is ($e$ is the charge of electron).

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:

$A$ particle of mass $m$ and charge $q$ moving with velocity $\vec v$ enters a region of uniform magnetic field $\vec B.$ Then:

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

Assertion : $A$ proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow{B}$ in a direction perpendicular to $\overrightarrow{B}$,will describe circular trajectories of equal radii.