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In 2+1 dimensions of spacetime, the electromagnetic field is made up of a vector electric field and a scalar magnetic field. At each point in space, there is a magnetic field value, which we can define as positive or negative (what we would describe as "into the page" and "out of the page" in 3D space). Current moving up along a vertical wire creates a constant-with-distance positive magnetic field to the left of the wire and a negative to the right. An infinitely long wire would fill half of the space with a positive magnetic field and the other half with a negative. The only permanent magnets would be stable loops of current (or spinning charges) and based on their handedness be permanent positive or negative magnets.

Now my question is this: How do these magnetic fields generated by currents and permanent magnets interact? I have read that there is no magnetic attraction or repulsion between magnets (The Planiverse, A.K. Dewdney) but no explanation as to why. Electron paths are bent by these magnets however, would wires be attracted/repulsed by magnets, or only have induced torques? And currents induced by magnets, do they work to nullify the field within the loops, or outside them? And would there be any significant difference in behaviour if one used a magnetic dipole from two monopoles stuck together instead of a wire?

Any explanation or links to other resources would be much appreciated, thanks.

EDIT: My assumption that a loop of current would act like a permanent magnet and posses a magnetic field was incorrect, they have a magnetic field inside them and none outside, and do not interact with wires. My question now is this: are any sort of permanent magnet even possible in 2D space? Would they be dipoles or monopoles? Do spinning charges not have a magnetic field? And if wires are the only generators of magnetic fields, do they attract/repel nearby wires, because don't the field they generate cause electrons in the other wire to feel a tangential force to their direction?

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The 2+1 dimensional Maxwell equations are like the 3+1 dimensional Maxwell equations but with $B_x, B_y, E_z$ and all $z$ derivatives set to zero. There is only one component of magnetic field that we might continue to call $B_z$. So it as though we are in 3 spatial dimensions but all fields are constant in the z-direction and there are no currents in the z-direction.

So what you are calling a permanent magnet in 2+1 dimensions acts a lot like an infinite solenoid in 3+1 dimensions. It has non-zero field $B_z$ inside, but there is no magnetic field outside. Just like parallel infinite solenoids, these 'permanent magnets' do not exert a force on each other, and they also do not exert a force on charges outside (although there will be an Aharanov-Bohm effect quantum mechanically).

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  • $\begingroup$ Why would the 2+1 magnet behave more like an infinite solenoid and not a single loop of current? The field outside a loop of current is not zero, and neither does it need to be for there to be a non-zero force on other magnets. Are you also suggesting that the 2+1 wire behaves more like a sheet of current in 3+1 space and not like a wire? $\endgroup$ Commented Feb 7 at 20:29
  • $\begingroup$ "The field outside a loop of current is not zero": The field around a loop of current in 3+1 dimensions involves the third dimension in an essential way, and it doesn't translate to 2+1 dimensions. Think of everything in 2+1 as extended trivially in the z-direction if you want to use your knowledge from 3+1 dimensional electromagnetism. So yes, the 2+1 wire behaves like a sheet of current in 3+1. $\endgroup$
    – octonion
    Commented Feb 7 at 20:41
  • $\begingroup$ I understood what you were saying, I just doubted it was true, as you made a jump in logic in your answer, between line 2 and 3 in your first paragraph, that I felt wasn't convincing. LPZ's answer satisfied that and now I'm convinced. I see loops of current are not a way to create the magnets I read in A.K. Dewdney's work, I guess they just presupposed the existence of magnetic monopoles. Is there any charge configuration you know of to create permanent magnets then? $\endgroup$ Commented Feb 9 at 17:48
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In 2D magnetostatics (and 2D+1 EM), you don't have magnetic monopoles. You can revert to the usual 3D magnetostatics by assuming that the currents are sheets in the $x,y$ directions invariant by translation along $z$. This forces $B$ to be only along the $z$ axis: $$ \begin{align} j &= j_x(x,y)e_x+j_y(x,y)e_y & B &=B_z(x,y)e_z \end{align} $$ The equations are therefore (setting $\mu_0=1$): $$ \begin{align} \partial_yB_z &= j_x & -\partial_xB_z &= j_y \end{align} $$ In 2D, magnetostatics, there is no analogue of Gauss' law, by construction, so there are no monopoles. From the 3D perspective, since $B$ is only along $z$ and there is no $z$ dependence, the divergence is automatically zero. Even if you were to use a magnetically charged line, the magnetic field would be radial and so it would not be 2D magnetostatics anymore.

You are probably rather alluding to flux lines. They are the singular solutions: $$ B_z = \phi\delta(x,y) $$ These are the this solenoid solutions that octonion is referring to. Indeed, the corresponding current is: $$ \begin{align} j_x &= \phi\partial_y\delta & j_y &= -\phi\partial_x\delta \end{align} $$ which is an infinitesimal current loop. Back to the 3D perspective, this is an infinitesimal current sheet loop, i.e. an infinitesimal solenoid. By superposition, you can construct any current with these solutions, and you can therefore interpret $B_z$ and the flux density.

Using the solenoid analogy, you can deduce that non overlapping current distributions do not interact magnetostatically. Since magnets are just currents in disguise, the same applies.

More formally, for a given $j$, you can solve for $B$ by rewriting: $$ \begin{align} \partial_xB_z &= -j_y & \partial_yB_z &= j_x & B_z = \int j_xdy-j_ydx \end{align} $$ and setting $B_z=0$ at infinity. The line integral is path independent thanks to the conservation of current: $$ \partial_xj_x+\partial_yj_y=0 $$ Thus, $B_z$ in every path connected region where $j=0$. In particular, the magnetic force on another current distribution is always zero. Indeed, the current distribution can be decomposed into magnetic dipoles and the force on a dipole requires a non zero magnetic gradient.

Once you understand what happens to current distributions, you know how to treat permanent magnets: they are just currents caused by magnetic dipoles. In the 3D perspective, the dipole moment are necessarily along $z$. The previous example of the infinitesimal solenoid can be recast as a uniformly magnetized line (magnetized along its axis). By superposition, you can construct any permanent magnet with these magnetized lines. Consequently, magnets do not apply forces between each other.

Note however that for AC currents, the result is not valid anymore thanks to induction and magnetic field is radiated away. Also, while mutual forces are zero, you still can have mutual inductances. They will not depends on the relative position of the currents in accordance to the principle of virtual work.

Note that you can consider 2D magnetism on other domains, not necessarily a plane. If they are topologically interesting it can accommodate monopoles. The monopole is not on the domain but by imagining the domain being extended to 3D space, you can “enclose” the monopole with the surface.

Explicitly, for the sphere: $$ \begin{align} j &= \frac{j_\theta(\theta,\phi)}{r^3}e_\theta+\frac{j_\phi(\theta,\phi)}{r^3}e_\phi & B &=\frac{B_r(\theta,\phi)}{r^2}e_r \end{align} $$ For the cylinder: $$ \begin{align} j &= \frac{j_\phi(\phi,z)}{\rho^2}e_\phi+\frac{j_z(\phi,z)}{\rho^2}e_z & B &=\frac{B_r(\phi,z)}{\rho}e_\rho \end{align} $$ For the flat torus just impose periodicity on top of the previous restriction for the plane.

Hope this helps.

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  • $\begingroup$ Ah, I see my initial assumption of loops of currents forming magnets was wrong. You say there are no monopoles, in that case is there any way to make a permanent magnet? What kind of current configuration would make a dipole? $\endgroup$ Commented Feb 9 at 17:41

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