An $\alpha - $ particle travels in a circular path of radius $0.45\, m$ in a magnetic field $B = 1.2\,Wb/{m^2}$ with a speed of $2.6 \times {10^7}\,m/\sec $. The period of revolution of the $\alpha - $ particle is
$1.1 \times {10^{ - 5}}\,\sec $
$1.1 \times {10^{ - 6}}\sec $
$1.1 \times {10^{ - 7}}\,\sec $
$1.1 \times {10^{ - 8}}\,\sec $
An electron moving towards the east enters a magnetic field directed towards the north. The force on the electron will be directed
An electron having a charge e moves with a velocity $v$ in positive $x$ direction. A magnetic field acts on it in positive $y$ direction. The force on the electron acts in (where outward direction is taken as positive $z$-axis).
Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $\mathrm{X}$ and $\mathrm{Y}$ is :
A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5$ $metre$ in a plane perpendicular to magnetic field $\overrightarrow B $. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ $metre$ in the same plane with the same $\overrightarrow B $ is........$keV$
An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicularly, Then