Question

A wire having a mass per unit length of 0.440 g/cm carries a 1.50-A current horizontally to the south. (a) What is the direction of the minimum magnetic field needed to lift this wire vertically upward?

Answer 1

B = 0.287 Tesla

According to thumb rule the direction of magnet is to the east.

Step-by-step explanation:

F = BiL sin∅

note that F = mg

B = magnetic field

i = current

therefore,

mg = BiLsin∅

Make B subject of the formula

B = mg/iLsin∅

m/L = mass per length of the wire and it is given in gram per centimeter . It should be converted to kg per meter.

m/L = 0.440 g/cm

convert to kg/m = 0.440 × 0.001 kg × 100 m = 0.044 kg/m

B = 9.8 × 0.044 / sin 90 × 1.50 × 1

B = 0.4312/ 1.50

B = 0.28746666666

B = 0.287 Tesla

According to thumb rule the direction of magnet is to the east.