An $\alpha – $ particle travels in a circular path of radius $0.45\, m$ in a magnetic field $B = 1.2\,Wb/{m^2}$ with a speed of $2.6 \times {10^7}\,m/\sec $. The period of revolution of the $\alpha – $ particle is

If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force, what will happen to proton and electron

A charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

A particle of mass $m$ and charge $q$ is thrown from origin at $t = 0$ with velocity $2\hat{i}$ + $3\hat{j}$ + $4\hat{k}$ units in a region with uniform magnetic field $\vec B$ = $2\hat{i}$ units. After time $t =\frac{{\pi m}}{{qB}}$ , an electric field  is switched on such that particle moves on a straight line with constant speed. $\vec E$ may be

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 $56 cm$.