$A$ proton and a deuteron $(q=+e, m=2.0 \ u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$,moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of the deuteron path to the radius $r_p$ of the proton path is:

Two ions of masses $4 \, amu$ and $16 \, amu$ have charges $+2e$ and $+3e$ respectively. These ions pass through a region of constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then:

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, Wb/m^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is

$A$ deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to

An electron falling freely under the influence of gravity enters a uniform magnetic field directed towards the south. The electron is initially deflected towards: