Loss cone opening angle

  • #1
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Homework Statement


Hi!
I need to find the opening angle of the loss cone at a given altitude, when the magnetic latitude is 65 deg.

Homework Equations


See below

The Attempt at a Solution


First, I used the following equation to calculate the magnetic field at a given altitude:
B=B0/[(r/RE))*sqrt(1+3sin^2(70deg)] ; where r= 2000 km (given altitude) , B0= 3.11*10^-5 nT for Earth and RE= 6378 km (Earth's radius).
Afterwards, I use: sin^2(alfa)=B0/B and the angle I get is too big. Please, help me with this one :)
 

Answers and Replies

  • #2
TSny
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I'm not familiar with the formula that you are using, but are you sure that r represents the altitude (i.e., distance above the surface of the earth)? What would the formula give for zero altitude?
 
  • #3
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That's exactly the problem, it should theoretically be 7 degrees, but I get something completely wrong. I don't know if this equation is valid or not
 
  • #4
TSny
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The formula B=B0/[(r/RE))*sqrt(1+3sin^2(70deg)] doesn't make sense if r represents altitude. On the surface of the earth, the altitude would be zero. But if you put r = 0 into your formula, what would you get?

So, if r is not the altitude, what do you think r actually represents? You must have gotten the formula from somewhere. Did your source of information define the symbols?

Your problem states that the latitude is 65 degrees. In your formula, you used 70 degrees. Is there a reason these are different?
 
  • #6
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I personally think that r should be the altitude plus the radius of Earth
 
  • #7
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I need to find the maximum field strength at this altitude and at 65 deg. I know what to do after that
 
  • #8
TSny
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I found this site helpful: https://farside.ph.utexas.edu/teaching/plasma/lectures1/node22.html

As you noted in post #6, r is the distance from the center of the earth, not the altitude.

Equation (123) in the link when written in your notation is B=B0(RE/r)3 sqrt[1+3sin2(θ)]. Note that the distances are cubed and the square root factor is not in the denominator.

I spent quite a while working with the formulas on this page, but was not able to generate an answer of 7o for the loss-cone angle. It might be helpful if you could state the question word for word as given to you.

[EDIT: If I'm understanding the equations, I get a loss-cone angle of about 7o for latitude 65o if the altitude is about 20,000 km rather than 2,000 km.]
 
Last edited:
  • #9
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I found this site helpful: https://farside.ph.utexas.edu/teaching/plasma/lectures1/node22.html

As you noted in post #6, r is the distance from the center of the earth, not the altitude.

Equation (123) in the link when written in your notation is B=B0(RE/r)3 sqrt[1+3sin2(θ)]. Note that the distances are cubed and the square root factor is not in the denominator.

I spent quite a while working with the formulas on this page, but was not able to generate an answer of 7o for the loss-cone angle. It might be helpful if you could state the question word for word as given to you.

[EDIT: If I'm understanding the equations, I get a loss-cone angle of about 7o for latitude 65o if the altitude is about 20,000 km rather than 2,000 km.]
So, it should be the radius of the Earth divided with the altitude that I need to use? How will it then be in the equatorial plane, since r would be equal to RE.
The thing with the 7 degrees might be wrong. The book says:

For example, at L = 4 if all particles whose mirror points are below an altitude of about 1,000 km (the exact altitude is not critical) are lost via collisions, then the distribution function of these particles at the magnetic equator (X = 0) has a loss cone starting at an approximate pitch angle of a\c = 7°

But that will depend on the L-value I believe. L is the surface of rotation of a field line about the magnetic dipole axis and you can calculate different L's , but I'm not doing it here.
 
  • #10
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By the way, when it comes to the equatorial plane, I only used
sin^2 (alfa)=(cos^6(lambda))/sqrt(1+3*sin^2(lambda)) , where lambda is 65 degrees. For the other height (I tried 1700 km), I used the equation you posed and got a value of 22 degrees. I might be correct, I have no idea.
 
  • #11
TSny
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So, it should be the radius of the Earth divided with the altitude that I need to use? How will it then be in the equatorial plane, since r would be equal to RE.
r is the distance from the center of the earth to the point of interest. If h is the altitude of the point, then r = RE + h.

The equatorial plane just refers to all points where the latitude angle is zero. r does not need to equal RE on the equatorial plane.

The parameter L for a particular magnetic field line is just req/RE, where req is the distance from the center of the earth to the point where the field line crosses the equatorial plane.
The thing with the 7 degrees might be wrong. The book says:

For example, at L = 4 if all particles whose mirror points are below an altitude of about 1,000 km (the exact altitude is not critical) are lost via collisions, then the distribution function of these particles at the magnetic equator (X = 0) has a loss cone starting at an approximate pitch angle of a\c = 7°
Using the equations in the link in post #8, you can verify the above. That is, for a magnetic field line corresponding to L = 4 and for mirror points at altitude of 1000 km, the loss cone angle on the equatorial plane is about 7o.

But that will depend on the L-value I believe. L is the surface of rotation of a field line about the magnetic dipole axis and you can calculate different L's , but I'm not doing it here.
Yes, it depends on L. L can be calculated if you know r and θ for some point on the magnetic field line that you are working with. From equation (122) in the link you can use r and θ to determine req. Then, L = req/RE.
 

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