Module 3 : Electromagnetism
Lecture 13 : Electric Current and Current Density
  Motion in a Magnetic Field - Quantitative
  Let the initial velocity of the particle be $\vec u$. we may take the direction of the component of $\vec u$perpendicular to $\vec B$as the $x-$direction, so that
 
\begin{displaymath}\vec u = (u_x, 0, u_z)\end{displaymath}
  Let the velocity at time $t$be denoted by $\vec v$
 
\begin{displaymath}\vec v = v_x\hat\imath + v_y\hat\jmath + v_z\hat k\end{displaymath}
  we can express the force equation in terms of its cartesian component
 
\begin{eqnarray*} m\frac{dv_x}{dt} = q(v_yB_z - v_zB_y) &=& qv_yB\\ m\frac{dv_y... ..._xB_z) &=& -qv_xB\\ m\frac{dv_z}{dt} = q(v_xB_y - v_yB_x) &=& 0 \end{eqnarray*}
  where we have used $B_x=B_y=0$and $B_z=B$ .
  The last equation tells us that no force acts on the particle in the direction in which $\vec B$acts, so that
 
\begin{displaymath}v_z = {\rm constant}\ = u_z\end{displaymath}
 
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