Question

A wire having a mass per unit length of 0.540 g/cm carries a 1.60 -A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

____?_____T

Answer 1

m/L = mass per unit length = 0.54 x 10^-3 x 100 kg/m = 0.054 kg/m

i = current = 1.60 A

for lifting the wire

weight of wire = magnetic force

mg = i BL

(m/L) g = i B

0.054 x 9.8 = (1.60) B

B = 0.331 T

using right hand rule , magnetic field must be in east direction