$A$ mixed beam of $He^+$ and $O^{2+}$ ions (mass of $He^+ = 4 \ amu$ and that of $O^{2+} = 16 \ amu$) passes through a region of a constant perpendicular magnetic field. If the kinetic energy of all the ions is the same,then:

$A$ $5 \text{ mg}$ particle carrying a charge of $5 \times 10^{-6} \text{ C}$ is moving with a velocity of $(3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}$ in a region having a magnetic field $\vec{B} = 0.1\hat{k} \text{ Wb/m}^2$. It moves a distance of $\alpha \text{ m}$ along the $\hat{k}$ direction when it completes $5$ revolutions. The value of $\alpha$ 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.

The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the Earth's magnetic field points down. The electron is deflected towards:

If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field,then

An electron has mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is moving with a velocity of $10^6 \ m/s$. It enters a region where a magnetic field exists. If it describes a circle of radius $0.10 \ m$,the intensity of the magnetic field must be: