The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

$A$ positive charge $q$ of mass $m$ is moving along the $+x$ axis with velocity $v$. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction,crossing the $y$ axis at a distance $d$. Then:

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

$A$ charged particle moving in a magnetic field $B$ has velocity components both along $B$ and perpendicular to $B$. The path of the charged particle will be:

Two identical charged particles enter a uniform magnetic field with the same speed but at angles $30^{\circ}$ and $60^{\circ}$ with the field. Let $a, b,$ and $c$ be the ratios of their time periods,radii,and pitches of the helical paths,respectively. Then:

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be: