An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: If oxygen ion $(O^{-2})$ and hydrogen ion $(H^{+})$ enter normal to the magnetic field with equal momentum,then the path of $O^{-2}$ ion has a smaller curvature than that of $H^{+}$.
Reason $R$: $A$ proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statement,choose the correct answer from the options given below.

An electron,a proton,and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e, r_p$,and $r_{\alpha}$ respectively in a uniform magnetic field $B$. The relation between $r_e, r_p$,and $r_{\alpha}$ is:

$A$ proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$,perpendicular to the plane of the motion,the proton will move in a circular path of radius

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)