$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

The figure shows a region of length $l$ with a uniform magnetic field of $0.3 \, T$ in it. $A$ proton enters the region with a velocity of $4 \times 10^{5} \, m/s$,making an angle of $60^{\circ}$ with the field. If the proton completes $10$ revolutions by the time it crosses the region shown,$l$ is close to....... $m$ (mass of proton $= 1.67 \times 10^{-27} \, kg$,charge of the proton $= 1.6 \times 10^{-19} \, C$)

$A$ particle of mass $m$ and charge $q$,moving with velocity $V$,enters Region $II$ normal to the boundary as shown in the figure. Region $II$ has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of Region $II$ is $\ell$. Choose the correct choice$(s)$.
Figure: $222707-q$
$(A)$ The particle enters Region $III$ only if its velocity $V > \frac{qB\ell}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $V < \frac{qB\ell}{m}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V = \frac{qB\ell}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$

An electron enters a chamber in which a uniform magnetic field is present as shown in the figure. Ignore gravity. During its motion inside the chamber:

The motion of a charged particle can be used to distinguish between a magnetic field and an electric field in a certain region by firing the charge:

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,