Nuclear Fusion Physics and Technology/Example page

As summarized in previous chapters, electromagnetic field is mathematical abstraction of two projections ${\displaystyle {\vec {E}}({\vec {r}}):\mathbb {R} ^{3}\rightarrow \mathbb {R} ^{3}}$ and ${\displaystyle {\vec {B}}({\vec {r}}):\mathbb {R} ^{3}\rightarrow \mathbb {R} ^{3}}$, which meets Maxwell equations

${\displaystyle rot{\vec {H}}={\vec {j}}+{\frac {\partial {\vec {D}}}{\partial t}}\qquad rot{\vec {E}}+{\frac {\partial {\vec {B}}}{\partial t}}=0}$
${\displaystyle div{\vec {B}}=0\qquad {\vec {D}}=\rho }$

and may be represented by field lines defined as

${\displaystyle {\frac {d{\vec {x}}}{ds}}=\alpha {\vec {B}}({\vec {r}})}$

Field line is open, when it is not closed in plasma.

Field line is closed, when it is closed in plasma.

Lets assume electromagnetic field ${\displaystyle {\vec {E}}({\vec {r}}),{\vec {B}}({\vec {r}})}$ with field lines. Then

${\displaystyle {\frac {dl_{x}}{B_{x}}}={\frac {dl_{y}}{B_{y}}}={\frac {dl_{z}}{B_{z}}}}$

Proof:
The theorem results from field line definition directly

${\displaystyle {\frac {d{\vec {x}}}{ds}}=\alpha {\vec {B}}({\vec {r}})\qquad /.{\frac {ds}{\vec {B}}}}$

${\displaystyle {\frac {d{\vec {x}}}{\vec {B}}}=\alpha .ds}$

which is a vector equation of three scalar equations

${\displaystyle {\frac {dl_{x}}{B_{x}}}=\alpha .ds\qquad {\frac {dl_{y}}{B_{y}}}=\alpha .ds\qquad {\frac {dl_{z}}{B_{z}}}=\alpha .ds}$

and thus may be written

${\displaystyle {\frac {dl_{x}}{B_{x}}}={\frac {dl_{y}}{B_{y}}}={\frac {dl_{z}}{B_{z}}}\qquad Q.E.D.}$