An electron, moving in a uniform magnetic field of induction of intensity $\vec B,$ has its radius directly proportional to

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6: 5$ and their respective masses ratio is $9: 4$. Then, the ratio of their charges will be.

A charged particle of specific charge $\alpha$ is released from origin at time $t = 0$ with velocity $\vec V = {V_o}\hat i + {V_o}\hat j$ in magnetic field $\vec B = {B_o}\hat i$ . The coordinates of the particle at time $t = \frac{\pi }{{{B_o}\alpha }}$ are (specific charge $\alpha = \,q/m$) 

A $2\, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5\, tesla$. The force on the proton is

A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to

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