Answer:
185 A
Explanation:
In order for the rod to float above the ground, the magnetic force between the two rods must be equal to the weight of the floating rod.
So we can write:
[tex]\frac{\mu_0 I_1 I_2 L_2}{2\pi r}=m_2 g[/tex]
where the term on the left is the magnetic force and the term on the right is the weight, and where:
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I_1 = I_2 = I[/tex] is the current in the two rods (they carry the same current)
[tex]L_2[/tex] is the length of the floating rod
r is the distance between the rods
[tex]m_2[/tex] is the mass of the floating rod
g is the acceleration due to gravity
Here we have:
[tex]L_2 = 1.1 m[/tex]
[tex]m_2=0.064 kg[/tex]
[tex]r=12 mm = 0.012 m[/tex]
[tex]g=9.8 m/s^2[/tex]
Therefore, solving for I, we find:
[tex]\frac{\mu_0 I^2 L_2}{2\pi r}=m_2 g\\I=\sqrt{\frac{2\pi r m_2 g}{\mu_0 L_2}}=\sqrt{\frac{2\pi(0.012)(0.064)(9.8)}{(4\pi \cdot 10^{-7})(1.1)}}=185.0 A[/tex]
