A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is $R$. When energy of particle is doubled, then new radius will be

At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v  = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that

If two protons are moving with speed $v=4.5 \times 10^{5} \,m / s$ parallel to each other then the ratio of electrostatic and magnetic force between them

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

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 (mass = $9.1 \times {10^{ - 31}}$ $kg$; charge = $1.6 \times {10^{ - 19}}$ $C$) experiences no deflection if subjected to an electric field of $3.2 \times {10^5}$ $V/m$, and a magnetic fields of $2.0 \times {10^{ - 3}} \,Wb/m^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius.......$m$