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I wan to calculate the voltage difference generated by sun EM Field on a wire placed on Moon or Mars surface, but I can't find consistent data about solar EM Field. I found different values:

The magnetic field at an average place on the Sun is around 1 Gauss, about twice as strong as the average field on the surface of Earth (around 0.5 Gauss).

https://www.windows2universe.org/sun/sun_magnetic_field.html

The Sun’s magnetic field During solar minimum, the magnetic field of the Sun looks similar to Earth’s magnetic field. It looks a bit like an ordinary bar magnet with closed lines close to the equator and open field lines near the poles. Scientist call those areas a dipole. The dipole field of the Sun is about as strong as a magnet on a refrigerator (around 50 gauss). The magnetic field of the Earth is about 100 times weaker.

https://www.spaceweatherlive.com/en/help/the-interplanetary-magnetic-field-imf

Magnetic field at Earth orbit

A video simulation of Earth's magnetic field interacting with the (solar) interplanetary magnetic field (IMF) The plasma in the interplanetary medium is also responsible for the strength of the Sun's magnetic field at the orbit of the Earth being over 100 times greater than originally anticipated. If space were a vacuum, then the Sun's magnetic dipole field, about 10^−4 teslas at the surface of the Sun, would reduce with the inverse cube of the distance to about 10−11 teslas. But satellite observations show that it is about 100 times greater at around 10^−9 teslas.

https://en.wikipedia.org/wiki/Interplanetary_magnetic_field

Edit: new source says IMF strength at Earth distance is around 6 nT

Does it exist a quantitative equivalent of this tipical picture, which is just qualitative?

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    $\begingroup$ EMF usually stands for electromotive force in the context of electromagnetism. Best to avoid it in your question. $\endgroup$ – user2705196 Feb 5 at 13:02
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Answer is:

  • Moon: ~6 nanoTesla
  • Mars: >1 nanoTesla

[Tesla] = $\frac{[V][s]}{[m^2]}$ = $\frac{[N]}{[A][m]}$

1 nT = $10^{-9}$ T


This plot from Explorer 35 spacecraft, which orbited Moon in 1967, explains answer: Interplanetary Magnetic Field around the Moon

  • Title: The magnetic fields of Mercury, Mars, and moon
  • Authors: Ness, N. F.
  • Journal: In: Annual review of earth and planetary sciences. Volume 7. (A79-37176 15-42) Palo Alto, Calif., Annual Reviews, Inc., 1979, p. 249-288. Bibliographic Code: 1979AREPS...7..249N

Value of "sun magnetic field", actually named "Interplanetary Magnetic Field - IMF", is around 6 nanoTesla, with its angles continuously changing (Theta = 0°/-40°, Phi = -30°/0°)

Paper also states that there is only slight modification on IMF performed by the Moon.

Subsequent studies revealed the presence of several "magnetic anomalies" in various regions of the Moon, where values of ~100 nT of local EMF were measured.

Further plots:

MEASUREMENTS OF THE PERTURBED-INTERPLANETARY MAGNETIC FIELD IN THE LUNAR WAKE bY Harold E. Taylor", K. W. Behannon and N. F. Ness


This plot shows instead the measured and modeled strength of IMF along the whole Solar System, based on Voyager data: IMF plot along solar system https://solarscience.msfc.nasa.gov/people/suess/Interstellar_Probe/IMF/IMF.html

HP (Helio Pause), BS (Bow Shock) and TS (Termination Shock) are depicted below: BS, TS and HP

This image shows instead planets distances from sun (beware: logarithmic scale!): Planets distances

Average distance of Mars from Sun is 1.5 AU, which means that IMF is > 1nT.

Juno spacecraft measurements:

Juno IMF measurements

The interplanetary magnetic field observed by Juno enroute to Jupiter - Jacob R. Gruesbeck


Further data:

It appears that Open Circuit Voltage (Voc) on a wire moving into an EM field is given by (source 1, source 2):

  • Voc = v * B * L

Hence current:

  • I = v * B * L / Rw

  • v = speed [m/s]

  • B = EM field [Tesla] = $\frac{[V][s]}{[m^2]}$
  • L = wire length [m]
  • Rw = wire resistance [ohm]

and

  • Rw = $\rho * L / A$
  • $\rho$ = resistivity (copper = 1.68E-8 Ohm*m)
  • L = length
  • A = section area

Hence:

$I = \frac{v B L}{\rho L / A}$

$I = \frac{v B A}{\rho}$

Earth+Moon speed around sun: ~30000 m/s (3E+04)

Mars speed around sun: ~24000 m/s (2.4E+04)

It has also to be taken into account the rotation speed of IMF, given by Sun rotation period of 25.38 days (= 2pi radians in 2192832 seconds):

$\omega = 2.865 × 10^{−6} \frac{rad}{s}$

Hence linear speed:

v = ω * R

  • Earth+Moon: R = 1 AU = 1.5E+11 m
  • Mars: R = 1.5 AU = 1.5 * 1.5E+11 m = 2,25E+11 m

$V_{earth} = 2.865 * 10^{-6} * 1.5*10^{11} = 4.3 * 10^{5} [m/s]$

$V_{mars} = 2.865 * 10^{-6} * 1.5*1.5*10^{11} = 6.45 * 10^{5} [m/s]$


Control data

For Italian Tethered Satellite System (TSS-1R on STS-75 Shuttle mision) there are these known data:

  • Satellite orbit height: 300 km
  • Relative speed between orbiter and plasma: 7300 m/s
  • Earth magnetic field: 20000-60000 nT
  • Tether Length = ~20000 m
  • Tether electric resistance: 2000 or 260 Ohm (?)
  • Tether diameter: 2.54 mm
  • Resulting DeltaV = 5000 V or 3500 V
  • Theoretical current achievable: 19A
  • Actual current achieved: 1 A
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