A charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$ . When electric field is switched off, the particle will move in a circular path of radius
$\frac {E}{BS}$
$\frac {ES}{B}$
$\frac {ES}{B^2}$
$\frac {E}{B^2S}$
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
An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.
(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )
A particle having a mass of $10^{- 2} \,kg$ carries a charge of $5 \times 10^{-8}\, C.$ The particle is given an initial horizontal velocity of $10^5\, m/s $ in the presence of electric field $E$ and magnetic field $B.$ To keep the particle moving in a horizontal direction, it is necessary that
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to
The electrostatic force $\left(\vec{F}_1\right)$ and magnetic force $\left(\vec{F}_2\right)$ acting on a charge $q$ moving with velocity $v$ can be written :