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What is the larmor radius of an electron in the inner van allen belt?

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the larmor radius of an electron in the inner van allen belt?
    This is for a General Astr class so I feel like I must be over complicating it.

    Electron Kinetic Energy in the inner VA belt: K≥30 [MeV] (much greater then its rest energy)
    VA Belt Radius: r~1.1-2.0 Earth Radii

    2. Relevant equations

    Larmor Radius: r=(mv)/(qB)
    where m is the mass of the particle, v is the component of its velocity perpendicular to the B field, q is the charge of the particle, and B is the B field magnitude at that point.

    Earth's B field strength as a function of r: B(r)=(B0*R^3)/r^3
    where B_o is the Earths B field strength at its surface, R is the radius of the earth, and r is the distance

    Kinetic energy in relativistic terms: K=gm0c2-m0c2
    where g is gamma, m is the particles rest mass and c is the speed of light

    g=1/[itex]\sqrt{1-(v^2/c^2)}[/itex]

    3. The attempt at a solution

    First I attempted to solve for B field strength in the inner VA belt

    B(1.1R) = (3.1*10^-5)/(1.1^3) = 2.33*10^-5[T]
    B(2.0R) = (3.1*10^-5)/(2.0^3) = 3.875*10^-6[T]
    So the field strength is between 2.33*10^-5 and 3.875*10^-6 [T]

    Next I used the electrons kinetic energy so solve for its velocity

    v = [itex]\sqrt{[1-(.511/30.511)^2]c^2}[/itex]
    ≈2.9996*10^8 [m/s]

    now I found the max larmor radius by first assuming that all the electrons velocity is perpendicular (I have no idea how to determine how much of v is in fact perpendicular)
    and using the weakest strength of B
    r=(9.109*10^-31*2.9996*10^8)/(1.602*10^-19*3.875*10^-6)
    ≈440[m]
    so the larmor radius must be ≤440[m] is my final conclusion

    some obvious problems:
    1:this doesnt really narrow down my answer very much
    2:this course dosent assume any knowlage of relativity (so I must be overcomplicating it)

    any input would be appriciated!!
     
  2. jcsd
  3. Apr 19, 2012 #2
    Never mind, that is the answer they were looking for.
     
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