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:

$A$ 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

$A$ beam of ions with velocity $2 \times 10^5 \ m/s$ enters normally into a uniform magnetic field of $4 \times 10^{-2} \ T$. If the specific charge of the ion is $5 \times 10^7 \ C/kg$,then the radius of the circular path described will be $....... \ m$.

An electron having mass $9.1 \times 10^{-31} \ kg$,charge $1.6 \times 10^{-19} \ C$ and moving with velocity of $10^6 \ ms^{-1}$ enters a region where a magnetic field exists. If it describes a circle of radius $0.2 \ m$,then the intensity of the magnetic field must be . . . . . . $\times 10^{-5} \ T$.

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .