The earth's magnetic field is roughly that of a dipole, tilted from the rotational axis by about 10 degrees at the moment, but it's actually somewhat lumpy/bumpy. I plotted the three components (x, y, z = North, East, Down) as a function of longitude, for a LEO at 300km above the equator using the World Magnetic Model WMM2015 from NOAA > NESDIS > NCEI (formerly NGDC) > Geomagnetism using the menu settings and script shown below. The field is roughly 0.3 Gauss pointing North, but there are "bumps" available in the other two directions of about +/- 0.1 Gauss. The full description of the model is here. I also plotted the fields from a dipole approximation.
You can read more about the Earth's magnetic field here for example.
The force on a current in a field follows the Laplace right hand rule (I never knew he had one!) so for a satellite in orbit, using North, East, and Down as the three axes, to boost an equatorial LEO satellite higher, I think we need an East-West force (accelerate the speed), which means either an Up-Down rod using the strongest North-South field, or - as you mention in the question - a North-South rod using the weaker Up-Down field.
The magnitude of the force (Newtons) is equal to the length of the rod (meters) times the field strength (Tesla). So 1 Amp in 0.3 gauss gives 3E-5 Newtons. I'm not sure how the ends of the rod work - but if you assume a 1.5 eV work function for the electron and ion guns (and otherwise no losses), that's 1E-05 Newtons per Watt, or 100kW per Newton - very roughly the same scale as plasma thrusters.
So in order to carry a current, the rods must emit negative charge (electrons or ions) from one end and positive charge (ions) from the other. But I have no idea where an Amp of ions can come from! If this technology is supposed to take advantage of gas that it encounters in its orbit, it will probably be microamps or nanoamps.
I have a sneaking suspicion that the drag from a given gas pressure may be worse than the thrust from a rod carrying a current produced by ionizing the gas at that pressure, but that's a different question.
If instead it carries it's own gas to ionize, then I wonder if maybe a conventional ion thruster is a better choice?
Let me know if this is helpful or if I've missed the point.
Some scrappy Python to give an idea what I did:
import numpy as np
import matplotlib.pyplot as plt
fnames = ["equator_X.csv", "equator_Y.csv", "equator_Z.csv"]
directions = ["North", "East", "Down"]
# http://www.ngdc.noaa.gov/geomag-web/#igrfgrid
fields = []
for fname in fnames:
with open(fname, "r") as infile:
lines = [line.rstrip('\n') for line in infile]
print lines[13]
lon, hkm, nT = np.array([[float(x) for x in line.split(',')]
for line in lines if "#" not in line]).T[2:5]
fields.append(nT * 1E-09 * 1E+04) # nT to T to gauss
fields = np.vstack(tuple(fields))
total = np.sqrt((fields**2).sum(axis=0))
fields = np.vstack((fields, total))
directions += ["Total"]
lon = np.array(lon)
zlon = np.zeros_like(lon)
# Now get simple dipole approximation
# Geomagnetic poles - axis of "best dipole approximation" to earth's real field
gmlat, gmlon = 80.3, -72.6
rads, degs = np.pi/180., 180./np.pi
mlon = lon - gmlon - 90
sinmlat = -np.sin(rads*mlon)*np.sin(rads*(90-gmlat))
mlat = np.arcsin(sinmlat)
plt.figure()
plt.plot(lon, degs*mlat)
plt.show()
def earth_dipole(r_km, maglat):
"""from: https://en.wikipedia.org/wiki/Dipole_model_of_the_Earth%27s_magnetic_field"""
a_km = 6371. # km
B0 = 3.12E-5 # T
R = r_km / a_km # in earth radii
Br = -2. * (B0 / R**3) * np.sin(maglat)
Bth = (B0 / R**3) * np.cos(maglat)
return Br, Bth
r_km = (6371. + 300)*np.ones_like(mlat)
Bz_dip, Bx_dip = earth_dipole(r_km, mlat)
fields = np.vstack((fields, 1E+04*Bx_dip, -1E+04*Bz_dip)) # Tesla to Gauss
directions += ["North(dipole)", "Down(dipole)"]
plt.figure()
for field, label in zip(fields, directions):
plt.plot(lon, field, label=label)
plt.plot(lon, zlon, '-k')
plt.legend(loc='center right', shadow=False, fontsize = 'large')
plt.xlim(lon.min(), lon.max())
plt.title("Magnetic field 300km above Earth Equator", fontsize=18)
plt.show()
