# 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?

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 $\nabla \times \mathbf{B} \neq 0$.

• Since my scalar function is $$a(\textbf{r}) = 0$$ Using the following equation: $$\nabla P - \rho \textbf{z} = 0$$ and since $$P = nk_bT$$ -my approximation is T(x,y,z) = Const - I don't get the exponential you have. – Billy Matlock May 24 '17 at 14:38
• *Couldnt edit my previous comment: I get the result $$\rho = \rho_o e^{-mgz/kT}$$ – Billy Matlock May 24 '17 at 14:44
• @BillyMatlock - That's fine. In this case the scale height is just $h = kT/mg$ and $\rho = \bar{m} \ n$, where $\bar{m}$ is the mean mass of the gas (i.e., roughly the ion mass here). – honeste_vivere May 24 '17 at 15:57