Starting with the momentum equation for a fluid element in ideal MHD (ignoring viscosity and gravity):

\rho \frac{d}{dt} \bold{u} + \rho (\bold{u} \cdot \bold{\nabla}) \bold{u} = - \bold{\nabla} P + \frac{1}{\mu}(\bold{j} \times \bold{B})

The 2nd term on the RHS is the Lorentz force (or more accurately force per unit volume). Since we don't follow charges or currents, we rewrite it as:

\bold{j}\times\bold{B}= (\bold{\nabla} \times \bold{B})\times \bold{B}

which using a vector identity is rewritten as:

(\bold{\nabla} \times \bold{B})\times \bold{B}=-\bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{1}{\mu}(\bold{B}\cdot \bold{\nabla}) \bold{B}

the first term on the RHS is the same form as thermal pressure and so we call it the magnetic pressure. The second term is generally called the magnetic tension. To see why this is so, we rewrite it as:

\frac{1}{\mu}(\bold{B}\cdot \bold{\nabla}) \bold{B} = \hat{\bold{b}}\hat{\bold{b}} \cdot \bold{\nabla}(\frac{B^2}{2 \mu}) + \frac{B^2}{\mu} \frac{\hat{\bold{n}}}{R_c}

The first term on the RHS is a unit tensor dotted with the gradient of the magnetic pressure. The tensor is formed by the outer product of \hat{\bold{b}} with itself. \hat{\bold{b}} is the unit vector along the magnetic field line. Thus, if the field line was oriented along z,

\hat{\bold{b}}\hat{\bold{b}} =

0	0	0
0	0	0
0	0	1

If the gradient of magnetic pressure was along the field line, that the first term on the LHS would be just the gradient of magnetic pressure. This would cancel with the other term for magnetic pressure. Thus we see that the magnetic pressure does not provide a force along field lines. It is thus does not provide an isotropic pressure force, but rather only acts perpendicular to the field lines, which is what we will consider next.

Say the gradient of magnetic pressure is perpendicular to the magnetic field lines, say for example the gradient is along y (field is pointing along z still). Then,

\hat{\bold{b}}\hat{\bold{b}} \cdot \bold{\nabla}(\frac{B^2}{2 \mu}) = \bold{0}