NE528_HW3_Fall_2009_with_worked_solutions - NE528 Fall 2009...

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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 / nn ra = , , and ( cos / ee o TT 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 du E JB dt ρσ p = +×− and keep the term E σ in the equation for a finite charge density, i.e. 0 0 and hence E Helpful vector relation : ( ) ( ) .. A BC A CB A BC ××= 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 + EJB p = ×− Taking the cross product of p JB E ∇= × + with B and using the vector relation ( ) ( ) ABC A . C . BA B C ( ) ( ) Bp B J B B EB B J B B σσ ×∇ = × × + × = + × E With B and J perpendicular to each other the second term ( ) . BJ B vanishes and solving for the current gives 22 2 2 B pB E B pE J B B BB B × × =− =+ The term 2 E B B × 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 , EBd r i f t J ×
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i.e., , which modifies the diamagnetic current for non- quasi neutral plasma.
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This note was uploaded on 12/08/2010 for the course PY 528 taught by Professor Bourham during the Fall '09 term at N.C. State.

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NE528_HW3_Fall_2009_with_worked_solutions - NE528 Fall 2009...

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