A uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3 R}{2}$ (region $2$ in the figure) pointing normally into the plane of the paper. A particle with charge $+Q$ and momentum $p$ directed along $x$-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 $B=\frac{8}{13} \frac{\mathrm{p}}{QR}$, the particle will enter region $3$ through the point $P_2$ on $\mathrm{x}$-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 $y$-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

If $\alpha $ and $\beta  – $ particles are moving with equal velocity perpendicular to the flux density $B$, then the radii of their paths will be

A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{ i }(v \neq 0)$ towards a large screen placed in the $Y – Z$ plane at a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is

A proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,\, MeV,$ the energy acquired by the alpha particle will be……$MeV$

A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled, then the radius of the circular path will be…..$cm$