Hall in Round Wire 1 REVIEWED

Hall Effect in Round Wire 1 PhL 3.23.14 This was my first attempt at the "radial Hall effect" problem, and this attempt is summarized in Summary of Hall in Round Wire REVIEWED.doc. Today is 3/31/14. Here I came up with (Plan B below) a very fancy ODE for η which when simplified gives the same result obtained now by a new method which needs no ODE at all. I was unaware of the Magnetic Ohm's Law when I wrote this so there were some seeming paradoxes. Plan A below is interesting: It shows that you cannot explain the radial Hall effect in terms of a line charge down the middle and a surface charge outside. This concept gives Er ~ 1/r instead of Er ~ r, and this leads to contradictions. This was my first approach to the radial Hall effect. Soon after this I realized I could have Maple handle all the dimensions as well as the numbers! You can see below all my painful work with dimension bugs which I eventually got all fixed. We have a round isolated conductor shown in red, and B are the mag field lines Now (ignoring polarity), the Lorentz force points everywhere toward the center. Let's try to solve this problem in its Hall aspect. We assume [ in this doc, e < 0 ] Jz(r) = n(r) e vz(r) (1) // ρ(r) = n(r)-n0 Dim check: dim(LHS) = amp/m2 dim(RHS) = m-3 Cou m/sec = Cou/sec/m2 = amp/m2 OK The Hall field is this Er(r) = vz(r) Bθ(r) (2) Dim check: dim(LHS) = volt/m dim(RHS) = m/sec*amp-henry/m2 = m/sec*amp-ohm-sec/m2 = volt/m OK just because at each point, we have to neutralize the q v x B force so there will then be no steady-state transverse currents. Finally, the divergence operator in this situation is div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) and then our third equation is Maxwell equation r-1∂r(rEr) = ρ/ε0 (3) Dim check: dim(LHS) = volt/m2 dim(RHS) = Cou/m3 * m/farad = volt/m2 OK where ρ would then be "some kind" of charge inside the wire. Normally of course ρ = 0 and everything is balanced, electrons against lattice ions. Note: What about curl E = 0 ? curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] or curl E = [ 0 - 0] + [0- ∂rEz] + [ 0 - 0 ] = - ∂rEz [ correct ] This tells us that ∂rJz(r) = 0 and thus Jz = constant. Now at least I know where that is coming from! Of course it also comes from my Chapter 2 result in the low ω limit. The following Plan A is interesting: It shows that you cannot explain the radial Hall effect in terms of a line charge down the middle and a surface charge outside. This concept gives Er ~ 1/r instead of Er ~ r, and this leads to contradictions. This was my first approach to the radial Hall effect. Plan A. Assume that n(r) = n0 so that ρ = 0 and also that Jz = constant. In this case, equation (3) says that ∂r(rEr) = 0 and then rEr(r) = K => Er(r) = K / r This is my "radial Hall effect" idea. This E field could be generated by a line charge along the axis of the cylinder. Recall that φ = ln(1/r) = -ln(r) and Er = -∂rφ = 1/r in this situation. So let us assume that this is just one "surface" of the induced Hall surface charge, and the other polarity is on the outside surface of the conductor. We also have this equation combining the above with (2) r vz(r) Bθ(r) = K . If Jz is non-uniform, then we don't really know Bθ(r) so we are looking here at two unknown functions vz(r) and Bθ(r). But in Plan A we are assuming that Jz = constant. Equation (1) then says we must have vz(r) = constant. In this case, since Jz = constant, we know that Bz(r) = Cr and then we have r vz(r) Bθ(r) = K → r vz Cr = K => inconsistent! Er(r) = vz(r) Bθ(r) → K/r = vzCr => inconsistent! OK, my "line charge radial Hall effect theory" does not allow vz(r) = constant . Plan B. Now assume only that Jz = constant. Let's now assume there does exist a charge density ρ. The electrons can shift radially relative to the ions. I think it is best to call this a free charge density and not a polarization charge density. So (3) then says ∂r(rEr) = r ρ(r)/ε0 . I think the best way to model this is to say that ρ(r) = e [n(r)-n0] so we might have either an excess or a deficiency of electron density and this makes ρ. Here are our three equations again, Jz(r) = n(r) e vz(r) (1) // ρ(r) = n(r)-n0 Er(r) = vz(r) Bθ(r) (2) r-1∂r(rEr) = ρ/ε0 (3) Since Jz = const we know that Bθ(r) = Cr. Then we are saying from (1), n(r) vz(r) = Jz/e Jz = constant Bθ(r) = Bθ(a) (r/a) From (2) we then have Er(r) = vz(r) Bθ(r) = Bθ(a) (r/a) = [Jz Bθ(a)/ea] [ r / n(r) ] Dim check: dim(LHS) = volts/m dim (RHS) = amp m-2 amp-henry-m-2 Cou-1 m-1 m m3 = amp amp-henry Cou-1 m-1 = amp-henry sec-1 m-1 = amp-ohm-sec sec-1 m-1 = amp-ohm m-1 = volt/m OK Now I suspect that n(r) could deviate only very slightly from n0, so I think it is best to define this variable as the one of interest, and which I suspect will be very small. The "fractional variation in n(r) " : η(r) ≡ [n(r)-n0]/n0 = [n(r)/n0 - 1] => n(r) = n0[1 + η(r)] ρ(r) = e [n(r)-n0] = en0 η(r) . Then we have Er(r) = = Dim check: LHS = volt/m RHS = amp/m2 amp-henry/m2 * Cou-1 m-1 m3 m = amp/m amp-henry * Cou-1 = Coul sec-1/m amp-ohm-sec * Cou-1 = 1/m amp-ohm = volt/m OK Now planning ahead, we know that ∂r = = - r2 η'(r) Dim OK Then ∂r(rEr) = ∂r = [ - r2 η'(r) ] Dim OK Dim check: LHS = volt/m RHS = amp/m2 * amp-henry/m2 * Cou-1 m-1 m3 m = = amp/m2 * amp-henry * Cou-1 m = amp/m2 * amp-ohm-sec * Cou-1 m = Coul/sec/m2 * volt-sec * Cou-1 = m–2 * volt = volt/m OK But equation (3) says ∂r(rEr) = rρ(r)/ε0 = (r/ε0) ρ(r) = (r/ε0) en0 η(r) = (en0/ε0) r η(r) dim check: LHS = volt/m RHS = Cou m-3 m/Farad *m = volt/m OK Equating our two expressions for ∂r(rEr) we then obtain this differential equation [ - r2 η'(r) ] = (en0/ε0) rη(r) dim check: LHS = volts/m2 m RHS = Cou m-3 m/farad * m = Cou m-1 1/farad = volt/m OK We can move all the constant stuff to the right side [ - r2 η'(r) ] = (en0/ε0) r η(r) = r η(r) Now define β ≡ so our ODE is then [ - r2 η'(r) ] = β r η(r) where β ≡ or [ - r η'(r) ] = β η(r) where β ≡ Notice that each term is dimensionless, and therefore we should have dim(β) = 1. Dimension check: dim(β) = Cou2 m m-6 / [ farad/m * amp/m2 * amp-henry/m2 ] = Cou2 m m-6 * m/farad * m2/amp * m2-amp-1 henry-1 = Cou2 * 1/farad * 1/amp *amp-1 henry-1 = Cou2 * 1/farad * amp-2 henry-1 = amp2 sec2 1/farad * amp-2 henry-1 = sec2 1/farad henry-1 = sec2 / [farad-henry ] = sec2 / [farad-ohm-sec ] = sec / [ farad-ohm ] = sec/sec = 1 Let's work on this constant β. We know that 2πa Hθ(a) = I => Hθ(a) = I/(2πa) => Bθ(a) = μ0I/(2πa) Dimension check: dimB = amp-henry/m2 dim[μ0I/(2πa)] = henry/m * amp m-1 = amp henry/m2 OK Jz = I/(πa2) Dimension check: dimJ1 = amp/m2 dim(RHS) = amp/m2 OK β = * * = * * = = 2π2 (ea2n0c/I)2 Dimension check: dim [ea2n0c/I] = Cou -m2 m-3 m/sec / Cou/sec =1 dim(β) = 1 Let's evaluate this β thing to see the ball park using Maple: So β = 0.32 x 1031. And here is our ODE [ - r η'(r) ] = β η(r) where β ≡ = 0.32 x 1031 OK, mult through [ - r η'(r) ] = β η(r) 2(1+η) - r η' = β η (1+η)2 r η' + β η (1+η)2 - 2(1+η) = 0 r η' + (1+η )[ β η (1+η) - 2] = 0 Right here I think we are forced to have (since β is so large) [ β η (1+η) - 2] ≈ 0 => β η ≈ 2 => η = 2/β ≈ very small OK, if we know that η << 1, then we can set 1+η = 1 to get r η' + [ β η - 2] = 0 r η' + β η - 2 = 0 // a pretty simple ODE! OK, this says η(r) = C1 r-β + 2/β I verified this solution on scratch. Now we need a boundary condition. I want the total charge over the cross section to be 0, so that says !Syntax Error, Irdr!Syntax Error, Idθ ρ(r) dz = 0 !Syntax Error, Irdr!Syntax Error, Idθ en0 η(r) dz = 0 !Syntax Error, Irdr η(r) = 0 But we have trouble now because the r-β term in η(r) diverges violently at r = 0. This seems to say that we must have C1 = 0 and then we get η(r) = 2/β. But since this is always positive, it cannot integrate to 0. [ OK, fix this up by having a surface charge of opposite sign on the wire surface, Hall-like ] Perturbation Attempt: How about a perturbation theory solution: η(r) = [2/β + ε u(r)] 1+η = [1 + 2/β + ε u(r)] η' = εu' Then r η' + β η (1+η)2 - 2(1+η) = 0 r εu' + β [2/β + ε u(r)] [1 + 2/β + ε u(r)]2 - 2 [1 + 2/β + ε u(r)] = 0 r εu' + β [2/β + ε u(r)] [1 + ε u(r)]2 - 2 [1 + ε u(r)] = 0 r εu' + [2 + εβ u] [1 + ε u]2 - 2 [1 + ε u] = 0 So we then have r εu' + ε (2+β)u + order(ε2) = 0 So the first order correction ODE is then r u' + (2+β)u = 0 or r u' + βu = 0 (*) So our first order solution is this problem child u(r) = C1 r-β and then we are back to our original solution. Let's then look at the ε2 situation: η(r) = [2/β + ε u(r) + ε2v(r)] 1+η = [1 + 2/β + ε u(r) + ε2v(r)] η(r) = [2/β + ε u + ε2v] 1+η = [1 + 2/β + ε u + ε2v] ≈ [1 + ε u + ε2v] Then we get r η' + β η (1+η)2 - 2(1+η) = 0 Write this stuff as η = x so x = [2/β + ε u + ε2v] r x' + β x (1+x)2 - 2(1+x) = 0 Here is what Maple says: The last terms for large b vanish so OK. The order ε equation is this ru' +6u + bu +12u/b = 0 => ru' +bu = 0 agrees with * Now what do we do for order e2 ? It says this 2bu2 + 6u2 + rv' + bv + 6v + 12v/b = 0 rv' + bv + 6v + 12v/b = -2bu2 - 6u2 rv' + bv = -2bu2 But I had to set C1 = 0 and that made u = 0 and then this v equation is the same as the u equation, and we have the same problem again! So perturbation is not working. Resume with Different Interpretation. Suppose we go with this as the solution η(r) = 2/β . Thus we have ρ(r) = e [n(r)-n0] = en0 η(r) = en02/β = ρ = a constant ! Evaluate this: ρ = en02/β = en02* [2π2 (ea2n0c/I)2]-1 = 2en0 * = Dim check" LHS = cou/m3 RHS = amp2 cou-1 m-4 m3 sec2 m-2 = amp2 cou-1 m-3 sec2 = amp2 cou-1 m-3 sec2 = cou2 sec-2 cou-1 m-3 sec2 = cou m-3 OK Maple says that ρ = 0.8 x 10-20 cou/m3. OK, then what is the total charge inside the wire slice of thickness dz ? Q/dz = ρ πa2 = 0.26 x 10-23 Coulombs/meter For neutrality, we must then have -Q/dz Coulombs/m on the outside surface! A Hall charge. So Qsurface = - ρ πa2 = - 0.26 x 10-23 Coulombs/meter What is the surface charge per meter? σ = Q/2πa = 0.41 x 10-22 Coul/m2 I wonder about div E = 0 near the surface, where we would write ∫V ρ dV = ∫S ε E dS Using the obvious gaussian box, we shrink it and it only picks up the surface charge and that then creates a radial Hall-created E field outside the wire. Wrong! True that box only has the surface charge, but this will be accounted for by the Er on the inside, leaving Er = 0 on the outside. This has to be because outside the ρ charge appears as a line charge and so does the equal and opposite surface charge, so outside we have to have Er ≡ 0. So what is my solution to this problem? In Plan B I use div E = 0, Lorentz balance, and Jz = nevz and nothing else! I find that: ρ(r) = en0 η(r) = ρ = en0 (2/β) where β ≈ 1031 Qs = -ρ πa2 = - en0 (2/β) πa2 n = Qs/2πa = -ρ πa2/2πa = -ρ a/2 = - en0 (2/β) a/2 linear surface charge Er(r) = r Bθ(a) = μ0I/(2πa) Jz = I/(πa2) Inserting these expressions gives Er(r) = r = (r/a) Dimension check henry-m-1 amp2 m-3 cou-1 m3 = henry-m-1 amp amp cou-1 = volt-sec-m-1sec-1 = volt/m Paradox: If Ohm's Law holds that Jr = σ Er , we get Jr(a) ≠ 0 which cannot be correct. However, from my Ohm's Law doc, we do know that Ohm's Law in the presence of a B field reads Jr = σ0( Er + τωcEθ) σ0 = (n|e|2τ/m) Jθ = σ0( Eθ - τωcEr) ωc(r) = - |e| B(r)/m Jz = σ0 Ez We could kill off Jr(a) by requiring there be some Eθ(a) ≠ 0, but that violates equipotential surface. Also, at low B we still have Jr = σ Er . So something is wrong with my solution!