The radius of the circular path of an electron when subjected to a perpendicular magnetic field is:

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

An electron is accelerated by a potential difference of $12000\, V$. It then enters a uniform magnetic field of $10^{-3}\, T$ applied perpendicular to the path of the electron. Find the radius of the path. Given: mass of electron $m = 9 \times 10^{-31}\, kg$ and charge on electron $q = 1.6 \times 10^{-19}\, C$.

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 ............ $m/s$.

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

An electron is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. The angle $\theta$ between $\vec{v}$ and $\vec{B}$ lies between $0^\circ$ and $\frac{\pi}{2}$. Its velocity vector $\vec{v}$ returns to its initial value in a time interval of: