Generals 1998 8

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[edit] Magnetohydrodynamics

We start with:

\mathbf{J}\times\mathbf{B}=\nabla p


and:

\mathbf{B}=B_{T}\left(R,Z\right)\hat{\mathbf{\theta}}


and:

p=p\left(R,Z\right)


With p vanishing at:

\left(R-R_{0}\right)^{2}+Z^{2}=a^{2}


We have:

\nabla\times\mathbf{B}=\mu_{0}\mathbf{J}


So that:

\mathbf{J}=\frac{1}{\mu_{0}R}\nabla\left(RB_{T}\right)\times\hat{\mathbf{\theta}}


Plugging in:

\mathbf{J}\times\mathbf{B}=\nabla p


Dotting with \mathbf{J}:

0=\left(\frac{1}{\mu_{0}R}\nabla\left(RB_{T}\right)\times\hat{\mathbf{\theta}}\right)\cdot\nabla p


Or:

0=\hat{\mathbf{\theta}}\cdot\nabla\left(RB_{T}\right)\times\nabla p


So that RBT must be a function of p:

RB_{T}=F\left(p\right)


Using this in the equation for \mathbf{J}:

\mathbf{J}=\frac{1}{\mu_{0}R}\nabla\left(F\left(p\right)\right)\times\hat{\mathbf{\theta}}=\frac{F^{\prime}\left(p\right)}{\mu_{0}R}\nabla p\times\hat{\mathbf{\theta}}


And in \mathbf{B}:

\mathbf{B}=\frac{F\left(p\right)}{R}\hat{\mathbf{\theta}}


We get from \mathbf{J}\times\mathbf{B}=\nabla p:

\left(\frac{F^{\prime}\left(p\right)}{\mu_{0}R}\nabla p\times\hat{\mathbf{\theta}}\right)\times\left(\frac{F\left(p\right)}{R}\hat{\mathbf{\theta}}\right)=\nabla p


Giving:

-\frac{F^{\prime}\left(p\right)}{\mu_{0}R}\frac{F\left(p\right)}{R}\nabla p=\nabla p


And:

\mu_{0}R^{2}=-F^{\prime}\left(p\right)F\left(p\right)


Since the right hand side has to be constant on the boundary (it is only a function of p, and p=0 on the boundary), but the left hand side varies on the boundary, this cannot be satisfied.

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