What is the effect of Earth’s magnetic field on airplanes? As an airplane accelerates through the earth’s magnetic field, it experiences a changing flux $d\phi \over dt$ and a potential difference is induced along its wings.
Given that the wings are made of metal and act as a continuous conductor, current flows along the wings in a direction perpendicular to the vertical component of the earth’s magnetic field. Does this mean that the airplane would experience a Lorentz force?
What other effects could the building up of charge have on an airplane, and what difference would it make if the airplane’s wings were made of an insulating material?
Also, what would happen to an airplane in a thunderstorm?

Update: I discovered that airplanes have something called “discharge wicks” (picture below), which are designed to counter this problem. However, the buildup of static charge seems to be more significant than the buildup of charge due to the effect I described, so these are probably used to dissipate static charges.

 A: Nothing would really happen that may be of any significant consideration at least for normal sized passenger planes from private jets to Boeing.
Lets see two major steps, I have considered landing and take off similar, and flight as separate.
First Lets consider take off/landing

The blue lines are supposed to be magnetic field lines of earth, they are not normally so straight , but this would be a good approximation.
Lets assume a $30^o$ angle between the field and plane's body, the earth's magnetic field being $0.65G$ at maximum that is a mere $6.5 \times 10^{-5} T$. While The largest wingspan of an aeroplane till now is $97.51m$ we'll consider is $100m$ just for our ease.
So even if an aeroplane had been a square of such a large side, its area would be $10^4m^2$, and a decent $30^o$ angle with magnetic field would give you a feeble $0.56weber$ of flux.
Even if it reaches this much change of flux in just one second which it doesn't, then we would have a potential difference of about $0.5V$ across the wings, now with the resistance the current would just become too negligible and already these numbers are too large because we considered the plane a square and a constant angle of $30^o$ that the area vector had with the horizontal while its not like that and is much over $45^o$
Thus concluding that at least while take-off and landing we do not need to care about any induced currents and voltages
Now lets consider the flight

This is a very typical route for a flight, But still at this point you can rotate the plane in all directions and still get a near $90^o$ angle between the magnetic field and area vector in nearly all places where aeroplanes can land and/or take off! Same calculations with just a difference of angles when applied in this case give even smaller and much less noticeable currents.
Now you might think of Antarctica where the lines are pretty much perpendicular, but even for that if you calculate for $90^o$ you will get negligible results only.
Now I am going to take an impossible scenario just to show that any real life flux changes are easily negligible. The minimum that a flux can be is $0weber$ and the maximum it can be even for the largest square sheet aeroplane is $0.65weber$, even if any plane makes this amount of flux change in a mere second the potential difference developed would only be $0.65V$ this would give negligible currents and hence can be neglected over all.
Now to the case of thunderstorms, the maximum trouble that any aeroplane would receive is well known to arise from air turbulence, during thunderstorms these would create a danger far greater than any lightning bolts etc can produce, lets take a look.
During a typical thunderstorm the potential difference between the clouds and earth is more than $1\times 10^8$, the total resistance of the plane would be negligible as compared to that of the air, the major current that runs between the cloud and earth to establish a channel for the discharge is of the order of $100A$, now lets calculate the energy density that will arise in the plane due to Joules heating,
The energy generated would be for the square sheet plane $U = I^2 \frac{\rho l}{A}$ and energy per unit area would be $$u = \frac{I^2 \rho l}{A^3}$$
For our square plane, where sides are equal it would result in $$u = \frac{I^2 \rho}{s^5}$$
Lets assume an impossible current of $1000A$ for our square plane of $100m$ side, and check again$$u=\rho \times 10^{-4}$$ clearly negligible energy density
For a small plane, like the drones we have nowadays or even any thing with wingspan less than $10m$ the energy density increases rapidly, but I don't think those planes are operational under such harsh weather conditions!
PS: I am not 100% sure about the thunderstorm analysis, but I am sure about the analysis for magnetic flux
Addendum : the "discharge wicks" you show in your picture are 100% for static discharge with as simple a reason as their other name is "static discharger". I won't increase the length of the answer explaining their working etc but you can easily check it out here or simply do a google search of "discharge wick".
