A uniform magnetic field $\vec B\,\, = \,\,{B_0}\,\hat j$ exists in a space. A particle of mass $m$ and charge $q$ is projected towards negative $x$-axis with speed $v$ from the a point $(d, 0, 0)$. The maximum value $v$ for which the particle does not hit $y-z$ plane is

charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then

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 proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha - $particle (mass = $4m$ and charge = $+ 2e),$ so that it can revolve in the path of same radius.......$MeV$

If a positive ion is moving, away from an observer with same acceleration, then the lines of force of magnetic induction will be

${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ ions having same kinetic energy pass through a region of space filled with uniform magnetic field $B$ directed perpendicular to the velocity of ions. The masses of the ions ${H^ + },\,H{e^ + }$and ${O^{ + + }}$ are respectively in the ratio $1:4:16$. As a result