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 $56 cm$.

An electron (charge $q$ $coulomb$) enters a magnetic field of $H$ $weber/{m^2}$ with a velocity of $v\,m/s$ in the same direction as that of the field the force on the electron is

An electron (mass = $9.0 × $${10^{ - 31}}$ $kg$ and charge =$1.6 \times {10^{ - 19}}$ $coulomb$) is moving in a circular orbit in a magnetic field of $1.0 \times {10^{ - 4}}\,weber/{m^2}.$ Its period of revolution is

When a charged particle enters a uniform magnetic field its kinetic energy

A particle having the same charge as of electron moves in a circular path of radius $0.5
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )

A charge $q$ moves in a region where electric field and magnetic field both exist, then force on it is