Question

A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? 

Answer 1

Answer

Force required to lift the wire

$IlB=mg\\B=\frac{mg}{Il}\\$

By putting the value

$B=\frac{(0.500)(9.8)}{2}=2.45T$

Direction should be inside perpendicular to the plane of paper