An electron and a proton enter region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius ${r_e}$ and ${r_p}$ respectively. Then

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

A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. If the direction of the magnetic field is along the outward normal to the plane of the paper, then the time spent by the particle in the region of the magnetic field after entering it at $C$ is nearly :-......$ns$

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 same radius.......$MeV$

A particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and magnetic field $\vec B = {B_0}\hat k$. Its velocity at $(x_0 , y_0 , 0)$ is ($(4\hat i + 3\hat j)$ . The value of $x_0$ is: 

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