Variations in the geomagnetic field penetrate quite deeply into the earth and sea and may be a feasible energy-harvesting source, especially in the case of another Carrington event.
But how feasible is it? The Halloween geomagnetic storm of 2003 provoked planetwide variations of some 10 μT, out of a total 25–65 μT, with larger variations toward the poles, while typical daily variation is about 25 nT, "with variations over a few seconds of typically around 1 nT".
A rate of change of 1 nT (1 nWb/m²) per second is one nanovolt per square meter inside a single-turn inductive loop; even if we have 1000 turns, that's still only a microvolt per square meter. You could imagine stepping that up to a usable 0.4 volts or so with a chain of transformers; three 80:1 step-up transformers would probably serve. If you were trying to get 1 μW, which is a challenging but achievable level for modern energy-harvesting machinery to survive from, you'd need 1 A m² at the 1000 turns level or 1000 A m² at the 1-turn level.
One problem is that you can't get an arbitrarily large amount of power out of an inductor in a varying magnetic field just by winding more turns around it; at some point the current that's being induced will cancel out most of the magnetic field that would otherwise exist, and you'll stop getting more power.
It turns out that the energy of the magnetic field is ½B²/μ, so in empty space the difference between 30 nT and 31 nT is 24 pJ/m³, and we probably can't capture more than half of that for impedance-matching reasons, so we're probably limited to a few picowatts per cubic meter. (I don't think using higher-permeability materials helps here; the μ is on the wrong side of that fraction.)
A further complicating factor is that, if you're using conventional
conductors, you probably need to use ridiculously thick wires.
Suppose your primary coil is 1000 m², the size of a big-box store
façade, and you're getting 100 μV out of 100 turns around it.
(Remember, unless you're near the poles, it has to be oriented
north-south.) To get 1 μW you need 10 mA, and if you want no more
than 90% of the energy to be lost in heating the primary coil, you
need a voltage drop of no more than 90 μV, 900 μV per turn. So your
130-meter-long coil needs to be no more than 90 mΩ per turn, which
requires 3-gauge copper wire (~awg(2*~circlearea(130 m /
copperconductivity 90 milliohm)) in units(1) gives about 3.3), which
is is normally used for carrying 150 amps or more and costs abut US$3
per meter.
(I'm not entirely sure but I think you might need to enclose a larger area to grab enough energy. This helps a little with the wire thickness because you can enclose a larger area per unit length of wire, but the wire is still ridiculously thick.)