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$$\vec{\nabla} P = \frac{1}{c} \vec{J} \times \vec{B}$$

Now, if ##P=const## and ##\vec{J}## (in MHD ##\vec{J}\propto\vec{\nabla}\times\vec{B}##) is parallel to ##\vec{B}##, we get that ##(\vec{\nabla} \times \vec{B}) \times \vec{B}=0##. Thus implies that:

$$(\vec{\nabla} \times \vec{B}) = \alpha(r) \vec{B}$$

That is the condition for a Free-Froce fields. So ... the question is, shouldn't be the curl of a vector always be orthogonal to the vector?