$A$ 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

$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

$A$ 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.

The velocity of certain ions that pass undeflected through crossed electric field $E = 7.7 \, kV/m$ and magnetic field $B = 0.14 \, T$ is.....$km/s$

$A$ particle of mass $m$ and charge $q$ moving with velocity $\vec v$ enters a region of uniform magnetic field $\vec B.$ Then:

Two charged particles of mass $m$ and charge $q$ each are projected from the origin simultaneously with the same speed $V$ in a transverse magnetic field $B$. If $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of the particles (with respect to the origin) at $t = \frac{\pi m}{qB}$, then the value of $\vec{r}_1 \cdot \vec{r}_2$ at that time is: