$A$ 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

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. They will move along circular paths.

$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{i}$ with an initial velocity $\vec{v} = v_0 \hat{j}$. The speed of the particle will become $2v_0$ after a time:

In the figure shown,a charge $q$ of mass $m$ moving with a velocity $v$ along the $x$-axis enters a region of uniform magnetic field $B$ directed into the page. If the particle is able to enter the region $x > b$,then the velocity $v$ must be greater than:

$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.

$A$ 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?