A particle moving in a magnetic field increases its velocity, then its radius of the circle

A particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with uniform magnetic field $B$ along the $\hat k$ direction. The particle will penetrate in this region in the $x$-direction upto a distance $d$ equal to

What is the behaviour of perpendicular electric field ${\rm{\vec E}}$ and magnetic field ${\rm{\vec B}}$ ?

A proton (mass $m$ ) accelerated by a potential difference $V$  flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be

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

A particle of mass $m$ and charge $q$, accelerated by a potential difference $V$ enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$, the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by