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\begin{document}
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\heading{Lorenz Force}
\auname{I.D. 303464531}
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{\bf The problem:}
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A particle with velocity $\vec v$, mass $m$ and charge $q$ is moving in magnetic field $\vec B$. The angle between $\vec v$ and $\vec B$ is $\theta$, so the movement is circular. Find:
\begin{enumerate}
\item the radius of the circle $R$.
\item the time period $T$
\item the distance the particle moved in $\vec B$ direction in time $T$.
\item the equations of motion
\end{enumerate}
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\Dn
{\bf The solution:}
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1. The force on the particle is $\vec F = q\vec V \times \vec B$. We'll separate $\vec V$ to it's parallel and perpendicular to $\vec B$ components
\begin{eqnarray}
V_\parallel=V \cos\theta \\
V_\bot=V\sin\theta
\end{eqnarray}
In the plane perpendicular to $\vec B$ the movement is circular. The magnetic force $F_B$ balances the centrifugal force
\begin{eqnarray}
\frac{{mV_ \bot ^2 }}{R} = qV_ \bot B
\end{eqnarray}
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which leads to
\begin{eqnarray}
R = \frac{{mV\sin \theta }}{{qB}}
\end{eqnarray}
2.The time period is
\begin{eqnarray}
T = \frac{{2\pi }}{\omega } = \frac{{2\pi R}}{{{\rm{V}}_ \bot }} = \frac{{2\pi m}}{{qB}}
\end{eqnarray}
3.the distance the particle moved in $\vec B$ direction in time $T$ is
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\begin{eqnarray}
P = {\rm{V}}_\parallel T = \frac{{2\pi mV\cos \theta }}{{qB}}
\end{eqnarray}
4.The equations of motion are
\begin{eqnarray}
z(t) - z_0 &=& \Delta z = V_\parallel t = Vt\cos \theta \\
x(t) &=& \Delta x = r\cos (\omega t) \\
y(t) &=& \Delta z = r\sin (\omega t)
\end{eqnarray}
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\end{document}