# Is there a magnetic field that satisfies $\nabla \times B = 0$?

In electrostatics Maxwell's equations for the magnetic field are

$\nabla \cdot B = 0$ and $\nabla \times B = \mu_0 J$

Now, take $B = xi-yj$, where $i$ and $j$ are the usual unit vectors, then one can show that

$\nabla \times B = 0$

which consequently means that $J=0$.

But in Maxwell's equations, isn't that $J$ supposed to be the source of the magnetic field B?

Then how come $J$, the source of the magnetic field, is zero yet $B$ is not zero? What am I missing?

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Maxwell's equation are local. That is the curl of $B$ at some point depends on the current at that point.

But "the current is the source of the magnetic field" is not local in that a current at $\vec{r}_1$ can create a magnetic field at $\vec{r}_2 \ne \vec{r}_1$.

So there is no conflict because the magnetic field can be caused by a current somewhere else, but restricted to rotation-free due to the lack of a local current.

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I want to get this right, please correct me if I am wrong. So rotation of a magnetic field is due to a local current, but that magnetic field of which we are calculating its rotation could be due to other currents. If what I wrote is write that would mean that the J in maxwell's equation is not the source of that magnetic field in the same equation, right? But now there is an equivalent equation to $\nabla \times B=\mu_0 J$, namely its integral form, Ampere's law. In Ampere's law the enclosed current in the equation is necessarily the source of the magnetic field under the closed intergral!! – Revo Dec 19 '11 at 20:35
So how come 2 equivalent equations, one in the integral form and the other is its differential form, in one equation J is not the source of B, while in the other I is the source of B? – Revo Dec 19 '11 at 20:37
Think a bit about the nature of the divergence and the curl. They are evaluated locally (i.e. $\nabla \times B(\vec{r}) = \mu_o J(\vec{r})$, but they impose constraints on the field even in places where the RHS vanishes. Take a single, isolated point charge and the divergence equation for the electric field sets the field in all space up to a gauge. Local equations, but because they apply at all points, global effects. – dmckee Dec 19 '11 at 21:03

Note that your magnetic field is unphysical. It goes to infinity as $x\to \infty$, for example.

Maxwell's equation do not uniquely specify the electromagnetic fields. In order to make a unique specification, you must also impose some boundary conditions.

If you say, for example, that there is a sphere of radius $R$ and that $B = 0$ on that sphere and $E = 0$ on it, and also that $J = 0$ in the sphere, then you will have $B = 0$ as the unique solution for the magnetic field in the sphere.

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But the magnetic field of infinite constant current sheet is constant above and below the plane, and it does not go to infinity when you move away from the current sheet. – Revo Dec 19 '11 at 5:50
What's your point? – Mark Eichenlaub Dec 19 '11 at 6:04
My point is that I do not understand your argument. One of the undergraduate classic examples of a constant magnetic field is a magnetic field due to infinite current sheet. That magnetic field does not go to 0 as you go to infinity, but still it is a magnetic field, that respects Maxwell's equations of magnetostatics. I am not talking about electromagnetic field, I am talking about magnetostatics only where the 4 coupled differential equations of Maxwell's becomes decoupled. – Revo Dec 19 '11 at 20:28
I didn't say all magnetic fields go to zero at infinity. I said that it's not a unique solution unless you impose boundary conditions. Even in your example of an infinite sheet of charge the magnetic field is not uniquely specified without boundary conditions. It so happens that those conditions are different here because you're talking about an unphysical current distribution. – Mark Eichenlaub Dec 19 '11 at 20:38