Question

A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive direction. What is the magnitude of the minimum magnetic field needed to levitate the rod

Answer 1

Magnetic field, B = 0.23 T

Step-by-step explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

`BIL=mg\\\\B=(mg)/(IL)\\\\B=(0.15* 9.8)/(13* 0.49)\\\\B=0.23\ T`

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.