1
$\begingroup$

I am a mathematician and have a paper which models a situation where a homogeneous magnetic field is applied to a moving electrically conducting fluid. There is a Lorentz force formula on which all the work is made:

$$\mathbf{F}= \sigma (\mathbf{E} + \mathbf{V} \times \mathbf{B}) \times \mathbf{B}.$$

But I think that the last $\times \ \mathbf{B}$ is not needed. Is this the correct formula for this case, or is the work not correct?

$\endgroup$

1 Answer 1

4
$\begingroup$

Let us start by summarising the governing equations of MHD. We have the reduced form of the Maxwell equations

$$\nabla \times \mathbf{B} = \mu \mathbf{J},$$ $$\nabla . \mathbf{J} = 0,$$

$$\nabla \times \mathbf{E} = - \partial \mathbf{B} / \partial t,$$ $$\nabla.\mathbf{B} = 0$$

and the auxillary equations

$$\mathbf{J} = \sigma(\mathbf{E} + u \times \mathbf{B}),$$ $$\mathbf{F} = \mathbf{J} \times \mathbf{B}.$$

From the last two equations you get your result, namely that

$$\mathbf{F} = \sigma(\mathbf{E} + u \times \mathbf{B}) \times \mathbf{B}.$$

Note that the above combine to give the induction equation.

I hope this helps.

$\endgroup$
2
  • $\begingroup$ Is it so I can't use the general formula of moving particle with charge mentioned in en.wikipedia.org/wiki/Lorentz_force anyhow? Are these both formulas somehow related? Is this force of MHD also called "Lorentz force"? Thank you! $\endgroup$
    – Gedrox
    Commented May 30, 2013 at 9:39
  • 1
    $\begingroup$ You cannot use the Lorentz force equation for particles, on fluids. So the short answer is no. This is due to the fact that plasmas are conducting fluids. The closest you can get to the Lorentz force law is to consider a fluid element of charge density $q$ and current density $\mathbf{j}$ then you can write the force per unit mass acting on that element as $\rho\mathbf{F} = q\mathbf{E} + \mathbf{j}\times\mathbf{B}$. No this is not commonly referred to as the 'Lorentz Force' as far as I am aware... $\endgroup$
    – MoonKnight
    Commented May 30, 2013 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.