$A$ 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

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

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:

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

An electron enters the space between the plates of a charged capacitor as shown. The surface charge density on the plates is $\sigma$. The electric field intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in the space,perpendicular to the direction of $E$. The electron moves perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$ without any change in direction. The time taken by the electron to travel a distance $l$ in the space is:

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be: