Physics Friday 123

Part 3: Vector Potential of a Solenoid

Consider the ideal, infinitely long solenoid of radius R, n turns per unit length, and carrying a current I. Let us use cylindrical coordinates (ρ, φ, z), with our z-axis on the axis of the solenoid, so that the field inside the solenoid is in the positive z direction (the current is in the positive φ direction). Then consideration of the symmetries and the right-hand rule indicates that the magnetic field is purely in the z direction. Consideration of Ampère’s Law in the integral form, , as described here over rectangular loops in a ρz plane indicates that the field is a uniform inside the solenoid, and vanishes outside the solenoid (see here). Now, what then is the magnetic vector potential (in the Coulomb gauge)?


We note that due to the cylindrical symmetry, the current density is independent of both φ and z, and as noted before, the current is directed in the direction of . Lastly, since the current is confined to the solenoid of radius R, the radial dependence of the current density is of the form of a Dirac delta ; thus , for a constant C (dependent on n and I). Thus, from last week, we see
.

Now, by our cylindrical symmetry, we can choose the point at which we measure the potential to lie on our x axis. Then . Similarly, we note that the direction of is dependent on φ‘; using , we see
.
Examining the integration over φ‘ for the two components, we see that the x component is zero, thanks to the sine term in the numerator, and so we see that only is nonzero. Thus, our symmetry means
.
Now, rather than integrate further, or try to solve the vector differential equation directly, we can use our final result from here. Choosing a circle around the z axis of radius r, we see that A is constant on the circle, and everywhere tangent to it, so that the line integral over this loop is just the magnitude of A times the length of the loop:
.
For r<R, we see that the magnetic field is constant on, and perpendicular to, the disc bounded by this circle, and so the magnetic flux through this loop is just the magnitude of the field times the area of the loop:
.
Since these two are equal, we have

For r>R, the flux is constant, that through a cross-section of the solenoid, , so
.
Taking the curl of the above, one can confirm that the above gives the field we expect.

We note that outside the solenoid, that while the magnetic field is zero, the vector potential is not. In quantum electrodynamics, the vector potential is significant, and can be seen to have an effect even in regions where the field is zero, such as outside a solenoid, as detected in the magnetic Aharonov-Bohm effect, where the non-zero A leads to a phase difference between charged particles travelling to either side of a solenoid.

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