posted by Justin on .
Suppose a straight 1.25 mm diameter copper wire could just "float" horizontally in air because of the force of the Earth's magnetic field B, which is horizontal, perpendicular to the wire, and of magnitude 5.00 multiplied by 10-5 T. What current would the wire carry?
I want to use the formula F=IlBsin(theta); but am confused as to what F would be, what does just "float" horizontally imply?
I won't be able to answer exactly, but we know that the net force on the wire would be zero. So it means that Force(gravity) = Force(magnetic)
Hey, I thought about it a little more and I have your answer here.
B = mu(naught) * I / (2pi * r)
You can solve for I in that equation. I hope it looks familiar.
mu(naught) = vacuum permeability
I = current
r = radius of wire