If a proton, deuteron, and $\alpha$-particle are accelerated by the same potential difference and enter perpendicular to a magnetic field, then the ratio of their kinetic energies 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 $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

Two charged particles $A$ and $B$ of masses $m$ and $2m$, charges $2q$ and $3q$ respectively, moving with the same velocity, enter a uniform magnetic field such that both particles make the same angle $( < 90^{\circ} )$ with the direction of the magnetic field. Then the ratio of the pitches of the helical paths of the particles $A$ and $B$ is

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

$A$ singly ionized magnesium atom $(A=24)$ ion is accelerated to kinetic energy $5\,keV$ and is projected perpendicularly into a magnetic field $B$ of magnitude $0.5\,T$. The radius of the path formed will be . . . . . . $cm$.

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .