Is there any Electric Current on (Solar) Coronal Loops? I'm studying Plasma Astrophysics and now I have came across the following problem, which I have been trying to solve on my own, and came to a "non-logical" conclusion?
Here is the Problem:
Let the Coronal Loop have the following Magnetic Field: 
$$ B_{x} = B_o e^{-kz}cos(kx)$$ and $$ B_z = -B_o e^{-kz}sin(kx) $$
for z>0 , and |x| < pi/2k
After finding the Magnetic field lines, it says find the Current and Density distribution on z dependence.
From $ J = \bigtriangledown x B$ 
One can find that $J=0$ ??
From Equilibrium Equation the Density has no dependence on z , except the gravitational field, which is weak?
Any help?  
 A: 
One can find that J=0 ??

Your math is okay.  What you are being asked to examine is a special case of a coronal loop constructed from what is called a potential field.  You often see the converse, namely a non-potential field, in discussions of dynamo theory.
It is perfectly fine, mathematically, to have $\nabla \times \mathbf{B} = 0$.  Whether that describes reality is another issue but it is okay mathematically.  One of the consequences is that this geometry describes a force-free field, i.e., one that satisfies $\mathbf{j} \times \mathbf{B} = 0$.
A general force-free field is defined by assuming:
$$
\nabla \times \mathbf{B} = \alpha \left( \mathbf{r} \right) \ \mathbf{B}
$$
where $\alpha \left( \mathbf{r} \right)$ is a scalar function of position/altitude.  Under these assumptions, a potential field is the limit where $\alpha \left( \mathbf{r} \right) \rightarrow 0$ while the non-potential field corresponds to $\alpha \left( \mathbf{r} \right) \neq 0$.  Using vector calculus and $\nabla \cdot \mathbf{B} = 0$, one can show that:
$$
\mathbf{B} \cdot \nabla \ \alpha \left( \mathbf{r} \right) = 0
$$

From Equilibrium Equation the Density has no dependence on z , except the gravitational field, which is weak?  Any help?

Under these conditions, one can describe the number density and pressure using hydrodynamics, similar to how one examines neutral atmospheres, i.e.:
$$
n\left( r \right) \propto e^{-r/h}
$$
where $h$ is the characteristic scale height and $r$ is the altitude.
Side Note
While I have not seen your specific example used, I have seen something similar given by:
$$
\begin{align}
  B_{x} & = B_{o} \ e^{- l \ z} \ \sin{\left( k x \right)} \\
  B_{y} & = B_{o} \ e^{- l \ z} \ \sin{\left( k x \right)} \\
  B_{z} & = B_{o} \ e^{- l \ z} \ \cos{\left( k x \right)} \\
\end{align}
$$
which has $\nabla \times \mathbf{B} \neq 0$.  As you probably already noticed, if you change the sign of your z-component from your example you would also have $\nabla \times \mathbf{B} \neq 0$.
