A positively charged $(+ q)$ particle of mass $m$ has kinetic energy $K$ enters vertically downward in a horizontal field of magnetic induction $\overrightarrow B $ . The acceleration of the particle is :-
$qB\sqrt {\frac{{2K}}{m}} $
$\frac{{qB\sqrt {2K} }}{{{{(m)}^{\frac{3}{2}}}}}$
$\frac{{2qB}}{{{{(m)}^{\frac{3}{2}}}}}\sqrt {2K} $
$2qB\sqrt {\frac{{2K}}{m}} $
A proton and an $\alpha - $particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25$ $\mu \, sec$ to make $5$ revolutions, then the periodic time for the $\alpha - $ particle would be........$\mu \, sec$
A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointing in the same direction
A particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$, The radius of the circular path described by it will be
In an experiment, electrons are accelerated, from rest, by applying, a voltage of $500 \,V.$ Calculate the radius of the path if a magnetic field $100\,mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19}\,C$ Mass of the electron $= 9.1 \times 10^{-31}\,kg$ ]
The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat z$ A particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{{mv}}{{2qB}}$ , the acceleration of the charged particle at the point of its emergence at the other side is