In a <magnetostatic equilibrium>, inertia is absent and all forces balance. A <force-free magnetic field> is the limiting case in which the <Lorentz force density> vanishes:
$$
\frac1c\mathbf j\times\mathbf B=0.
$$
Thus the <current density> is parallel to the <magnetic field>, so for some scalar <force-free parameter> $\alpha(\mathbf r)$,
$$
\mathbf j=\alpha\mathbf B.
$$
Using <Ampère's law> in magnetostatics gives
$$
\boxed{\nabla\times\mathbf B=\frac{4\pi}{c}\alpha\mathbf B}.
$$
Taking the <divergence> and using both the <divergence of a curl is zero> and $\nabla\mathbin\cdot\mathbf B=0$ yields
$$
0=\nabla\mathbin\cdot(\alpha\mathbf B)
=\mathbf B\mathbin\cdot\nabla\alpha,
$$
so
$$
\boxed{\mathbf B\mathbin\cdot\nabla\alpha=0}.
$$
The <force-free parameter> is therefore constant along every <magnetic field line>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 1 / a / Solution

{title2=$\nabla\mathbin\cdot(\nabla\times\mathbf F)=0$}

For a twice continuously differentiable <vector field>, commuting partial derivatives and antisymmetry of the Levi-Civita symbol give
$$
\nabla\mathbin\cdot(\nabla\times\mathbf F)=0.
$$

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