An electron is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along:

Ionized hydrogen atoms and $\alpha$-particles with same momenta enter perpendicular to a constant magnetic field $B$. The ratio of the radii of their paths $r_{H}: r_{\alpha}$ will be

$A$ 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:

$A$ particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The speed of the particle is $v = 10^7 \, m/s$. The magnetic field is directed along the inward normal to the plane of the paper. The particle enters the field at $C$ and leaves at $D$. Then the angle $\theta$ must be :-