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

$${\partial B_{y}\over \partial t} = - {\partial\over \partial x}\left (v_{x}B_{y}\right ).$$ (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

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

The flux is

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

Figure 4.4: The flux between $x=a$ and $x=b$.

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,

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

Hence, we may write

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

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

$${\partial B_{y}\over \partial t} + v_{x}{\partial B_{y}\over \partial x} + {\partial v_{x}\over \partial x} B_{y} = 0.$$

Thus,

$${\partial B_{y}\over \partial t} - {v_{0}x\over L}{\partial B_{y}\over \partial x} - {v_{0}\over L}B_{y} = 0.$$ (4.11)

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

$$B_{y} = X(x)T(t),$$

so that (4.11) may be expressed as

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

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

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

Consider the $x$-dependence first of all.

$${1\over X}{dX\over dx} = \lambda {L\over v_{0}x}.$$

Integrating we obtain

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

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

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

Hence, we obtain

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

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

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

Therefore,

$$B_{y} = B_{0}e^{(v_{0}t/L)}.$$

Thus, the flow carries in the field lines and makes the magnetic field strength grow exponentially in time.