100%(1)1 out of 1 people found this document helpful
This preview shows pages 1–3. Sign up to view the full content.
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 by2BpJB×∇=. Now, consider a plasma where both the density and temperature have radial profiles and vary as()0cos/nnra=, , and (cos/eeoTTra=)4eiTTa=, where ais the plasma radius and ris 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 uEJBdtρσp=+×− ∇and keep the term Eσin the equation for a finite charge density, i.e. 0 0and henceEσσ≠≠Helpful vector relation: ()()..ABCAC BA B C××=−Where A, B and Care vectors. Worked Solution: Single fluid equation is: pBJEdtud∇−×+σ=ρThe steady-state equilibrium condition is: 0+ EJBpσ=×−∇Taking the cross product of pJBEσ∇=×+with B and using the vector relation ()()ABCA.C.BA B C××=−()()..BpBJBBEB B JB JBBσσ×∇=××+×=−+×EWith B and J perpendicular to each other the second term ().B JBvanishes and solving for the current gives 2222BpBEBpEJBBBBBσσ×∇××∇×=−=+The term2EBBσ×represents a drift current because σis the charge density of the fluid, and2EBB×is the drift velocity, thus this term is , EB driftJ×
This preview
has intentionally blurred sections.
Sign up to view the full version.