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

The radius of curvature of the path of a charged particle moving in a static uniform magnetic field is

A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is  $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If $r_{d}$ and $r_{p}$ are the radii of their circular paths respectively, then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt{ x }: 1$ where $x$ is ..........

When a charged particle moving with velocity $\vec V$ is subjected to a magnetic field of induction $\vec B$ , the force on it is non-zero. This implies the

If two protons are moving with speed $v=4.5 \times 10^{5} \,m / s$ parallel to each other then the ratio of electrostatic and magnetic force between them