$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes three times the original,the new radius will be

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is

An $\alpha$-particle,a proton,and a deuteron enter a uniform transverse magnetic field $B$ with the same accelerating potential $V$. Find the ratio of the radii of the paths followed by these particles.

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

$A$ 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 $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.