Topic is that last mentioned in the so far underwhelmingly responded to entry here Referencing mainly to fig.9 and accompanying in the article http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2009_Griffiths.pdf , it will be helpful to work from a Cartesian coordinate system xyz, with unit vector x horizontal to the right, y vertically up, z out of the page. let's consider what happens when the rectangularish loop current shown in fig.9 there (height w, length l) is made to precess at some constant angular velocity ω about an axis along y and coinciding with the center of loop left vertical leg. We have from (16) there the general result phid = 1/c2m×E for a magnetic moment m immersed in a uniform E field, it being understood the moving charges comprising loop current m must feel the applied E. That rules out an ordinary conducting loop current. At the instant phid acts along axis y (fig.9) there must be exerted on the loop a force Fhid equal to minus that rate-of-change-of-phid, given by Fhid = -dphid/dt = -ω×phid = -ω×(1/c2m×E) = z1/c2ωwlIE - (1), and thus a torque ζhid acting about y ζhid = r×Fhid = (1/2lx)×(z1/c2ωwlIE) = -y1/2w(l/c)2ωIE - (2) That factor of 1/2 owes to that Fhid acts through the center of the loop. Naturally one expects conservation of angular momentum holds, requiring an equal and opposite torque to be exerted by some other internal agency. There is indeed an opposing internally sourced torque, seen by moving into the instantaneous rest frame S' of the right vertical leg. In S', moving at velocity v = lω×y = -zlω relative to rest frame S, E seen in S is transformed into an E', B' given by E' = γE, B' = -1/c2v×E' = -x1/c2lωE' - (3) This B' acts on I' = -yγI in that leg to give a magnetic Lorentz force Fem = (I'w)×B' = I'w(-y)×(-x1/c2lωE') = -zI'w1/c2lωE' = -γ2Fhid - (4) As Fem acts along the direction of relative motion, it transforms back into S unchanged. The resulting torque ζem exerted on the loop owes to Fem acting over the full moment arm l = xl, and by inspection is therefore ζem = r×Fem = (xl)×(-yγ2Fhid) = -2γ2ζhid - (5) Net torque acting is thus ζ = ζhid + ζem = ζhid(1-2γ2) ~ -ζhid (v<<c) - (6) At this point it must be asked whether there is an overall conservation of angular momentum. The one 'reasonable' place left to look is the source of E, which we can take to be two very large uniformly charged circular capacitor plates spaced equally above and below the precessing loop. That any equal and opposite back-reaction torque exists on the sum of these charged plates can be quickly ruled out. That's because the loop sweeps out the form of a toroidal solenoid each revolution. There can be no requisite circular E field generated by such a toroid - rotating or not, unless one wishes to challenge validity of ∇×E = -∂B/∂t. Hence there is zero net torque exerted on the source charges giving rise to the applied E = yE. Sure looks like conservation of angular momentum is in serious trouble! Returning to the loop itself, a non-zero ζ means a mechanical angular power generation W = ζ.ω = 1/(2c2)ω2wl2IE(2γ2-1) - (7) The sign of W can be changed at will given we are free to alter independently ω, E, and sense of I. Can we at least rescue conservation of energy here - there surely must be a power-drain in the circulating I exactly off-setting the mechanical W in (7)? The one seeming possibility is that E fields acting in the loop's two vertical legs are effectively different, and to the right degree. This is not evident in lab frame S, where E is uniform, and what's more the charge flow rates must be equal. The Lorentz force law F = q(E+v×B) explicitly forbids any y-acting 'boost' in one leg owing to it's relative velocity in S. (it does not however prevent existence of a transverse x-acting reaction force in supporting tube - necessary in order to explain Fem in (4)) Hence in S we have no net power drain in circulating current I owing to E. In S' E' is a factor gamma greater, but the resulting forces qE' transform back in S to give just qE. Nope, looks a lot like conservation of energy, and thus in any other frame, energy-momentum, is in deep pooh too. Goodness me, can someone point out my obvious blundering here? A simple sign error maybe, or worse, just wrong-headed thinking? Well judging by the neg-response in that other thread, this might be a long wait. Maybe the topic is just too boring for the erudite readership here at PF. Or something. : Assuming E generates all the pressure differential between upper and lower horizontal legs, Fhid is the sum of uniform precessional force densities that are opposed and weighted in favor of the upper leg. The center of Fhid thus acts half-way between the two vertical legs. We ignore any phid induced in the right vertical leg owing to centripetal acceleration from rotation at ω about y as it makes no contribution to the x-acting phid of interest here. : There are perhaps two other conceivable places to look: a) - Precession of SFM (stored field momentum) of density ε0E×B, which as shown in eq'n (57c) in above cited article, is numerically equal and opposite to phid in the situations covered there. But what can one say here? Where is this precessing SFM able to physically manifest? We have applied the usual expressions involving Lorentz force and SR transformations, and there is no room at all for any 'cancellation' of Fhid by precessing SFM. Hence an unphysical quantity; at least in this situation. b) - Self-interaction between charges flowing in the loop. Any such interactions must go as the square of charge density for a given mean flow speed, hence as the square of I. Whereas the expressions for force and torque above found are directly proportional to I. So that is ruled out on functional dependence grounds. And ruled out completely if our model was based on an 'incompressible' charged fluid for which any EM self-interactions would be identical to that with no imposed E, which is an essential for phid etc.