Joule heating for a fluid

Say you have a conductive liquid with a changing magnetic field going right through it, causing an electric current. How exactly does the electric current travel and how could you calculate the effect of Joule heating on the liquid?

1 Answer

You are looking for something called the generalized Ohm's law, which is given by: $$\mathbf{E} + \mathbf{v} \times \mathbf{B} \approx \frac{ \mathbf{j} \times \mathbf{B} }{ n \ e } - \frac{ \nabla}{ n \ e } \cdot \left( \mathcal{P}_{e} + \frac{ m_{e} }{ m_{i} } \mathcal{P}_{i} \right) + \eta \ \mathbf{j} + \frac{ m_{e} }{ n \ e^{2} } \frac{ d \mathbf{j} }{ d t } \tag{1}$$ where $\mathbf{j}$ is the total current density, $n$ is the total number density (assuming quasi-neutrality, i.e., $n_{e} = n_{i}$), $e$ is the fundamental charge, $\mathcal{P}_{s}$ is the pressure tensor of species $s$, $m_{s}$ is the mass of species $s$ ($s$ can be $e$ for electron or $i$ for ion), and $\eta$ is the scalar electrical resistivity.

The Joule heating term is the $\eta \ \mathbf{j}$. In linear circuits, one often assumes that Equation 1 reduces down to something akin to: $$\mathbf{E} \approx \eta \ \mathbf{j} \tag{2}$$ and then one can use a relationship from Poynting's theorem which relates the rate of change of electromagnetic energy per unit volume into mechanical energy per unit volume (e.g., heat and/or particle kinetic energy). This term is given by $\mathbf{E} \cdot \mathbf{j}$ or approximated as $\eta \ j^{2}$.

In your specific example, Equation 2 would probably include the Hall term as well, i.e., the $\mathbf{j} \times \mathbf{B}$ term.

• Number density of electrons? And does "of species" just mean, "of the fluid in question?" Commented Jul 25, 2017 at 3:16
• Oh nevermind, a better question than the second one I asked is, "what exactly is 'e' and 'i'?" Commented Jul 25, 2017 at 3:19
• @A.Frasch - e = electron and i = ion. Electrons are generally the current carrying particles but ions can contribute as well in some cases. If you mean a neutral conducting liquid like mercury, then electrons will be the current carriers and Equation 1 will reduce to Equation 2 (possibly with the Hall term as well if under and external magnetic field). Commented Jul 25, 2017 at 12:58
• So in a liquid metal like iron, there wouldn't be any ions, right? Commented Jul 26, 2017 at 3:49
• The nuclei that make up the liquid are effectively the ions, but in its neutral state, yes, one can treat this as an electron-only current. Commented Jul 26, 2017 at 13:27