An electron (mass $m$) is accelerated through a potential difference of $V$ and then it enters a magnetic field of induction $B$ normal to the field lines. The radius of the circular path is ($e$ = electronic charge).

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

At $t = 0$,a charge $q$ is at the origin and moving in the $y$-direction with velocity $\vec{v} = v\hat{j}$. The charge moves in a magnetic field that is for $y > 0$ out of the page and given by $B_1\hat{k}$ and for $y < 0$ into the page and given by $-B_2\hat{k}$. The charge's subsequent trajectory is shown in the sketch. From this information,we can deduce that:

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

$A$ particle of mass $M$ and charge $q$ moves with a constant velocity $V$ in the positive $x$-direction. $A$ constant magnetic field $B$ exists in the negative $z$-direction between $x = a$ and $x = b$. Find the minimum value of $V$ such that the particle enters the region $x > b$.