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Solutions to the problems in Chapter 27

Solutions to the problems in Chapter 27

Solutions to the problems in Chapter

The problems of 6th week (“superconductivity”) 26 Solutions to the problems in Chapter 27 27.1 Superconducting sphere The free-energy density of a type-I superconductor is determined by the critical field H c .Since ⃗ B = ⃗0 throughout the volume of the type-I superconductor, when surface effects are neglected (λ L ≈ 0 compared to the dimensions of the sample) ⃗B i = H ⃗ i +4πM ⃗ = ⃗0 ⇒ M ⃗ = − 1 H 4π ⃗ i (1) Assuming the sample to be a thin needle along the direction of the applied magnetic field H ⃗ 0 (along z), the demagnetization field H ⃗ d = −N zM ⃗ ≈ ⃗0. In this case, the magnetic field within the sample is H ⃗ i = H ⃗ 0 and the magnetization is M ⃗ = −H ⃗ 0 /(4π). According to eq. (2.6) in the note “Magnetic energy and domains”, the magnetic energy density is then, ∫ [∫ 1 V F d⃗r M = G M = − sample V ⃗M · δ ⃗ H 0 ] = 1 ∫ H0 ⃗H 0 ′ · dH 4π ⃗ 0 ′ = H2 0 0 8π [This energy density corresponds to G in Marder because H ⃗ 0 is the independent variable, see problem 27.2]. The total energy density is the sum of G M and the field-independent condensation-energy density, G S . Since the two contributions just outbalance each other at the critical field H 0 = H c ,wehaveameasureforG S , G = G S + G M , G S = − H2 c (3) 8π If the sample has instead the shape of a sphere, then (1) and (3) still apply, but ⃗H i = ⃗ H 0 + ⃗ H d = ⃗ H 0 −N z ⃗ M = ⃗ H0 + D ⃗ H i , D ≡ N z 4π = 1 3 (2) (sphere) (4) implying H ⃗H i = ⃗ 0 1 − D , G = G S + G M = −H2 c 8π + H0 2 (5) 8π(1 − D) The internal field H i is larger than the applied field and becomes equal the critical one, when the applied field is H 0 =(1− D)H c = 2 3 H c . This condition is not necessarily critical, since the global free energy of the superconducting sphere is still negative (as long as H 0 < √ 1 − DH c ). A detail discussion of the intermediate state of the superconductor, when (1 − D)H c

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