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Next: Summary Up: MHD Equilibrium Structures Previous: Potential Problems leading to

Non-potential current flowing inside a cylindrical magnetic loop

Assume that ${\bf J} = j_{0}(R){\bf e}_{z}$, then there is a magnetic field induced by the current having the form ${\bf B} =
B_{\phi}(R){\bf e}_{\phi}$ such that ( $\nabla \times {\bf B} =
\mu_{0}{\bf J}$)
\begin{displaymath}
{1\over R}{d\over dR}\left (R B_{\phi}\right ) = \mu_{0} j_{0}(R).
\end{displaymath} (4.16)

This may be integrated to give $B_{\phi}$ as

\begin{displaymath}
B_{\phi}(R) = {1\over R}\int_{0}^{R}\mu_{0}s j_{0}(s) ds,
\end{displaymath}

where $s$ is a dummy variable for the integration. As an example, consider a uniform current flowing inside a loop of radius $a$, with zero current outside. Hence, we have

\begin{displaymath}
j_{0}(R) = \left \{
\begin{array}{ll}
j_{0}, & R < a, \\
0, & R > a.
\end{array} \right.
\end{displaymath}


\begin{displaymath}
{d\over dR}\left (RB_{\phi}\right ) = \left \{
\begin{arra...
... \mu_{0}j_{0}R, & R < a, \\
0, & R > a.
\end{array} \right.
\end{displaymath}


\begin{displaymath}
RB_{\phi} = \left \{
\begin{array}{ll}
\mu_{0}j_{0}{R^{2}\over 2}, & R < a, \\
C, & R > a,
\end{array} \right.
\end{displaymath}

where $C$ is a constant. Here we have used the boundary condition $B_{\phi}(0) = 0$. To determine the value of the constant we use continuity of $B_{\phi}$ at $R=a$. Thus, from

\begin{displaymath}
B_{\phi}(R) = \left \{
\begin{array}{ll}
\mu_{0}j_{0}{R\over 2}, & R < a, \\
{C\over R}, & R > a.
\end{array} \right.
\end{displaymath}


\begin{displaymath}
B_{\phi}(a) = \mu_{0}j_{0}{a\over 2} = {C\over a},
\end{displaymath}

and so

\begin{displaymath}
C = \mu_{0}j_{0}{a^{2}\over 2}.
\end{displaymath}

Therefore, the final form for $B_{\phi}$ is

\begin{displaymath}
B_{\phi}(R) = \left \{
\begin{array}{ll}
\mu_{0}j_{0}{R\o...
...\
{\mu_{0}j_{0}a^{2}\over 2R}, & R > a.
\end{array} \right.
\end{displaymath}

Since the current is no longer zero everywhere and because ${\bf J}$ is not parallel to ${\bf B}$, there will be a non-zero Lorentz force. Thus,

\begin{displaymath}
{\bf J}\times {\bf B} = {\nabla \times {\bf B}\over \mu_{0}...
... 2}{\bf e}_{R}, & R < a, \\
0, & R > a.
\end{array} \right.
\end{displaymath}

As mentioned earlier in this section, the Lorentz force can be expressed in terms of a magnetic pressure force and a magnetic tension force. We can now calculate each of the effects.

The magnetic tension force is, for $R<a$,

\begin{displaymath}
{1\over \mu_{0}}\left ({\bf B}\cdot \nabla\right ){\bf B} =...
...i}^{2}\over \mu_{0}R} = -{\mu_{0}\over 4}j_{0}^{2}R{\bf e}_{R}
\end{displaymath}

while for $R>a$ we get

\begin{displaymath}
-{\mu_{0}j_{0}^{2}a^{4}\over 4R^{3}}{\bf e}_{R}.
\end{displaymath}

The magnetic pressure force is, for $R<a$,

\begin{displaymath}
-\nabla \left ({B^{2}\over 2\mu_{0}}\right ) = - {d\over dR...
...right ){\bf e}_{R} = -{\mu_{0}\over
4}j_{0}^{2}R{\bf e}_{R},
\end{displaymath}

and for $R>a$,

\begin{displaymath}
{\mu_{0}j_{0}^{2}a^{4}\over 4 R^{3}}{\bf e}_{R}.
\end{displaymath}

From these results we see that the two forces add together inside the magnetic loop ($R<a$) but the two forces cancel outside where $R>a$.

Because the Lorentz force acts towards the centre of the loop, there must be another force inorder to keep the loop in equilibrium if there is no flow. This must be due to a gas pressure gradient. If the pressure is $p(R)$ then the Lorentz force is balanced by

\begin{displaymath}
\nabla p = {\bf J}\times {\bf B}.
\end{displaymath}

Thus,

\begin{displaymath}
{dp\over dR} = -{d\over dR}\left ({B_{\phi}^{2}\over
2\mu_{0}}\right ) - {B_{\phi}^{2}\mu_{0}R}.
\end{displaymath}


next up previous
Next: Summary Up: MHD Equilibrium Structures Previous: Potential Problems leading to
Prof. Alan Hood
2000-11-06