$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

$A$ charge $+Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

$A$ charged particle is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. For the magnetic force on it to be maximum,which of the following is correct?

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.

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?