$A$ particle of charge $q$ moves with a velocity $\vec{V} = a \hat{i}$ in a magnetic field $\vec{B} = b \hat{j} + c \hat{k}$,where $a$,$b$,and $c$ are constants. The magnitude of the force experienced by the particle is:

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: If oxygen ion $(O^{-2})$ and hydrogen ion $(H^{+})$ enter normal to the magnetic field with equal momentum,then the path of $O^{-2}$ ion has a smaller curvature than that of $H^{+}$.
Reason $R$: $A$ proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statement,choose the correct answer from the options given below.

$A$ 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 $^{\circ}$)

An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \, m/s$. It enters a region of uniform electric field of $300 \, V/cm$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:

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).