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 F G 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 / G m = × ) kg 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 F G r = 2 ( ) Magnetic Gravitational F q v B Mm G r = + × ±²³²´ ±²³²´ 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) 0 : ........................... : x x y y y x z x y y x y x dv m q v B qv B dt dv m q v B dt dv m dt dv B dy qv v v y dt m dt dv d y m qv B dt dt Thus d y qv B y y oscillatory dt Solution y Mm Mm G G r r Mm M G G r r M G r M G r ω ω ω ω ω = × = = + × = = = = = = = = + = = 1 2 1 2 2 cos sin cos sin K t K t dy K t K t dt ω ω ω ω ω ω ω + + = − +
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