An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicularly, Then

A particle with charge to mass ratio, $\frac{q}{m} = \alpha $ is shot with a speed $v$ towards a wall at a distance $d$ perpendicular to the wall. The minimum value of $\vec B$ that exist in this region perpendicular to the projection of velocity for the particle not to hit the wall is

Ionized hydrogen atoms and $\alpha$ -particles with same momenta enters perpendicular to a constant magnetic field $B$. The ratio of their radii of their paths $\mathrm{r}_{\mathrm{H}}: \mathrm{r}_{\alpha}$ will be

A charge $Q$ moves parallel to a very long straight wire carrying a current $l$ as shown. The force on the charge is

A proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,\, MeV,$ the energy acquired by the alpha particle will be......$MeV$

Electron of mass $m$ and charge $q$ is travelling with a speed along a circular path of radius $r$ at right angles to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius