$A$ proton and a deuteron $(q=+e, m=2.0 \ u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$,moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of the deuteron path to the radius $r_p$ of the proton path is:

$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

$A$ rectangular region $ABCD$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. $A$ narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $BC$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$,the magnetic field has to be

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$.

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

$A$ 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: