When an electron beam passes through an electric field,it gains kinetic energy. If the same beam passes through a magnetic field,then:

$A$ uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3R}{2}$ (region $2$ in the figure) pointing normally into the plane of the paper. $A$ particle with charge $+Q$ and momentum $p$ directed along the $x$-axis enters region $2$ from region $1$ at point $P_1(y=-R)$. Which of the following option$(s)$ is/are correct?
$[A]$ For $B > \frac{2}{3} \frac{p}{QR}$,the particle will re-enter region $1$.
$[B]$ For $B = \frac{8}{13} \frac{p}{QR}$,the particle will enter region $3$ through the point $P_2$ on the $x$-axis.
$[C]$ When the particle re-enters region $1$ through the longest possible path in region $2$,the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from the $y$-axis is $p/\sqrt{2}$.
$[D]$ For a fixed $B$,particles of same charge $Q$ and same velocity $v$,the distance between the point $P_1$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle.

$A$ 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

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$. The radius of the circular path described by it will be:

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?