An electron has mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is moving with a velocity of $10^6 \ m/s$. It enters a region where a magnetic field exists. If it describes a circle of radius $0.10 \ m$,the intensity of the magnetic field must be:

$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$,then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)

$A$ stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field $E$ (vertically downwards) and a uniform magnetic field $B$ (out of the plane of the figure) as shown. Then:

An electron is orbiting in a circular orbit under the influence of a constant transverse magnetic field of strength $B$. Assuming that Bohr's postulate regarding the quantisation of angular momentum holds good for this electron,find the kinetic energy of the electron in the $n^{th}$ orbit. ($m_e =$ mass of electron,$e =$ magnitude of charge of electron)

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is