An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

An electron (mass = $9.0 × $${10^{ - 31}}$ $kg$ and charge =$1.6 \times {10^{ - 19}}$ $coulomb$) is moving in a circular orbit in a magnetic field of $1.0 \times {10^{ - 4}}\,weber/{m^2}.$ Its period of revolution is

A beam of ions with velocity $2 \times {10^5}\,m/s$ enters normally into a uniform magnetic field of $4 \times {10^{ - 2}}\,tesla$. If the specific charge of the ion is $5 \times {10^7}\,C/kg$, then the radius of the circular path described will be.......$m$

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

As shown in the figure, the uniform magnetic field between the two identical plates is $B$. There is a hole in plate. If through this hole a particle of charge $q$, mass $m$ and energy $E$ enters this magnetic field, then the particle will not collide  with the upper plate provided

A proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to direction of magnetic field. If both the particles move around magnetic field in circles of equal radii, the ratio of momentum of proton to alpha particle $\left( {\frac{{{P_p}}}{{{P_\alpha }}}} \right)$ is