$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

An electron is travelling in the east direction and a magnetic field is applied in the upward direction. In which direction will the electron deflect?

$A$ 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?

An $\alpha$-particle (mass $4 \text{ amu}$) and a singly charged sulfur ion (mass $32 \text{ amu}$) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region,the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_s$,respectively. The ratio $(r_s / r_\alpha)$ is:

$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

If the velocity of an electron is $(2 \hat{i} + 3 \hat{j}) \text{ m s}^{-1}$ and it enters a magnetic field of $4 \hat{k} \text{ T}$,then . . . . . . .