$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .

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

$A$ proton with kinetic energy of $1 \; MeV$ moves from south to north. It experiences an acceleration of $10^{12} \; m/s^2$ due to an applied magnetic field (directed from west to east). The value of the magnetic field is: ....... $mT$ (Rest mass of proton is $1.6 \times 10^{-27} \; kg$)

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is:

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. They will move along circular paths.