NE528_HW2_Fall_2009_with_worked_solutions

NE528_HW2_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#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
Image of page 1

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

View Full Document Right Arrow Icon
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 + + = − +
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern