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hw5solutionextra - Problem 5.17. [Magnetic systems.) (a)...

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Unformatted text preview: Problem 5.17. [Magnetic systems.) (a) Faraday’s law relatEs the hack-emf in the coil (against which we must do work) to the time rate of change of the magnetic flux. At any moment the magnetic flux is <11 3 = NAB, where A is the cross-sectional area of the coil. According to Faraday’s law, thereforeT the magnitude of the back-emf is dilifi dB Efllfrg —~ W — and so the power that we must supplyr is {LB dB 033 since ’H = NI/L. To obtain the total energy (work) required for an infinitesimal change in the current:I we integrate the power over time to obtain work = HV/gég—dt :- HVdB. (1:1) From the definition of H, we can write 3 = are + MW), and hence as = New + ear/V). The result of part (a) is therefore work = VHhu{dH + dM/V) : ,uflVH (H! + ,ufl’H dM. In the first term we can write Hd'H = d[%?1f2), so this term is the change in the quantity ELEVHE. If there were no specimen inside the solenoid, this term would give the change in the vacuum field energy; with the specimen, H is the same as without, so this term represents the work we would have to do to increase the field if there were no specimen. If we define the work done on the “system” to exclude this term but include everything else, then ' W = work done on system = pO’H dM. (c) The work done on a mechanical system is ——P dV. Apparently, the analogous term for a magnetic system is +M0H dM. The thermodynamic identity for a magnetic system should therefore be dU = TdS + poHdM. (d) The magnetic analogue of the enthalpy would be Hm = U _._HOHM, in analogy with the ordinary enthalpy H = U + PV. An infinitesimal change in Hm can then be written mm = dU -— “OH dM u. MOM d’H = TdS —— MOM dH, where I‘ve used the thermodynamic identity for U in the last step. Interpretation? Apparently the quantity Hm is less than the “system” energy (at least for our situ- ation), and is the more natural “energy” function to use when a process takes place at constant H. To obtain the magnetic analogue of the Gibbs free energy, we can subtract TS just as for a mechanical system: Gm = Hm ~— TS. Under an infinitesimal change in conditions, dGm = de —-- TdS —— SdT = -—SdT—- non’H. Presumably, Gm is the energy that can be extracted as work when the system is held at constant T and constant ’H (whereas the Helmholtz free energy, F = U —- TS , would give the available work in a process at constant T and M [The references given in the text provide further interpretations of the various energy functions for a magnetic system.] ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 404 taught by Professor Anlage during the Fall '11 term at Maryland.

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hw5solutionextra - Problem 5.17. [Magnetic systems.) (a)...

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