Magnetizing Mars
... sorry, no
Superconductors do not make a stupid idea into a super idea.
Some talk about terraforming Mars, turning this dead ball of rock into a miniature Earth. Mars has no magnetic field, so an artificial atmosphere (somehow) would get stripped away by solar UV and the much weaker gravity. How would a terraformed Mars 2 compared to Earth and Mars 1 today?

units 
Earth 
Mars 1 
Mars 2 
Semimajor Axis 
AU 
1.0 
1.52 
1.52 
Mass 
kg 
5.972e24 
6.42e23 
6.42e23 
Radius 
km 
6371 
3389.5 
3389.5 
Gravity 
m/s² 
9.81 
3.71 
3.71 
Planet Density 
kg/m³ 
5510 
3930 
3930 
Escape Velocity 
m/s 
11200 
5030 
5030 
Escape Energy 
MJ/kg 
63 
12.6 
12.6 
Dipole moment 
Am² 
7.94e22 
~0 
3.4e22 
Iron Core radius (est) 
km 
3400 
1800 
1800 
Atmosphere @surface: 

Mass 
kg 
5.15e18 
2.5e16 
4e18 
Temperature 
K 
288 
210 
288 
Density 
kg/m³ 
1.2 
0.015 
1.2 
Pressure 
kPa 
101 
0.64 
101 
Pressure Scale Height 
km 
8.5 
11.5 
22.8 
Temperature Lapse Rate 
K/km 
6.5 
2.2 
? 
Primary Components 

N₂, O₂ 
CO₂ 
N₂, O₂ 
Molecular Weight 
g/mol 
30.5 
44.0 
30.5 
Molecular Weight 
kg 
1.54e24 
2.23e24 
1.54e24 
Thermal Velocity √3kT/2m 
m/s 
485 
345 
485 
(V_{esc}/V_{thrm})² 

533 
213 
108 
CO₂ mass 
kg 
3.2e15 
2.4e16 
? 
The most disturbing problem with Mars 2 is that a hotter, lowermass gas atoms are more likely to escape in the weak Mars gravity field ... (V_{esc}/V_{thrm})² is halved compared to the colder, higher molecular weight current Mars 1 atmosphere.
But we'll ignore that, and also ignore how we can create enough greenhouse effect to heat Mars to Earth temperatures. Terraforming in science fiction novels is done with a few keystrokes. 
The goal is making an Earthequivalent magnetic field with an equatorial coil.
The shielding effect of the magnetic field is not due to surface field strength, but the dipole of the planetary magnetosphere compared to the solar wind pressure. Presume the solar wind density is inversesquare  (1/1.52)² or 0.433 less dense than at Earth, so the artificial Martian dipole may be 0.433*7.94e22 Am² or 3.4e22 Am².
The potentially magnetic core of Mars is small; it will reduce the reluctance of the path a bit, but the magnetic field we are concerned with goes far into space, and most of the field will be close to the equatorial coil, not the core of Mars.
The dipole field of a coil is the circumferential current times the area; the martian equatorial radius is R = 3396200 m, and the area of the coil is the \pi R^2 = 3.63e13 m², so the current is 940 megaAmps. By Ampere's law for a round wire with radius r , the B field at the edge of the wire is ~ B = \mu_0 I / 2 \pi r ~ where ~ \mu_0 = 4e7 \pi ~, hence B = 2e7 I/r or r = 2e7 I/B . For a maximum field B at the edge of the wire of 20 Tesla, r = 2e7 * 9.4e8 / 20 = 9.4 meters. Our superconducting wire (which may be a thin hollow cylinder of superconductor around a structural core) is about 19 meters (62 feet) in diameter. The cross section might be oval shaped; however, the field will concentrate around the tightest radius, so the total surface area must be larger to limit the maximum field at the radius to 20 Tesla.
20 Tesla is the highest field ever generated in stacked magnetic fields in small lab experiments for a few days at very cold temperatures. In Real Life, The 4.3 km diameter Large Hadron Collider runs magnets at 7.7 Tesla in 96 tonnes of 1.9 Kelvin liquid helium; one small weld failure released as much energy as a high speed train collision, and took the collider offline for 14 months.
20 Tesla on the inside surface of the coil, almost that much on the outside; the magnetic pressure on the coil surface will be B^2/2 \mu_0 = 160 MPa, or the pressure at 1.6 kilometers deep in the ocean.
How much energy is stored in the field? Using an online calculator, the inductance is 27.5 Henries. The stored energy is ½ L I² = 0.5 * 27.5 * (9.4e8)² = 1.2e19 Joules or 3.4e6 GWhrs or 2900 megatonnes explosive equivalent.
Solar insolation on Mars is 590 W/m², and a 40% efficient solar panel on the Martian equator might collect an average of 60 W/m² of electricity. An earth year is 31.56 Msec. so a square meter produces about 1.89e9 J per year. A solar array area of 1000 square kilometers (60 GW average, 230 GW peak noontime at Mars perigee), will produce enough energy to charge the coil in about 6 years.
However  if even a tiny portion of the superconductor surface fails, that area will stop conducting and make I²R resistive heat instead. The superconductor system will need massive cooling systems, and a high thermal and electrical conductivity substrate to bypass the current and conduct away the resistive heat. If that fails ... a complete failure and a GYNORMOUS electrical arc, dissipating as much energy as 2900 megatonnes of nuclear weapons, starting near the disruption, with a shock wave launched around the planet in a few seconds. The incandescent arc will climb rapidly into the sky, like a nuclear fireball ringing the planet, probably emitting high energy xrays.
This isn't terraforming  it is hellonMarsforming. Something that huge running for that many years must be perfect, forever.
Science fiction writers can type any numbers they want into their stories, and their science and engineering ignorant readers will accept them if the story is entertaining. However, entertainment won't terraform Mars.