The Earth\'s magnetic field is modeled by a magnetic dipole with magnetic moment

Question

The Earth's magnetic field is modeled by a magnetic dipole with magnetic moment pointing in the direction near the South Pole. In the magnetic equatorial plane the strength of the magnetic field is given as 4rr with A,3.1210-5T and R. = 6370 km. The Earth is surrounded by belts of ener- getic charged particles, e. the Van Allen belts. The inner belt created by cosmic rays extends from 1.2 Ro to 3 Ro above the Earth atmosphere, and consists of pro- tons with energies around 100 MeV and electrons with energies of 1 MeV. The outer belt created by the solar wind ex- tends from 4 R to 11 R above the Earth atmosphere and consists of electrons with energies in the range of 0.1-10 Mev and protons with energies around 100 MeV Considering the charges gyration motion in the equatorial plane, which parti- cle is the most susceptible to be lost in the Earth's atmosphere? Hints: Calculate the gyration (Larmor) radius and associ- ate particle energy with perpendicular speed

Explanation / Answer

given

for, inner belt
radius range = 1.2Re to 3Re ( where RFe is radius if earth)
consists of protons of energies, Ep = 100 MeV
electron energies, Ee = 1 MeV

For outer belt
radius range = 4Re to 11 Re above the surface of earth
consists of electrons with energies E'e = 1 MeV
Proton energies E'p = 0.1 - 10 MeV

now for a particle of mass m in magnetic field B, gyration radius is r and speed is v
then
qvB = mv^2/r
r = mv/qB = sqrot(2m^2v^2/2)/qB = sqroot(2mE)/qB ( where E is the energy)
also, B = Bo(Re/r)^3
Bo = 3.12*10^-5 T
Re = 6370,000 m

hence
r = sqroot(2mE)(r/Re)^3/qBo

for inner belt, for proton
r = sqroot(2*1.6*10^(-27)*100*10^6*1.6*10^-19)(1.2)^3/1.6*10^-19*3.12*10^-5 = 61921.88245384 m
for electron
r = sqroot(2*9.1*10^-31*10^6*1.6*10^-19)(1.2)^3/1.6*10^-19*3.12*10^-5 = 186.79482 m

for outer belt
for proton
r = sqrot(2*1.6*10^(-27)*10*10^6*1.6*10^-19)(5)^3/1.6*10^-19*3.12*10^-5 = 1416479.9302 m
for electron
r = sqrt(2*9.1*10^-31*10^6*1.6*10^-19)(5^3)/1.6*10^-19*3.12*10^-5 = 13512.3573349679 m

hence we see that the highest gyration radius is that of proton is outer layer, hence protons in outer layer can easily get lost in the atmosphere of earth

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