A charge particle of charge $q$ and mass $m$ is accelerated through a potential diff. $V\, volts$. It enters a region of orthogonal magnetic field $B$. Then radius of its circular path will be
$\sqrt {\frac{{Vm}}{{2q{B^2}}}} $
${\frac{{2Vm}}{{q{B^2}}}}$
$\sqrt {\frac{{2Vm}}{q}} \left( {\frac{1}{B}} \right)$
$\sqrt {\frac{{Vm}}{q}} \left( {\frac{1}{B}} \right)$
An electron beam passes through a magnetic field of $2 \times 10^{-3}\,Wb/m^2$ and an electric field of $1.0 \times 10^4\,V/m$ both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively
An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the $+x$ direction and a magnetic field along the $+z$ direction, then
An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be
If a charged particle enters perpendicularly in the uniform magnetic field then