Final answer:
The altitude of the Sun at noon on December 21 from the Tropic of Cancer is 44°, considering the declination angle. Magnetic forces affect aircraft with static charges, and the Hall Effect results in a potential difference across the aircraft's wingspan when flying through Earth's magnetic field.
Step-by-step explanation:
The question you've asked pertains to the observation of degrees and altitude for a magnetic course, focusing mainly on celestial navigation and magnetic force effects on aircraft. On December 21, as seen from a place on the Tropic of Cancer, the altitude of the Sun at noon would be 44°. This is because the Sun is located about 23° S of the celestial equator on that date, and the difference in declination between the zenith at the Tropic of Cancer and the Sun's position is 46°, leading to a Sun altitude of 90° - 46° = 44°. Regarding the effect of Earth's magnetic field on an aircraft, a supersonic jet with a static charge flying over the Earth's south magnetic pole in a magnetic field directed straight up would experience a magnetic force, the direction and magnitude of which depend on the charge, velocity, and magnetic field strength and orientation, as per the Lorentz force law.
In terms of the Hall Effect with an airplane flying due north, the west end of the wingspan would be positively charged because the magnetic force acting on the electrons in the wingspan would push them towards the east end, making it negatively charged. The Hall emf can be calculated using the width of the wingspan, the velocity of the aircraft, and the component of the magnetic field perpendicular to the motion of the conductive rod that represents the wingspan.