= Solution
Use the source's <Gaussian units> and neglect the <displacement current>, as in nonrelativistic <magnetohydrodynamics>. Write $\Delta p=p_\perp-p_\parallel$ for the <pressure anisotropy>. The <pressure tensor> is $\mathsf P=p_\perp\mathsf I-\Delta p\,\mathbf b\mathbf b$, so its <force density> is
$$
-\nabla\cdot\mathsf P=-\nabla p_\perp+\nabla\cdot(\Delta p\,\mathbf b\mathbf b).
$$
<Ampère's law> and the <solenoidal magnetic-field constraint> give the <Lorentz force> decomposition
$$
\frac{\mathbf j\times\mathbf B}{c}
=\frac{(\nabla\times\mathbf B)\times\mathbf B}{4\pi}
=-\nabla\frac{B^2}{8\pi}+\frac{(\mathbf B\cdot\nabla)\mathbf B}{4\pi}
=-\nabla\frac{B^2}{8\pi}+\nabla\cdot\frac{\mathbf B\mathbf B}{4\pi}.
$$
Adding these two <force densities> and using $\mathbf B\mathbf B=B^2\mathbf b\mathbf b$ yields
$$
\boxed{\rho\frac{D\mathbf u}{Dt}=-\nabla\left(p_\perp+\frac{B^2}{8\pi}\right)+\nabla\cdot\left[\left(\Delta p+\frac{B^2}{4\pi}\right)\mathbf b\mathbf b\right].}
$$
The first term is the <gradient> of the modified <magnetohydrodynamic total pressure>; the second is the anisotropic <Maxwell stress tensor> contribution together with the directional <pressure force>. Their effective coefficient is the <pressure-anisotropy-modified magnetic tension>, $\mathcal T=B^2/(4\pi)+\Delta p$. At fixed <magnetic field>, positive <pressure anisotropy> increases resistance to bending, while negative <pressure anisotropy> reduces it. When $\mathcal T$ becomes negative, bending experiences an amplifying <force> rather than a restoring <force>. Gradients of $\Delta p$ can also drive motion, so the curvature interpretation refers specifically to the bending contribution.
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