An electron $(e^-)$ is moving in the North direction and the magnetic field at this place is in the upward direction. Then,the electron $(e^-)$ will be deviated towards:

The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

$A$ current-carrying long solenoid is placed on the ground with its axis vertical. $A$ proton is falling along the axis of the solenoid with a velocity $v$. When the proton enters into the solenoid,it will

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

$A$ positive,singly ionized atom of mass number $A_M$ is accelerated from rest by a voltage $V = 192 \text{ V}$. Thereafter,it enters a rectangular region of width $w$ with a magnetic field $B_0 = 0.1 \hat{k} \text{ T}$,as shown in the figure. The ion finally hits a detector at a distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $= (5/3) \times 10^{-27} \text{ kg}$,charge of the electron $= 1.6 \times 10^{-19} \text{ C}$.]
Which of the following option$(s)$ is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ ion is $4 \text{ cm}$.
$(B)$ The value of $x$ for an ion with $A_M = 144$ is $48 \text{ cm}$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$,the minimum height $(x_1 - x_0)$ of the detector is $52 \text{ cm}$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $28 \text{ cm}$.