By Hiroshi Maeda
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Additional info for Bismuth-based High-temperature Superconductors (Applied Physics Series , No 6)
Note: There are two flaws in the answer m / πh 2 quoted in the text. First, the area A is missing, meaning the quoted answer is a density per unit area. This should not be a major issue. Second, the h should be replaced by = . (b) N = 2 ⋅ πk F2 2 (2π / L) => ns = N / A = k F2 / 2π L m where ns is the 2D sheet density. For a square sample, W=L, so: W ns e 2τ 2π m Rs = 2 2 and using =k F / m = vF : kF e τ h 1 2π= Rs = = 2 2 e kF A k F vF e τ (c) Rs = 17-2 CHAPTER 18 1. Carbon nanotube band structure.
The induced dipole moment on the atom at the origin is p = αE, where the electric −3 field is that of all other dipoles: E = 2 a 3 ∑ p n = 4p a 3 ∑ n ; the sum is over ( ) ( )( ) positive integers. We assume all dipole moments equal to p. The self-consistency condition is that p = α(4p/a3) (Σn–3), which has the solution p = 0 unless α ≥ (a3/4) (1/Σn–3). 202; it is the zeta function ζ(3). 16-3 CHAPTER 17 1. (a) The interference condition for a linear lattice is a cos θ = nλ. The values of θ that satisfy this condition each define a cone with axis parallel to the fiber axis and to the axis of the cylindrical film.
12-2 5a. U ( θ ) = K sin 2 θ − Ba M s cos θ Kϕ2 − Ba M s 1 2 ϕ , for θ = π + ϕ 2 and expanding about small ϕ . 1 Ba M s . ). For minimum near ϕ = 0 we need K > b. If we neglect the magnetic energy of the bidomain particle, the energies of the single and bidomain particles will be roughly equal when M s d 3 ≈ σ w d 2 ; or d c ≈ σ w M s . 2 2 For Co the wall energy will be higher than for iron roughly in the ratio of the (anisotropy 2 constant K1)1/2, or 10. Thus σ w ≈ 3 ergs cm . For Co, Ms = 1400 (at room ° temperature), so M s ≈ 2 ×106 erg cm .
Bismuth-based High-temperature Superconductors (Applied Physics Series , No 6) by Hiroshi Maeda