A magnetostatic <force-free magnetic field> is in equilibrium when its electric current exerts no <Lorentz force density>:
$$
\mathbf J\times\mathbf B=0.
$$
Thus $\mathbf J$ is pointwise parallel to $\mathbf B$, so for some scalar $\alpha(\mathbf r)$,
$$
\mathbf J=\alpha\mathbf B.
$$
Using the magnetostatic <Ampère's law>, $\nabla\times\mathbf B=(4\pi/c)\mathbf J$, gives
$$
\boxed{\nabla\times\mathbf B=\frac{4\pi}{c}\alpha\mathbf B}.
$$
Taking the divergence and using $\nabla\cdot\mathbf B=0$ yields
$$
\boxed{\mathbf B\mathbin\cdot\nabla\alpha=0}.
$$
The <force-free parameter> is therefore constant along each magnetic field line.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-314/1/a/solution