$A$ 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 $^{\circ}$)

$A$ particle having a mass of $10^{-2} \, kg$ carries a charge of $5 \times 10^{-8} \, C$. The particle is given an initial horizontal velocity of $10^5 \, m/s$ in the presence of an electric field $\vec{E}$ and a magnetic field $\vec{B}$. To keep the particle moving in a horizontal direction,it is necessary that:
$(1)$ $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity.
$(2)$ Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity.
$(3)$ Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.
$(4)$ $\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity.
Which one of the following pairs of statements is possible?

$A$ particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction,extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

$A$ 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 is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along:

An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$