Ionized hydrogen atoms and $\alpha$-particles with same momenta enter perpendicular to a constant magnetic field $B$. The ratio of the radii of their paths $r_{H}: r_{\alpha}$ will be

The magnetic field is uniform for $y > 0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y < 0$. $A$ proton,denoted by a filled circle,leaves $y = 0$ in the $-y$-direction with some speed,as shown below. Which of the following best denotes the trajectory of the proton?

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

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:

An electron,a proton,a deuteron,and an $\alpha -$ particle enter a uniform magnetic field normally with the same velocity. Which particle will have the maximum rotational frequency?