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Charged particle in magetic field problem

  1. Jun 15, 2006 #1
    Hello everyone, I have a concept problem im having a problem understanding

    Assume the Earth's magnetic dipole moment is aligned with the earth's rotational axis, and the earth's magnetic field is cylindrically symmetric.
    If i have a proton traveling in orbit to the earth at a height of 7950000 meter where the Earth's magetic field has an intensity of 5.45 X 10^-8 T at a speed of 41890384.6154 m/s

    How do I determine the way the proton's velocity is traveling?

    I have the choices of

    1. an equatorial orbit from East to West.
    2.a polar orbit.
    3. an equatorial orbit from West to East.

    I believe it would be an equatorial orbit as it is perpendicular to the Earth's magnetic field, however I am not complete sure and even if I was right, how would I determine if it orbits west to east or east to west?

    Any help or tip would be appreciated, thx
     
  2. jcsd
  3. Jun 15, 2006 #2

    Andrew Mason

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    If it is travelling in a circular orbit, you know that the combined force of gravity and the Lorentz force are constant. Since gravity is constant, the Lorentz force must be constant. So either the Lorentz force is non-zero and constant or it is 0.

    In order to determine which it is, work out the gravitational force and determine if it is sufficient to keep the proton in that orbit. If it is, the Lorentz force must be 0 (polar orbit). If it is not, determine whether gravity is greater or less than the centripetal force to determine the direction of the Lorentz force, and hence the direction of the proton.

    AM
     
  4. Jun 15, 2006 #3
    Thanks for the speedy tip AM

    I believe the question implied that the proton is to be traveling at the speed that I said ealier and kept there in constant to be in orbit and cancel the gravity to 0, hence polar orbit.

    However that turned out to be the wrong answer, what do you think I did wrong here and could u provide futher explaination as how to determine which way the orbit would be if the centripetal force is greater or lesser
     
  5. Jun 16, 2006 #4

    Tide

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    HINT: What is the direction of [itex]\vec g \times \vec B[/itex]?
     
  6. Jun 16, 2006 #5

    Andrew Mason

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    If you work it out, the force of gravity is not sufficient to provide the centripetal force for such an orbit.

    So the magnetic force has to be non-zero. What kind of orbit does that imply? The magnetic force has to be in what direction?

    Since [itex]\vec{F} = q\vec{v}\times\vec{B}[/itex], what direction must v be in order to provide that centripetal force?

    AM
     
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