Question

 Imagine a wire with linear mass density 2.40 g/m encircling the Earth at its magnetic equator, where the field is modeled as having the uniform value 28.0 mT horizontally north. What magnitude and direction of the current in the wire will keep the wire levitated immediately above the ground?

Answer 1

Answer

The magnetic and gravitational forces must balance.Therefore, it is necessary to have

$IlB=mg$

$I=\frac{mg}{lB}$

$I=\frac{(2.40\times 10^{-3})(9.8) }{28\times10^{-6}}\\I=840A$

Direction of current should be eastward.