$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon:

Two electrons,$e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the kinetic energy of $e_1$ is double than that of $e_2$. The relation of their frequencies of rotation,$f_1$ and $f_2$ is

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.

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix

$A$ proton, a deuteron, and an $\alpha$-particle enter a region of a uniform magnetic field perpendicular to their velocities with the same kinetic energy. If $r_p, r_d,$ and $r_\alpha$ are the radii of the circular paths of these particles, respectively, then:

$A$ charge of $1\,C$ is moving in a magnetic field of $0.5\,T$ with a velocity of $10\,m/s$ perpendicular to the field. The force experienced is.....$N$.