Why Is the Magnetic Field Inside the Cylinder Negative?

[stunner5000pt]

Homework Statement

An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization. parallel to the axis, $M = ks\hat{z}$ where k is a constant and s is the distance from the axis; there is no free current anywhere. Find the magnetic field inside and outside the cylinder

Homework Equations

$$J_{b} = \nabla \times M$$
$$K_{b} = M\times \hat{n}$$
$$\oint B \cdot dl = \mu_{0} I_{enc}$$

The Attempt at a Solution

Here $$J_{b} = -k\hat{\phi}$$
and $$K_{b} = kR \hat{\phi}$$

so the field inside s<R
$$B \cdot 2\pi s = \mu_{0} \int J_{b} da = \mu_{0} \int -k s'ds' d\phi'$$
so i get $$B = -\mu_{0} ks \hat{z}$$
but the answer is supposed to be positive...
why is that? Am i supposed to include the surface current density to find the field? But for a question in the past (for a cylinder with magnetization $M = ks^2 \hat{\phi}$.. however that time the enclosed current in the enitre (s>R) cylinder was zero - there was symmetry between the two surface currents. The amperian loop was a circlular loop within the cylinder...

Is this question to be solved differently because there is no symmtery between the surface and volume current densities?

 

[kuruman]

If $\mathbf{J}_b=-k \mathbf{\hat \phi}$ and you want to write the current as $I_b=\int\mathbf{J}_b\cdot \mathbf{da}$, then in what direction should the normal to area element $da$ be?

You wrote $da=s' ds'd\phi'$ which implies that $\mathbf{da}=s' ds'd\phi'\mathbf{\hat z}$. In that case $I_b=0$.