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\heading{Fields inside a rolling charged cylinder}
\auname{I.D. 310159025}
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{\bf Problem}
A cylinder of a radius $R$, a length $L$, charded with a surface density charge $\sigma$, is rolling with a linear velocity $u \ll c$ without sliding.
Find the electric and magnetic fields inside the cylinder.
\Dn
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{\bf Solution}
Let $S$ be the rest system of the cylinder.
The laboratory system is $S'$. The rolling direction is $-\hat x$.
the current in the S system is:
\begin{eqnarray}
I&=&\frac{q}{T}=\frac{\sigma2\pi{RL\omega}}{2\pi} \\
\omega &=& \frac{u}{R} \\
I &=& \sigma{uL}
\end{eqnarray}
where $\omega$ is the angular velocity.
From the Ampere's law we can find out the magnetic field in the $S$ system:
\begin{eqnarray}
\oint{\vec B \cdot d\vec r}&=&4\pi K I \\
B&=&{4\pi}K\sigma{u}\hat{z}
\end{eqnarray}
The electric field in the $S$ system is zero:
\begin{eqnarray}
\mbox{div}\vec E={4\pi}{k\rho}\Rightarrow{\rho=0}\Rightarrow{E=0}
\end{eqnarray}
Since $u \ll c$, then $\gamma \simeq 1$ and $\sigma \simeq \sigma'$.
Using the Lorentz transformations we can find the fields in the lab system.
The $S'$ system moves with the velocity $u$ to the $x$ direction
Electric field $E'$
\begin{eqnarray}
E'_\| &=& E_\|=0 \\
E'_\perp &=& \gamma_0{E_\perp+{u}\times{B_\perp}}
=\gamma_0{uB\perp} \simeq {4\pi}{K\sigma{u^2}\hat{y}}
\end{eqnarray}
Magnetic field $B'$
\begin{eqnarray}
B'_\|&=&B_\|=0 \\
B'_\perp&=&\gamma_0{B_\perp+{u}\times{E_\perp}}=\gamma_0{B_\perp}
\simeq {4\pi}{K\sigma{u}\hat{z}} = B_\perp
\end{eqnarray}
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\end{document}