An electron gun is placed inside a long solenoid of radius $\mathrm{R}$ on its axis. The solenoid has $\mathrm{n}$ turns/length and carries a current $I$. The electron gun shoots an electron along the radius of the solenoid with speed $v$. If the electron does not hit the surface of the solenoid, maximum possible value of ${v}$ is (all symbols have their standard meaning)
$\frac{\mathrm{e} \mu_{0} \mathrm{nIR}}{\mathrm{m}}$
$\frac{\mathrm{e} \mu_{0} \mathrm{nIR}}{2 \mathrm{m}}$
$\frac{2 \mathrm{e} \mu_{0} \mathrm{nIR}}{\mathrm{m}}$
$\frac{\mathrm{e} \mu_{0} \mathrm{nIR}}{4 \mathrm{m}}$
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
A particle of charge $q$ and velocity $v$ passes undeflected through a space with non-zero electric field $E$ and magnetic field $B$. The undeflecting conditions will hold if.
A proton of mass $1.67 \times {10^{ - 27}}\,kg$ and charge $1.6 \times {10^{ - 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X - $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y - $ axis, the path of proton is
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 carrying $1\, Me V$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an $\alpha -$ particle to describe a circle of same radius in the same field ?........$MeV$