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:

$A$ particle of charge $16 \times 10^{-16} \, C$ moving with velocity $10 \, ms^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \, Vm^{-1}$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $\vec{B}$ is:

The figure shows a region of length $l$ with a uniform magnetic field of $0.3 \, T$ in it. $A$ proton enters the region with a velocity of $4 \times 10^{5} \, m/s$,making an angle of $60^{\circ}$ with the field. If the proton completes $10$ revolutions by the time it crosses the region shown,$l$ is close to....... $m$ (mass of proton $= 1.67 \times 10^{-27} \, kg$,charge of the proton $= 1.6 \times 10^{-19} \, C$)

An $\alpha$-particle,a proton,and a deuteron enter a uniform transverse magnetic field $B$ with the same accelerating potential $V$. Find the ratio of the radii of the paths followed by these particles.

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:

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a uniform magnetic field,speed and energy remain the same for a moving charged particle.
Reason $(R)$: $A$ moving charged particle experiences a magnetic force perpendicular to its direction of motion.