Module 3 : Electromagnetism
Lecture 13 : Electric Current and Current Density
 

As in the earlier case, we can solve the equations by differentiating one of the equations and substituting the other,

\begin{displaymath}m\frac{d^2v_x}{dt^2} = qB\frac{dv_y}{dt} = -\frac{q^2B^2}{m}v_x\end{displaymath}

which, as before, has the solution

\begin{displaymath}v_x = A\sin\omega_ct\end{displaymath}

with $\omega_c = qB/m$. Substituting this solution into the equation for $v_y$, we get

\begin{displaymath}v_y = a\cos\omega_ct + C\end{displaymath}

Since $v_y =0$at $t=0$, the constant $C=-A$, so that

\begin{displaymath}v_y = A(\cos\omega_ct-1)\end{displaymath}

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