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Module 3 : Electromagnetism

Lecture 13 : Electric Current and Current Density

Motion in a crossed electric and magnetic fields

The force on the charged particle in the presence of both electric and magnetic fields is given by

$\begin{displaymath}\vec F = q(\vec E + \vec v\times\vec B)\end{displaymath}$

Let the electric and magnetic fields be at right angle to each other, so that,

$\begin{displaymath}\vec E\cdot\vec B =0\end{displaymath}$

If the particle is initially at rest no magnetic force acts on the particle. As the electric field exerts a force on the particle, it acquires a velocity in the direction of $\vec E$ . The magnetic force now acts sidewise on the particle.
For a quantitative analysis of the motion, let $\vec E$ be taken along the x-direction and $\vec B$ along z-direction. As there is no component of the force along the z-direction, the velocity of the particle remains zero in this direction. The motion, therefore, takes place in x-y plane. The equations of motion are

$\displaystyle m\frac{dv_x}{dt}$	$\textstyle =$	$\displaystyle q(\vec E + \vec v\times\vec B)_x = qE + qBv_y$	(1)
$\displaystyle m\frac{dv_y}{dt}$	$\textstyle =$	$\displaystyle q(\vec E + \vec v\times\vec B)_y = - qBv_x$	(2)