$A$ particle with $10^{-11} \, C$ of charge and $10^{-7} \, kg$ mass is moving with a velocity of $10^8 \, m/s$ along the $y$-axis. $A$ uniform static magnetic field $B = 0.5 \, T$ is acting along the $x$-direction. The force on the particle is:

$A$ $10 \; eV$ electron is circulating in a plane at right angles to a uniform magnetic field of magnetic induction $10^{-4} \; Wb/m^2$ $(1.0 \; \text{gauss})$. The orbital radius of the electron is ........ $cm$.

Given below are two statements:
Statement $I$: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.
Statement $II$: The electric force accelerates the positively charged particle perpendicular to the direction of the electric field. The magnetic force accelerates the moving charged particle along the direction of the magnetic field.
In the light of the above statements,choose the most appropriate answer from the options given below:

$A$ 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:

$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):

$A$ particle is moving in a uniform magnetic field,then