An $\alpha -$ particle of $1\,MeV$ energy moves on circular path in uniform magnetic field. Then kinetic energy of proton in same magnetic field for circular path of double radius is......$MeV$
$1$
$4$
$2$
None
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let $r_p, r_e$ and $r_{He}$ be their respective radii, then
An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of
Two protons $A$ and $B$ move parallel to the $x$-axis in opposite directions with equal speeds $v$. At the instant shown, the ratio of magnetic force and electric force acting on the proton $A$ is ( $c=$ speed of light in vacuum)
A charge $q$ is released in presence of electric $(E)$ and magnetic field $(B)$ then after some time its velocity is $v$ then
Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p, r_d$ and $r_{\alpha}$ Which one of the following relation is correct?