An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

The charge on a particle $Y$ is double the charge on particle $X$. These two particles $X$ and $Y$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

$A$ proton of mass $1.67 \times 10^{-27} \, kg$ and charge $1.6 \times 10^{-19} \, C$ is projected with a speed of $2 \times 10^6 \, m/s$ at an angle of $60^\circ$ to the $X$-axis. If a uniform magnetic field of $0.104 \, T$ is applied along the $Y$-axis,the path of the proton is:

$A$ particle with charge $q$,moving with a momentum $p$,enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$,where $d < \frac{p}{Bq}$. The particle is deflected by an angle $\theta$ in crossing the field.

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

$A$ particle of mass $m$ and charge $q$ moving with velocity $\vec v$ enters a region of uniform magnetic field $\vec B.$ Then: