The structure of the magnetic field of an ℓ = 1 toroidal torsatron

Abstract

The authors study the structure of the magnetic field in an C = 1 toroidal torsatron. Experiments on the Vint-20 device and numerical calculations on a computer have shown that closed magnetic surfaces with high parameters exist in an { = 1 torsatron. The forms of the magnetic surfaces are given together with graphs showing how their parameters vary within the working volume under different operating conditions. It has been established that the magnetic surface parameters can be adjusted within wide limits by changing the vertical field. It is shown that this type of system is characterized by high inhomogeneity of the magnetic field in the working volume and that this inhomogeneity can be substantially reduced by applying a vertical field or by changing some parameters of the helical winding.

Recent stellarator research

“…7, the equivalent field is represented by so-called 'torsatron-type' coil system. The properties of torsatron are studied in Refs [25][26][27][28][29][30][31][32][33][34][35]. As the number of helical conductors is half that of the standard configuration, the torsatron type is more convenient to form a helical diverter (see Fig.…”

“…There is no exact and simple equilibrium solution for the toroidal stellarator. When the plasma pressure is finite, a Pfirsch-Schliiter current given by 4TT 1 3p '• cos 6 i B o 3r (35) flows. When the distributions of pressure and rotational transform are p(r) = p 0 (1 -(r/a) m ) and t(r) = i o (r/a) 2fi ~4' respectively, then the Pfirsch-Schluter current yields a vertical field B i:…”

Effect of alternating electric field on diffusion of weakly collisional plasma in ℓ = 1 Vint-20 torsatron

“…Very characteristic in this respect is the magnetic configuration of the £ = 1 Vint-20 torsatron [8]. Table I presents the main characteristics of the two devices and, from a comparison, it is evident that the magnetic-field inhomogeneities are much greater in the Vint-20 torsatron: eh = 0.65, et = 0.25.…”