$A$ positively charged particle $q$ of mass $m$ is passed through a velocity selector. It moves horizontally rightward without deviation along the line $y = \frac{2mv}{qB}$ with a speed $v$. The electric field is vertically downwards and the magnetic field is into the plane of the paper. Now,the electric field is switched off at $t = 0$. The angular momentum of the charged particle about the origin $O$ at $t = \frac{\pi m}{qB}$ is:

$A$ 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 $I$

$A$ metallic block carrying current $I$ is subjected to a uniform magnetic induction $\overrightarrow{B}$ as shown in the figure. The moving charges experience a force $\overrightarrow{F}$ given by ........... which results in the lowering of the potential of the face ........ Assume the speed of the carriers to be $v$.

$A$ particle is moving in a uniform magnetic field,then

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is