Answer:
[tex]i_2 = 7.6 A[/tex]
Explanation:
As we know that the force per unit length of two parallel current carrying wires is given as
[tex]F = \frac{\mu_o i_1 i_2}{2\pi d}[/tex]
here we know that
[tex]F = 3.6 \times 10^{-5} N/m[/tex]
[tex]i_1 = 0.52 A[/tex]
d = 2.2 cm
now from above equation we have
[tex]3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}[/tex]
[tex]3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2[/tex]
[tex]i_2 = 7.6 A[/tex]