# MHD problem

1. Jan 14, 2009

### Logarythmic

1. The problem statement, all variables and given/known data
Two of the MHD equations can be formulated as

$$\vec{E} + \vec{v} \times \vec{B} = \eta \vec{J}$$
$$\nabla \times \vec{B} = \mu_0 \vec{J}$$

where [itex]\eta[/tex] is the resistivity of the plasma.

a.) Derive an equation for the magnetic field at very high resistivity and describe the corresponding consequences for the behaviour of the magnetic field lines.

b.) Derive an equation for the magnetic field av very low resistivity and describe the corresponding consequences for the behaviour of the magnetic field lines.

2. Relevant equations
The set of Maxwell equations?

3. The attempt at a solution
I have no clue. This has been a problem for a week or so, and it's the only one I got left. The only thing I know is that when the resistivity is low, the field is frozen-in to the plasma.

2. Jan 14, 2009

### gabbagabbahey

Hmmm... I'm not too familiar with Magnetohydrodynamics, but it seems to me that you might need at least one more equation...

I'd start by eliminating J from your equations, giving you a single realtion involving E and B.

In the limit of extremely high resistivity, you may be able to treat $$\frac{1}{\eta}(\vec{E} + \vec{v} \times \vec{B})$$ as effectively zero.

In the opposite limit, I think you need another equation in order to eliminate E.....Does $$\vec{\nabla} \times \vec{E}=\frac{-\partial \vec{B}}{\partial t}$$ hold true in MHD? If so, I'd use that to eliminate E.

3. Jan 14, 2009

### Logarythmic

Yeah Faraday's law holds, so maybe

$$\frac{\eta}{\mu_0} \nabla^2 \vec{B} - (\nabla \cdot \vec{v})\vec{B} - \frac{\partial \vec{B}}{\partial t} = 0$$

But what about the behaviour of the field lines? Could one say that for high resistivity the current is zero and the magnetic field is static, and that for low resistivity the field lines are frozen-in?

4. Jan 14, 2009

### gabbagabbahey

I think you are missing a couple of terms; shouldn't it be:

$$\frac{\eta}{\mu_0} \nabla^2 \vec{B} +(\vec{B} \cdot \vec{\nabla})\vec{v}-(\vec{v} \cdot \vec{\nabla})\vec{B}- (\vec{\nabla} \cdot \vec{v})\vec{B} - \frac{\partial \vec{B}}{\partial t} = 0$$

And since you are talking about the low resisitivity limit, you can probably neglect the first term:

$$(\vec{B} \cdot \vec{\nabla})\vec{v}-(\vec{v} \cdot \vec{\nabla})\vec{B}- (\vec{\nabla} \cdot \vec{v})\vec{B} - \frac{\partial \vec{B}}{\partial t} = 0$$

Well, I'm not really sure what the field lines look like in the low resistivity limit, but you could try plotting the solution to the above PDE to see.

EDIT: You can probably make some assumptions about the velocity field of the plasma in the high resistivity limit that will simplify the PDE further.

As for the high resistivity limit, don't you have

$$\vec{\nabla} \times \vec{B} \approx 0$$

That should tell you something about the field lines.