Imagine a very long, uniform wire that has a linear mass density of 0.0085 kg/m
ID: 1310266 • Letter: I
Question
Imagine a very long, uniform wire that has a linear mass density of 0.0085 kg/m and that encircles the Earth at its equator. Assume the Earths magnetic dipole moment is aligned with the Earths rotational axis. The Earths magnetic field is cylindri-cally symmetric (like an ideal bar magnetic). The acceleration of gravity is 9.8 m/s" and the magnetic field of the earth is 1 X 10-5 T. What is the magnitude of the current in the wire that keeps it levitated just above the ground? Answer in units of A The current in the wire goes in the same direction as the Earths spinning motion (West to East). opposite direction as the Earths spinning motion (East to West).Explanation / Answer
The net force on a 1 meter segment of the wire would be the upward force due to the current, countered by the downward force due to gravity.
The upward force on a 1 meter length of the wire will be BIL = (1 x 10-5)(A)(1.0)=(1 x 10-5)A.
The downward force due to gravity will be mg = (0.0085)(9.8) = 0.000867 N. These forces will be equal when the wire "floats" so:
A = 0.000867/0.00001 = 86.7 Amps (approx)
The current is in same direction as the Earth's spinning motion.