An electron having mass $m$ and kinetic energy $K$ enter in uniform magnetic field $B$ perpendicularly, then its frequency will be
$\frac{e K}{q v B}$
$\frac{e B}{2 \pi m}$
$\frac{2 \pi m}{e B}$
$\frac{2 m}{e B K}$
A charge particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,m\,T$ at right angle to the direction of field. The charge particle completes semicircle of radius $3\,cm$ inside magnetic field. The mass of the charge particle is $........\times 10^{-18}\,kg$.
A particle moving with velocity v having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram......$^o$
A particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly in a region of uniform magnetic field $B$ and executes circular arc of radius $R$, then $\frac{q}{m}$ equals
As shown in the figure, the uniform magnetic field between the two identical plates is $B$. There is a hole in plate. If through this hole a particle of charge $q$, mass $m$ and energy $E$ enters this magnetic field, then the particle will not collide with the upper plate provided
An $e^-$ is moving parallel to a long current carrying wire as shown. Force on electron is