An electron is moving along the positive $x$-axis. To make it move in an anticlockwise circular path in the $x-y$ plane,a magnetic field must be applied:

$A$ negatively charged particle projected towards east is deflected towards north by a magnetic field. The direction of the magnetic field is

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

$A$ charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is

$A$ horizontal conductor is oriented north-south and carries some current. $A$ positively charged particle located vertically above it and having a velocity directed northward experiences an upward force. What is the direction of the force if this charged particle were located to the east of the conductor and had a velocity directed towards the conductor?

$A$ $5 \text{ mg}$ 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 . . . . . . .