$A$ charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a

$A$ 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 $= 1.6 \times 10^{-19} \ C$)

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels:

$A$ particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ describes a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B$. The work done by the field when the particle completes one full circle is

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

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 is tripled,the radius will become