A very long straight wire carries a current $I$. At the instant when a charge $ + Q$ at point $P$ has velocity $\overrightarrow V $, as shown, the force on the charge is

A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ – 27}}\,kg$ and charge $ = 1.6 \times {10^{ – 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be

A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. the particle leaves the magnetic field at point $D,$ then the distance $CD$ is :-

A proton of mass $1.67\times10^{-27}\, kg$ and charge $1.6\times10^{-19}\, C$ is projected with a speed of $2\times10^6\, m/s$ at an angle of $60^o$ to the $X-$ axis. If a uniform magnetic field of $0.104\, tesla$ is applied along the $Y-$ axis, the path of the proton is

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 electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below