Text_Template

Module 3 : Electromagnetism

Lecture 13 : Electric Current and Current Density

Motion of a Charged Particle in a Uniform Magnetic Field

Let the direction of the magnetic field be taken to be z- direction,

$\begin{displaymath}\vec B = B\hat k\end{displaymath}$

we can write the force on the particle to be

$\begin{displaymath}\vec F_m = m\frac{d\vec v}{dt} = q\vec v\times\vec B\end{displaymath}$

The problem can be looked at qualitatively as follows. We can resolve the motion of the charged particle into two components, one parallel to the magnetic field and the other perpendicular to it. Since the motion parallel to the magnetic field is not affected, the velocity component in the z-direction remains constant.

$\begin{displaymath}v_z(t) = v_z(t=0) = u_z\end{displaymath}$

where $\vec u$ is the initial velocity of the particle. Let us denote the velocity component perpendicular to the direction of the magnetic field by $v_\perp$ . Since the force (and hence the acceleration) is perpendicular to the direction of velocity, the motion in a plane perpendicular to $\vec B$ is a circle. The centripetal force necessary to sustain the circular motion is provided by the Lorentz force

$\begin{displaymath}\frac{mv_\perp^2}{R} = \mid q\mid v_\perp B\end{displaymath}$