$A$ 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 the same radius?

An electron revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$,then

An alpha particle moving with a certain speed towards the east enters a uniform magnetic field directed vertically up. The alpha particle will then move in:

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

Two particles of masses $m_{a}$ and $m_{b}$ and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius $r_{a}$ and $r_{b}$ such that $r_{a} > r_{b}$. Then which is true?

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