An electron moving with a speed $u$ along the positive $x-$axis at $y = 0$ enters a region of uniform magnetic field $\overrightarrow B = - {B_0}\hat k$ which exists to the right of $y$-axis. The electron exits from the region after some time with the speed $v$ at co-ordinate $y$, then

A proton (mass $m$ ) accelerated by a potential difference $V$  flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be

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......$MeV$

An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again it is projetced with velocity $v_0$ perpendicular to a uniform magnetic field $B/$ If $r_1$ is initial radius of curvature just after entering in the electric field and $r_2$ is initial radius of curvature just after entering in magnetic field then the ratio $r_1:r_2$ is equal to 

A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be

If a charged particle goes unaccelerated in a region containing electric and magnetic fields