A wire having a mass per unit length of 0.600 g/cm carries a 1.90 A current horizontally to the south.

(a) What is the direction of the minimum magnetic field needed to lift this wire vertically upward?

(b) What is the magnitude of the minimum magnetic field needed to lift this wire vertically upward?

a) the direction of the magnetic field is east.

b) The magnitude = 0.310 T

Explanation:

Step 1: Data given

Mass per unit length = 0.600 g/cm

current = 1.90 horizontally to the south

Step 2:

The force of gravity = magnetic force

m*g = i*L*B

m/L * g = i*B

B = ((m/L)*g) /i

⇒ with m/L = the mass per length unit = 0.600 g/cm = 0.0600 kg /m

⇒ g = 9.81 m/s²

⇒ i = the current = 1.90 A

B = (0.0600 * 9.81) / 1.90

B = 0.310 T

According to the right hand rule, the direction of the magnetic field is east.

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