By Yukikazu Iwasa
Designed for graduate scholars in mechanical engineering, this textbook discusses the elemental thoughts of superconducting magnet know-how. very important themes lined contain box distribution, magnets, strength, thermal balance, dissipation, and security. To support the scholars excel within the box, every one bankruptcy comprises educational difficulties, observed via recommendations, using solenoidal magnets as examples.
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Additional info for Case Studies in Superconducting Magnets: Design and Operational Issues (Selected Topics in Superconductivity)
8) That is, the magnetic pressure is equivalent to magnetic energy density. For B o equal to 1 T, Eq. 98 × 10 5 Pa or ~4atm, from which it follows that for B o = 50 T, a magnetic pressure of ~10 GPa is reached. Fig. 2 Top view of a differential element for a “thin-walled” solenoid (thickness δ) of an average diameter 2a. The element is ∆ z high in the z -direction (out of the paper). d. d. 3 defines the winding cross section and important parameters. The differential magnetic field at the center, d Hz (0, 0), generated by a currentcarrying ring of differential cross section dA located at (r, z) is given by a modified form of Eq.
Field Fig. 9 Power dissipation vs resistivity for induction heated cylindrical shell. 7: Eddy-current loss in a metallic strip In this problem an expression for eddy-current loss in a metallic strip subjected to a time-varying magnetic field is derived. It is useful in computing eddy-current heating in copper matrix superconductor strips. ) One relevant example where eddy-current heating is a nuisance as a source of extraneous heat is in high-field, low-temperature (millikelvin range) experiments.
43. d) In reality the magnetic flux in the shielding material must be kept below the material’s saturation flux, µ o M sa . 44) e) Draw field lines for the case µ/µo >> 1. Fig. 4 Spherical magnetic shell in uniform magnetic field. 3 a) The problem is divided into three regions: region 1 (r ≥ R ), region 2 (the shell), and region 3 (r ≤ R – d) . 1c) Note that φ 1 → H 0 r cos θ for r → ∞ and that φ 3 remains finite as r → 0. 2c) Boundary Conditions 1) At r = R, the tangential component of (H θ ) is continuous: φ 1 = φ 2 .