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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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