$A$ particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly into a region of uniform magnetic field $B$ and executes a circular arc of radius $R$. Then,$\frac{q}{m}$ equals:

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

The charge on a particle $Y$ is double the charge on particle $X$. These two particles $X$ and $Y$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

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:

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix