2020
DOI: 10.1029/2020ja028153
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Euler Potentials for the Earth Magnetic Field With Field‐Aligned Currents

Abstract: Euler potentials of the Earth's dipole magnetic field with field‐aligned currents are found. The results generalize ones obtained earlier for systems of plane currents parallel to an ambient uniform magnetic field. A real current system corresponding to an observed increase of the magnetic field measured by a low‐altitude spacecraft is used in this modeling. Parameters of charged particle trajectories, such as length and period, are determined for typical and strongly disturbed conditions. A particle distribut… Show more

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Cited by 5 publications
(9 citation statements)
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“…Romashets and Vandas (2020) treated as an example a superstorm case of a Carrington‐Event caliber (Gonzalez et al., 2011; Tsurutani et al., 2003) with an estimated total current I 0 = 30 MA. Here are some other estimated values for the Carrington Event: the average speed of the ejectum about 2,400 km s −1 , E y ≈ 200 mV m −1 , D st ≈ −1,100 nT, and Δ H ≈ 1,600 nT.…”
Section: Resultsmentioning
confidence: 99%
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“…Romashets and Vandas (2020) treated as an example a superstorm case of a Carrington‐Event caliber (Gonzalez et al., 2011; Tsurutani et al., 2003) with an estimated total current I 0 = 30 MA. Here are some other estimated values for the Carrington Event: the average speed of the ejectum about 2,400 km s −1 , E y ≈ 200 mV m −1 , D st ≈ −1,100 nT, and Δ H ≈ 1,600 nT.…”
Section: Resultsmentioning
confidence: 99%
“…The parameters of the RC yield the value of Δ H ≈ 20 nT as a contribution to the equatorial horizontal surface field. Our g function corresponds to slightly disturbed conditions with the interplanetary electric field E y = 4 mV m −1 (Korth et al., 2010; Romashets & Vandas, 2020) and the total current I 0 ≈ 4.5 MA (as the total current we take an average of unsigned upward and downward currents). For our ordinary case, we consider E y = 2 mV m −1 which is characterized by I 0 = 2 MA (Korth et al., 2010), so our g function is scaled by 0.45 to yield approximately this value.…”
Section: Resultsmentioning
confidence: 99%
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