Three ions $H^+$,$He^+$,and $O^{2+}$ having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. Then:

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

$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

An electron enters the space between the plates of a charged capacitor as shown. The surface charge density on the plates is $\sigma$. The electric field intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in the space,perpendicular to the direction of $E$. The electron moves perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$ without any change in direction. The time taken by the electron to travel a distance $l$ in the space is:

$A$ proton, a deuteron, and an $\alpha$-particle enter a region of a uniform magnetic field perpendicular to their velocities with the same kinetic energy. If $r_p, r_d,$ and $r_\alpha$ are the radii of the circular paths of these particles, respectively, then: