In Thomson's experiment,if the value of $q/m$ is the same for all positive ions striking the photographic plate,then the trace would be

$A$ particle of charge $q = -16 \times 10^{-18} \, C$ moving with velocity $v = 10 \, m/s$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis,and an electric field of magnitude $E = 10^4 \, V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $B$ is:

$A$ $10 \; eV$ electron is circulating in a plane at right angles to a uniform magnetic field of magnetic induction $10^{-4} \; Wb/m^2$ $(1.0 \; \text{gauss})$. The orbital radius of the electron is ........ $cm$.

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

$A$ 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 $I$

$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):