Comments on the electromagnetic fields of a time - dependent solenoid

The electromagnetic fields associated with a time-dependent infinitely long solenoid are analysed. I t is proved that if the magnetic field vanishes at the solenoidal outer region then its flux is a linear function of the time. It is also shown that a time-dependent solenoid does not satisfy the field-free requirement of the Aharonov-Bohm effect. An infinite solenoid is frequently used in a theoretical analysis of electromagnetic interactions. A solenoid with a time-dependent flux belongs to this category. The electromagnetic potential and fields of this system are the subject of this comment. Several authors have recently discussed systems that contain a time-dependent infinite solenoid [ 1-61. In these works, the z component of the magnetic field associated with this device is written in the following form: E,(r, t) = B (?)e(R-r) (1) where r denotes the cylindrical radial coordinate, R is the solenoidal radius and O(x) is the unit step function which vanishes for a negative argument and is unity for a positive one. The solenoidal axis coincides with the z axis of cylindrical coordinates. Units used in this work are those where the speed of light c = 1. All calculations are carried out for the vacuum and the fields D and H are omitted from the discussion. (Some authors use a related expression where the solenoidal radius shrinks to zero in a manner which conserves its magnetic flux.) The main discussion of the present work shows that (1) holds only if B (t) is a linear function of the time. Using Maxwell equations and the symmetries of the solenoid, one proves that E, = E, = 0. Relying on this result, the following analysis shows the validity range of (1). Assume that (1) holds. Consider a circle of radius r > R, which is a part of the x, y plane and whose centre is at the origin. Using vector analysis, Maxwell's equations and (l) , one finds for this circle E * d l =-b(t) S f where S denotes the area of the solenoidal cross section. The cylindrical symmetry of the problem shows that Eq =-B (t) S / 2 ~ r .