# Capacitor and resistor in a flowing liquid

1. Feb 14, 2007

### Saketh

1. The problem statement, all variables and given/known data

A parallel-plate capacitor with cross-sectional area S and plate separation d is put into a stream of conducting liquid with resistivity $\rho$. The liquid moves parallel to the plates with a constant velocity v, and the whole system is located in a uniform magnetic field of induction B, vector $\textbf{B}$ being parallel to the plates and perpendicular to the stream direction. The capacitor plates are interconnected by means of an external resistance R. What amount of power is generated in that resistance?

2. Relevant equations

$$\rho J = E$$
Maxwell's equations

3. The attempt at a solution

I have no intuitive grasp of what's going on in this problem. My first guess was that the charges in the fluid are being pushed by the magnetic field, but what does that have to do with the current flowing through the resistor? I'm not sure why current is flowing through the resistor at all.

I can't really show work because I'm not sure where to start. Once I figure out what's going on, I can do the rest of this problem myself.

Last edited: Feb 14, 2007
2. Feb 14, 2007

### Staff: Mentor

Cool question. It's related to the field of magnetohydrodynamics:

http://en.wikipedia.org/wiki/Magnetohydrodynamics

You are correct that the moving charges in the fluid will experience a force perpendicular to the magnetic field. This will effectively pump charges from one capacitor plate to the other, which will create a voltage which causes a current through the external resistor.

So if you can calculate what the plate-to-plate flux of charges is due to the force from the magnetic field, then you will be able to calculate the current flow through the resistor. The only thing I'm not sure about is whether you need to also account for a reverse leakage current back through the conductive liquid (as appearing like another resistor in parallel with the explicit load resistor)..... I'm not sure about that aspect without giving it more thought.

3. Feb 14, 2007

### Saketh

If the magnetic field is exerting a force on the charges, I thought that the fluid would be moving upwards, in which case it would not be moving parallel to the plates as said earlier.

This confuses me. How can the charges move from one plate to the other when the magnetic force will push them in an arc?
My guess is that the resistance of the liquid is negligible because it acts as a "current source" instead of a pure "voltage source." But I am the one who's confused about this question, so take that with a grain of salt.

4. Feb 14, 2007

### Staff: Mentor

The "conducting liquid" is flowing straight through the capacitor, I would think, under the influence of some hydraulic pressure source. The fact that the free conductors in the fluid are experiencing forces orthogonal to the fluid flow should not alter the fluid's direction substantially. The fluid flowing is what moves the conductors along perpendicular to the B field, which is what imparts a force on them to make them tend to flow from plate-to-plate.

I guess I'd start by ignoring the fluid aspect of the question, and just think about the charges as being free and flowing in a stream. I'd just think about the electrons and ignore the fact that there are matching protons somewhere in the soup to keep everything electrically neutral.

I'm not sure about the next steps, but I think I would assume that a voltage difference develops across the plates, which generates an opposing E field to the B field Lorentz force, in order for the current flow to reach equilibrium. I think that it is the balance between the E and B field forces on the electrons which will result in a steady upward motion of the electrons instead of a curved trajectory like they would follow in the absence of the capacitor plates.

Maybe explore that angle some -- see if you can get to where the electrons are travelling upward at some angle, and calculate the net flux of electrons between the plates from that. Fun problem.

5. Feb 14, 2007

### Saketh

Yes, it is a fun problem. It's from the Irodov book -- 3.265, to be exact.

I'll think about it more tomorrow, and report back if I have any breakthroughs.