When a charged particle enters a uniform magnetic field,its kinetic energy

$A$ particle of specific charge $q / m = \pi \text{ C kg}^{-1}$ is projected from the origin towards the positive $X$-axis with a velocity $10 \text{ ms}^{-1}$ in a uniform magnetic field $\vec{B} = -2 \hat{k} \text{ T}$. The velocity $\vec{v}$ of the particle after time $t = \frac{1}{12} \text{ s}$ will be (in $\text{ ms}^{-1}$):

$A$ proton moving with a velocity $2.5 \times 10^7 \ m/s$ enters a magnetic field of intensity $2.5 \ T$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is

Statement-$1$: The path of a charged particle may be a straight line in a uniform magnetic field.
Statement-$2$: The path of a charged particle is decided by the angle between its velocity and the magnetic field acting on it.

$A$ charged particle is moving in a uniform magnetic field. It penetrates a layer of lead and thereby loses half of its kinetic energy. What happens to the radius of curvature of its path?

$A$ proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is