$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will 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:

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

$A$ charged particle of charge '$q$' is accelerated by a potential difference '$V$' and enters a region of uniform magnetic field '$B$' at right angles to the direction of the field. The charged particle completes a semicircle of radius '$r$' inside the magnetic field. The mass of the charged particle is

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ proton and an $\alpha$-particle enter a magnetic field perpendicularly with the same velocity. If the proton takes $25 \mu \text{sec}$ to complete $5$ revolutions,what is the time period of the $\alpha$-particle in $\mu \text{sec}$?