$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is:

$A$ velocity selector is to be constructed to select ions with a velocity of $6 \,km \,s^{-1}$. If the electric field used is $400 \,V \,m^{-1}$, then the magnetic field to be used is

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$: $A$ charged particle is moving at a right angle to a static magnetic field. During the motion,the kinetic energy of the charge remains unchanged.
Statement $2$: $A$ static magnetic field exerts a force on a moving charge in the direction perpendicular to the magnetic field.

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

$A$ $10 \; eV$ electron is circulating in a plane at right angles to a uniform magnetic field of magnetic induction $10^{-4} \; Wb/m^2$ $(1.0 \; \text{gauss})$. The orbital radius of the electron is ........ $cm$.