A copper wire of radius 0.5 mm at the Earth's equator carries a current.Assuming that the magnetic field of the earth of 0.5 G at the equator is parallel to the surface of the Earth and that the current flows to the east,what is the current necessary in the wire to allow it to levitate?

Question

Quizplus · Accepted Answer

The magnetic force must equal the gravitational force for levitation. Using the formula F = BIL (magnetic force) and equating it to mg (gravitational force), and given the Earth's magnetic field (B = 0.5 G = 0.5 x 10^-4 T), the mass per unit length (m/L) can be calculated using the density of copper and the wire's cross-sectional area. Solving for the current (I) gives approximately 1370 A.

A Copper Wire of Radius 0