$A$ charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$. When the electric field is switched off,the particle will move in a circular path of radius:

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis,under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off,the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

$A$ particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ describes a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B$. The work done by the field when the particle completes one full circle is

$A$ 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 accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

Two ions of masses $4 \, amu$ and $16 \, amu$ have charges $+2e$ and $+3e$ respectively. These ions pass through a region of constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then: