# diffusion of magnetic field and Joule heating effects

Since the magnetic induction equation is

$${\frac {\partial {\boldsymbol {B}}}{\partial t}}=\nabla \times ({\boldsymbol {v}}\times {\boldsymbol {B}})+\eta \nabla ^{2}{\boldsymbol {B}}$$

I want to ask why the second term (diffusion of magnetic field) is caused by Joule heating effects?

Thank you very much!

• My answers here and here may provide some context to the $\nabla^2B$ term, but I don't think they answer your question. – Kyle Kanos May 18 '19 at 16:43

This equation, which describes the magnetic field in a conducting fluid, is derived from the system of Maxwell equations and Ohm's law. Consider three equations (the law of Ampere, Ohm, and Faraday respectively) $$\nabla \times \vec {B}=\mu \vec {j}$$ $$\vec {j}=\sigma (\vec E+\vec {v}\times \vec {B})$$ $$\frac {\partial \vec {B}}{\partial t}=-\nabla \times \vec {E}$$ Combining equations we find $$\frac {\partial \vec {B}}{\partial t}=\nabla \times(\vec {v}\times \vec {B})+\eta \nabla ^2 \vec {B}$$ with $$\eta = 1/\mu\sigma$$. Thus, the diffusion of the magnetic field is associated with the electrical conductivity of the fluid and with Ohm's law. Joule heating is also linked to conductivity and Ohm’s law. Both effects are related by the equation $$\frac {\vec {j}^2}{\sigma }=\frac {\eta }{\mu }(\nabla \times \vec {B})^2$$