$A$ proton and an electron both moving with the same velocity $v$ enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that

$A$ 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.

If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field,then

An $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.

An electron enters with a velocity $\vec{v} = v_0 \hat{i}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the $xy$-plane. Suggest a configuration of fields $\vec{E}$ and $\vec{B}$ that can lead to this.

$(a)$ $A$ monoenergetic electron beam with electron speed of $5.20 \times 10^{6} \;m s^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} \;T$ normal to the beam velocity. What is the radius of the circle traced by the beam,given $e/m$ for electron equals $1.76 \times 10^{11} \;C \;kg^{-1}$?
$(b)$ Is the formula you employ in $(a)$ valid for calculating the radius of the path of a $20 \;MeV$ electron beam? If not,in what way is it modified?