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How many turns in the field?

  1. Feb 22, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the solar wind has velocity ##450## km s##^{−1}##, i.e. it is slow solar wind, and if the heliopause is ##150## AU from the Sun, how many turns are wound into each field line between the Sun and the heliopause? Assume the magnetic field is ##6##nT, and the rotational period of the Sun is ∼27 days.

    2. Relevant equations


    3. The attempt at a solution
    Not sure where the orbital period of the sun comes into it. And I assume the ##450## km s##^{−1}## are the velocity of solar wind particles radially, but I'm not 100% on that.

    What I thought was if it takes ##\frac{150AU}{450 \times 10^3 ms^{-1}} = 5 \times 10^7s## to get to the heliopause, then in that time since a particle in the magnetic field will move in a helix with frequency ##\frac{eB}{m}##, and with a frequency and a time I could work out how many times a particle would loop the field line.

    Problem is I haven't been give a mass and I feel like the number of turns in the field line shouldn't depend on the mass, it should be a property of the field. Even if that's not true and it should depend on particle type, I haven't been given a particle type to go look up the mass of. And this answer doesn't involve the rotational period of the sun. So I'm stuck! Thanks for any help! :)
     
    Last edited: Feb 22, 2017
  2. jcsd
  3. Feb 23, 2017 #2

    andrevdh

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    I think the 450 km/s is the forward speed. The rotation of the sun gives it a sideways component. Not sure how this comes into getting to the answer though, but in order to make it turn around the field lines the particles need to have a sideways component.
     
    Last edited: Feb 23, 2017
  4. Feb 23, 2017 #3

    gneill

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    Staff: Mentor

    I think they are asking for how many times the field line is wrapped around by the Sun's rotation. Start by determining how many day's travel is it for the solar wind to travel from the Sun to the heliopause.
     
  5. Feb 24, 2017 #4

    andrevdh

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    I would really like to see where this is going, so please Kara do as gneill suggests, although I cannot see the connection.
    I know that the field lines are wound up in the pasma (worked at an observatory long ago and did some spectroscopy investigations on the sun), but did not know it is due to the sun's rotation. Think they are wound in opposite directions in the two hemispheres?
     
  6. Feb 24, 2017 #5
    The solar wind takes 577 days to reach the heliopause just using speed = distance/time. In that time the sun rotates 21.4 times so from gneill's suggestion there are 21.4 turns in the field. Although I would never have known to interpret the question like that. :)
     
  7. Feb 24, 2017 #6

    andrevdh

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    So what is happening is that the field lines are wound up in the plasma as the sun rotates!
    No wonder that the magnetic field of the sun has an 11 year cycle during which time the field gets weaker and weaker as the field lines are wound up more and more!
    After that it starts over with a reversed polarity.
     
    Last edited: Feb 24, 2017
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