A charge $q$ is released in presence of electric $(E)$ and magnetic field $(B)$ then after some time its velocity is $v$ then
$v\, \propto \,E$
$v\, \propto \,\frac{1}{E}$
${v^2}\, \propto \,B$
$v\, \propto \,B^o$
At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that
A charged particle initially at rest at $O$,when released follows a trajectory as shown alongside. Such a trajectory is possible in the presence of
An electron having a charge e moves with a velocity $v$ in positive $x$ direction. A magnetic field acts on it in positive $y$ direction. The force on the electron acts in (where outward direction is taken as positive $z$-axis).
A proton, a deuteron and an $\alpha$ particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is.......... and their speed is.................. in the ratio.
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