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Homework Statement
A wire of mass 3.0g and length 60cm is placed on a table parallel to the East-West direction. If a current flows through the wire from West to East, how large must the current be to cause the wire to lift off the surface of the table? (The Earth's magnetic field has strength 5.3x10^(-5) Tesla.)
Homework Equations
F = ILBsin(q) (magnetic force on a current carrying conductor).
The Attempt at a Solution
The magnitude of the force generated by the interaction between the current and the magnetic field must be at least as large as the force due to gravity acting on the wire. The right-hand rule tells me that if the current flows through the wire from West to East, the magnetic force must act vertically upwards. Thus I'm looking for the value of the current, I, that satisfies
ILBsin(90) > mg
Rearranging this gives me
I > mg / LB = (2.94x10^(-2)) / (0.60 x 5.3 x 10^(-5))
I calculate this to be 924.5 Amps. That seems like a huge current. Where am I going wrong?