An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

$A$ uniform electric field and a uniform magnetic field are produced,pointing in the same direction. An electron is projected with its velocity pointing in the same direction.

Two particles of masses $m_{a}$ and $m_{b}$ and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius $r_{a}$ and $r_{b}$ such that $r_{a} > r_{b}$. Then which is true?

$A$ 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:

$A$ 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 an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2 \text{ mT}$ perpendicular to the direction of the field. After some time,it is found that the velocity of the proton has changed in direction by $90^{\circ}$. Then,at this time,the angle between the velocity vectors of the proton and the $\alpha$-particle is (in $^{\circ}$)