Ampère's law states that

$$ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} \tag{1}$$

Simultaneously, from Ohm's law, we know that


When equating both currents $\mathbf{J}$, and applying Faraday's law, and knowing that the magnetic field $\mathbf{B}$ is solenoidal, one arrives at the transport equation of the magnetic field

$$ \frac{\partial\mathbf{B}}{\partial t}=\nabla\times(\mathbf{u}\times\mathbf{B)}+\frac{1}{\sigma\mu_0}\nabla^2\mathbf{B}\tag{3}$$

In the low magnetic Reynolds numbers limit $Re_m=\mu_0\sigma u L \ll 1$, the externally imposed magnetic field does not change and equation $(3)$ is not explicitly solved, because the diffusional term becomes negligible. In this so-called one-way coupled magnetohydrodynamics (MHD) (valid for most liquid metals), the electric field $\mathbf{E}$ has a potential $\Psi$, which is calculated from the divergence free ($\nabla\cdot\mathbf{J}=0$) condition, using equation $(2)$, with a Lorentz force $\mathbf{J}\times\mathbf{B}$.

My question is how should one look at Ampère's law in the context of one-way coupled, or low $Re_m$ MHD? Specifically considering the following two examples

Example 1 Suppose a situation with an insulating boundary (with $\mathbf{n}=\mathbf{\hat{y}}$). This means that $J_y=0$. Furthermore, the imposed magnetic field is constant in time, but varies in space. From $(1)$

$$J_y = \frac{dB_x}{dz}-\frac{dB_z}{dx},$$

which, for an arbitrary, non-uniform, steady magnetic field, would be non-zero.

Example 2 Suppose the externally imposed magnetic field $\mathbf{B}(x,y,z,t)=\mathbf{B}_0$. According to $(1)$, $\mathbf{J}=\mathbf{0}$, and thus there would be no Lorentz force?

What is exactly happening at the interface, and am I applying Ampère's law correctly. What am I overlooking?

  • $\begingroup$ "the externally imposed magnetic field does not change", does that mean you are interested in steady state case only (ignore the time dependence)?? $\endgroup$ Nov 28, 2013 at 10:28
  • $\begingroup$ @Gotaquestion Steady state would probably be enough for me to understand the issue (especiall considering the magnetic field). Adding time-dependence of the magnetic field would make it more difficult, I suppose. $\endgroup$
    – Bernhard
    Nov 28, 2013 at 10:36
  • $\begingroup$ I agree. One more question please, in your first example you are discussing a boundary condition, in your second example you are assuming a constant uniform magnetic field everywhere (not a boundary condition) Am I right? $\endgroup$ Nov 28, 2013 at 10:38
  • $\begingroup$ @Gotaquestion Yes, that is right. But I think they have the same answer. $\endgroup$
    – Bernhard
    Nov 28, 2013 at 11:33
  • $\begingroup$ I am not fully sure that I understand the question quite well; I will assume it is about where Amperes law comes into play in equation 3. Please have a look at my answer $\endgroup$ Nov 28, 2013 at 22:24

1 Answer 1


Let's start with a few issues considering Amperè's law. Amperè's law describes the magnetic field generated by current. The current can be localized at a certain point in space, but its magnetic field is spread everywhere. Applying Amperè's law to a certain point where current is zero does not generate a magnetic field. However, it doesn't necessarily mean the magnetic field is zero at that point: there could be a magnetic field at that point that is generated by a current some distance away.

In general (excluding permanent magnets - which comes from the magnetic dipole moment) all magnetic fields are generated by currents. Whether it is important to know the current that generated a magnetic field or it is enough to know the magnetic field distribution, depends on the problem being investigated.

In this question, there are two magnetic fields: First, the one externally imposed. We don't care about the current that generated it, all what we care about is the fields value which we assume we know. Second, the self-generated magnetic field in the considered medium (we must know the current generating this field in order to know the field distribution using Ampère's law).

In low $Re_m$ MHD, this induced, self-generated field, following from the currents inside the domain, is low compared to the external field.

Plasma For plasma's, the story is different. Think of it this way, if you suddenly applied a magnetic field into plasma described by MHD that sudden change (time derivative) is going to generate an electromotive force according to Farady’s law. Since the medium described by MHD is conducting, the electromotive force will generate current which will generate a magnetic field according to Amperes law, which will be imposed to the externally applied magnetic field. The direction of the self generated magnetic field weakens the external magnetic field, such that the TOTAL magnetic field appears to be diffusing in the medium. The diffusion continues until the magnetic field inside the medium is zero. Having zero total magnetic fields means all the energy of the external field has been dissipated in generating internal current.

From the physical picture I explained, Amperes law is used to describe the internal magnetic field resulting from internal currents. If you redo the derivation of equation 3 you will see that the diffusion term in equation 3 comes from Amperes law, which is consistent with physics I described above.

Getting back to equation 3, we know that magnetic Reylonds number is proportional to the conductivity of the medium, which means having zero conductivity (insulating medium) sets Reynolds number to zero. However, that breaks the whole system of equations and the applicability of MHD. In other words, if a medium is insulating it can’t be described by MHD. There has to be a minimum conductivity that ensures the MHD assumption holds.

With respect to your first example, I will take this picture from Chen’s book with adding y axis to it: enter image description here

Assuming an insulating interface contradicts the validity of MHD assumption, so the simple answer to your example is that MHD doesn’t hold for such a situation. If we assumed that the medium is conducting, there still will be a zero Jy component, but the other two components are not. So Amperes law would still give rise to magnetic field.

For your second example, it is unphysical. You can’t have a uniform magnetic field within MHD. From diffusion equation solution you see that one might START (at time = 0) with a uniform magnetic field, but eventually it is going to settle to zero. There will be no Lorentz force.

I advise you to have a look at Chen’s book chapter 6 and Piels book chapter 5 for more information.

  • $\begingroup$ Thanks for your answer, but it is still not completely what I am looking for. I am specifically asking for the low $Re_m$ limit, where the magnetic field is imposed, and does not change. SO this would be in the liquid metal range rather than plasma's. I am sorry if this was unclear. $\endgroup$
    – Bernhard
    Nov 29, 2013 at 6:25
  • $\begingroup$ With respect to the second example: this is actually often done in liquid metal MHD. Forced flow and imposed magnetic field do cause electromagnetic forces. $\endgroup$
    – Bernhard
    Nov 29, 2013 at 7:13
  • $\begingroup$ My background in MHD comes from plasma obviously. My answer is based on low Magnetic Reynolds number case. I am not sure how the liquid metal case differed from plasma, but since they are described by the same equations they behave similarly. If they are physically different they wouldn't be described by the same set of equations, don't you agree?? Can you please give me a reference to a case where uniform static magnetic field is imposed in MHD?? Equation 3 implies that only can be true if the magnetic field is 0 inside MHD. Outside MHD medium it is not necessarily so @Bernhard $\endgroup$ Nov 29, 2013 at 9:26
  • $\begingroup$ I understood the different background :) Well, in liquid metal MHD, equation 3 is never explicitly used, but that does not mean that Ampere's law is not valid. Example (assuming you have access) journals.cambridge.org/action/… MHD flow in a duct. Can we move this discussion maybe to chat? $\endgroup$
    – Bernhard
    Nov 29, 2013 at 10:40
  • $\begingroup$ That will be better, how do you do it? I have never done it before $\endgroup$ Nov 29, 2013 at 10:55

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