proof

Try to prove that

$$\nabla \cdot \frac{(\mathbf{B}_0 \times \nabla \psi)}{B^2_0} = - 2\ensuremath{\boldsymbol{\kappa}} \cdot \left( \frac{\mathbf{B}_0 \times \nabla \psi}{B_0^2} \right) .$$ (357)

Proof:

$$\nabla \cdot \frac{(\mathbf{B}_0 \times \nabla \psi)}{B^2_0} = \frac{1}{B_0^2} \nabla \cdot (\mathbf{B}_0 \times \nabla \psi) + (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2}$$

$$= \frac{1}{B_0^2} \nabla \psi \cdot \nabla \times \mathbf{B}_0 + (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2}$$

$$= \frac{1}{B_0^2} \nabla \psi \cdot {\textmu}_0 \mathbf{J}_0 + (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2}$$

$$= 0 + (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2} .$$ (358)

Using

$$(\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2} = ... ...B_0^2} (\mathbf{B}_0 \times \nabla \psi) \cdot \ensuremath{\boldsymbol{\kappa}}$$ (359)

(proof of this is given in Sec. 9.9), Eq. (358) is written as

$$\nabla \cdot \frac{(\mathbf{B}_0 \times \nabla \psi)}{B^2_0} = - ... ...B_0^2} (\mathbf{B}_0 \times \nabla \psi) \cdot \ensuremath{\boldsymbol{\kappa}}$$ (360)