$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

$A$ 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 :-

The ratio of time periods of $\alpha$-particle and proton moving on circular path in a uniform magnetic field is . . . . . . .

The motion of a moving electron is not affected by:

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then: