Modified Faraday Disc: EMF Calculation

[sebqas]

The EMF for the Faraday disc is calculated as
\epsilon = \int_0^R \textbf{v} \times \textbf{B} \cdot d\textbf{l}
the integration is between the disc axis (0) and its edge (R is the disc radius).
As for uniformly rotating disc \boldmath v = \omega \times r and B-field is orthogonal to the disc plane, one gets famous result:
\epsilon = 1/2 \omega B R^2
However such calculation works only if the integrand (v x B) is curl-free, then the integration does not depends
on the contour and the EMF is well defined. But anybody can imagine that instead of solid disc we have a
conducting fluid flowing out from the center and rotating. In this case v has radial component, curl(v x B) is
non-zero and the integral for EMF depends on the shape of the contour.

So then, what is the actual EMF ?

 

[MS La Moreaux]

In another thread, I recently posted the answer to your question before you posed it. https://www.physicsforums.com/showthread.php?t=430457&page=4"

Mike

 

[kcdodd]

I think that you have to assume your fluid has a finite resistivity. If it did not then the curl of the emf would induce an infinite current around the axis of fluid source, which really means the fluid could not flow out. It would be frozen by the magnetic field. The resistivity + azimuthal current must exactly cancel the azimuthal emf which means the integration doesn't gain any potential by considering the non-radial paths, at least assuming cylindrically uniform resistivity.

 

[sebqas]

In another thread, I recently posted the answer to your question before you posed it. https://www.physicsforums.com/showthread.php?t=430457&page=4"

Mike

I can't see where exactly, its a long thread.

 

[sebqas]

I think that you have to assume your fluid has a finite resistivity. If it did not then the curl of the emf would induce an infinite current around the axis of fluid source, which really means the fluid could not flow out. It would be frozen by the magnetic field. The resistivity + azimuthal current must exactly cancel the azimuthal emf which means the integration doesn't gain any potential by considering the non-radial paths, at least assuming cylindrically uniform resistivity.

sorry, what you mean by the "the curl of the emf" , the electromotive force (EMF) is a scalar
defiend by the above inetgral. Do you mean the Lorentz force (E+ v x B) ?

It is commonly claimed in the textbooks the EMF is only a matter of B-field and the movement of the conductor described by \vec{v}. The EMF exists independently of the resistivity and
the possible current (any current appears only if the circuit is closed). For instance, the EMF for a battery is well defined even if the bulb is not connected to it.

The problem is that the EMF is equivalent to the potential difference (see Griffith p.293) but the Lorentz force is not conservative force and then it cannot be used (without any additional constrains) to derive the EMF:

\nabla \times (\vec{v} \times \vec{B}) = \vec{v} (\nabla \cdot \vec{B}) - \vec{B} (\nabla \cdot \vec{v})

the first term is zero, whereas the second not in general.

An example of real fluid: static radial flow with expansion (cylindrical coordinates r,\phi,z),
the velocity has only the radial component:
\vec{v} = \alpha r \hat{r}
the density decreases radialy:
\rho = \beta/r^2.

the continuity equation \nabla \cdot (\rho \vec{v}) + \partial_t \rho = 0 is fulfilled.

Any non-radial part of the path gives arbitrary large contribution to the integral \int_0^R \vec{v} \times \vec{B} \cdot d\vec{l}

 

[kcdodd]

Sorry, yes I mean the curl of the vector force.

If you stick two probes in this fluid you can measure a potential difference between the probes. Clearly, this must be well defined. However, that measurement will certainly depend on not only the "pure emf", which seems undefined to me, but also ohmic considerations. Different possible paths will have different total "emf", resistance, and current densities, but all should give the same potential difference for given two points.

If I were designing this machine to generate power then I would think it terms of the placement of the probes as giving me an output voltage, or emf, of the generator. Now, this is well defined with zero external load just like the battery, even though there are lots of internal currents.

Also if this is an incompressible fluid then the density can not change, which means the radial velocity will vary instead.

 

[MS La Moreaux]

sebqas,

Sorry for the delay. It was my last post in the thread of the link, #62.

Mike