9780030839931, Chapter 34, Problem 1P

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Chapter 34, Problem 1P Problem Thermodynamics of the Superconducting State The equilibrium state of superconductor in uniform magnetic field determined by the temperature and the magnitude of the field (Assume that the pressure fixed, and that the superconductor is long cylinder parallel to the field so that demagnetization effects are The thermodynamic identity conveniently written the terms of the Gibbs free energy G: dG (34.37) Where Sis the entropy and the total magnetization (IR where Mis the magnetization density). The phase boundary between the superconducting and normal states the H- plane given by the critical field Hc(T) (Figure 34.3) (a) from the fact that continuous across the phase that (34.38) (where the subscripts and indicate values in the superconducting and normal phase respectively). (b) Using the fact that the superconducting state displays perfect diamagnetism (B while the normal state has negligible diamagnetism (M show from the entropy discontinuity across the phase boundary is and thus the latent when the transition occurs in is (c) Show that when the transition occurs at zero field (i.e., at the critical point) there is a specific heat discontinuity given by Step by solution Step Solution The thermodynamic identity for the Gibbs free energy G given by dG (1) Here S the T the the total magnetization For process constant we will have dG (2) The normal state of most superconductors paramagnetic and the magnetization small compared with that the superconducting state Therefore we can neglect the normal state magnetization Consequently from equation (2) dG or we can state that Gibb's function in the normal state is not changed by the application of the magnetic field. (3) Step (a) with dT =0 at constant MV dH (Since, MV) VdH (4) H VdH Or on integrating (as function of magnetic VdH (5) Step At the critical the free energies of the normal and superconducting states must be equal for equilibrium of two phases, (6) Substitute for in the equation (8) From equation dG we -dG Differentiating the above equation with respect T keeping pressure constant G constant). dT Step (b) Entropy: Let us now calculate the difference in entropy of two phases. For solids given by, Differentiating equation (8) with respect T we get, (9) Step Latent heat: The latent heat defined as, obtain the latent heat when the transition occurs field, substitute dT the equation So, Q=-TV (10) dT Step of 6 (c) Specific Heat: The specific heat dT dT The equation with zero field will reduce dT Differentiating the above equation with respect to T and multiplying by we get, dT dT dT =0 and