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Magnetic field lines and the Sun

  1. Jan 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Hey!
    I have trouble with an exercise that I need help with. Here it is:

    The magnetic field lines for the average interplanetary magnetic field (IMF) follow Archimedean spirals.

    i) Find the heliocentric distance r in Astronomical Units (AU), where a field line has wrapped itself around the Sun once. Assume that the solar wind speed is Usw = 400 km/s.

    ii) Also determine the number of times the magnetic field has wound around the Sun by a heliocentric distance of 100 AU.

    2. Relevant equations
    See below

    3. The attempt at a solution

    i) The plasma flows radially from the Sun as I've understood it. Therefore, the flow of the plasma shall have a constant radial velocity Usw with respect to the heliocentric distance.
    We have to look at the velocities in polar coordinates I believe, so vφ = −ΩR in the φ-direction and Vr=Usw. Here, Ω is the angular rotation rate of the Sun at its equator and R is the radial distance from the Sun.
    Further, an equation for Archimedian sprials is: R − R⊙ = − VSW Ω (φ − φ0) , where R⊙ is the solar radius and φ0 is the angular position of a particular magnetic field line at the surface of the Sun.
    Now, how should I apply this to the problem? It is said that the field line has wrapped itself around the Sun once, is that related to the angular rate (Ω maybe)?

    ii) I actually have no idea here. I think that we need to consider that it has a circular move, so something with the angular velocity I guess, where the radius is involved maybe.

    Hope that you can help me
     
  2. jcsd
  3. Jan 19, 2016 #2

    SteamKing

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    If you go around a circle one revolution, is that related to the angular velocity or the angular displacement?
    You should get a pitcher of an Archimedean Spiral and study it. Here is one:

    a_1529.gif

    Note that this spiral begins at the origin (0,0). For the magnetic field line erupting from the sun, you need to take into account the radius of the sun, which is why the formula for the spiral above is written the way it is. The equation of this spiral is r = a ⋅ θ. Remember, θ is measured in radians.
     
  4. Jan 19, 2016 #3
    Thanks, I'll give it a try tomorrow :) is a=9.8 m/s^2?
     
  5. Jan 19, 2016 #4

    SteamKing

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    If you are talking about the formula for the sample spiral I showed you, a is just a constant. If you make a small, you get a tight spiral.

    Remember, 9.8 m/s2 is the acceleration due to gravity only for the earth. This value has nothing to do with acceleration due to gravity on the sun.
     
  6. Jan 20, 2016 #5
    Okay, a constant then. Am I supposed to give it a value or change the formula to find it? Also, can I say that θ. is 2*pi ?
     
  7. Jan 20, 2016 #6
    I assume that I will need to insert the equation you gave me in R − R⊙ = − VSW Ω (φ − φ0) , is it for Ω? Or would it replace φ − φ0 instead? Sorry, but this problem is a little tricky.
     
  8. Jan 20, 2016 #7
    I know that the Sun has a sidereal rotation period at the equator of ∼25 days, which should be Ω (give 25 days in radians). How can I implement your equation in the one that I mentioned from the beginning? R − R⊙ = − VSW Ω (φ − φ0) --> a*θ -R⊙ = − VSW Ω ? But then I don't have the heliocentric distance R, just the Sun's radius
     
  9. Jan 20, 2016 #8

    SteamKing

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    You're not supposed to do anything with it. It was intended to be just an example of how a spiral equation looks expressed in polar coordinates.

    Again, you don't need to do anything with my example spiral. You were given an equation to use in the OP, along with certain other information to allow you to solve this problem.
     
  10. Jan 20, 2016 #9
    Okay, now I get it. The thing is that I don't know what to do with φ − φ0, I'm stuck here
     
  11. Jan 20, 2016 #10

    SteamKing

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    You still haven't answered this question. If you go around something once, is that an angular velocity or an angular displacement?
    All they want you to do here is to calculate how many times around the sun the mag field line wraps until it is 100 AU from the solar surface.

    You're trying to make this a more difficult problem than it's supposed to be, for some reason. It's more about geometry than solar physics.
     
