Nuclear Fusion Physics and Technology/Example page

Note: Electromagnetic field
As summarized in previous chapters, electromagnetic field is mathematical abstraction of two projections $$\vec{E}(\vec{r}):\mathbb{R}^3 \rightarrow \mathbb{R}^3$$ and $$\vec{B}(\vec{r}):\mathbb{R}^3 \rightarrow \mathbb{R}^3$$, which meets Maxwell equations $$rot \vec{H} = \vec{j} + \frac{\partial \vec{D}}{\partial t} \qquad rot \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$ $$div \vec{B} = 0 \qquad \vec{D} = \rho $$ and may be represented by field lines defined as $$\frac{d\vec{x}}{ds} = \alpha \vec{B}(\vec{r})$$

Definition: Open field line
Field line is open, when it is not closed in plasma.

Definition: Closed field line
Field line is closed, when it is closed in plasma.

Theorem: Magnetic field line equation
Lets assume electromagnetic field $$\vec{E}(\vec{r}),\vec{B}(\vec{r})$$ with field lines. Then $$\frac{dl_x}{B_x} = \frac{dl_y}{B_y} = \frac{dl_z}{B_z}$$  Proof:  The theorem results from field line definition directly $$\frac{d\vec{x}}{ds} = \alpha \vec{B}(\vec{r}) \qquad /. \frac{ds}{\vec{B}}$$ $$\frac{d\vec{x}}{\vec{B}} = \alpha. ds$$ which is a vector equation of three scalar equations $$\frac{dl_x}{B_x} = \alpha. ds \qquad \frac{dl_y}{B_y} = \alpha. ds \qquad \frac{dl_z}{B_z} = \alpha. ds$$ and thus may be written $$\frac{dl_x}{B_x} = \frac{dl_y}{B_y} = \frac{dl_z}{B_z} \qquad Q.E.D.$$