Using stokes theorem to find magnetic field

[indie452]

Homework Statement

magnetic field is azimuthal B(r) = B(p,z) [tex]\phi[/tex]
current density J(r) = Jp(p,z) p + Jz(p,z) z
= p*exp[-p] p + (p-2)*z*exp[-p] z

use stokes theorem to find B-filed induced by current everywhere in space

Homework Equations

stokes - {integral}dS.[curl A] = {closed integral}dl.A
curl B(r) = J(r)

The Attempt at a Solution

={integral}dS.[curl B(r)] = {closed integral}dl.B(r)
={integral}dS.J(r) = {closed integral}dl.B(p,z) [tex]\phi[/tex]

{integral}dS.J(r) = {integral}dS.p*exp[-p] p + (p-2)*z*exp[-p] z = {closed integral}dl.B(p,z) [tex]\phi[/tex]

dl = pd[tex]\phi[/tex] [tex]\phi[/tex]
dS = pd[tex]\phi[/tex]dz p + pd[tex]\phi[/tex]dp z

So:
{closed integral}pd[tex]\phi[/tex].B(p,z) [tex]\phi[/tex] - with limits 0-2pi
= B(p,z)*2pi*p

{integral}dS.p*exp[-p] p + (p-2)*z*exp[-p] z
do in 2 parts:
{integral}pd[tex]\phi[/tex]dz.p*exp[-p] p - with limits 0-2pi and 0-R
= 2pi*p²*R*exp[-p]

{integral}pd[tex]\phi[/tex]dp.(p-2)*z*exp[-p] - with limits 0-2pi and 0-r
= -2pi*r²*exp[-r]

so B(p,z)*2pi*p = 2pi*p*R*exp[-p] - 2pi*r²*exp[-r]
= B(p,z) = p*R*exp[-p] - r²/p*exp[-r]

is this answer right?

 

[vela]

What's the difference between R, r, and p (by which I think you mean ρ)? That's the first thing you need to clear up.

You didn't say what surface S you're integrating over. In any case, I think your evaluation of

[tex]\oint_S (\nabla\times\textbf{B})\cdot d\textbf{S}[/tex]

is incorrect.