$A$ proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of the radii of their circular paths is

$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon:

$A$ 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:

An electron has mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is moving with a velocity of $10^6 \ m/s$. It enters a region where a magnetic field exists. If it describes a circle of radius $0.10 \ m$,the intensity of the magnetic field must be:

Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)