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

If an electron and a proton having same momenta enter perpendicular to a magnetic field,then

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2} \, T$. If the value of $e/m$ is $1.76 \times 10^{11} \, C/kg$,the frequency of revolution of the electron is

$A$ singly ionized magnesium atom $(A=24)$ ion is accelerated to kinetic energy $5\,keV$ and is projected perpendicularly into a magnetic field $B$ of magnitude $0.5\,T$. The radius of the path formed will be . . . . . . $cm$.

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)