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In the Magnetohydrodynamics the electric current is given by the Ampere's law is:

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

where $\mathbf{B}$ is the magnetic field and $\mathbf{J}$ is the electric current.

Now, consider Ohm's law:

$$\mathbf{E}+\mathbf{V}\times\mathbf{B}=\mathbf{J}/\sigma$$

where $\mathbf{E}$ is electric field and $\mathbf{V}$ is the fluid velocity.

The question is if there is a uniform external magnetic exerting in the absence of an electric field, Ampere's law says $\mathbf{J}=0$ while the Ohm's law says $\mathbf{J}\neq 0$, why?

Kyle Kanos
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Danial Rezai
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2 Answers2

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If the external magnetic field is uniform, total field $\mathbf B$ need not be uniform; there is contribution to total field due to current in the medium. If total field $\mathbf B$ is somehow made uniform, then current density vanishes. Then $\mathbf E + \mathbf V \times \mathbf B = 0$, so electric field is determined by the magnetic field.

Ján Lalinský
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Contrary to what you've asserted, assuming $\mathbf{E}=0$ and $\mathbf{B}=\text{const}$, Ohm's law does not make the claim that $\mathbf{J}\neq0$, it makes no claim on the value of $\mathbf{J}$ as there is an additional component you've neglected: the velocity (i.e., you have two unknowns & 1 equation). So by itself, it can only make the claim that the current density is orthogonal to the velocity & magnetic field, $$ \mathbf{v}\times\mathbf{B}=\mathbf{J} $$

It is only by using additional information, in the form of Ampere's law wherein we find that $\mathbf{J}=0$, that Ohm's law reduces to, $$\mathbf{v}\times\mathbf{B}=0$$ which means that the velocity $\mathbf{v}$ must be parallel to the magnetic field.

Kyle Kanos
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