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.
The magnetic force acting on the wire (assuming the wire is perpendicular to the magnetic field) is [tex]F=ILB[/tex] 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: [tex]F=mg[/tex]
Equalizing the two forces, we have [tex]mg=ILB[/tex] that we can solve to find I, the minimum current in the wire required to levitate it: [tex]I= \frac{mg}{LB}= \frac{(0.63 kg)(9.81 m/s^2)}{(2.8 m)(0.94 T)} =2.35 A [/tex]
