An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

Three ions $H^+$,$He^+$,and $O^{2+}$ having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. Then:

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is: