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

$\frac{dl_x}{B_x} = \frac{dl_y}{B_y} = \frac{dl_z}{B_z} \qquad Q.E.D.$ Last modified on 21 March 2012, at 15:48