An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

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

$A$ proton carrying $1\, MeV$ kinetic energy is moving in a circular path of radius $R$ in a uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of the same radius in the same field? ........$MeV$

An electron moving with a velocity $\vec{V_1} = 2\,\hat{i}\,\text{m/s}$ at a point in a magnetic field experiences a force $\vec{F_1} = -2\hat{j}\,\text{N}$. If the electron is moving with a velocity $\vec{V_2} = 2\,\hat{j}\,\text{m/s}$ at the same point,it experiences a force $\vec{F_2} = +2\,\hat{i}\,\text{N}$. The force the electron would experience if it were moving with a velocity $\vec{V_3} = 2\hat{k}\,\text{m/s}$ at the same point is

$A$ proton enters a magnetic field of flux density $1.5 \,Wb \,m^{-2}$ with a velocity of $2 \times 10^7 \,ms^{-1}$ at an angle of $30^{\circ}$ with the field. The force on the proton will be

$A$ beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20 \ V m^{-1}$ and $0.5 \ T$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons in $m s^{-1}$?