If a proton, deutron and $\alpha - $ particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energies is

A particle of mass $m$ and charge $q$, accelerated by a potential difference $V$ enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$, the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by

What is the radius of the path of an electron (mass $9 \times 10^{-31}\;kg$ and charge $1.6 \times 10^{-19} \;C )$ moving at a speed of $3 \times 10^{7} \;m / s$ in a magnetic field of $6 \times 10^{-4}\;T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. ( $\left.1 eV =1.6 \times 10^{-19} \;J \right)$

A charge having $q/m$ equal to $10^8\, C/kg$ and with velocity $3 \times 10^5\, m/s$ enters into a uniform magnetic field $0.3\, tesla$ at an angle $30^o$ with direction of field. The radius of curvature will be ......$cm$

Mark the correct statement

The motion of a charged particle can be used to distinguish between a magnetic field and electric field in a certain region by firing the charge