An electron beam is moving between two parallel plates having an electric field of $1.125 \times 10^{-6} \, N/C$. $A$ magnetic field of $3 \times 10^{-10} \, T$ is also applied so that the beam of electrons does not deflect. The velocity of the electron is ............ $m/s$.

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will be

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

$A$ 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 $mT$)

$A$ uniform electric field and a uniform magnetic field are produced,pointing in the same direction. An electron is projected with its velocity pointing in the same direction.