How to Prove P+B²/8π in MHD Equilibrium?

[QuarkDecay]

Homework Statement

Prove that in an MHD equilibrium state in a pinch with flux and B_θ(r) and B_z(r), the equation for pressure is;
d/dr(P + B²/8π) + B_θ²/4πr - ρ( u²_θ/r)= 0

B is the total Magnetic field, u_θ(r) the velocity of plasma, ρ= density, and solving in a (r,θ,z) system

Homework Equations

MHD equilibrium equations ( solution for z and θ pinches)
(1) ∇P= (J x B)/c
(2) ∇ x B= (4π/c)J

Edit;

The MHD equlibrium equation when there's flux is

ρ(u∇)u= (j x B)/c -∇P
That I apparently have to use instead of (1)
3. The Attempt at a Solution

I know how to solve the problem with the θ-pinch, that has a solution for Pressure; d/dr( P(r) + B_z²(r)/8π) = 0

We calculate the J first, from equation (2), by calculating the ∇ x B, with ∇= ∂/∂r r^{^} + 1/r(∂/∂θ) θ^{^} + ∂/∂z z^{^} and B only in the z^{^} axis, B-> B(r) z^{^}= B_z(r)

Then we calculate the J x B ( J_θ= -c/4π (dB_z/dr ),and equate it from (1) like;
∇P = (JxB)/c, and here ∇P=dP_r/dr
dP_r/dr= (JxB)/c = -1/4π (dB_z/dr =>
=> d/dr( P(r) + B_z²(r)/8π) = 0

 

[NateD]

But I don't know how to calculate the pressure of a θ-pinch from the MHD equilibrium equation.I have to assume Bθ2(r)/4πr, but I don't understand why and from what equation that part comes from.