Suppose a straight 1.00-mm-diameter copper wire could just “float” horizontally in air because of
the force due to the Earth’s magnetic field, which is horizontal, perpendicular to the wire, and of magnitude 5.03*10^-5T .What current would the wire carry? Does the answer seem feasible? Explain briefly. (ρCU = 8.9*10^3Kg/ m^3).

Question

Suppose a straight 1.00-mm-diameter copper wire could just “float” horizontally in air because of 
the force due to the Earth’s magnetic field, which is horizontal, perpendicular to the wire, and of magnitude 5.03*10^-5T .What current would the wire carry? Does the answer seem feasible? Explain briefly. (ρCU = 8.9*10^3Kg/ m^3).

jamesj · Answer

What is the formula for the force on a wiring carrying a current $ I $ perpendicular to a magnetic field $ B $?

jamesj · Answer

If you don't know, you will probably enjoy watching the first 20 minutes of this lecture where this is explained and demonstrated: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-11-magnetic-field-and-lorentz-force/

anonymous · Answer

formula? i think F=ILB sin(tetha)

jamesj · Answer

Yes (and if you're not sure, watch the video!).  Now you need to find the current in the wire such that the Lorentz force is equal and opposite to the gravitational force.

jamesj · Answer

Figured this out?