Current required to lift a wire in a magnetic field

[uibble]

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?

 

[TSny]

Welcome to PF!

Your work looks correct. Yes, it would take a very large current.

Your calculation assumes that the Earth's B field is parallel to the ground where this wire is located. I think that's what you were supposed to assume. But in most places on the Earth the Earth's magnetic field is not parallel to the ground. It has a "dip" angle that gets larger as you move toward the Earth's magnetic poles.

 

[uibble]

Yes, that observation about the direction of the magnetic field was given in the question and I should have stated it in the template. Sorry!

I'm new to all of this and I'm surprised that it takes such a large current to move such a light wire. I'm used to 13A fuses and so forth on domestic appliances here in the UK and so a 1000A current seems enormous to me.

Thanks for your help.

 

[TSny]

Yes, 1000 A is not common. But neither is levitating a wire using the Earth's magnetic field