  12. Jan 20, 2016 #11

    SteamKing

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    The quantity φ0 is there just to make the formula more general, as in the case where you were, say, dealing with multiple field lines. Presumably, each field line would erupt from the solar surface at a different angle as measured around the solar equator.

    Since we are dealing with a single field line for this problem, you can assume that φ0 = 0.
     
  13. Jan 20, 2016 #12
    Oh, sorry. It is angular displacement, not velocity.
    I know, I think that I'm trying to make it more difficult : / All I can think of is that φ − φ0 should be 2*pi
     
  14. Jan 20, 2016 #13
    Then φ should be 2*pi radians, for one revolution
     
  15. Jan 20, 2016 #14

    SteamKing

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    For 1 trip around the sun, sure.
     
  16. Jan 20, 2016 #15
    Great then. I believe that I should use 2*pi , they always begins with the "most obvious" assumptions.
     
  17. Jan 20, 2016 #16
    Thanks buddy, you've helped me a lot!!
     
  18. Feb 14, 2016 #17
    Hi again,

    I got the wrong answer here :/ Anyone that can help?
     
  19. Feb 14, 2016 #18

    SteamKing

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    Wrong answer doesn't tell us much. Please show your calculations.
     
  20. Feb 15, 2016 #19
    Ok, here are my calculations:

    I assumed that the radial velocity of the plasma , V_SW, is constant since it's an Archimedian spiral. This is with respect to the heliocentric distance. The plasma flow in the ecliptic plane has an apparent velocity in polar coordinates of v_R = V_SW in the R-direction and vφ = −ΩR in the φ-direction
    I also know that it takes about T=25 days for the Sun to rotate about its axis , so I can use : Ω=2π/T. R is the radial distance from the Sun.
    The two magnetic field components must also be considered and these can be : B_R/B_φ = v_R/v_φ = − V_SW /ΩR.

    The overall shape of any given magnetic field line can be determined by rewriting the equation above as: d_R/d_φ = − V_SW/Ω. We can write v_R has as dR/dt and vφ as R *(dφ/dt) . Since VSW and Ω are constants, the obvious solution to this differential equation, with boundary conditions applied, is the equation of an Archimedean spiral: R − R⊙ = (− V_SW/Ω) *(φ − φ0) (*) (R⊙ is the solar radius and φ0 is the angular position of a particular magnetic field line at the surface of the Sun.).

    I am looking for R, so I can change (*) , by inserting Ω in (*) I get : R − R⊙ = (− V_SW/2π)*T*(φ − φ0) (**)
    --> R=R⊙−( V_SW/2π)*T*(φ − φ0) (**)
    φ=2π and φ0=0 --> insert in (**) --> R=R⊙−V_SW*T (***).
    I convert T to seconds and insert V_SW and R⊙ in (***), divide it by 1 AU, which gives me R=6AU . This seems to be wrong.

    b) I calculated the angle by using: tanΘ=((R-R⊙)*Ω*sinΘ))/V_SW. <-- equation my book. Θ= angle between the magnetic field direction and radial direction. I guessed that it was 90 deg (equatorial plane). Also, the typical velocity for solar winds is V_SW=400 km/s. Θ=90 degrees , so the equation can also be written as:

    tanΘ=(R*Ω)/V_SW. R=6 AU, which I got earlier. I got about 81 degrees, which is wrong.

    c)What is the magnitude of the field, |B|, at this point if the source surface is located at Rs = 10R⊙ and if the magnetic field at Rs is Bs = 10^-66 T?

    I used:
    |B|=B_s*(R⊙/R)^2*(1+((Ω^2(R-R⊙)^2)/(V_sw^2))*sin2Θ) B_s = 2,000 nT (field strength here, from the book). Wrong answer here as well..

    d) Number of times the magnetic field has wound around the Sun at 100 AU.
    n=100AU/(diameter of Sun*π) , which is totally wrong!

    Hope that my solutions are clear enough
     
    Last edited: Feb 15, 2016
  21. Feb 15, 2016 #20
    I believe that I can get a)-c) right if I know what values φ − φ0 should have. I chose φ =2π and φ0=0. Can I use R(t)=R⊙+R⊙*T for the heliocentric distance, where T is about 25 days?
    d) is a little tricky, even if it might be the easiest one
     
    Last edited: Feb 15, 2016
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