A charge particle of charge $q$ and mass $m$ is accelerated through a potential diff. $V\, volts$. It enters a region of orthogonal magnetic field $B$. Then radius of its circular path will be

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

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2}\, T $. If the value of $e/m$ is  $1.76 \times 10^{11}\, C/kg $, the frequency of revolution of the electron is

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

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