$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

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:

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

$A$ particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly into a region of uniform magnetic field $B$ and executes a circular arc of radius $R$. Then,$\frac{q}{m}$ equals:

$A$ proton accelerated by a potential difference of $500 \ kV$ flies through a uniform transverse magnetic field of $0.1 \ T$. The field is spread over a region of $1.0 \ cm$ thickness. The angle through which the proton gets deviated from its original direction is (Proton mass $= 1.6 \times 10^{-27} \ kg$ and charge of proton $= 1.6 \times 10^{-19} \ C$) (in $rad$)