An electron and a proton of equal linear momentum enter in a direction perpendicular to a uniform magnetic field. If the radii of their circular paths are $r_e$ and $r_p$ respectively,then $\frac{r_e}{r_p}$ is equal to - (mass of electron $= m_e$,mass of proton $= m_p$)

$A$ $10 \; eV$ 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 ........ $cm$.

$A$ positive,singly ionized atom of mass number $A_M$ is accelerated from rest by a voltage $V = 192 \text{ V}$. Thereafter,it enters a rectangular region of width $w$ with a magnetic field $B_0 = 0.1 \hat{k} \text{ T}$,as shown in the figure. The ion finally hits a detector at a distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $= (5/3) \times 10^{-27} \text{ kg}$,charge of the electron $= 1.6 \times 10^{-19} \text{ C}$.]
Which of the following option$(s)$ is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ ion is $4 \text{ cm}$.
$(B)$ The value of $x$ for an ion with $A_M = 144$ is $48 \text{ cm}$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$,the minimum height $(x_1 - x_0)$ of the detector is $52 \text{ cm}$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $28 \text{ cm}$.

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that

An electron having a charge $e$ moves with a velocity $\vec{v}$ in the positive $x$-direction. $A$ magnetic field $\vec{B}$ acts on it in the positive $y$-direction. The force on the electron acts in (where the outward direction is taken as the positive $z$-axis):