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A now-deleted answer to this recent question prompted me to wonder about this and I can't find a clear answer in the top layer of google results, so I thought I'd ask here.

What are the possible magnetic fields with constant magnitude?

That is to say, suppose that $\mathbf B: \mathbb R^3 \to \mathbb R^3$ is

  • solenoidal, so $\nabla \cdot \mathbf B = 0$, and
  • with constant magnitude $|\mathbf B(\mathbf r)| \equiv B_0$.

What can be said about $\mathbf B$? Is the solenoidality condition strong enough to imply that $\mathbf B(\mathbf r)$ must be a constant vector field? Or is it possible for the direction of the vector field to change from point to point? If so, can a general description of this class of fields be formulated?

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  • $\begingroup$ When you say "constant" you mean no change in time or no change in space? I think the answer by @knzhou assume you are asking about time variation. Which I think is the more usual meaning. A field is uniform if the value is the same everywhere (at least in some region). $\endgroup$ – nasu Dec 10 '18 at 15:06
  • $\begingroup$ @nasu This is a constant-magnitude requirement on top of a magnetostatic (i.e. constant in time) framework. The mathematical specification is unambiguous. $\endgroup$ – Emilio Pisanty Dec 10 '18 at 15:16
  • $\begingroup$ @EmilioPisanty I think one approach here might be to write $B=\nabla \times A$ where $\nabla\cdot A = 0$. Requiring that $\nabla(|B^2|)=0$ will then give you a condition on $A$ although I'm not sure its pretty. $\endgroup$ – jacob1729 Dec 10 '18 at 15:19
  • $\begingroup$ I assume that you're ruling out a sheet of current in a plane because the field isn't regular at the plane? $\endgroup$ – Ben Crowell Dec 10 '18 at 17:40
  • $\begingroup$ @Ben I was ruling it out primarily because I hadn't thought of the possibility ;-). But yes, it's probably a good idea to restrict this to regular fields. $\endgroup$ – Emilio Pisanty Dec 10 '18 at 18:35
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Partial answer: if there are no currents, all such magnetic fields must be constant.

In the absence of currents, we have $$\nabla \cdot \mathbf{B} = 0, \quad \nabla \times \mathbf{B} = 0.$$ The curl-free condition is equivalent to $\partial_i B_j = \partial_j B_i$, as is clear by writing it in terms of differential forms. As a result, the Laplacian of any field component vanishes, $$\partial^2 B_i = \partial_j \partial_j B_i = \partial_j \partial_i B_j = \partial_i (\partial_j B_j) = 0.$$ The Laplacian of the magnitude squared is hence $$\partial^2 |\mathbf{B}|^2 = 2B_i \partial^2 B_i + 2 (\partial_j B_i)(\partial_j B_i) = 2 (\partial_j B_i)^2.$$ Since $|\mathbf{B}|^2$ is constant, the left-hand side is zero and so is every term on the right-hand side. But then $\partial_j B_i = 0$, so $\mathbf{B}$ is constant.

When there are currents, we pick up an extra term, $$\partial^2 B_i \sim (\nabla \times \nabla \times \mathbf{B})_i \sim (\nabla \times \mathbf{J})_i.$$ Hence the argument also goes through if $\nabla \times \mathbf{J} = 0$. I'm not sure what the answer is for general $\mathbf{J}$.

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  • $\begingroup$ I believe that the second line (and therefore the conclusion) still holds so long as $\vec{\nabla} \times \vec{J} = 0$. This can be shown by noting that in general, $\partial_j B_i = \partial_i B_j + \mu_0 \epsilon_{ijk} J_k$. (I may not quite have the sign right, though.) $\endgroup$ – Michael Seifert Dec 10 '18 at 15:07
  • $\begingroup$ Also, the line after the last equation should read "Since $|\mathbf{B}|^2$ is constant, the left-hand side is zero..." $\endgroup$ – Michael Seifert Dec 10 '18 at 15:13
  • $\begingroup$ @MichaelSeifert Agreed on both counts, thanks! $\endgroup$ – knzhou Dec 10 '18 at 15:14
  • $\begingroup$ ....How can the div be equal to the curl when the curl is a pseudovector and the div is a scalar? $\endgroup$ – jpmc26 Dec 10 '18 at 19:07
  • $\begingroup$ @jpmc26 It's a standard abuse of notation. $0$ can stand for zero or the zero vector. $\endgroup$ – knzhou Dec 14 '18 at 13:08
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The energy of a magnetic field is [Jackson, Classical Electrodynamics (1975) p. 216, converted to SI units]

$$ W = \frac{1}{2}\int \mathbf{H}\cdot\mathbf{B} d^3 x. $$

In a vacuum,

$$ \mathbf{H} = \mathbf{B} / \mu_0, $$

so

$$ W = \frac{1}{2}\int B^2 d^3 x. $$

Integrated over all space, the energy of a field with constant $B^2$ is infinite unless $B=0$. Thus, the only possible magnetic field with constant magnitude is identically zero. Note that the direction of $\mathbf{B}$ doesn't matter because it is in a dot product with itself.

In a diamagnetic or paramagnetic body, replace $\mu_0$ by $\mu$ and you get the same result. Not that there are any infinite diamagnetic or paramagnetic bodies!

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  • $\begingroup$ This doesn't answer the question. $\endgroup$ – Ben Crowell Dec 10 '18 at 17:36
  • $\begingroup$ It says the possible magnetic fields with constant magnitude are identically zero. How does that not answer the question? $\endgroup$ – A. Newell Dec 10 '18 at 17:39
  • $\begingroup$ This answers the question in the title but not the one in the body. I suppose it is technically correct, but I think it was clear from the question that the spirit is "what do Maxwell's equations say about a constant magnitude field? $\endgroup$ – Javier Dec 10 '18 at 18:06
  • $\begingroup$ The negative reactions seem strange to me, and another example of what a hostile environment the Stack Exchange is. It's the second time that I have provided a correct answer and been voted down for it. If I can demonstrate that a field is identically zero, what else is worth saying? $\endgroup$ – A. Newell Dec 10 '18 at 18:10
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    $\begingroup$ You didn't show that the magnetic field is zero, you showed that if the magnetic field is constant over infinite space and the energy of the magnetic field must be finite, then the magnetic field must be zero. Those are not assumptions is it likely the OP accepts. A similar argument would say that the universe can't be isotropic. $\endgroup$ – Acccumulation Dec 10 '18 at 18:31

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