NE528_HW3_Fall_2009_with_worked_solutions

NE528_HW3_Fall_2009_with_worked_solutions - NE528 Fall 2009...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
NE528 Fall 2009 HW#3 (total 80 points) Due Wednesday, October 7 by class time 1. (40 points) We derived the expression for the current density “the diamagnetic current” at equilibrium condition and have shown that it is given by 2 B p J B ×∇ = . Now, consider a plasma where both the density and temperature have radial profiles and vary as ( ) 0 cos / n n r a = , , and ( cos / e eo T T r a = ) 4 e i T T a = , where a is the plasma radius and r is the radial distance. Obtain the expression of the diamagnetic current density at equilibrium condition under the above conditions and discuss your result. Use the single fluid equation d u E J B dt ρ σ p = + × − ∇ and keep the term E σ in the equation for a finite charge density, i.e. 0 0 and hence E σ σ Helpful vector relation : ( ) ( ) . . A B C AC B A B C × × = Where A , B and C are vectors. Worked Solution: Single fluid equation is: p B J E dt u d × + σ = ρ The steady-state equilibrium condition is: 0 + E J B p σ = × −∇ Taking the cross product of p J B E σ = × + with B and using the vector relation ( ) ( ) A B C A.C . B A B C × × = ( ) ( ) . . B p B J B B E B B J B J B B σ σ ×∇ = × × + × = + × E With B and J perpendicular to each other the second term ( ) . B J B vanishes and solving for the current gives 2 2 2 2 B p B E B p E J B B B B B σ σ ×∇ × ×∇ × = = + The term 2 E B B σ × represents a drift current because σ is the charge density of the fluid, and 2 E B B × is the drift velocity, thus this term is , E B drift J ×
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon