Question

A square loop of wire, 15 cm on a side, carries a current of 0.50 A. It stands vertically, such that its normal is parallel to a uniform horizontal magnetic field that exists in a region in which half of the loop is immersed, with the other half below the region. The mass per unit length of the wire is 5.6 g/m. What is the strength of the magnetic field if the loop simply hangs there?

Answer 1

BIL= mg or
B*0.50*0.15= 0.0056*0.15*4*9.81 or
B = (0.0056*4*9.81)/0.50 = 0.439 or 0.44 T

Answer 2

your calc is ok