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

$A$ proton with kinetic energy of $1 \; MeV$ moves from south to north. It experiences an acceleration of $10^{12} \; m/s^2$ due to an applied magnetic field (directed from west to east). The value of the magnetic field is: ....... $mT$ (Rest mass of proton is $1.6 \times 10^{-27} \; kg$)

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

$A$ 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 $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

Two particles $A$ and $B$ of masses $m_A$ and $m_B$ respectively,having the same charge,are moving in a plane. $A$ uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_A$ and $v_B$ respectively,and the trajectories are as shown in the figure. Then: