Summary of Hall in Round Wire REVIEWED

Summary of Hall Effect in a Round Wire PhL 3.24.14 Reviewed this doc in detail on 3/31/14, I think all is resolved in "Appendix N". We assume azimuthal symmetry and an infinitely long wire so this is a 2D problem. Here are the governing equations: Jz(r) = n(r) e vz(r) // current equation e<0 (1) Er(r) = vz(r) Bθ(r) // Hall effect equation (2) r-1∂r(rEr(r)) = ρ(r)/ε0 // divE=0 equation (3) - ∂rEz(r) = 0 // curlE = 0 equation (4) Paradox Right off the Bat. If vz = constant, we know Bθ(a) ≠ 0 and then from (2) Er(a) ≠ 0. At least for small B this means Jr(a) ≠ 0 [ wrong because of Magnetic Ohm's Law! ] and this is a contradiction because this current has nowhere to go (vacuum outside the wire). So something is wrong with equation (2). [ no, but something is wrong with Ohm's Law!] Comments: 1. In any differential operator, we can set ∂z[...] = 0 and ∂θ[...] = 0 and Eθ = 0. The first since everything is uniform in z, the second because nothing changes with θ, and the third due an expected symmetric solution for the E field. Certainly Eθ = 0 on the surface since it is an equipotential. With these assumptions we find that div E = r-1∂r(rEr) and curl E = - ∂rEz from which (3) and (4) above derive. [ correct ] 2. In the Jz = nevz equation, n is the density of electrons in the classical electron cloud inside the conductor, e < 0 is the electron's charge, and vz is the average drift velocity of the flowing electrons. If we think of Jz > 0 flowing out of the plane of paper (the z direction), then electrons flow into the plane of paper and vz < 0. Of course n(r) is positive. [ correct ] 3. This same vz appears in the Hall effect equation which says there must be a local E field to offset the Lorentz force e v x B acting on an electrons. Consider this picture, The B field is CCW by the right hand rule. We show one electron heading into paper at the cross, which indicates vz < 0 . The direction of v x B is radial outward. The comments in the picture are self explanatory, and we end up with a radial E field pointing inward to balance the Lorentz force. We shall find below that Er(r) = - K (r/a) where K is a positive constant, and so the E field does in fact point radially inward. Notice in the equation (2) that Er(r) = vz(r) Bθ(r) < 0 since Bθ(r) > 0 and vz < 0, so we have the sign right in equation (2) as well as in equation (1). [ correct ] 4. None of the functions can depend on z or θ due to our geometry. We allow all functions then to possibly vary in r, even though they may end up being independent of r. [ ok ] 5. Normally one thinks of the free charge ρ(r) being 0 inside a conductor because the electrons offset the positive ion charges, but we allow that in this situation ρ(r) might not be 0. In fact, we know that ρ(r) = e [n(r)-n0] (5) where n0 is the number density of electrons in the absence of current I in the wire. [ ok ] We now restate the above four equations: [ all correct, note missing Ohm's Law ] Jz(r) = n(r) e vz(r) // current equation e<0 (1) Er(r) = vz(r) Bθ(r) // Hall effect equation (2) r-1∂r(rEr(r)) = ρ(r)/ε0 // divE=0 equation (3) - ∂rEz(r) = 0 // curlE = 0 equation (4) The last equation says that Ez(r) = constant. Since Jz(r) = σEz(r) [ an assumption, but turns out to be correct], this says Jz is a constant. [ correct ] This is consistent with our conclusions of Chapter 2 in the low-frequency limit. We no longer need the last equation [ correct ], and our equation set then becomes, using (5) for ρ(r), Jz = n(r) e vz(r) // current equation e<0 (6) Er(r) = vz(r) Bθ(r) // Hall effect equation (7) r-1∂r(rEr(r)) = (e/ε0) [n(r)-n0] // divE=0 equation (8) We still have several functions to worry about: n(r), vz(r) and Er(r). At least we know Bθ(r) as follows using Ampere's law with our uniform Jz : [ correct ] 2πr Hθ(r) = I(r2/a2) => Hθ(r) = [I/(2πa2)] r => Bθ(r) = [μ0I/(2πa2)] r = C r C = [μ0I/(2πa2)] (9) so Bθ(r) is a simple linear function inside the conductor. [ correct ] At this point, we replace function n(r) by another function η(r) as follows η(r) = [n(r)-n0]/n0 = n(r)/n0 - 1 => n(r) = n0 ( η(r)+1) (10) [ Comment: Suppose n(r) - n0 = 1 electron. Then η = 1/n0 = 1/[8.5 x 1028] = 1.1 x 10-29 . ] This dimensionless function η(r) is then the fractional deviation of n(r) from its at-rest value of n0 . We can then say ρ(r) = e [n(r)-n0] = e n0 η(r) [ ok ] (11) Our three equations are now Jz = n0 ( η(r)+1) e vz(r) // current equation e<0 (12) Er(r) = vz(r) Bθ(r) // Hall effect equation (13) r-1∂r(rEr(r)) = (e/ε0) n0 η(r) // divE=0 equation [ ok ] (14) and our unknown functions are vz(r), Er(r) and η(r) . We can solve (12) for vz(r) to get vz(r) = Jz /(e n0) * 1/( η(r)+1) (15) and then insert this into (13) to get Er(r) = Jz/(e n0) * 1/( η(r)+1) * Cr = (JzC/en0) (16) and so r Er(r) = (JzC/en0) We then compute, ∂r = = - r2 η'(r) and therefore ∂r(rEr(r)) = (JzC/en0) [ - r2 η'(r)] But according to (14) we have ∂r(rEr(r)) = (e/ε0) n0 r η(r) Setting these last two expressions equal and dividing by r we get (JzC/en0) [ - r η'(r)] = (e/ε0) n0 r η(r) or [ - r η'(r)] = η(r) We define this constant to be β so we then have [ - r η'(r)] = β η(r) β = Notice that β is dimensionless since all terms in the ODE here are dimensionless. [ I think all the above math regarding η and the final ODE is correct, have not checked details ] But we know that C = [μ0I/(2πa2)] and Jz = I/(πa2) so β = * * = * * = But 1/(ε0μ0) = c2 (c = speed of light) so then β = = 2π2 [ ea2n0c/ I ]2 This number β is quite large for normal parameters. For example, consider, So we then have this ODE [ - r η'(r)] = β η(r) β ~ 1031 We multiply through by (1+η)2 to get [2(1+η) - r η'] = β η(1+η)2 or r η' + β η(1+η)2 - 2(1+η) = 0 or r η' + (1+η) [ βη(1+η) - 2] = 0 This is a standard issue non-linear ODE for the function η(r) with variable r. Because β is so large, it must be approximately true that [ βη(1+η) - 2] ≈ 0 or η(1+η) ≈ 2/β = exceedingly small = η + η2 Therefore η itself must be very small, and certainly η << 1. We can then approximate 1+η = 1 and then our ODE becomes r η' + [ βη - 2] = 0 which has the following solution η(r) = C1 r-β + 2/β In order to avoid an astronomical singularity at r = 0, we must set C1 = 0 and then η(r) = 2/β << 1 [ I think this is correct, and η = 10-21 in the last App N example] which is that same as η(1+η) ≈ 2/β if η << 1. We conclude then that η(r) is a very small constant. Recall now (15) and (16), vz(r) = Jz /(e n0) * 1/( η(r)+1) (15) Er(r) = Jz/(e n0) * 1/( η(r)+1) * Cr = (JzC/en0) (16) We then see that vz(r) is a constant we will call vz and vz = Jz /(e n0) Er(r) = (JzC/en0) r = (JzCa/en0) (r/a) = K (r/a) [ K now called - Es ] But recall that C = [μ0I/(2πa2)] and Jz = I/(πa2) so so (JzCa/en0) = * * = ≡ K [ K = -Es see App N (N.7.13) ] A few extra lines of Maple, So we can write Er(r) = Er(a) (r/a) Er(a) = - 0.47 x 10-11 volts/m I = 1 amp, a = 1 cm For a 1 mm radius wire, the field will be 103 times larger so Er(r) = Er(a) (r/a) Er(a) = - 0.47 x 10-8 volts/m I = 1 amp, a = 1 mm Meanwhile we have vz = Jz /(e n0) = I/(πa2en0) = -0.23 x 10-6 m/sec I = 1 amp, a = 1 cm = -0.23 x 10-4 m/sec I = 1 amp, a = 1 mm The charge density ρ(r) is given by ρ(r) = e n0 η(r) so it is also a constant, ρ = e n0 η = e n0 2/β = - 0.83 x 10-20 coulombs/m3 The total charge inside a slice of thickness dz of the wire is then Q = dz * πa2 ρ = dz πa2 e n0 2/β An equal and opposite charge forms on the surface of the wire. Outside the wire these charges cancel and there is no external electric field. Conclusions (1) the basic idea of Jz = constant and vz = constant in a round wire carrying current I is born out. ok (2) there is a small radial inward-directed "Hall effect" electric field inside the wire which is required to offset the Lorentz force acting on the electrons . It has the form Er(r) = K (r/a). ok (3) inside the wire there is a small and constant amount of negative free charge ρ. This is what deflects the electrons curving toward the center line. ok (4) the surface of the wire then has a small positive surface charge. ok (5) outside the wire there is no radial electric field due to this Hall effect. ok (6) it is not very clear how one might measure this internal electric Hall field. ok Question: How do these facts relate to r-1∂r(rEr(r)) = ρ(r)/ε0 // divE=0 equation (3) Answer: Er(r) = K (r/a) so r-1∂r(rEr(r)) = (K/a) r-1∂r(r2) = (2K/a) = ρ/ε0 OK d2x' Does this radial E field create some kind of measurable magnetic field outside the wire? I could use my 2D Biot-Savart to answer that question [ wrong wrong! Outside field cancels and is 0. ] H(x,y) = ∫d2x' J(x') x R R ≡ x - x' (B.2.24) If you put point (x,y) on the right side x axis and consider contributions to this integral from two mirror points about the x axis, you find that the two contributions are equal, but of opposite sign since the cross product has opposite sign for the two pieces. Thus these two pieces contribute 0, and that applies to all pieces, so there is no external H field created by the Hall current. Question: If there is a radial Er field inside the conductor, isn't there a radial current Jr, and if there is, doesn't that violate the assumption of no transverse currents? [ resolved in Magnetic Ohm's Law! ] Answer #1: Ohm's Law has the form J = σE - D grad ρ But you would think that grad ρ = 0 if ρ is a constant, and then we are back to J = σE , so what gives here? A fly in the ointment! I will have to ponder this later. Answer #2: OK, this is resolved now in Section 8 of "Ohms Law and Magnetic Fields".