Answer:
Magnetic field, B = 0.23 T
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 :
[tex]BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T[/tex]
So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.
