If a proton, deuteron, and $\alpha$-particle are accelerated by the same potential difference and enter perpendicular to a magnetic field, then the ratio of their kinetic energies is:

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be:

Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $\theta$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then,they will meet again for the first time at a point whose distance from the point of projection is

$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is

$A$ proton is moving along the $Z$-axis in a magnetic field. The magnetic field is along the $X$-axis. The proton will experience a force along

An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.