Asked Dec 22, 2019

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?


Expert Answer

Step 1


Mass per unit length (m/L) = 0.5 g/cm.

Current in wire towards south direction (I) = 2 A.

Step 2

The gravitational force on wire should be equal to the magnetic force applied on wire in upward direction.


Image Transcriptionclose

mg = B.I.L (m/ L)g = B.I (0.5 g/cm)x9.81 m/s² = B × 2 A 0.5 x9.81 B = B = 2.45 T

Step 3

The current is in south direction and to keep wire in vertically upward the angle between current direction and magnetic field should be 90˚...

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