An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec B$ . If kinetic energy is doubled and magnetic induction tripled, the radius will become 

A proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,\, MeV,$ the energy acquired by the alpha particle will be......$MeV$

A charged particle moves through a magnetic field perpendicular to its direction. Then

Mixed $H{e^ + }$ and ${O^{2 + }}$ ions (mass of $H{e^ + } = 4\,\,amu$ and that of ${O^{2 + }} = 16\,\,amu)$ beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then

A charged particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path described by the particle is proportional to

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,