Answer:
Current, I = 2.45 T
Explanation:
It is given that,
Magnetic field, B = 0.92 T
Length of wire, l = 2.6 m
Mass, m = 0.6 kg
We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.
[tex]Ilb=mg[/tex]
[tex]I=\dfrac{mg}{lB}[/tex]
[tex]I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}[/tex]
I = 2.45 A
So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.
