$A$ singly ionized magnesium atom $(A=24)$ ion is accelerated to kinetic energy $5\,keV$ and is projected perpendicularly into a magnetic field $B$ of magnitude $0.5\,T$. The radius of the path formed will be . . . . . . $cm$.

$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

An electron at rest is accelerated by a potential $V_{1}$ and then enters a uniform magnetic field,experiencing a force $F_{1}$. When the potential is changed to $V_{2}$,the force experienced by the electron becomes $2F_{1}$. The ratio of $V_{1}$ to $V_{2}$ is:

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.

$A$ beam of well-collimated cathode rays travelling with a speed of $5 \times 10^6 \, m/s$ enters a region of mutually perpendicular electric and magnetic fields and emerges undeviated from this region. If $|B| = 0.02 \, T$,the magnitude of the electric field is: