$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

$A$ beam of electrons,initially at rest,is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating potential is increased to $V^{\prime}$ and the force experienced by the electrons in the same magnetic field becomes $2F$. The ratio $\frac{V}{V^{\prime}}$ is:

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

$A$ $5 \text{ mg}$ particle carrying a charge of $5 \times 10^{-6} \text{ C}$ is moving with a velocity of $(3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}$ in a region having a magnetic field $\vec{B} = 0.1\hat{k} \text{ Wb/m}^2$. It moves a distance of $\alpha \text{ m}$ along the $\hat{k}$ direction when it completes $5$ revolutions. The value of $\alpha$ is . . . . . . .

$A$ particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform electric and magnetic fields, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along