$A$ charge '$q$' moves with a velocity $2 \ m/s$ along the $x$-axis in a uniform magnetic field $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$. The charge will experience a force:

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively,in a uniform magnetic field $\vec{B} = B_0 \hat{i}$,each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

$A$ particle of mass $m$ and charge $q$ is incident on the $XZ$ plane with velocity $v$ in a direction making an angle $\theta$ with a uniform magnetic field applied along the $X$-axis. The nature of motion performed by the particle is . . . . . . .

$A$ proton and an alpha particle of the same kinetic energy enter a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and the proton is ....