A proton (or charged particle) moving with velocity $v$ is acted upon by electric field $E$ and magnetic field $B$. The proton will move undeflected if
$E$ is perpendicular to $B$
$E$ is parallel to $v$ and perpendicular to $B$
$E$, $B$ and $v$ are mutually perpendicular and $v = \frac{E}{B}$
$E$ and $B$ both are parallel to $v$
Assertion : A proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow B $ in a direction perpendicular to $\overrightarrow B $, will describe circular trajectories of equal radii.
Two ions having masses in the ratio $1 : 1$ and charges $1 : 2$ are projected into uniform magnetic field perpendicular to the field with speeds in the ratio $2 : 3$. The ratio of the radii of circular paths along which the two particles move is
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
A rectangular region $A B C D$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. A narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $B C$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$, the magnetic field has to be
A particle having some charge is projected in $x-y$ plane with a speed of $5\ m/s$ in a region having uniform magnetic field along $z-$ axis. Which of the following cannot be the possible value of velocity at any time ?