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\begin{document}
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\heading{Displacement Current}
\auname{I.D. 043423755}
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{\bf The problem:}
A capacitor made of circular plates of the area $A$ is connected to the voltage source $V$, the distance between the two plated is $d$.
The two plates are moving to each other with the constant speed $u$.
\begin{enumerate}
\item What is the magnetic field between the plates?
\item What is the magnetic field between the plates if the capacitor is disconnected but charged?
\end{enumerate}
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\Dn
{\bf The solution:}
\Dn
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1. The Maxwell equation
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\begin{eqnarray}
\oint \vec B \cdot d\vec l = \mu_0 I + \mu_0 \epsilon_0 \frac{d}{dt} \int \vec{E}\cdot d\vec{a}
\end{eqnarray}
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Since there is no current between the plates we are left with
\begin{eqnarray}
\oint \vec B \cdot d\vec l = \mu_0 \epsilon_0 \frac{d}{dt} \int \vec{E}\cdot d\vec{a}
\end{eqnarray}
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where the displacement current is
\begin{eqnarray}
I_d = \epsilon_0 \frac{d}{dt} \int \vec{E}\cdot d\vec{a}
\end{eqnarray}
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The electrical field as a function of time when the distance between the plates decreases is:
\begin{eqnarray}
\vec{E} = \frac{V_0}{d-ut} \hat z
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\end{eqnarray}
where $z$ is the symmetry axis of the capacitor.
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The magnetic field is tangential
\begin{eqnarray}
\oint B\cdot d\vec{l} &=& B(r) 2 \pi r \\
\oint E\cdot d\vec{a} &=& \frac{V_0}{d-ut} \cdot \pi r^2
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\end{eqnarray}
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Therefore:
\begin{eqnarray}
B = \frac12 \mu_0 \epsilon_0 V_0 \frac {u}{(d-ut)^2}r \hat{\theta}
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\end{eqnarray}
\Dn
2. if the capacitor is disconnected but charged:
\begin{eqnarray}
\vec E = 4 \pi k \sigma \hat{z} = 4 \pi k \frac{Q}{A} \hat{z}
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\end{eqnarray}
The electrical field is time-independent. Hence, there is no magnetic field.
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