Question

A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.94 t. if the wire has a length of 2.8 m and a mass of 0.63 kg, determine the minimum current needed to levitate the wire.

Answer 1

The magnetic force acting on the wire (assuming the wire is perpendicular to the magnetic field) is

`F=ILB`
where I is the current in the wire, L its length, B the intensity of the magnetic field.

In order to levitate the field, this force must be at least equal to the weight of the wire, which is:

`F=mg`

Equalizing the two forces, we have

`mg=ILB`
that we can solve to find I, the minimum current in the wire required to levitate it:

`I= (mg)/(LB)= ((0.63 kg)(9.81 m/s^2))/((2.8 m)(0.94 T)) =2.35 A`