An $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

An electron with kinetic energy $5 \ eV$ enters a region of uniform magnetic field of $3 \ \mu T$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$,so that the electron moves along the same path,is . . . . . $N C^{-1}$.
(Given: mass of electron $= 9 \times 10^{-31} \ kg$,electric charge $= 1.6 \times 10^{-19} \ C$)

$A$ charged particle of charge $q$ and mass $m$ is accelerated through a potential difference of $V$ volts. It enters a region of orthogonal magnetic field $B$. The radius of its circular path will be: