charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then
$\overrightarrow {\;v} $=$\;\frac{{\left( {\vec BX\vec E} \right)}}{{{E^2}}}$
$\overrightarrow {\;v} = \frac{{\left( {\vec EX\vec B} \right)}}{{{B^2}}}$
$\overrightarrow {\;v} $=$\;\frac{{\left( {\vec BX\vec E} \right)}}{{{B^2}}}$
$\;\overrightarrow {\;v} = \frac{{\left( {\vec EX\vec B} \right)}}{{{E^2}}}$
A particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with uniform magnetic field $B$ along the $\hat k$ direction. The particle will penetrate in this region in the $x$-direction upto a distance $d$ equal to
A uniform magnetic field $\vec B\,\, = \,\,{B_0}\,\hat j$ exists in a space. A particle of mass $m$ and charge $q$ is projected towards negative $x$-axis with speed $v$ from the a point $(d, 0, 0)$. The maximum value $v$ for which the particle does not hit $y-z$ plane is
A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is ....
Mixed $H{e^ + }$ and ${O^{2 + }}$ ions (mass of $H{e^ + } = 4\,\,amu$ and that of ${O^{2 + }} = 16\,\,amu)$ beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then
In a certain region static electric and magnetic fields exist. The magnetic field is given by $\vec B = {B_0}\left( {\hat i + 2\hat j - 4\hat k} \right)$. If a test charge moving with a velocity $\vec v = {v_0}\left( {3\hat i - \hat j + 2\hat k} \right)$ experiences no force in that region, then the electric field in the region, in $SI\, units$, is