# MHD Flow in a channel with and external B field and circuitry determining E. Different circuits and Ohm's law

I was wondering if someone can explain why E is the way it is in cases 2,3,4 in page 9 of these notes ?

In case 2 "Short Circuit", do I just have to assume that for a perfect conductor E = 0 for short circuit ?

I don't really get why current density is negative in this case (or the 3rd case for that matter) and what it means physically, for current density to be negative...

Any extra comments on what is going on in there would also be appreciated.

• You should include the author/university/publisher information about the notes in case of link rot. May 14, 2014 at 17:52
• It appears to me that case 2 the assumption is a perfect short. Current density being negative in case 3 just defines it is flowing towards the source and in case 4 it is flowing away from the source. May 14, 2014 at 17:59
• I would also like a slightly bigger discussion of what you already tried. I don't want to read 10 pages first. But $j_y<0$ follows more or less directly from the equaion above 4.26 May 14, 2014 at 17:59
• Apologies, have not posted in SE for a good while. The course is University of Glasgow, MHD. Thanks for the replies! I do not get why the electric field is the way it is... That is, how I would know that, without having been given the information in the 2-4 cases. That is, following the equations GIVEN the electric field is trivial, I just don't understand the origin of E-field in each case...
– leb
May 14, 2014 at 18:47
• @user60150 in case 2 it is assumed perfect short so there is no field. In case 3 it comes from the kinetic energy generating an electrical potential and in case 4 it is probably chemical because an external battery which adds a field is connected. Is there supposed to be a diagram of the circuit somewhere? May 14, 2014 at 22:37

The relevant equation in the notes is just above (4.26): $$j_y=\sigma\left(E_0-uB_0\right)$$
• For Case 2, $E_0=0$ meaning that the above becomes $$j_y=-\sigma uB_0<0$$ (assuming that $\sigma,u,B>0$).
• For Case 3, $uB_0>E_0$ which means, again that $j_y<0$ because the second term (which is negative) is larger in magnitude than the first term.
Current density is a vector term, so the negative sign of $j_y$ in the above two cases indicates that it is flowing opposite the positive direction, from $+y$ to $-y$.