Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a uniform magnetic field,speed and energy remain the same for a moving charged particle.
Reason $(R)$: $A$ moving charged particle experiences a magnetic force perpendicular to its direction of motion.

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.

$A$ uniform magnetic field $B$ exists in a region. An electron of charge $q$ and mass $m$ moving with velocity $v$ enters the region in a direction perpendicular to the magnetic field. Considering Bohr angular momentum quantization, which of the following statement$(s)$ is/are true?

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

$A$ charged particle is moving in a circular orbit of radius $6 \, cm$ with a uniform speed of $3 \times 10^6 \, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4} \, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be: