The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the Earth's magnetic field points down. The electron is deflected towards:

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

When an electron beam passes through an electric field,it gains kinetic energy. If the same beam passes through a magnetic field,then:

$A$ particle of charge $16 \times 10^{-16} \, C$ moving with velocity $10 \, ms^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \, Vm^{-1}$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $\vec{B}$ is:

$A$ proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be