A $2\, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5\, tesla$. The force on the proton is

An $\alpha$-particle (mass $4 amu$ ) and a singly charged sulfur ion (mass $32 amu$ ) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_5$, respectively. The ratio $\left(r_s / r_\alpha\right)$ is. . . . .$(4)$

A charged particle moves in a magnetic field $\vec B = 10\,\hat i$ with initial velocity $\vec u = 5\hat i + 4\hat j$ The path of the  particle will be

A particle moving with velocity v having specific charge $(q/m)$ enters a region of  magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram......$^o$ 

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p, r_d$ and $r_{\alpha}$ Which one of the following relation is correct?

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is