An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3}\,T$. The radius of path of the circle will approximately be :

Ionized hydrogen atoms and $\alpha$ -particles with same momenta enters perpendicular to a constant magnetic field $B$. The ratio of their radii of their paths $\mathrm{r}_{\mathrm{H}}: \mathrm{r}_{\alpha}$ will be

A proton of velocity $\left( {3\hat i + 2\hat j} \right)\,ms^{-1}$ enters a magnetic field of  $(2\hat j + 3\hat k)\, tesla$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times10^8\,Ckg^{-1}$)

When a charged particle moving with velocity $\vec V$ is subjected to a magnetic field of induction $\vec B$ , the force on it is non-zero. This implies the

An electron, moving along the $x-$ axis with an initial energy of $100\, eV$, enters a region of magnetic field $\vec B = (1.5\times10^{-3}T)\hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2\, cm$. The electron is detected at the point $Q$ on a screen placed $8\, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6\times10^{-19}\, C$, mass of electron $= 9.1\times10^{-31}\, kg$)

A positive charge $'q'$ of mass $'m'$ is moving along the $+ x$ axis. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then