As shown in the figure, the uniform magnetic field between the two identical plates is $B$. There is a hole in plate. If through this hole a particle of charge $q$, mass $m$ and energy $E$ enters this magnetic field, then the particle will not collide with the upper plate provided
$B > \frac{{2mE}}{{qd}}$
$B > \frac{{\sqrt {2mE} }}{{qd}}$
$B < \frac{{2mE}}{{qd}}$
$B < \frac{{\sqrt {2mE} }}{{qd}}$
A charged particle is moving in a circular orbit of radius $6\, cm$ with a uniform speed of $3 \times 10^6\, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4}\, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is
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
A particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$, The radius of the circular path described by it will be
If a charged particle goes unaccelerated in a region containing electric and magnetic fields
Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :.......$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )