If a proton enters perpendicularly a magnetic field with velocity $v$, then time period of revolution is $T$. If proton enters with velocity $2 v$, then time period will be

Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $q$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then, they will meet again for the first time at a point whose distance from the point of projection is

A proton and an electron both moving with the same velocity $v$ enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that

A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{ i }(v \neq 0)$ towards a large screen placed in the $Y - Z$ plane at a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e,r_p$ and ${r_\alpha }$ respectively in a uniform magnetic field $B$. The relation between $r_e,r_p$ and $\;{r_\alpha }$ is

An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall