$A$ particle is moving in a uniform magnetic field,then

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

$A$ uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3R}{2}$ (region $2$ in the figure) pointing normally into the plane of the paper. $A$ particle with charge $+Q$ and momentum $p$ directed along the $x$-axis enters region $2$ from region $1$ at point $P_1(y=-R)$. Which of the following option$(s)$ is/are correct?
$[A]$ For $B > \frac{2}{3} \frac{p}{QR}$,the particle will re-enter region $1$.
$[B]$ For $B = \frac{8}{13} \frac{p}{QR}$,the particle will enter region $3$ through the point $P_2$ on the $x$-axis.
$[C]$ When the particle re-enters region $1$ through the longest possible path in region $2$,the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from the $y$-axis is $p/\sqrt{2}$.
$[D]$ For a fixed $B$,particles of same charge $Q$ and same velocity $v$,the distance between the point $P_1$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle.

An electron moves with a speed of $2 \times 10^5\, m/s$ along the $+x$ direction in a magnetic field $\vec{B} = (\hat{i} - 4\hat{j} - 3\hat{k})\,T$. The magnitude of the force (in newton) experienced by the electron is (charge on electron $= 1.6 \times 10^{-19}\,C$).