A superconducting current loop could perhaps do the trick. Let's consider a very simplified analysis for creating a protective magnetic field for Venus.
We're going to be essentially imitating earth's magnetic field.
I don't know much about the solar wind, so I'll assume the strength varies per the inverse square law. Further, I assume the magnetic field strength required for planetary protection is directly proportional to solar wind strength.
Venus orbits at 0.72 AU from the sun. Then, in order to have the same protection on Venus against the solar wind as Earth does now, the magnetic field should be $(1/0.72)^2 = 1.93$ times as strong as Earth's field.
The earth's magnetic field is usually approximated as a dipole. The magnetic field at a given radius R and magnetic latitude theta is given by
$\left|B\right| = \frac{B_0}{R^3} \sqrt{1 + 3\sin^2 \theta }$ Tesla
Planetary magnetic fields vary across radius and latitude. For the purpose of this analysis, let us take as reference the value of the magnetic field at the pole. In other words, the target is to get Venus' polar magnetic field strong enough, and hope the rest of the field is also strong enough, since that kinda works for earth.
Plugging $B_0 = 3.12 \times 10^{-5} T $, $R=1$ and $\theta = 90 °$, we get a neat figure of $B_{pole-earth} = 6.24 \times 10^{-5} T$. The required $B_{pole-venus}$ then comes out to be $1.2 \times 10^{-4} T$.
Again, to simplify calculations, I'm going to assume a single equatorial loop of superconductor resulting in the magnetic poles coinciding with the rotational poles. The radius of Venus is 6052 km, and thus the length of the superconducting cable is around $38000 km$.
Now the magnetic field of a single circular loop of current, at a point on the axis of the circle is given by
$$B = \frac{\mu_0}{4 \pi} \frac{2 \pi R^2 I}{(z^2 + R^2)^{3/2}} T$$
Since the pole is at ground level, the expression simplifies quite a bit to yield
$$B_{pole-venus} = \frac{\mu_0}{4 \pi} \frac{pi I}{R} T$$
Solving for current, we get
$$I = B_{pole-venus} \frac{R}{\pi} \frac{4 \pi}{\mu_0} A$$
yielding a value of 2.3 GA . Huge, but surely not universe wrecking.
The inductance of a coil of wire does not have an accurate closed-form expression even in the ideal scenario, so we will use Kirchhoff's approximation given by
$$L = 4 \pi R (\ln{\frac{8 R}{p}} - 1.75)$$
Let us assume a generous conductor of $p = 3 m$ radius. We get an inductance of $L = 415 MH$ - an enormous value, but entirely expected given the planetary scale.
Finally, the energy in a loop of current is given by $E = \frac{1}{2} L I^2$ , giving a value of $1.12 \times 10^{27} J$. This is a pretty big number - obviously we would want to generate this from Venus' abundant solar energy.
Venus' insolation, another figure that definitely has an inverse square law variation, should be around ${1/0.72}^2 = 1.93$ times the earth's insolation which is the solar constant, equal to $1362 W/m^2$. This gives us a very encouraging $P = 2627 W/m^2$ as Venus' insolation, and $P \pi R^2 = 3.023 \times 10^{17} W$ as Venus' total solar budget. Assuming 100% efficiency shows us that it would require around 37 years to fully charge up the magnetic field. Not trivial at all, but certainly practically doable on a timescale of millenia.
Addendum : Since both solar wind (presumably) and insolation vary as the inverse square of distance to the sun, the number of years of capturing solar energy should be independent of distance from the sun. The inductance of the loop however is proportional to $R \, \text{ln}R$, and current required is proportional to $R$, so dependence of energy required on the radius ($0.5 L I^2$)is more than cubic. As suggested by RiskyScientist below, running the numbers for Callisto shows it would require $2.4 \times 10^{22} J$ or $0.26 y = 95$ days for solar charging of the magnetosphere generator.
The major assumptions here are :
- Dipole approximation
- Polar magnetic field strength
- Inverse square law variation in solar wind
- Necessity/sufficiency of earth's magnetic field for protection against atmospheric stripping
- Single current loop
Most of these serve to decrease the required energy, so there should be some hope here.
The magentic field at the surface of the conductor is $154 T$ which is far greater than any human-created field. The superconducting cable would occupy a volume of at least $38000 \pi p^2 = 1.075 km^3$. weighing around $8.6 \times 10^{12} kg$ (assuming density of $8000 kg/m^3$).
However, such a loop offers a critical advantage should industrial civilisation ever arise : electric storage AND transmission across the planet, with basically unlimited discharge and charge rates.
