magnetic potential vs. solid angle It is an old result, maybe going back all the way to Ampere or Poisson that the locus of points having the same solid angle when viewing a current loop is a (magnetic) equipotential surface corresponding to the current. The proof that can be found in, for example, Sommerfeld's Electrodynamics, p118, uses standard properties of Green's Theorem, but it is a derivation that I do not find particularly illuminating.
My question is if there is an intuitive explanation as to why this is (must be) true.
 A: I believe this beautiful theorem (in fact the slightly stronger one that the magnetic potential is everywhere proportional to the solid angle subtended by the loop) is due to Gauss. It certainly appears in his book on Terrestrial Magnetism. To derive it we need to make a couple of assumptions: 1) at sufficiently large distances the magnetic field of a small planar current loop is indistinguishable from the field of an ordinary (small) magnet i.e. a magnetic dipole, with moment equal to the current times the area of the loop (multiplied by a suitable constant, depending on your units); 2) if two current loops, each carrying the same current, are placed next to each other so that part of loop A coincides with part of loop B, with the current in A going the opposite way to the current in B, in that shared part, then the combined field of the two loops is the same as the field of a single larger loop made by combining A and B and cutting out their shared part. (In plain language, the opposite currents cancel out. Note that the loops are not necessarily planar.)
Now, outside a magnet the force on a (hypothetical) magnetic monopole is 'conservative' which means that it is derivable from a potential function. Far away from the magnet the potential can be regarded as the sum of the potential due to a North pole and the potential due to a South pole, very close together (that is how you make a dipole). The potential due to a monopole is (a constant times) $1/r$. The potential due to a dipole is the $z$-derivative of this:
$$-d/dz(1/r) = z/r^3 = cos\theta/r^2.$$
Hence the potential at a point P due to a small current loop is 
$$IAcos\theta/r^2.$$
But $A cos\theta/r^2$ can be interpreted as the solid angle subtended by the small loop at the point P. Note that it is zero when $\theta=\pi/2$ (loop seen from edge on) and negative when $\theta\gt\pi/2$. The negative sign goes with the fact that the sense of the current (as seen from P) has changed from clockwise to anticlockwise (or whatever!). So the potential due to the little loop is I times the solid angle subtended by the little loop.
Now, given some loop L, not necessarily small or planar, carrying a current I, it can be broken down into many small planar loops as follows. Fill in the loop with a surface S, so that L is the boundary of S. Then draw a roughly rectangular curvilinear grid on S, dividing it into many small compartments. Then imagine that each little square of the grid is a little current loop, carrying that same current I. The currents cancel out everywhere except at the boundary of S, so the field of the original loop L is the sum of the fields of all the little square loops.
Consider the potential at P. This is the sum of the potentials due to the little square loops, each of which is I times the solid angle subtended at P by the little square. So their sum is I times the solid angle subtended by the whole loop L. Q.E.D.
Now (says Gauss) consider how the potential changes as you traverse a loop which links with the current loop. Suppose you start just above the loop. This is like the solid angle subtended by France as seen from the top of the Eiffel tower; it takes up nearly half of all the available directions, so the solid angle is nearly half the surface area of a unit sphere, so nearly $2\pi$. Then you go round on your loop. The solid angle drops to zero, then goes into the negative. As you come up to the loop from below, the solid angle approaches the value $-2\pi$; as you go through the loop, back to your starting point, it goes to a value just less than $-2\pi$ (this is assuming the solid angle changes continuously). I think you will be able to convince yourself that the change in solid angle during the complete circuit is precisely $-4\pi$; so the change in the potential is I times this.
This is remarkable because the area of the loop has dropped out: the change in potential (round a complete circuit) depends only on the current. If you went on a different circuit, not linking the current loop, the change in potential would be zero. It is only interested in current that goes through the circuit. This is essentially the rule that the circuit integral of $\mathbf{H.dl}$ equals the current going through the circuit, which is called, unaccountably, Ampère's Law. I'm fairly sure it ought to be called Gauss' Second Law of Magnetism (the first being that $\mathbf{divB}=0$). 
Note: most of the above is rehash of material from Maxwell's Treatise on Electricity and Magnetism, Vol II
