I was trying to derive the cross-product of vectors:
$\vec{B} = B_\theta$
$\vec{J} = -\frac{1}{\mu_0} \frac{d}{dr}B_z(r)$
substituted on
$$\nabla p = \vec{J} \times \vec{B}$$
getting
$$\frac{d}{dr}\Bigg(p+\frac{B_\theta ^2}{\mu_0}\Bigg) + \frac{B^2_\theta}{\mu_0 r} = 0$$
but the text says it is
$$\frac{d}{dr}\Bigg(p+\frac{B_\theta^2}{2\mu_0}\Bigg) + \frac{B^2_\theta}{\mu_0 r} = 0$$
Where does the 1/2 factor come from in term 2?
Thanks!