$A$ proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

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.

Two infinitely long straight wires $A$ and $B$, each carrying current $I$, are placed on the $x$ and $y$-axes, respectively. The current in wires $A$ and $B$ flows along $-\hat{i}$ and $\hat{j}$ directions, respectively. The force on a charged particle having charge $q$, moving from position $r = d(\hat{i} + \hat{j})$ with velocity $v = v\hat{i}$ is:

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?

$A$ magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region,a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. $A$ narrow beam of (single species) charged particles all accelerated through $15 \;kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{5} \;V \,m^{-1}$,make a simple guess as to what the beam contains. Why is the answer not unique?

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels: