$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the kinetic energy of the particle is increased to three times,then the new radius will be

$A$ proton and an $\alpha$-particle enter a magnetic field perpendicularly with the same velocity. If the proton takes $25 \mu \text{sec}$ to complete $5$ revolutions,what is the time period of the $\alpha$-particle in $\mu \text{sec}$?

$A$ 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:

The figure shows the face of a cathode-ray oscilloscope tube,as viewed from the front. The electron beam is coming out normally from the plane of the paper. The electron beam passes through a region where there are electric and magnetic fields directed as shown. The deflections of the spot from the center of the screen produced by the electric field $E$ and the magnetic field $B$ separately are equal in magnitude. Which one of the diagrams below shows a possible position of the spot on the screen when both fields are operating?

$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = E\hat i$ and a magnetic field $\vec B = B\hat i$ with an initial velocity $\vec v = v_0\hat j$. The speed of the particle will become $2v_0$ after a time:

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)