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\frac{\nabla V\times\mathbf{B}}{B^2} = -\frac{2\pi I_{pol}^d}{\langle B^2\rangle}\mathbf{e}_\theta + \frac{2\pi I_{tor}}{\langle B^2\rangle}\mathbf{e}_\phi~. | \frac{\nabla V\times\mathbf{B}}{B^2} = -\frac{2\pi I_{pol}^d}{\langle B^2\rangle}\mathbf{e}_\theta + \frac{2\pi I_{tor}}{\langle B^2\rangle}\mathbf{e}_\phi~. | ||
</math> | </math> | ||
The above expressions adopt very simple forms for the 'vacuum' field, i.e. one with <math>\nabla\times\mathbf{B} = 0</math>. In this case <math>I_{tor} = 0</math> and <math>\tilde{\eta} = 0</math> leaving, e.g. | |||
:<math> | |||
\mathbf{B} = \frac{I_{pol}^d}{2\pi}\nabla\phi,\quad (\text{for a vacuum field)} | |||
</math> | |||
In a low-<math>\beta</math> stellarator the equilibrium magnetic field is approximatelly given by the vauum value. |
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