An electron is moving along the positive $x$-axis. To make it move in an anticlockwise circular path in the $x-y$ plane,a magnetic field must be applied:

$A$ particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then

$A$ 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?

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

The magnetic field is uniform for $y > 0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y < 0$. $A$ proton,denoted by a filled circle,leaves $y = 0$ in the $-y$-direction with some speed,as shown below. Which of the following best denotes the trajectory of the proton?

An electron is moving in the north direction. It experiences a force in the vertically upward direction. The magnetic field at the position of the electron is in the direction of