Acurrent-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

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:


Equalizing the two forces, we have

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

No it doesnt because it can be any number.

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way