Awire having a mass per unit length of 0.500 g/cm carries a2.00-a current horizontally to the south. what are the directionand magnitude of the minimum magnetic field needed to lift thiswire vertically upward?

Explanation:

Given that,

The mass per unit length of the wire,

Current in the wire, A = 2 A (due south)

To find,

The direction and magnitude of the minimum magnetic field.

Solution,

When the wire needed to lift this wire vertically upward, the magnetic force is balanced by its weight.

B = 0.245 T

The direction of magnetic field is given by the right hand rule and it is in clockwise direction around the wire.