NE528_HW2_Fall_2009_with_worked_solutions

NE528_HW2_Fall_2009_with_worked_solutions - NE528 Fall 2009...

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NE528 Fall 2009 HW#2 (total 80 points) Due Wednesday, September 23 by class time 1. (30 points) A fully ionized, uniform, hydrogen plasma has a density (n e = n i = n) in orthogonal uniform gravitational and magnetic fields, where the gravitational field is along -y direction g = (0, - g, 0), and the magnetic field is along the z direction B = (0, 0, B). (a) Calculate the drift velocities of each species for the following situations: (i) In the Earth's magnetosphere at r = 2 R Earth , assuming the surface field strength is 10 -4 Tesla. (ii) Near the surface of the Sun in an active region where B = 10 -2 Tesla. (iii) In the magnetosphere of a neutron star at r = 2 R Sun assuming the neutron star has the same mass of the sun and that the surface field strength is 10 8 Tesla at a distance equals to 10 6 cm from the center of the neutron star. (b) Check, for each case, the gravitational and Lorentz force acting on the plasma and see if they cancel out. Useful Information: Radius of the Earth R Earth = 10 8.8 cm Radius of the Sun R Sun = 10 10.84 cm Mass of the earth M Earth = 10 27.78 gram Mass of the Sun M Sun = 10 33.3 gram Surface gravity of the Earth g Earth = 10 2.99 cm.s -1 Surface gravity of the Sun g Sun = 10 4.44 cm.s -1 Universal gravity equation 2 2 ( , . . / ) g in Newtons i e kg m s Mm FG r = , where , M is the mass of the object and m is the mass of the electron, r is the distance between the electron and the object ( -11 3 2 6.67 10 / Gm ) k g s Worked Solution a) Calculate the drift velocity for each case. G B y x z v
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The case is very similar to the motion of electron in steady electric and magnetic fields. Universal gravity equation: 2 g Mm FG r = 2 () Magnetic Gravitational Fq v B Mm G r =+ × ±²³²´ 2 2 2 2 2 2 22 2 2 0 : ........................... : x xy y yx z x yy x y x dv mq v B q v B dt dv v B dt dv m dt dv Bd y qv v v y dt m dt dv dy v B dt dt Thus q v B y y oscillatory dt Solution y Mm Mm GG rr Mm M M G r M G r ωω ω = × =− = == = = = += = 12 2 cos sin cos sin Kt dy dt ++ + Assume the particle starts at the origin with 0 ox oy vv = = 0 at = , hence 21 0; (1 cos ) KK y t M G r M G r
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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_HW2_Fall_2009_with_worked_solutions - NE528 Fall 2009...

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