Two ions of masses $4 \, amu$ and $16 \, amu$ have charges $+2e$ and $+3e$ respectively. These ions pass through a region of constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then:

When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution 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:

$A$ stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field $E$ (vertically downwards) and a uniform magnetic field $B$ (out of the plane of the figure) as shown. Then:

$A$ positively charged particle $q$ of mass $m$ is passed through a velocity selector. It moves horizontally rightward without deviation along the line $y = \frac{2mv}{qB}$ with a speed $v$. The electric field is vertically downwards and the magnetic field is into the plane of the paper. Now,the electric field is switched off at $t = 0$. The angular momentum of the charged particle about the origin $O$ at $t = \frac{\pi m}{qB}$ is: