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Effect of ${\bf v}$ on ${\bf B}$

For a field ${\bf B} = B_{y}(x,t){\bf j}$ and flow ${\bf v} =
v_{x}(x,t){\bf i}$, (4.2) becomes
\begin{displaymath}
{\partial B_{y}\over \partial t} = - {\partial\over \partial x}\left
(v_{x}B_{y}\right ).
\end{displaymath} (4.10)

Now consider the magnetic flux between two lines at $x=a$ and $x=b$, which are moving with the plasma. Thus, $a = a(t)$ and $b = b(t)$ such that

\begin{displaymath}
{da\over dt} = v_{x}(a,t), \qquad {db\over dt} = v_{x}(b,t).
\end{displaymath}

The flux is

\begin{displaymath}
\hbox{Flux } = \int_{a(t)}^{b(t)} B_{y}(x,t)dx.
\end{displaymath}

Figure 4.4: The flux between $x=a$ and $x=b$.
\includegraphics [scale=0.5]{fundfig42.eps}

Notice that the flux is a function of time only. The spatial dependence has been integrated out. Flux changes in time due to two effects, namely because $B_{y}$ changes in time and because the end points move at speeds $v_{a}$ and $v_{b}$ reducing or increasing the range of integration. On differentiating with respect to time we must remember to differentiate not only the integrand but also the limits. Thus,

\begin{displaymath}
{D\over Dt}\int_{a}^{b}B_{y}dx = \int_{a}^{b}{\partial B_{y...
...rtial t} dx + {db\over dt}B_{y}(b,t) - {da\over dt}B_{y}(a,t).
\end{displaymath}

Hence, we may write

\begin{displaymath}
{D\over Dt}\int_{a}^{b}B_{y}dx = \int_{a}^{b}{\partial B_{y...
...} + {\partial \over \partial x}\left
(v_{x}B_{y}\right ) dx.
\end{displaymath}

By (4.10) this is zero. In other words, the amount of magnetic flux remains constant, and we say that it is frozen to the plasma.

Example 4. .1Consider the effect of a flow $v_{x} = - v_{0}x/L$ on an initially uniform field, $B_{y}(x,0) = B_{0}$. Equation (4.10) becomes

\begin{displaymath}
{\partial B_{y}\over \partial t} + v_{x}{\partial B_{y}\over \partial
x} + {\partial v_{x}\over \partial x} B_{y} = 0.
\end{displaymath}

Thus,
\begin{displaymath}
{\partial B_{y}\over \partial t} - {v_{0}x\over L}{\partial
B_{y}\over \partial x} - {v_{0}\over L}B_{y} = 0.
\end{displaymath} (4.11)

Look for a seperable solution to this linear equation of the form

\begin{displaymath}
B_{y} = X(x)T(t),
\end{displaymath}

so that (4.11) may be expressed as

\begin{displaymath}
X T^{\prime} - {v_{0}x \over L} X^{\prime}T - {v_{0}\over L} XT = 0.
\end{displaymath}

Dividing by $XT$ and collecting all the functions of time onto one side of the equation and all the functions of $x$ onto the other we get

\begin{displaymath}
{T^{\prime}\over T} - {v_{0}\over L} = {v_{0}x\over
L}{X^{\prime}\over X} = \lambda = \hbox{ constant.}
\end{displaymath}

Consider the $x$-dependence first of all.

\begin{displaymath}
{1\over X}{dX\over dx} = \lambda {L\over v_{0}x}.
\end{displaymath}

Integrating we obtain

\begin{displaymath}
\log X = \lambda {L\over v_{0}}\log x + C \qquad \Rightarrow \qquad
X(x) = A x^{\lambda L/v_{0}}.
\end{displaymath}

Similarly we may integrate the $t-$dependent terms,

\begin{displaymath}
{dT\over dt} = \left (\lambda + {v_{0}\over L}\right ) T, \qquad
\Rightarrow \qquad T(t) = C e^{(\lambda + v_{0}/L)t}.
\end{displaymath}

Hence, we obtain

\begin{displaymath}
B_{y} = X(x)T(t) = D x^{\lambda L/v_{0}}e^{(\lambda + v_{0}/L)t}.
\end{displaymath}

In theory, we need to obtain the values for the separation constant $\lambda$ and add together all the possible solutions. However, this example is particularly simple. Satisfying the initial condition gives

\begin{displaymath}
B_{y}(x,0) = B_{0}, \qquad \Rightarrow \qquad \lambda = 0 \hbox{
and } D = B_{0}.
\end{displaymath}

Therefore,

\begin{displaymath}
B_{y} = B_{0}e^{(v_{0}t/L)}.
\end{displaymath}

Thus, the flow carries in the field lines and makes the magnetic field strength grow exponentially in time.
next up previous
Next: MHD Equilibrium Structures Up: MHD Equations Previous: MHD Equations
Prof. Alan Hood
2000-11-06