An electron moving with a velocity $\vec{V_1} = 2\,\hat{i}\,\text{m/s}$ at a point in a magnetic field experiences a force $\vec{F_1} = -2\hat{j}\,\text{N}$. If the electron is moving with a velocity $\vec{V_2} = 2\,\hat{j}\,\text{m/s}$ at the same point,it experiences a force $\vec{F_2} = +2\,\hat{i}\,\text{N}$. The force the electron would experience if it were moving with a velocity $\vec{V_3} = 2\hat{k}\,\text{m/s}$ at the same point is

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

$A$ particle of mass $M$ and positive charge $Q$,moving with a constant velocity $\vec{u}_1 = 4\hat{i} \text{ m/s}$,enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region,the particle emerges on the other side after $10 \text{ ms}$ with a velocity $\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j}) \text{ m/s}$. The correct statement$(s)$ is (are):
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction.
$(C)$ The magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

$A$ proton and an $\alpha$-particle enter a magnetic field perpendicularly with the same velocity. If the proton takes $25 \mu \text{sec}$ to complete $5$ revolutions,what is the time period of the $\alpha$-particle in $\mu \text{sec}$?

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

$OABC$ is a current-carrying square loop. An electron is projected from the center of the loop along its diagonal $AC$ as shown. The unit vector in the direction of initial acceleration will be