Why are Ohm's law and Ampere's law contrasting in Magnetohydrodynamics? 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?
 A: 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.
A: 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.
