$A$ $10 \; eV$ electron is circulating in a plane at right angles to a uniform magnetic field of magnetic induction $10^{-4} \; Wb/m^2$ $(1.0 \; \text{gauss})$. The orbital radius of the electron is ........ $cm$.

$A$ 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

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

$A$ particle of mass $m$ and charge $q$ is incident on the $XZ$ plane with velocity $v$ in a direction making an angle $\theta$ with a uniform magnetic field applied along the $X$-axis. The nature of motion performed by the particle is . . . . . . .

If a proton of kinetic energy $8.35 \text{ MeV}$ enters a uniform magnetic field of $10 \text{ T}$ at right angles to the direction of the field,then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27} \text{ kg}$ and charge of proton $= 1.6 \times 10^{-19} \text{ C}$)

$A$ 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?