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Hi David, Practically speaking, B1 = B2. This is because when the electrons collide with the lattice, they pick up momentum. The rate at which momentum is transfered is a function of the resistance (being a measure of how frequently the collisions occur). However, the transfer of momentum is nearly instantaneous compared to any mechanical acceleration you are likely to impose on the ring, and so you can consider that the electrons are moving at the same speed as the lattice. Therefore, the magnetic field is due to the excess charge moving with the ring as it spins. There is really no difference between this and a stationary ring with a current, at least as far as magnetics is concerned. Granted, when you first accelerate the ring, the lattice will move first, and subsequently the electrons will move, and so you can consider that there will be a *very* brief period when the system is not in equilibrium, and the excess charge would be *stationary* until the electrons catch up with the lattice. Let's investigate that for a second. The energy of the electrons, would be E = (N m v^2)/2 which is just (1/2)mv^2 times the number of electrons in the ring. m is the mass of an electron, and v is the velocity -- v is assumed to be the same for all electrons. It is an approximation, sure, but it won't harm us here. Likewise, the current moving through any point in the ring is I = A p q v where A is the cross sectional area of the ring, p is the electron density (I can't write rho easily here), q is the charge per electron, and v is the speed they are travelling. We can use the second equation to get: v = I / (A p q) and plug that into our expression for energy: E = (N m / 2) (I / (A p q))^2 Then notice that the energy of the electrons is changing due to the collisions with the lattice which is a function of resistance like you noted. We can use the relationship for the power dissipated by a current traveling through a resistance, and the fact that the power dissipated is the derivative of the energy with respect to time. P = I^2 R = dE/dt So that gives (notice that I is a function of time and so we have to use the chain rule): I^2 R = (N m / (2 (A p q)^2)) 2 I dI/dt So you get: dI/dt = ( R (A p q)^2 / (N m) ) I = k I This is just an exponential decay with a time constant k: I = e^(kt) => dI/dt = k e^(kt) = k I We can simplify k a little bit, and we had better do so because it looks at first glance as though it is dependent on the geometry of the ring. Notice that N is the total number of conducting electrons in the ring, so N = A L p Where A is still the cross section, L is the length (so A L is the volume in the limit that the cross section is much less than the length -- if not you'll have to integrate), and p is the electron density again. k = R A^2 p^2 q^2 / ( A L p m) so: k = (A / L) ( R p q^2 / m) But the resistance of the whole ring (R) is a function of the shape of the ring too. The standard formula for connecting the resitance (R) of a rod to the resistivity (p) with cross section A and length L is: p = A R / L We're not using a rod, but again, so long as L is much greater than A, this is a good approximation. (Otherwise, integrate.) k = p^2 q^2 / m And this is just a function of the resistivity. This is somewhat intuitive because we would expect that adding more copper to the ring would add more electrons, and so the behavior aught not change. In our case we're using copper, so you can look in the CRC Handbook, (Lide, David R, 77th Edition) The resistivity of copper at 300 K is about 1.7 * 10^-8 ohm-m (Table 12-40) The electric charge is 1.6 * 10^-19 C Electron mass is 9.1 * 10^-31 kg So k is really small, like 10^-24 seconds, if I didn't make a goof. That is to say, the time scale over which the electrons pick up momentum is MUCH shorter than the timescale over which a pressure wave travels through the copper. (About 4*10^3 m/s per http://hypertextbook.com/physics/waves/sound/). So that means that you can't possibly accelerate the crystal fast enough to get a differential between the electrons and the lattice which is big enough to measure. Hence, practically speaking, B1 = B2. Hope this helps! Zack

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