Answer
$i_1 = 23.4~mA$
Work Step by Step
We can write an expression for the magnetic field due to a current in a straight wire:
$B = \frac{\mu_0~i}{2\pi~r}$
On the graph, we can see that $u_B = 0$ when $x = 20.0~cm$
Then the net magnetic field at the origin when $x = 20.0~cm$ is $B_{net} = 0$
Thus wire 1 must be a distance of $~~\frac{20.0~cm}{3} = 6.67~cm~~$ from the origin since $\frac{i_1}{i_2} = \frac{1}{3}$
The curve has an asymptote of $u_B = 1.96\times 10^{-9}~J/m^3$
We can find the magnetic field:
$u_B = \frac{B^2}{2~\mu_0} = 1.96\times 10^{-9}~J/m^3$
$B^2 = (2~\mu_0) (1.96\times 10^{-9}~J/m^3)$
$B = \sqrt{(2~\mu_0) (1.96\times 10^{-9}~J/m^3)}$
$B = \sqrt{(2)~(4\pi\times 10^{-7}~H/m) (1.96\times 10^{-9}~J/m^3)}$
$B = 7.019\times 10^{-8}~T$
As $x \to \infty$, the magnetic field due to the current in wire 2 approaches zero.
We can find $i_1$:
$B = \frac{\mu_0~i_1}{2\pi~r}$
$i_1 = \frac{2\pi~r~B}{\mu_0}$
$i_1 = \frac{(2\pi)~(0.0667~m)~(7.019\times 10^{-8}~T)}{4\pi\times 10^{-7}~H/m}$
$i_1 = 23.4~mA$
