Answer:
299.11 A
Explanation:
You first equal the magnetic force on the upper wire with the gravitational force:
[tex]F_E=F_g\\\\i_1LB=Mg[/tex] ( 1 ) (first term is the magnetic force produced by the magnetic force of the second wire)
i1: current of the upper wire
L: length
M: mass of the upper wire
B: magnetic field generated by the second wire
Next, you calculate the magnetic field produced by the other wire:
[tex]B=\frac{\mu_o i_2}{2\pi r}[/tex] (2)
i2: current of the second wire
r: distance between wires
mo: magnetic permeability of vacuum = 4pi*10^-7 T/A
Next, you replace the expression (2) into the expression (1) in order to obtain an expression for i2:
[tex]i_1L(\frac{\mu_oi_2}{2\pi r})=Mg\\\\i_2=\frac{M}{L}\frac{2\pi r g}{\mu_oi_1}[/tex]
M/L = is the mass per unit length of the first wire
Finally, you replace the values of the parameters:
[tex]i_2=(2.6*10^{-4}kg/m)\frac{2\pi (0.054m)(9.8m/s^2)}{(4\pi*10^{-7}T/A)(2.3A)}\\\\i_2=299.11A[/tex]
hence, the current in the second wire must be 299.11A
