Q:

Magnetic flow meters use Faraday induction to produce a voltage across a channel in which a conductive fluid is flowing. The basic relationship (using convenient geometry so that the cross products simplify) isVoltage = k･v･B･Lwhere k is a calibration constant, v is the fluid flow speed, B is the constant external magnetic field strength and L is the distance across the channel. My first question is, other than specifics of the electronic design and exact instrument configuration, how does the constant k depend on the conductivity of the fluid? How does one calculate the effective charge and current densities? I am particularly interested in seawater as the conducting fluid.Secondly, given a voltage set up between the electrodes, presumably work can be down by connecting the output to a load. The situation I am envisioning is that the EMF set up by the flow through a magnetic field will be aligned along a conducting cable (oriented in the appropriate direction, of course). However, I am concerned about internal resistance of the cable "robbing" the available energy of the system. Can you explain how this set up would work (or not!)?

- randy (age 61)

Palos Verdes CA

- randy (age 61)

Palos Verdes CA

A:

It turns out that the conductivity of the fluid doesn't enter into that calibration constant. Here's one way to think about it. Let's say, for simplicity, that the flow velocity is uniform. Then let's look at the whole situation from the reference frame at rest with respect to the fluid. In that reference frame, that has a pure magnetic field in the lab beomes in part an electric field. In nice cgs units, the electric field E in that frame is E=(v/c)B, where B is the lab magnetic field, to lowest order in v/c. So that means that in these units, with c= speed of light, :

V= (v/c)BL.

Here V is in statvolts, B is in Gauss, and L is in centimeters.

Conversion to SI units or others isn't hard.

V (in volts)= v(in meters/second) * B (in tesla) * L (in meters).

In a real instrument, the fluid velocity isn't constant throughout the tube, being largest in the middle, so the effective "L" may be a little different than the simple geometrical distance, if for "v" you use the average velocity.

If you use this voltage to drive an external circuit, that means that some current is flowing at right angles to the magnetic field and to the fluid flow. The magnetic field exerts a force on that current, in the direction opposing the fluid flow. So whatever work is done by the current has to come from work done by whatever is pushing the fluid flow, just as you suspected.

Mike W.

*(published on 03/30/2015)*