Van Allen Belt

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MoreLater

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Estimating Energy Stored in the Geomagnetic Field

The magnetic field of the earth is complicated, not a simple dipole, but it can be approximated that way.

B_r = - 2 B_0 \left( { R_E } \over r \right)^3 \cos \theta

B_{\theta} = - B_0 \left( { R_E } \over r \right)^3 \sin \theta

dE = { { {B_r}^2 + {B_{\theta}}^2 } \over { 2 {\mu}_0 } } ~ dV

dV = 2 \pi r^2 \sin \theta ~ dr ~ d\theta

dE = { { \left( 2 B_0 \left( R_E \over r \right)^3 \cos \theta \right)^2 + \left( B_0 \left( R_E \over r \right)^3 \sin \theta \right)^2 } \over { 2 {\mu}_0 } } ~ 2 \pi r^2 \sin \theta ~ dr ~ d\theta

dE = { { \pi {B_0}^2 {R_E}^6 } \over { { {\mu}_0 } r^4 } } ( 3 \cos^2 \theta + 1 ) \sin \theta ~ dr ~ d\theta

The total energy of the magnetic field is the double integral from \theta = 0 to \theta = \pi and from R = R_0 to infinity. Assuming the earth's mantle has a permeability of {\mu}_0 and the earth's iron core has an infinite permeability (and thus no appreciable magnetic energy storage), R_0 is the radius of the core, not the earth's surface, about 3500 km.

Since r and \theta and the scaling terms are independent, we separate the integrals:

E = \left( { \pi {B_0}^2 {R_E}^6 } \over { {\mu}_0 } \right) ~ \left( \int^{\infty}_{R_0} { dr \over { r^4 } } \right) ~ \left( \int_0^{\pi} ( 3 \cos^2 \theta + 1 ) \sin \theta ~ d\theta \right)

E = \left( { \pi {B_0}^2 {R_E}^6 } \over { {\mu}_0 } \right) ~ \left( 1 \over { 3 {R_0}^3 } \right) ~ \left( 3 \times { 2 \over 3 } + 2 \right)

E = { { 4 \pi {B_0}^2 {R_E}^6 } \over { 3 {\mu}_0 {R_0}^3 } } = 2 \left( { 1 \over { 2 {\mu}_0 } } \left( B_0 \left( {R_E} \over {R_0} \right)^3 \right)^2 \right) \left( { { 4 \pi } \over 3 } {R_0}^3 \right)

The B field at the surface of the core is B_0 \left( {R_E} / {R_0} \right)^3 . The total energy is twice the energy density at the surface of the core { B^2 / { 2 {\mu}_0 } } times the volume of the core { { 4 \pi } \over 3 } {R_0}^3 .

• B_0 = 3.12E-5 Tesla

• R_E = 6378000 meters

• R_0 = 3500000 meters

• \mu_0 = 4e-7 π meter³ Tesla² Joule^-1

Above the core at 3500km, E = 5.1E18 Joules = 160 gigawatt-years. Above the earth's surface, E = 8.4E17 Joules = 26 gigawatt-years.

The earth's magnetic field is decreasing at about 6% per century, so the energy is decreasing 0.12% per year, about 200 megawatts above the core.

In 1960, Dessler and Vestine estimated that the maximum stored energy of particles in the van Allen belt must be less than 6e15 Joules. If this amount of energy was expended filling and emptying the van Allen belt once per year, that would also be about 200 megawatts. It is interesting that these are about the same magnitude. Both are 1.6 parts per billion of the solar energy absorbed and reradiated by the earth, and about 1/3 of the power of the solar wind times the earth's area.

I would like to learn the actual numbers for the stored energy and the particle power flux of the van Allen belt.

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Gyration Frequency at the Equator

B = - B_0 \left( { R_E } \over r \right)^3

freq = B q / 2 \pi m . . . . . for v << c

freq = B q / 2 \pi { m_0 } ( 1 + { E_k } / {E_0} ) . . . . . E_k is the particle energy in electron volts (not sure about energy scaling)

The rest energy E_0 of a proton is 938 MeV and particle energies range from 100 keV to over 400 MeV. The rest energy of an electrom is 511 keV with particle energies up to 10 MeV.