$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon:

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

$A$ particle moving in a magnetic field increases its velocity,then its radius of the circle