$A$ proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to the direction of the magnetic field. If both the particles move around the magnetic field in circles of equal radii,the ratio of the momentum of the proton to the alpha particle $\left( \frac{P_p}{P_\alpha} \right)$ is

$A$ 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 $^{\circ}$)

$A$ proton and an alpha particle moving with equal speeds enter normally into a uniform magnetic field. The ratio of times taken by the proton and the alpha particle to make one complete revolution in the magnetic field is

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?