A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

A particle with charge $+Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be 

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

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

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. $e/m$ for it will be