A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage $192 V$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $B_0=0.1 \hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.

[Given: Mass of neutron/proton $=(5 / 3) \times 10^{-27} kg$, charge of the electron $=1.6 \times 10^{-19} C$.]

Which of the following option($s$) is(are) correct?

$(A)$ The value of $x$ for $H^{+}$ion is $4 cm$.

$(B)$ The value of $x$ for an ion with $A_M=144$ is $48 cm$.

$(C)$ For detecting ions with $1 \leq A_M \leq 196$, the minimum height $\left(x_1-x_0\right)$ of the detector is $55 cm$.

$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M=196$ is $56 cm$.

A proton with a kinetic energy of $2.0\,eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\,T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $……….cm$ (Take, mass of proton $=1.6 \times 10^{-27}\,kg$ and Charge on proton $=1.6 \times 10^{-19}\,kg)$

An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the $+x$ direction and a magnetic field along the $+z$ direction, then

A charge $+ Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

Derived force on moving charge in uniform magnetic field with velocity $\overrightarrow {{v_d}} $.